MCAT Organic Chem/Gen Chem

The Stobbe condensation mechanism shares a similar reaction scheme with the Claisen condensation. Which of the following correctly identifies a step in the Claisen condensation? **what happens during the Stobbe condensation? A. Alkoxy group elimination results in generation of an ester. B. The enolate anion attacks the carbonyl through a nucleophilic acyl addition mechanism. C. The overall reaction is exergonic. D. Dehydration takes place via an E2 mechanism.

**an enolate anion attacks the carbonyl through a nucleophilic acyl addition mechanism B is correct. Step 1: An α-proton is removed by a strong base, resulting in the formation of an enolate anion, which is made relatively stable by the delocalization of electrons. Step 2: The carbonyl carbon of the (other) ester is attacked by the enolate anion/nucleophile. Step 3: The alkoxy group is then eliminated, and the alkoxide removes the newly formed doubly α-proton to form a new, highly resonance-stabilized enolate anion. Aqueous acid is added in the final step to neutralize the enolate and any base still present. The newly formed β-keto ester or β-di-ketone is then isolated. A: The elimination does not result in an ester. C: The reaction requires a stoichiometric amount of base as the removal of the doubly α-proton thermodynamically drives the otherwise endergonic reaction (i.e., Claisen condensation does NOT work with substrates having only one α-hydrogen because of the driving force effect of deprotonation of the β-keto ester in the final step). D: Claisen condensation, much like aldol condensation, occurs via a modified mechanism different from E1 and E2.

hemiketal

-OR, -OH, two alkyls

electrophiles

-electron loving -positive charge/positive polarization can accept an electron pair when forming new bonds with a nucleophile

600 mL of 1.5 M benzene-1,4-dicarboxylic acid (C8H6O4), one of the primary reagents used in the manufacture of Kevlar, is diluted with 2.4 L of water. What is the final molarity of the solution? A. 3.0 × 10-3 M B. 0.20 M C. 0.30 M D. 0.38 M

. For dilutions, use M1V1 = M2V2, keeping in mind that V2 must include the 2.4 L of water as well as the original 0.6 L of solute. M2 = (M1V1) / V2 = (0.6 L × 1.5 M) / 3 L = 0.30 M. Note also that this solution is being diluted to five times its original volume, so its new concentration must be one-fifth its initial value. **Remember with dilutions to add the values of the liquid together for the second, to get the total volume

what is percent dissociation?

. Remember that HF is a weak acid, meaning that it does not fully dissociate in water. In many weak acid problems, you must use an ICE table and the Ka of the acid to calculate the pH of an acidic solution. However, this question gives us pH outright. Remember, pH = -log[H+], so [H+] = 10-pH. Here, [H+] = 10-4 M. Now, we must understand what is meant by "percent dissociation." The percent dissociation of HF is simply the percent of the original acid concentration that has dissociated into H+ and F- ions. This value is equal to [H+]/[HF] x 100%. Original [HF] = 2 moles / 0.100 L = 20 M [H+]/[HF] = (10-4 M H+) / (20 M HF) = 5 x 10-6 % dissociation = 5 x 10-6 x 100% = 5 x 10-4 % dissociation = 5 x 10-4% = 0.0005%

The disruption of which membrane component is most likely to result in cellular traffic complications similar to those seen in gap junction disorders? A. Cholesterol B. Glycoproteins C. Glycolipids D. Phospholipids

. The cell membrane is composed of several different components, each responsible for different functions. Membrane transport is most likely to be affected if the disruption occurs in components that span the entire membrane. Transmembrane proteins (many of which are glycoproteins) are the only component listed that pass all the way through the cell membrane and facilitate membrane transport. A: Cholesterol does not facilitate membrane traffic, nor does it extend across the entire membrane. C, D: Phospholipids and glycolipids are located on the surface of cell membranes and typically do not extend though the entire bilayer. Glycolipids act to provide energy and also serve as markers for cellular recognition. Phospholipids are a structural component of the membrane and are not involved in traffic/transport.

elemental O2's oxidation state?

0

Regarding saponification, how many equivalents are there for every oxygen atom?

1 equivalent for every oxygen atom

calculation for ion product

1. first convert each ion concentration to molarity 2. use those concentrations in the solubility formula and compare final product with the Ksp 3. if greater, there will be precipitation

how do you find the molarity of a substance that is titrated, given the titration curve, the molarity of the substance used and the amount of substance added?

1. use the titration curve to find the number of mL used at the equivalence point (where acid = base) 2. divide the titrant molarity by the volume used = moles titrant 3. moles titrant = moles acid 4. divide moles acids by the volume used

how do you convert eV to joules?

1.6 * 10^-19 joules = 1 eV

where is pKa = pH

1/2 equivalence point

pressure/volume conversions *what do Pa =? *how many L per m^3? *what is a J equal to?

100 kPa = 100,000 Pa = 100,000 N/m2 = 105 N/m2 100 J/L x (1000 L/m3) = 105 J/m3 = 105 Nm/m3 = 105 N/m2

what is the Paul exclusion principle?

2 electrons cannot have the same quantum state

what does it take for 2 sugars to be anomers? which carbon is the anomer carbon?

2 sugars need to have different stereochemistry at solely anomeric carbon = carbonyl carbon

STP

273 K, 1 atm = 760mmHg

number of stereoisomers

2^n

saponification -what type of reaction? what is utilized for nucleophilic attack?

3 fatty acid salts and 1 glycerol (hydrolysis) reverse reaction between fatty acids and glycerol = triglycerides utilizes OH- for nucleophilic attack on carbonyl carbons

VSEPR theory

3D molecular geometry is determined by the electronic repulsions between its bonding and nonbonding electron pairs

Cr(OH)^2

4x^2 when determining molar solubility or Ksp

how much space will 3 moles of gas occupy?

67.2 L

What aspects separate single-crossover events from double-crossover events?

A double-crossover event is one in which chromosomal arms of homologous chromosomes cross over in two different places along the arm. This results in a section in the middle of each chromosome being exchanged. A simplified schematic of a single- vs. a double-crossover event is shown below.

Based on the information shown below, under which conditions in step 1 will 1-methylpiperazine react with diethylcarbamyl chloride to form diethylcarbamazine? A. 1M NaOH B. 1M NaN3 C. 1M HCl D. 1M NaCl

A is correct. 1-methylpiperazine must act as a nucleophile, attacking the carbonyl carbon of diethylcarbamyl chloride and displacing the chloride group. For this to occur, the -NH group on 1-methylpiperazine must be deprotonated, which would best be accomplished by a strong base like NaOH. See the diagram below for the reaction mechanism. B: The azide ion is much more reactive than 1-methylpiperazine. It would act as a nucleophile and attack the carbonyl carbon of diethylcarbamyl chloride itself, resulting in a different compound from that which is desired. C: Adding an acid would not deprotonate 1-methylpiperazine. Acids protonate other species; they do not deprotonate them. D: Sodium chloride is a neutral salt and would not react with 1-methylpiperazine at all.

Refractive myopia is corrected by the use of a diverging lens of appropriate optical power. When compared to a healthy eye, a myopic, but otherwise normal, eye's near point is most likely: A. smaller than that of a healthy eye. B. larger than that of a healthy eye. C. equal to that of a healthy eye. D. dependent upon the nature of the light incident on either eye.

A is correct. A diverging lens decreases the optical power of an instrument by increasing its effective focal length. Such a lens is corrective for the myopic eye, where its optical power exceeds that required for the axial length of the eye. As a result of this increased optical power, the myopic near point should be smaller than that found in a healthy eye. This is due to the myopic eye's capacity to focus light of great divergence on the retina.

According to the results of the experiment, which of the following blood vessels is LEAST likely experiencing turbulent flow? A. The aorta B. The femoral artery C. The saphenous vein D. All of these vessels are likely experiencing turbulent flow.

A is correct. According to the equation in the passage, a vessel will experience turbulent flow once a minimum critical velocity is reached. We are given the recorded velocities in each vessel. Next, we must compared them to the vessels' predicted vc values, and if vrecorded > vc, there will likely be turbulent flow. All of the vessels carry the same fluid, so we can ignore the effects of fluid viscosity and density. Thus, vc is proportional to Re / D. For the aorta, vrecorded = 48 cm/s; vc = Re / D = 380/2.5 = 152. 48 (our vrecorded value) is NOT greater than 152 (our vc value), so this vessel is unlikely to experience turbulent flow.

Perchloric acid is a superacid used in the production of rocket fuel. If two moles of perchloric acid are mixed with four liters of water, the normality of the resulting solution will be closest to: A. 0.5 N. B. 1 N. C. 2 N. D. 4 N.

A is correct. As a strong acid, perchloric acid is a compound that you should be familiar with for the MCAT. It has a formula of HClO4, making it a monoprotic acid. For this reason, its normality is the same as its molarity, or (2 moles HClO4) / (approximately 4 L total solution) = 0.5 M = 0.5 N.

A chemistry student notices that, at 273 K and 1.01 × 105 Pa of pressure, the Ksp value for copper (II) hydroxide is 2.2 × 10-20. If he wished to decrease this value, he could: **does Ksp respond to pressure? temperature? concentration? I. add more Cu(OH)2 to his reaction vessel. II. add sodium hydroxide, a strong base, to the solution. III. change the temperature of his flask to 220 K. IV. lower ambient pressure to 1 atm. A. III only B. I and III only C. I, II, and III only D. I, II, III, and IV

A is correct. As an equilibrium constant, Ksp only responds to changes in temperature. Specifically, if the student decreased the temperature of his solution, Ksp would decrease as well. I, II: Neither changes in ion concentration nor the common ion effect can alter the Ksp of a compound. IV: 1.01 × 105 Pa and 1 atm are actually equivalent pressures. In other words, this change in conditions is not a change at all.

Based on the data in Figure 2, what is the most likely mechanism for I- inhibition? A. I- binds to an external site on β-galactosidase, altering the enzyme conformation. B. I- competes with pNP-gal for the active site of β-galactosidase. C. I- binds to the enzyme-substrate complex. D. I- binds to β-galactosidase monomers, preventing formation of the functional tetramer.

A is correct. Based on the data in Figure 2, the Km of the enzyme doesn't change under I- inhibition, while the Vmax drops from roughly 32 µmol/min to 16 µmol/min. This is characteristic of noncompetitive inhibition, which choice A describes well.

In which of the following aqueous carbohydrate solutions would a given vesicle have the greatest buoyancy? A. 0.15 M sucrose B. 0.15 M glucose C. 0.20 M fructose D. 0.25 M galactose

A is correct. Buoyancy is the force that results from the displacement of fluid when an object is submerged. The magnitude of the buoyant force is equal to the weight of the fluid displaced. Therefore, the greater the density of the solution, the larger the buoyant force. Glucose, fructose, and galactose are all monosaccharides with the molecular formula C6H12O6, so their relative densities will depend on their concentrations. The most dense of those three choices, then, is choice D. However, choice A is sucrose, a disaccharide with the formula C12H22O11. Since sucrose has a molecular weight almost twice that of galactose, a 0.15 M sucrose solution would have a density approximately equivalent to that of a 0.3 M galactose solution. Choice D is not this highly concentrated, so we can eliminate it and choose option A.

Determine the empirical formula of an unknown sugar weighing 120 g that is combusted at high pressure, yielding 176 g CO2 and 72 g H2O. A. CH2O B. C4H8O4 C. CHO D. C3H6O3

A is correct. First, remember that all of the carbon in the original compound ends up in the form of carbon dioxide, while all of the hydrogen will be present as part of the water. We can then find the masses of C and H from the given masses of CO2 and H2O. 176 g CO2 × (1 mol CO2 / 44 g CO2) × (1 mol C / 1 mol CO2) = 4 mol C = 48 g. 72 g H2O × (1 mol H2O / 18 g H2O) × (2 mol H / 1 mol H2O) = 8 mol H = 8 g. Finally, the amount of oxygen can be found by taking 120 g - 48 g - 8 g = 64 g O = 4 mol. Last of all, find the mole ratio: for carbon, 4 mol C / 4 mol = 1; for hydrogen, 8 mol H / 4 mol = 2, and for oxygen, 4 mol O / 4 mol = 1. We are left with a 1:2:1 ratio, giving us CH2O.

Given the passage information about FLP, which biological reaction would they be most useful in catalyzing? A. Reduction of NAD+ B. Oxidation of glucose C. Phosphorylation of serine D. Dephosphorylation of glucose 6-phosphate

A is correct. From our reading of the passage, we know that FLP are useful for catalyzing the hydrogenation of molecules without expensive materials. Thus, we should look for an answer choice that involves hydrogenation. Because hydrogenation usually decreases the oxidation number of the atom the H becomes attached to, it is generally considered reduction. NAD+ accepts an H+ and two electrons to form NADH, so it is reduced and hydrogenated at the same time. Thus, choice A is correct.

OX-LDL most likely forms through the reaction of LDL with:

A is correct. From the passage, we can infer that OX-LDL is formed through the oxidation of LDL. Free radicals are molecules or atoms that contain one unpaired valence electron. As such, they typically serve as highly reactive oxidizing agents. B: These species are the reduced forms of flavin adenine dinucleotide (FAD) and nicotinamide adenine dinucleotide (NAD). Since they are reduced (and are carrying electrons), they tend to serve as reducing agents and would not oxidize LDL.

While the Nernst equation can be written in a variety of ways, one particularly biologically relevant iteration is shown below: Here, Vm represents the equilibrium membrane potential (in mV) for a particular ion, while z represents the valence of that ion. In neurons, the overall equilibrium potential is determined almost solely by potassium cations. Which set of conditions will yield a membrane potential closest to its actual resting value in humans? A. [K+outside] = 10 mM; [K+inside] = 149 mM B. [K+outside] = 149 mM; [K+inside] = 10 mM C. [K+outside] = 13 mM; [K+inside] = 100 mM D. [K+outside] = 100 mM; [K+inside] = 13 mM

A is correct. In humans, the resting membrane potential of a typical neuron is approximately -70 mV. For this reason, and since K+ has a valence of +1, we know that the log term in the above equation must be negative. Therefore, [K+inside] needs to be larger than [K+outside]. Furthermore, if the internal potassium concentration were exactly ten times the external [K+], the value for Vm would be -61 mV. As we are looking for a potential that is even more negative, [K+inside] must be relatively larger. Choice A displays a [K+inside] value that is around 15 times greater than [K+outside], which resembles values found in vivo. B, D: These choices contradict what we know about neuronal K+ concentrations. The intracellular potassium concentration is much smaller than the extracellular [K+]. C: This set of conditions is close, but would yield a Vm less negative than -61 mV.

When the outer hair cells vibrate, they generate a standing wave in the closed organ of Corti, which amplifies the frequency of interest at particular locations along the basilar membrane. How will the length of the enclosed space within the organ of Corti vary along the length of the basilar membrane to explain this? A. The length of the enclosed space in the organ of Corti will increase linearly from the base to the apex of the cochlea. B. The length of the enclosed space in the organ of Corti will decrease linearly from the base to the apex of the cochlea. C. The length of the enclosed space in the organ of Corti will decrease exponentially from the base to the apex of the cochlea. D. There should be no change in the length of the enclosed space of the organ of Corti, as length has no impact on the frequency of standing waves.

A is correct. In order for the standing wave generated by the outer hair cells to amplify the resonant frequency of the basilar membrane, it will have to match it to produce constructive interference. Thus, the fundamental frequency of the standing waves at each location along the organ of Corti should match the basilar membrane resonant frequency at that same location. Figure 2 shows that the resonant frequency decreases from the base to apex along the basilar membrane with a linear relationship, so, in order for the standing wave to produce constructive interference, its frequency will also have to decrease linearly along the membrane from base to apex. If the frequency must decrease, then the wavelength has to increase. The wavelength for a standing wave in an enclosed space is based on whether the ends of the space are closed-closed, open-open, or open-closed, and based on the length of the space. The question states that the organ of Corti is a closed space, and the wavelength of a standing wave in a closed-closed or open-open space is given by the following equation: λ = 2L / n where λ = wavelength, L = length and n = harmonic. The equation can also be derived from the standing waves as shown in the diagram below: Thus, in order to increase the wavelength, λ, it would be necessary to increase the length, L, of the enclosed space within the organ of Corti. B, C: These choices are opposite; the length should increase. D: The length does impact the wavelength and, thus, the frequency.

Which of the following characteristic changes in the IR spectrum would indicate the conversion of a fatty acid to an ester? A. Disappearance of a broad peak in the 2500 to 3300 cm-1 region only B. Appearance of a broad peak in the 2500 to 3300 cm-1 region only C. Appearance of a broad peak in the 2500 to 3300 cm-1 region, disappearance of an intense sharp band in the range 1730-1750 cm-1 D. Disappearance of a broad peak in the 2500 to 3300 cm-1 region, disappearance of an intense sharp band in the range 1730-1750 cm-1

A is correct. In the conversion of a fatty acid to an ester, we lose a hydroxyl group, and neither gain nor lose any other relevant functional groups. Hydroxyl OH bonds in carboxylic acids sound in the range of 2500 to 3300 cm-1, making answer choice (A) the correct answer. B: This is the opposite of the correct answer choice. An ester would be missing the -OH group that corresponds to a a broad peak in the 2500 to 3300 cm-1 region, whereas a fatty acid does have that hydroxyl group as part of the -COOH group. C, D: The intense sharp band in the range 1730-1750 cm-1 corresponds to the carbonyl (C=O) bond, which is present in both esters and fatty acids.

Which of the following best describes the variables plotted in Figure 1? A. The inverse of the tube diameter is the independent variable, plotted on the abscissa, and the critical velocity is the dependent variable, plotted on the ordinate. B. The critical velocity is the independent variable, plotted on the abscissa, and the inverse of the tube diameter is the dependent variable, plotted on the ordinate. C. The inverse of the tube diameter is the independent variable, plotted on the ordinate, and the critical velocity is the dependent variable, plotted on the abscissa. D. The critical velocity is the independent variable, plotted on the ordinate, and the inverse of the tube diameter is the dependent variable, plotted on the abscissa.

A is correct. In this case, the independent variable (the variable that is manipulated by the researcher) is the inverse of the diameter, while the dependent, or measured, variable is the critical velocity. By convention, the independent variable is plotted on the abscissa (x-axis) and the dependent variable is plotted on the ordinate (y-axis).

Of the choices below, the only one that is LESS reactive than isobutyric anhydride is: A. methyl propanoate. B. the conjugate base of ammonia. C. tert-butoxide anion. D. propionyl chloride.

A is correct. Isobutyric anhydride is an acid anhydride, one of the more reactive of the carboxylic acid derivatives. Esters, in contrast, are moderately unreactive, since their leaving groups are extremely unstable in solution. Methyl propanoate is an ester, as is evident from its nomenclature. B, C: Neither of these ions are carboxylic acid derivatives. In fact, both are fairly strong bases, as they are the conjugates of extremely weak acids. Bases like these are highly reactive, as they need to gain a proton to restore stability. D: From the name of this compound, we can see that it is an acyl halide; these molecules are highly reactive due to the presence of a good leaving group. Give feedback on this question

IMPERATIVE TO HAVE SIN/COS MEMORIZED!!!! write them down rn. While playing, a 40-kg child is accidentally pushed directly forward off a ledge, causing him to fall and hit the ground at a 60° angle. If the child has 720 J of kinetic energy at the instant before impact, at what velocity was he pushed? Assume negligible air resistance. A. 3.0 m/s B. 4.2 m/s C. 5.2 m/s D. 6.0 m/s

A is correct. Let's begin with the information that we have, most notably the kinetic energy value. Remember, KE = ½ mv2, so we can use this value to calculate the child's total velocity immediately before impact. 720 J = ½ (40 kg)(v2) 720 J = 20 kg (v2) 36 = v2 v = 6 m/s Note that this is the child's total final velocity, so it includes both a horizontal and a vertical component. Since air resistance is negligible and gravity acts only in the vertical direction, the child's final horizontal velocity is the same as his horizontal velocity at the moment of the push. This value can be found using the cosine of the given angle (if in doubt, draw out the situation)! vhorizontal = vtotal cos(60°) vhorizontal = 0.5vtotal = 0.5 (6 m/s) = 3 m/s

A mixture of propylamine and butanoyl chloride should create: A. N-propylbutanamide. B. N-butylpropanamide. C. aryl butanamide. D. a primary amine.

A is correct. Mixing an acyl halide with an amine will promote a nucleophilic attack, yielding an amide as the product. Here, the two reagents will combine to form N-propylbutanamide, a seven-carbon amide. B: This represents an incorrect naming of the proper answer. As the butyl chain is longer, it should form the suffix, while "propyl" should serve as part of the prefix. C: "Aryl" denotes a cyclic structure. Our product should not contain a ring. D: The amide that results from this reaction will be secondary, as its nitrogen atom will be attached to both the carbonyl carbon and its own original hydrocarbon chain.

The capacitance of a nerve membrane can be increased by: A. decreasing the width of the membrane. B. decreasing the surface area of the membrane. C. decreasing the charge stored across the membrane. D. increasing the voltage difference across the membrane.

A is correct. Remember that capacitance is coulombs per volt. So decreasing the charge (coulombs) would decrease the capacitance. That lets us eliminate choice C. Increasing the voltage would also decrease the capacitance, letting us eliminate choice D. For the MCAT, you should also remember that capacitance is directly proportional to area but inversely proportional to the distance between the two sides of the capacitor. Thus, decreasing the width of the membrane (choice A) would actually increase capacitance.

In aqueous solution, copper ion can react with water molecules to form a vibrant blue complex. This takes place according to the reaction below: Cu2+ + 6 H2O → [Cu(H2O)6]2+ In this reaction, H2O serves as: A. a Lewis base. B. a Lewis acid. C. the conjugate base of a Brønsted-Lowry acid. D. the conjugate acid of a Brønsted-Lowry base.

A is correct. Since this reaction involves the formation of a bond, we are dealing with the transfer of electrons, not the movement of a proton from one molecule to another. For this reason, the above reaction involves the Lewis definition of acids and bases, not the Brønsted-Lowry one. Specifically, water (which has two available lone pairs) is able to donate its electrons to copper (which, as a cation, is able to accept electrons). Electron-donating species are Lewis bases.

The following reaction is useful for the decarboxylation of esters under certain synthetic circumstances. Treatment of the starting material with sodium hydride would most immediately yield: A. a resonance-stabilized, tertiary carbanion. B. a resonance-stabilized, secondary carbanion. C. a primary carbanion. D. a secondary alcohol.

A is correct. Sodium hydride is most notable as a strong base. The most acidic site on the starting molecule is the tertiary carbon between the two carbonyl groups, a position that serves as the alpha carbon for two distinct carbonyl functionalities. Resonance stabilization of the conjugate base renders this site much more acidic than would normally be expected of a carbon atom. Sodium hydride would be more than adequate for the abstraction of a proton from this position.

Biological testing was conducted of blood serum from patients whose treatment regimen included taking one or both of the two taxane pharmaceuticals. One such test included an affinity chromatography technique in which it was attempted to determine how the two drugs affected liver functioning. When applying this technique, it could be expected that: I. migration of different enzymes would depend upon their respective affinities for the solid phase. II. migration of different enzymes would depend on their respective size and molecular mass. III. migration of different enzymes would depend on whether the enzymes' primary structures comprise more amino acids versus fewer amino acids. A. I only B. II only C. I and II only D. I, II, and III

A is correct. Statement I clearly describes the reason for different migration speeds in an affinity chromatography procedure — as its name implies, this technique is based on varying affinities for the stationary phase, as shown below. In such a procedure, the stationary phase often includes antibodies or other specific molecules.

Which of the following choices accurately represent(s) the Henderson-Hasselbalch equation?

A is correct. The Henderson-Hasselbalch equation can be written as either pH = pKa + log([A-]/[HA]) or pOH = pKb + log([HA]/[A-]). II: The log term should not be subtracted here. This would wrongly imply that more conjugate acid produces a lower pOH. In addition, this choice incorrectly mixes pOH (the log of a concentration) with Kb (an equilibrium constant). When pOH is used, pKb should be used as well. III: This equation mistakenly uses log[H+], when pH is always equal to -log[H+].

Which of the following compounds will have the highest retention time when run in a gas-liquid chromatography chamber? A. Methyl cyclohexane (highest boiling point = longest time to turn into a gas) B. Hexane C. Methylene chloride D. Heptane

A is correct. The compounds listed above are nonpolar, so they will separate mainly based on their boiling points. In gas-liquid chromatography, the sample is volatilized in a hot chamber. The higher the boiling point temperature, the more time the compound will spend in the beginning of the chamber waiting to get volatized. Retention time is the time it takes for the sample to reach the detector. Therefore, the compound with the highest boiling point, in this case methyl cyclohexane, will have the highest retention time. B, C, D: These compounds all have lower boiling point temperatures than that of methyl cyclohexane. Thus, their retention times are smaller than that of methyl cyclohexane.

A physical chemist is researching potential uses for the ion below. **what the differences between kinetic and thermodynamic? Whist one is more stable? less? This structure represents: A. the kinetic enolate. B. the thermodynamic enolate. C. neither the kinetic nor the thermodynamic enolate. D. it is impossible to decide from the information given.

A is correct. The kinetic product is the one that is less thermodynamically stable, but is easier to synthesize due to a lower activation energy. In general, kinetic enolates are less substituted than their thermodynamic counterparts. Here, the double bond has formed in the position that is less sterically hindered (between carbons 1 and 6). B: If this were the thermodynamic enolate, it would have the double bond between carbons 1 and 2 (the more substituted position adjacent to what formerly was the carbonyl carbon).

Based on the information presented in the passage, what is the limiting reagent for the preparation of hydroxyapatite? A. Calcium ion B. Phosphate ion C. Hydroxide ion D. Water

A is correct. The limiting reagent is the reactant that would produce the least amount of product, based on the stoichiometry of the reaction. Thus, the safe way to find this reagent is to determine the amount of product formed by the complete reaction of each of our two main reactants: calcium and phosphate ion. For calcium ion: 40 mL x [1 L/1000 mL] x 0.32 mol Ca2+/L x [1 Ca5(PO4)3OH / 5 Ca2+] = 0.00256 moles Ca5(PO4)3OH produced This calculation can be simplified by using scientific notation and the following approximations: 4 x 101 x 10-3 x 3 x 10-1 x (1/5) ~ (12/5) x 10-3 ~ 2 x 10-3 moles For phosphate ion: 60 mL x [1 L/1000 mL] x 0.19 mol PO43-/L x [1 Ca5(PO4)3OH / 3 PO43-] = 0.0038 moles Ca5(PO4)3OH produced This calculation can also be simplified by using scientific notation and the following approximations: 6 x 101 x 10-3 x 2 x 10-1 x (1/3) ~ (12/3) x 10-3 ~ 4 x 10-3 moles The amount of calcium available will thus react to form less product, making calcium ion our limiting reagent.

Radiation oncologists use machines called gamma knives to perform radiosurgery. Lead vests are used to protect non-cancerous areas and medical personnel from radiation exposure. A typical gamma knife uses a photon beam with a frequency of 1.6 x 1019 Hz. Suppose a stray photon hits a lead atom and ejects an electron and a lower-frequency photon of 1.599 x 1019 Hz. What is the maximum kinetic energy of the ejected electron? (Lead has a work function of 4.14 eV, Planck's constant is 6.626 x 10-34 J∙s, and 1 eV = 1.6 x 10-19 J.) A. 6.0 x 10-18 J B. 6.6 x 10-18 J C. 1.0 x 10-14 J D. 0 J

A is correct. The photons used for radiation therapy are so high-energy that they cannot be absorbed by an atom all at once. Instead, they display a scattering behavior in which a collision produces a second, lower-energy photon that moves off at an angle. Because of the law of conservation of energy, the energy of the initial photon must equal the sum of the energies of the emitted photon and the electron. However, we are given only frequencies. Since frequency and energy are directly proportional, we can use E = hf to find the energies required to solve. First, though, we must find the frequency of the electron, which is equal to the difference between the two photon frequencies. 1.6 x 1019 Hz - 1.599 x 1019 Hz = 1016 Hz E = hf = (6.6 x 10-34 J∙s)(10 16 Hz) = 6.6 x 10-18 J This quantity is NOT equal to the kinetic energy of the electron, as some energy is required simply to liberate the electron from lead's poisonous grasp. The work function represents this amount of energy. We must first convert lead's work function to joules, then subtract the work function from our calculated energy to get the kinetic energy. To make this calculation easier, use 4 instead of 4.14: WPb = (4 eV)(1.6 x 10-19 J) = 6.4 x 10-19 J KE = Etotal - WPb = 6.6 x 10-18 J - 6.4 x 10-19 J = 6 x 10-18 J

During the initial reduction reaction shown in Figure 2, FAD receives how many protons and how many electrons from cholesterol? A. 1 proton and 1 electron B. 1 proton and 2 electrons C. 2 protons and 1 electron D. 2 protons and 2 electrons

A is correct. The question asks about the effect of cholesterol on FAD. Go to Figure 2. In reducing FAD to FADH, one electron and one hydrogen ion—one proton—were transferred to FAD from cholesterol. Remember that reduction is the gain of an electron. While we often use the movement of H atoms to spot REDOX reactions, they are defined by the transfer of electrons.

In the human body, two FPP molecules are combined in a reaction that directly leads to which class of compounds? A. Cholesterol B. Sphingolipids C. Triacylglycerols D. Purines

A is correct. The structure of FPP is highly suggestive of its combination with a partner and transformation into the fused 4-ring system common to cholesterol. B: Sphingolipids, or glycosylceramides, are a class of lipids containing a backbone of sphingoid bases, a set of aliphatic amino alcohols. There is nothing in their structure to suggest 2 FPP molecules will lead to such a molecule. C: Triacylglycerols are esters derived from glycerol and three fatty acids. D: Purines are heterocyclic aromatic organic compounds. They consist of a pyrimidine ring fused to an imidazole ring (both of which contain nitrogen).

If researchers failed to take into account the effect of air resistance on the pitch, how would it impact their measurements of the efficiency of energy transfer from the arm to the baseball? A. It would be lower than the actual efficiency, as there was a higher initial velocity at release. B. It would be higher than the actual efficiency, as there was a higher initial velocity at release. C. It would be lower than the actual efficiency, as there was a lower initial velocity at release. D. It would be higher than the actual efficiency, as there was a lower initial velocity at release.

A is correct. This question requires us to examine the experimental setup and determine what effects air resistance would have on the scientists' calculations. Air resistance would decrease the velocity of the ball as it travels from the mound to home plate, where the velocity was recorded. Thus, the measured velocity should be lower than the velocity at release. The decreased velocity will result in a decreased calculated energy for the baseball, leading to a decreased calculation of efficiency for the energy transfer from the arm to the ball.

The molecule shown below is treated with a peroxyacid. Assuming that the starting material is completely enantiopure, subsequent treatment with potassium tert-butoxide would yield: A. a mixture of two diastereomers. B. a mixture of more than two diastereomers. C. a racemic mixture of products. D. a single, unique product.

A is correct. This reaction first involves the formation of an epoxide. Such reactions proceed without stereochemical bias, yielding an equal mixture of stereoisomeric products with respect to the site of epoxide formation. The carboxylic acid is then deprotonated and is subsequently able to open the epoxide ring by attacking the less sterically hindered side. Here, the opening of the epoxide leaves an intact chiral center that was not present in the original molecule. Since the epoxidation resulted in the formation of two diastereomers, and since the ring-opening reaction did not change any stereochemistry, the final products are best described as a mixture of two stereoisomers. Note that the single chiral center originally present in the molecule remains untouched.

Based on the information in the passage, what is the theoretical yield of hydroxyapatite? A. 1.29 g B. 1.91 g C. 3.02 g D. 502 g

A is correct. To save time, we can use the information obtained in the previous question, concerning the limiting reagent calculation for calcium. (While the MCAT typically does not ask questions that "depend" on previous questions, time can be saved by keeping clear track of your calculations, especially with regard to stoichiometry.) From that previous question, the amount of Ca5(PO4)3OH produced is 0.00256 moles. Multiplying by the gram formula mass of hydroxyapatite gives the theoretical mass of product: 0.00256 moles Ca5(PO4)3OH x 502.3 g/mol = 1.29 g

Administration of a muscle relaxant causes a vessel to increase radius by 10%. If all else is held constant, the volume flow rate will change to approximately what percent of the original rate? A. Increase to 150% of the original B. Remain at 100% of the original C. Decrease to 65% of the original D. Cannot be determined

A is correct. Volume flow rate is proportional to the radius raised to the fourth power. An increase of 10% means the rate of flow will increase to 1.14 of the original rate. We need not do any calculations here, since only choice A says that the flow rate will increase. B: If the rate of flow remained constant, then some other factors would also have to change. The question says that only radius increased and no other factors changed. C: This would be the result of a 10% decrease in radius: 0.94. D: This change can be determined.

What are the units for k in the following rate law: rate = k[A]2[B]? Note that the concentration unit is mol/L. A. L2mol-2s-1 B. s-1 C. s D. L mol-1s-1

A is correct. We can see that the reaction is third order overall, and reaction rates are measured in M/s, where M = mol/L. Thus, we can set up our equation: M/s = k (mol/L)3 = k (mol3/L3) k = (M/s) (L3/mol3) = (mol/L•s) (L3/mol3) k = L2/s•mol2 = L2mol-2s-1

The reaction between propanal and cyanide forms a: A. cyanohydrin. B. primary amine. C. secondary amine. D. enamine.

A is correct. When cyanide reacts with propanal (an aldehyde), it forms a cyanohydrin. Interestingly, it is the carbon - not the nitrogen - atom that acts as a nucleophile to attack the carbonyl carbon. The final cyanohydrin product consists of the former carbonyl carbon bound to -OH, -H, the original -R group from the aldehyde, and -C≡N.

Which of these compounds contains the most acidic alpha proton? A. Propanal B. 2-hexanone C. 2-propanone D. Benzaldehyde

A is correct. When dealing with aldehydes and ketones, remember that the alpha protons (those bound to the carbon atom(s) adjacent to the carbonyl carbon) are especially acidic. Of the choices, propanal includes the alpha proton with the conjugate that is least destabilized by nearby electron-donating groups. B, C: These molecules are ketones, which have two alkyl groups bound to the carbonyl carbon. Since alkyl groups are electron-donating, these add excess electron density to the carbonyl carbon. This added "partial negative" charge destabilizes the also-negative conjugate bases of these species, making their alpha protons less acidic. D: Benzaldehyde does not have an alpha hydrogen.

What is the electron configuration of Os3+? A. [Xe] 4f145d5 B. [Xe] 6s24f145d6 C. [Xe] 6s24f145d9 D. [Xe] 6s24f115d6

A is correct. When determining the electron configuration of a cation, remember that the highest-energy electrons should be removed first. As the 6s subshell possesses the highest principal quantum number, its electrons must also have the most energy of those involved. Thus, the two electrons in the 6s shell are removed first. Next, a single electron from the 5d shell must be taken off as well, as it has higher energy than the 4f electrons.

what do impurities do to melting point range?

A melting point range includes the temperature when the first crystal of a compound starts to melt and the temperature when the compound is entirely melted. For a pure compound, the melting point range is narrow. Therefore, melting point determination is sometimes used to identify an unknown pure compound. However, the presence of impurities in a compound lowers and broadens the melting point range. According to the passage, the melting point of pure caffeine is 235°C, making choice A the best answer.

Beer's Law -what does it measure, what does each part stand for? **what are the units for path length?

A=ebc A= absorbance e= absorptivity/extinction constant b = path length (cm) c = solution concentration

Assuming a membrane protein folds according to Equation 1 and has no intermediates, which of the following equations gives the rate of formation of unfolded proteins?

According to Equation 1, as ku increases, the rate of formation of the product (U) should increase. The rate is dependent only on the concentration of the native protein.

what does a similar wavelength in UV/Vis spectra mean?

According to the passage, the value of λmax for compound when performing UV/Vis spectroscopy indicates the presence of a specific combination of chromophoric groups. The close resemblance in the λmax values of cefixime and C1-C3, then, suggests that they contain a similar specific combination of chromophoric groups.

what does aldosterone do?

Aldosterone is released from the adrenal cortex in response to low blood pressure. Its primary function is to increase sodium reabsorption in the distal tubule and collecting duct. Aldosterone upregulates the sodium-potassium pumps along the lining of the nephron, pumping three sodium ions OUT of the nephron lining (and toward the blood) for every two potassium ions it pumps IN (toward the nephron and away from the blood). Since we have a net solute movement out of the nephron, aldosterone also increases the gradient that favors water reabsorption. The human nephron is shown below, with the distal tubule (a primary target of aldosterone) labeled. plus loss of potassium in the blood

Is it reasonable to expect that individuals with certain alleles of the gene discussed in the passage might be less likely to develop cancer in the first place?

Alleles for this gene either include or don't include an effective error-correcting machinery made up of proofreading enzymes that snip and fix mistakes in the DNA-copying process. This has an effect on the spread of cancer and, thereby, life expectancy after diagnosis. However, it makes sense that individuals who do not have cancer will be more or less likely to develop depending on their alleles for the relevant gene. There are two answer choices that say yes, but only A includes the correct reasoning in the answer choice. Metastasis occurs only after cancer has been developed, so it has nothing to do with cancer prevention.

Which of the following reagents would be involved in the first synthetic step to chemically derive an amide from oleic acid? A. Phosphorous pentachloride B. Aminobutane C. Octadecanol D. 2-aminopentane

Amides are the most stable (least reactive) of the carboxylic acid derivatives because the amino substituent is a very poor leaving group. In order to synthesize amides however, one cannot simply use the amino substituent as a nucleophile; as a relatively strong base, it will instead preferentially participate in an acid-base neutralization reaction with the acidic hydrogen of the organic acid. Instead, one must first create an acid chloride (or an acid anhydride, which is not an option in this question) to eliminate the confounding acidic hydrogen. Afterwards, one can proceed with a normal nucleophilic substitution. Aminobutane and 2-aminopentane are both suitable as amino substituents but are therefore unsuitable as a first step. The addition of octadecanol would be a poor choice of a first step; absent rigorous reaction conditions, it's unlikely anything productive would happen at all.

what happens to osmotic pressure when someone eats glucose? What does this ultimately do to the cell?

An increased concentration of circulating blood glucose, to which cell membranes are relatively impermeable, causes the osmotic pressure of the extracellular fluid to rise, in turn causing water to leave the cell. This movement of water promotes cellular dehydration.

The figure below depicts two anomeric forms of a sugar. These isomers can best be described as:

Anomers are a form of epimer. Epimers are sugars that differ only at one stereocenter, with all other stereocenters being exactly the same. Thus, epimers are a type of diastereomer, and choice C is correct. The stereocenter at which these structures differ is circled below. Note that the structure on the left is the alpha anomer, since the -OH group on its anomeric carbon points in the opposite direction (down from the ring) from the CH2OH group (which points upward). In contrast, the structure on the right, which has its -OH group pointing upward, is the beta anomer. A: Enantiomers are mirror images, which these two figures are not. B: Conformers can freely convert between forms with no bond breaking (such as the chair and boat forms of cyclohexane). Here, the hydroxyl group cannot simply rotate into the equatorial "up" position without breaking bonds. D: Constitutional or structural isomers have fundamentally different connectivity between the atoms. These molecules are stereoisomers, not constitutional isomers.

Aspartame (N-L-alpha-aspartyl-L-phenylalanine 1-methyl ester) is a very well-known artificial sweetener found in the large majority of non-sugar containing food products. This compound is classified as a(n):

Aspartame contains two amino acids as mentioned above in the formula; thus, it is a dipeptide. Aspartame (shown below) is the methyl ester of the dipeptide containing aspartic acid and phenylalanine. Of course, you do not need to have any outside knowledge about the structure of aspartame, but you should be able to recognize the structure below as a dipeptide.

Given the goal of calculating ΔG°w N-U, the researchers choose to study proteins with measurable folded and unfolded concentrations at equilibrium. This is most likely because the protein must have measurable folded and unfolded concentrations at equilibrium, and:

At equilibrium, the equilibrium constant is given by Keq = [products]/[reactants]. If the concentrations of the folded (reactant) and unfolded (product) protein are known while the reaction is at equilibrium, Keq can be calculated. Once Keq and temperature are known, ΔG°w N-U can be calculated with the following equation: ΔG°w N-U = -RTln(K eq).

High levels of LDL and OX-LDL increase the proportion of cholesterol in cell membranes. If the trisomy 21 data in Figure 2 can be attributed to the effect of cholesterol on these membranes, which of the following statements is most likely true? **what does cholesterol do for the cell membrane under normal conditions? how does it do this?

At moderate to high temperatures (including normal physiological temperature), cholesterol increases the rigidity of cell membranes by attracting adjacent phospholipid tails. Thus, a more rigid membrane appears to correlate with a higher incidence of trisomy 21. From Figure 3, however, we see that ethanol appears to counteract the trisomy-inducing effects of LDL and OX-LDL. It is therefore reasonable to conclude that ethanol decreases the rigidity (or increases the fluidity) of cell membranes under the conditions in this study.

what happens when a ride is at the top of the loop?

At the top of the loop, the gravitational and the normal forces (if any) point downward toward the center of the loop. Therefore, the net force causing centripetal acceleration is the sum of the gravitational force and the normal force. When the centripetal force is the minimum amount needed for the ride to stay on track, the normal force is zero. Note: At the top, the normal force can be zero but the gravitational force cannot. Let's designate the initial height where the ride starts as H. At the very beginning, the energy of the ride is the gravitational potential energy. Therefore, Ei = mgH. At the top of the loop, the ride includes both gravitational and potential energy. Therefore, Ef = mgH + ½ mv2 = mg (2r) + ½ mv2. The net acceleration at top (centripetal acceleration) = v2/r. Since we concluded that the normal force at the top is zero, the net acceleration at the top is the gravitational acceleration, g. Therefore, v2 = gr. Substituting this equation into the 'Ef' equation, we obtain the following: Ef = 2mgr + ½ mgr = (5/2) mgr To obey conservation of energy, Ei = Ef. Therefore: mgh = (5/2) mgr. We conclude that H is 5r/2. Since r is 8 m, H is 5(8 m)/2 = 20 m.

A scientist builds a MHC using NAD+ and ubiquinone, a vitamin-like substance found in most eukaryotic mitochondria. Given the data below, which of the following setups will result in a cell capable of running a heart monitor? NAD+ + H+ + 2e- → NADH E° = -320 mV Ubiquinone + 2H+ + 2e- → Ubiquinone (H2) E° = +100 mV A. Ubiquinone anode, NAD+ cathode; electrons will flow from ubiquinone to NAD+ B. Ubiquinone cathode, NAD+ anode; electrons will flow from NAD+ to ubiquinone C. Ubiquinone cathode, NAD+ anode; electrons will flow from ubiquinone to NAD+ D. Ubiquinone anode, NAD+ cathode; electrons will flow from NAD+ to ubiquinone

B

The development of PDUS technology alleviated a key disadvantage of classic Doppler ultrasound imaging. Which of the following best describes this disadvantage? A. Classic Doppler ultrasound gives a reading of zero for non-moving tissue. B. In classic Doppler ultrasound, readings become less accurate as cos θ approaches 0. C. In classic Doppler ultrasound, readings become less accurate as cos θ approaches 1. D. Classic Doppler ultrasound requires use of very high-frequency sound waves.

B

A student combines 364.6 g of HCl with 80 g of NaOH in 5 L of water. What additional volume of H2O must be added to this mixture to yield a solution with a pH of 1? Note that the molar mass of HCl is 35.46 g/mole, while that of NaOH is 40 g/mole. A. 8 L B. 75 L C. 80 L D. 120 L

B is correct. 364.6 g of HCl represents 10 moles of this strong acid. In contrast, 80 g of NaOH is only 2 moles of this strong base. When mixed in solution, the base will fully neutralize 2 moles of HCl according to the 1:1 stoichiometric ratio between these reagents. 8 moles of HCl will remain in 5 L of water; however, we need a total H+ concentration of 0.1 M to yield a pH of 1. This concentration can best be obtained by adding enough water to establish a total volume of 80 L. C: The question asks for the additional volume of water, not the final volume overall! Remember, we already had 5 L of water at the beginning of the setup.

What will be the difference in hematocrit between a person living at a higher elevation and a person living at a lower elevation? A. The person living at a higher elevation will have the higher hematocrit because of the decreased atmospheric pressure. B. The person living at a higher elevation will have the higher hematocrit because there is less oxygen in the air. C. The person living at a lower elevation will have the higher hematocrit because of the higher atmospheric pressure. D. The person living at a lower elevation will have the higher hematocrit because there is more oxygen in the air.

B is correct. A person living at a higher elevation will produce more blood cells because there is less oxygen in the air. The kidney has receptors that read the oxygen levels in the blood and produce erythropoietin in order to produce more blood cells. This will increase the hematocrit, making choice B correct. A, C: It is not the change in atmospheric pressure that would directly affect a person's hematocrit. D: At a lower elevation, more oxygen is present than at a higher elevation.

Under what set of conditions would an air embolism be MOST likely to form in blood? A. 42°C and PO2 = 300 mmHg B. 42°C and PO2 = 120 mmHg C. 13°C and PO2 = 120 mmHg D. 13°C and PO2 = 300 mmHg

B is correct. According to the first and second paragraphs, an air embolism is caused when a bubble of air is formed in the blood and blocks a vessel. If a gas is insoluble in a fluid, it can escape as a bubble (think of warm soda going flat). Gases generally have greater solubility in cooler liquids. Also, the solubility of a gas increases as the partial pressure of the gas increases. This is primarily due to the hydrogen bonding effects of aqueous fluids that produce cavities within the fluid; these cavities can accommodate the gas molecules. Thus, we would expect an air embolism to form when gas is insoluble, which occurs in high-temperature, low-pressure systems.

What is the formula of hydroxyapatite that is present in the unit cell of the crystalline material? A. Ca5(PO4)3OH B. Ca10(PO4)6(OH)2 C. Ca3(PO4)2 D. Ca(OH)2

B is correct. According to the information presented in the second paragraph, "The hexagonal unit cell ... contains two formula units." Since the formula unit for hydroxyapatite is Ca5(PO4)3OH, doubling this formula gives choice B.

What is the voltage generated by an electrochemical cell at equilibrium? A. The voltage must be positive. B. The voltage must be zero. C. The voltage must be negative. D. All REDOX reactions are at equilibrium, irrespective of their voltage.

B is correct. At equilibrium, the free energy change is zero. Therefore, the REDOX potential is zero as well, from ΔG = -nFE. When considering Equation 2, if Q = Keq, then the -0.06/n log K term cancels the E°. Choice B is the best answer. A, C: Since the free energy change at equilibrium equals zero, the voltage generated by a cell at equilibrium will also equal zero. D: It is not true that all REDOX reactions are at equlibrium.

In order for the image produced by the eyepiece to be erect, where must the image focused by the objective lens fall? A. Between the objective lens and the focal point of the eyepiece B. At a distance from the eyepiece smaller than the eyepiece's focal length C. At a distance from the objective less than the objective's distance from the sample D. At the focal point of the objective lens

B is correct. By "point the same direction," this question implies that the image created by the eyepiece is not inverted with respect to the image formed by the objective. In other words, the image is "upright" and virtual. (Be careful in cases of compound microscopes — if the first image was upside-down, the second image will be upside-down as well, since it is not inverted. It will still be a virtual image.) To create a virtual image with a converging lens, the object distance must be smaller than the focal distance of the lens in question. Here, this means that the object must fall within the focal length of the eyepiece lens.

List these four species in order of increasing pH, assuming that each is present in 0.10 M aqueous solution: H3PO4, H2PO4-, HClO4, PO43-. A. PO43- < H2PO4- < H3PO4 < HClO4 B. HClO4 < H3PO4 < H2PO4- < PO43- C. H3PO4 < HClO4 < H2PO4- < PO43- D. PO43- < H2PO4- < HClO4 < H3PO4

B is correct. Don't miss the inclusion of perchloric acid in the midst of three phosphorus-containing examples! HClO4 is a strong acid and thus has the lowest pH in solution of all four choices (assuming, of course, that starting concentrations are equal). Next, H3PO4 is a triprotic acid with a fairly stable conjugate base, although phosphoric acid certainly is not strong. Subsequent deprotonation yields stronger and stronger conjugate bases, making PO43- the most basic of the listed species. This tendency is very normal; intact polyprotic acids are always more acidic than their deprotonated conjugates.

Given a 5.0-ampere current running for a total of four hours and ten minutes, what will be the amount of hydrogen gas produced in an electrolysis reaction? (Faraday's constant = 96,485 C per mole of electrons) A. 0.19 moles B. 0.39 moles C. 0.78 moles D. 1.6 moles

B is correct. For questions like these, begin with the information given. We know that the current is 5.0 A, or 5.0 coulombs/s. To find coulombs, we simply need to multiply by seconds. Four hours and ten minutes is equal to 250 minutes, or 15,000 seconds. (5 C/s)(15,000 s) = 75,000 C of charge transmitted (75,000 C) / (96,485 C/mol) ~ 0.8 moles of electrons transferred (as 75/96 is slightly larger than 75/100, or 0.75) However, we are not quite finished. According to the reduction half-reaction, only one molecule of H2 is created for every two electrons. Thus, the amount of hydrogen gas produced must be approximately 0.4 moles.

The equivalence point of the titration of tolbutamide with NaOH was reached by adding 50 mL of NaOH. Which of the following correctly describes the solution during this process? A. The solution had a pH less than 7.1 at the equivalence point. B. After addition of 25 mL of NaOH, the pH of the solution was greater than 5. C. The concentration of the charged form of tolbutamide was greater than neutral tolbutamide during the titration. D. The concentration of the charged form of tolbutamide was less than neutral tolbutamide during the titration.

B is correct. If 50 mL of NaOH represents the volume of titrant required to reach the equivalence point, then 25 mL is the volume added at the half equivalence point. At this point, one half of the original tolbutamide present will have been converted to its conjugate base, and their concentrations will be equal (eliminate choices C and D). According to the Henderson-Hasselbach equation, when the values of protonated acid and conjugate base are equal, pH = pKa + log 1 = pKa + 0 = pKa. At the half-equivalence point, pH of solution equals the pKa of the analyte. According to paragraph 3, tolbutamide's pKa is 5.3. A: The titration of a strong base and a weak acid results in the formation of a basic salt, the hydrolysis of which causes the pH at the equivalence point to be greater than 7. C, D: The relative concentrations of the charged versus neutral forms of tolbutamide changed throughout the titration.

The distance between the ears for an average adult male is 20 cm. If a wave with a frequency of 17,000 Hz hits both the right and left ear and has a phase difference of 5π radians between the two, what is the difference in distance between the origin of the sound and each ear (velocitysound in air = 343 m/s)? what information does the phase difference give you? what is the equation for difference in distance (how is this calculated?)? A. 2 cm B. 5 cm C. 8 cm D. 20 cm

B is correct. If the phase difference between the sound wave hitting the right and the left ear is 5π, then the difference in distance from the origin of the wave to the right and the left ear is 2½ wavelengths, as (5π radians) / (2π radians / λ) = 2 ½ λ. Thus, the wavelength is needed in order to determine the difference in distance. The wavelength can be found by using the frequency and velocity in the following equation: v = λf. Rearranging for wavelength gives: λ = v/f. Plugging in velocity and frequency gives: λ = 343 m/s / 17000 Hz. In order to estimate, round 343 to 340 and use scientific notation: λ = 3.40 x 102 / 1.7 x 104 = 2 x 10-2 m = 2 cm Thus, the difference in distance = 2.5λ = (2.5)(2 cm) = 5 cm. "Difference in distance" = phase difference * wavelength

If a second experiment is run in which the velocity is kept constant, which of the following would increase the probability of turbulence? I. An increase in the diameter in the tube II. A decrease in the viscosity of the fluid III. A decrease in the density of the fluid A. I only B. I and II only C. II and III only D. I, II, and III

B is correct. In order to increase the likelihood of turbulence, we would need to lower the threshold at which turbulence occurs, thus making it more likely that the calculated critical velocity will be lower than the constant velocity used in the new experiment. In examining Equation 1, we can see that the critical velocity is inversely related to the diameter of the tube. Therefore, increasing the diameter should decrease the critical velocity (I is true, eliminate choice C).

A patient presents in the emergency department having ingested a large quantity of tolbutamide. Intravenous administration which of the following compounds is most likely to increase the rate of urinary excretion of the drug? A. KCl B. NaHCO3 C. NH4ClO4 D. NaCl

B is correct. In order to increase the percentage of drug excreted in the urine, it is necessary to decrease the fraction of tolbutamide capable of reabsorption, or diffusion out of the lumen of the nephron. Paragraph 3 says that for weakly acidic drugs, the uncharged state is capable of diffusion through membranes much more than the charged form. Thus, to prevent reabsorption, we must maximize the charged form of tolbutamide. We can do this to a weak acid by deprotonating the drug (rendering it negatively charged) via administration of a base. This will increase blood and urinary pH (i.e. decreasing [H+]) and increase the fraction of ionized drug present. Urinary alkalization can be accomplished by administration of a basic salt, like NaHCO3 (the salt of the conjugate base of carbonic acid). A, D: NaCl and KCl are salts that are very close to neutral (neither acidic nor basic). Because Cl- is the conjugate base of a strong acid (HCl), it will only be able to act as a base when pH is extremely low. C: ClO4- is also the conjugate base of a strong acid (HClO4), so it will also fail to act as a base under biological conditions. Additionally, NH4+ is acidic. Because of this, NH4ClO4 is acidic overall, not basic.

Which of the following is (are) present in a mixture of NH3 and BF3? I. FLP II. Brønsted-Lowry acid III. Brønsted-Lowry base A. I only B. III only C. II and III only D. I, II, and III

B is correct. Our strategy for RN questions should involve starting with the RN that appears in the most answer choices (in this case, RN III). BL bases are species that accept protons, and BL acids are species that donate protons. NH3 has a lone pair on the N that accepts protons to form NH4+, or ammonium. Thus, it is a BL base and RN III is correct. I: The passage states that FLP are mixtures or compounds that contain Lewis acids and Lewis bases that are sterically hindered, such as Compound 1. Though NH3 and BF3 are Lewis bases and acids, they are not sterically hindered, so they are not FLP. II: NH3 is a BL base, but BF3 is not a BL acid because it does not have a proton to donate.

Which of the following species, if present, will most likely result in the MFC undergoing fermentation at low voltages? A. AlCl3 B. CN- C. NO3- D. PO43-

B is correct. Paragraph 4 states "If no sulfate, nitrate or other electron acceptors are present, fermentation will be the main process when the anode potential remains low." This means that, for fermentation to predominate, we need a lack of available electron acceptors. Lewis bases are species that donate an electron pair. Thus, they are most likely to cause the MFC to undergo fermentation. Only choice B is a Lewis base.

The data in Table 1 show the effect of increasing pH on the relative rates of reduction of CO2 to CO compared with the reduction of CO to CH4. The researchers would be most likely to conclude that: A. at a low pH, the reduction of H2O outpaces the reduction of CO2. B. the production of H2 gas shows first-order kinetics. C. proton concentration has no effect on the reduction of CO to CH4. D. the rate of reduction of the CO2 radical anion slows at high pH.

B is correct. Since pH is defined as -log[H+] a pH of 1, the concentration of H+ is 0.1 M (or 1 x 10−1 M), while at a pH of 3, it is 0.001 M (or 1 x 10−3 M). This corresponds to a difference of two orders of magnitude, which is also approximately the difference in the production of H2 gas between the two pH conditions (1.32 mL/min vs. 0.01 mL/min). This corresponds to first-order kinetics, in which the reaction rate is dependent on the concentration of a single reactant (in this case, H+). A: We are given no direct information about the reduction of H2O, so cannot compare it to the rate of CO2 reduction. C: Because the rate of carbon monoxide reduction to methane decreases at high pH, it must be that the absence of protons confound the reductive process, having a demonstrable effect. D: The carbon dioxide radical anion is not specifically reduced during the reaction mechanism, but rather it is converted in an acid-base reaction into the metal-bound carboxy-hydroxyl intermediate.

Given the relationship between force per cross-sectional area and break frequency described in the passage, it is reasonable to expect: A. finely-boned women to be less liable to a fracture while running. B. those runners with the briefest interval of ground contact per step to be more liable to breaks. C. flat-footed runners to have a higher risk of fracture than individuals who run on their toes. D. compactly massive (heavily-muscled) runners to have a smaller energy loss due to air resistance.

B is correct. Since the passage states that greater force per cross-sectional area (pressure) correlates with higher break frequency, it's logical to expect that either smaller cross-sectional areas for the same force or larger forces for the same cross-sectional area will lead to more fractures. Runners who only contact the ground briefly are likely to have a smaller cross-sectional area actually hit the ground, leading to a large force per area value. Also, if you are familiar with the concept of impulse, you will remember that I = force x time. Assuming that I is a set value per step (which is true), the smaller the time of impact, the larger the force experienced.

A woman is standing two meters away from a speaker. The speaker is turned up to 11 on its dial, and the power of the sound hitting her tympanic membrane increases 100-fold. How far would the woman need to move to reduce the sound back to its original decibel level? A. 10 m B. 18 m C. 20 m D. 50 m

B is correct. The decibel level of sound is given by dB = 10log (I/Io), where I denotes intensity and Io is the reference intensity, which represents the lowest intensity audible to the human ear. In order to have the same decibel level, the same intensity, I, must reach the tympanic membrane. The intensity of sound is given by the equation I = P/A, where P = power and A = area. If the power increases 100-fold, then the area would need to increase by a factor of 100 to maintain the same intensity. The area, A, of the sound wave is given by the surface area of a sphere as the sound travels away from the speaker. The surface area of a sphere is proportional to the radius squared, or r2 (A = 4πr2). So, in order to increase the area 100-fold, the radius, r, must increase by a factor of 10 (102 = 100). Thus, if the woman originally stands two meters away then to maintain the same decibel level, she needs to stand 2 x 10 = 20 m away from the speaker. Standing 2 m away, the woman must move 18 m to reach a total final distance of 20 m (choice C is a clever trap).

Which of the following molecules has the highest predicted melting point? A. H2O B. CaCl2 C. Ne D. CH3COCH3

B is correct. The greater the intermolecular attractions between molecules, the higher the melting point of a sample of those molecules. Calcium chloride is a salt, meaning that it participates in ionic bonding. While ionic bonds are unusual in that they are technically chemical (or intramolecular) bonds, they do confer extremely high melting points on their associated solids. A: Water can experience hydrogen bonds, which are relatively strong for intermolecular attractions. However, water does not possess the ionic bonds, so its melting point is lower than that of CaCl2. Note also that water is clearly a liquid at room temperature, while calcium chloride is a solid! C: Neon is a noble gas and can only exert London forces, a very weak form of intermolecular attraction. D: CH3COCH3, or acetone, exerts dipole-dipole interactions. While these attractions are stronger than London forces, they are weaker than both hydrogen and ionic bonds.

If a signal is transmitted along a segment of nerve axon measuring 5 x 10-4 m, how much time is required for the signal to reach the end of the segment, assuming maximal transmission velocity? A. 2.5 x 10-6 ms B. 2.5 x 10-3 ms C. 1 x 10-1 ms D. 1 x 10-1 s

B is correct. The passage tells us that signals can be transmitted up to 200 m/s, which would be the maximal velocity, and the question tells us that the distance traveled is 5 x 10-4 m. Dividing distance by the speed gives us time: 5 x 10-4 m / 2 x 102 m/s = 2.5 x 10-6 s Converting to milliseconds, we get 2.5 x 10-3 ms.

If the tympanic membrane has three times the diameter of the round window (to amplify sound as it is transmitted through the ossicles) how would it impact the results shown in Figure 2 if the same experiment was performed on a mouse on whom the tympanic membrane had not been bypassed? A. The higher intensity would shift the values shown in Figure 2 to the left. B. The higher intensity would have no impact on the curve shown in Figure 2. C. The higher intensity would increase the wavelength of the sound waves, shifting the values in Figure 2 down and to the right. D. The higher intensity would shift the values shown in Figure 2 up.

B is correct. The question asks about how an increase in intensity would change the results shown in Figure 2. The values in Figure 2 were determined by changing the frequency of waves entering the cochlea and measuring the location of maximum amplitude along the basilar membrane. From paragraph 3, it is also clear that the frequency of resonance or maximum amplitude is based on the mass and stiffness of the particular portion of basilar membrane. So, changing the intensity of the incoming sound wave should have no impact on where the maximum amplitude occurs (i.e., what frequencies will vibrate certain locations) but will instead only increase the amplitude of vibration at those locations.

The auto-ionization of water is an endothermic process. With this in mind, if the temperature of a beaker of pure water is increased: A. the water will become acidic. B. the pH of the water will decrease. C. Kw will remain unchanged. D. the pOH of the water will increase.

B is correct. Think of auto-ionization in terms of equilibrium: heat + 2 H2O → H3O+ + OH-. The reaction will thus shift toward the product side in response to increased temperature. Interestingly, both [H+] and [OH-] will rise as a result, decreasing the pH and pOH to the same extent. A: Although pH decreases, which typically makes us think of an acidic solution, pOH also decreases. As long as proton and hydroxide concentrations are equal, the solution must be neutral. Remember, a pH of 7 only denotes a neutral solution when T = 25°C. C: Kw, like other equilibrium constants, is temperature-dependent. D: pOH decreases, not increases. The additional dissociation of water will yield more protons and more hydroxide ions.

If you wished to convert octanoic acid into a significantly more reactive molecule, you would most likely: A. expose it to PCl4. B. expose it to SOCl2. C. add it to aqueous solution. D. react it with isopropanol.

B is correct. Thionyl chloride (SOCl2) is a reagent commonly used to convert carboxylic acids into acyl halides. This particular reaction would produce octanoyl chloride, which (like other acyl halides) is highly reactive due to its halogen leaving group. A: PCl4 is not a stable molecule, a fact that can be discerned by attempting to draw its Lewis structure. PCl3 and PCl5, however, would accomplish the described goal. C: If anything, water would simply attack the carbonyl carbon of octanoic acid, producing the exact same molecule as the original species. D: Reaction of a carboxylic acid with an alcohol forms an ester. Esters are nearly identical to carboxylic acids in terms of reactivity.

In order to measure the ∆G, researchers needed to denature the proteins. Which of the following steps would be LEAST suitable for this procedure? A. Increasing temperature B. Irradiating the protein with non-ionizing radiation (does non-ionizing radiation break bonds?) C. Adding a concentrated chaotropic agent D. Lowering pH

B is correct. This answer choice specifically says that the protein will be irradiated with nonionizing radiation, which does not break bonds. Therefore, this method would not be able to denature the proteins. Irradiating the proteins with X-rays or gamma rays certainly can denature them, but these are forms of ionizing radiation. A: Heat will cause the protein to unfold and will eventually denature it. C: A chaotropic agent is any molecule in aqueous solution that can disrupt the hydrogen bonding network including the bonds between water molecules. As such, these agents (urea is one example) are capable of denaturing proteins by disrupting the hydrogen bonds that stabilize their configurations. D: Lowering the pH will protonate residues and potentially disrupt the structure of the protein, causing it to unfold.

A pharmaceutical researcher must choose a compound with the worst possible leaving group to prevent unwanted side reactions in a certain synthetic procedure. Of the following, he should use: A. propionic anhydride. B. methyl ethanoate. C. butanoyl chloride. D. acetyl iodide.

B is correct. This molecule's "-oate" suffix alerts us to its identity as an ester. The leaving group on an ester is an alkoxide ion, or RO-; this species is extremely unstable in solution. For this reason, esters are notably unreactive. A: This is an acid anhydride, which contains a very good, resonance-stabilized leaving group. C, D: Both of these molecules are acyl halides. Of the carboxylic acid derivatives, these contain the best leaving groups (halogen atoms). Note that within this subset, acyl fluorides possess the worst leaving groups (F-) while acyl iodides contain the best (I-).

The quantum number that denotes the subshell and the angular momentum of an electron is the: A. principal quantum number. B. azimuthal quantum number. C. magnetic quantum number. D. spin quantum number.

B is correct. This value tells us whether a particular electron is found in an s, p, d, or f subshell. It is also said to describe the shape of a particular orbital. A: The principal quantum number corresponds to an electron's shell and energy level. This number can be found by looking at the element's row in the periodic table. C: The magnetic quantum number narrows down the electron's location to a particular orbital within a subshell.

The researchers want to use narrow-spectrum LEDs to make their lamp more efficient. Assuming that the energy of a photon absorbed by porfirmer is transferred without loss to oxygen, what wavelength of light should the researchers select? (Note: Planck's constant is 6.626 x 10-34 J∙s) A. 1000 nm B. 1250 nm C. 2500 nm D. 3000 nm

B is correct. To answer this question, we need to work backwards by calculating the amount of energy required to excite one oxygen atom from the triplet state to the singlet state. That energy is given in the passage as 94.6 kJ/mol. We can divide that number by Avogadro's number to figure out the energy required for 1 atom to transition: (9.46 x 104 J/mol) / (6.02 x 1023 mol-1) = 1.57 x 10-19 J. To make that calculation doable without a calculator, use 9 instead of 9.46 and 6 instead of 6.02. By simplifying the fraction, you get that 9/6 = 3/2 = 1.5. Then you can deal with the exponents (4 - 23 = -19). Since the question says that no energy is lost in the transition from porfirmer to oxygen, we can assume that 1.5 x 10-19 J is the energy of the photon that we need. But how can we get the wavelength of the photon? Recall that E = hf relates the energy of the photon to its frequency, and v = fλ relates frequency, wavelength, and speed. The speed of the photon will be the speed of light (c = 3 x 108 m/s). Rearranging the second equation for frequency gives f = v/λ. Now we can substitute for f in the first equation, giving E = hv/λ. Rearranging this equation for wavelength gives λ = hv/E. λ = (6.6 x 10-34 J∙s)(3 x 108 m/s)/(1.5 x 10-19 J) λ = 13.2 x 10-7 m λ = 1.32 x 10-6 m λ = 1320 nm We should look for an answer that is slightly lower than that, since we estimated the energy of the photon to be lower than it actually is. Only choice B fits.

The proline in the PTS domain of mucin has a unique cyclical side chain and cannot form a standard α-helix. As a result, it is often used when mucin needs to make a tight turn in the polypeptide chain. This is likely to support the generation of which mucin structure? A. Quaternary B. Tertiary C. Secondary D. Primary

B is correct. Turning or bending of a polypeptide chain is the tertiary structure of a protein. A: Quaternary structure is the joining of multiple peptide chains, which was not mentioned by the question or the passage when discussing the PTS domain. C: Secondary structure is the formation of β-pleated sheets or α-helices, which the question states that proline cannot do. Thus, proline does not aid in the generation of secondary structure.

Towards the end of a game, a team's pitcher begins to lose his focus and shortens his stride, uncoupling the kinetic chain at his hips. How will this impact his pitch? A. It will make his shoulder more injury-prone as a result of the increased energy being dissipated during the follow-through from the inefficient kinetic chain. B. It will decrease the velocity of the pitch due to a decrease in energy transfer through the torso. C. It will decrease the velocity of the pitch as a result of the increased energy being dissipated during the follow-through from the inefficient kinetic chain. D. It will decrease the velocity of the pitch due to a decrease in the ability to generate force in the legs.

B is correct. Uncoupling the kinetic chain at the hips will decrease the efficiency of the energy transfer from the legs and hips to the lower torso and result in overall less energy transfer through the upper body to the upper extremities and ball. This will decrease the velocity of the pitch. A, C: The shoulder would be more injury-prone if there was a decrease in efficiency of the energy transfer from the upper body to the upper extremities, as the shoulder would need to dissipate more energy. However, the uncoupling described in the question stem occurs between the hips and lower torso, before reaching the shoulders. Thus, it would result only in an overall decrease in energy reaching the upper body. D: Uncoupling the kinetic chain will result in less energy transfer, not less force generation. The pitcher will still generate a large amount of force in his legs, but less of that energy will be transferred up the torso to the shoulders.

Experiments on the hydrolysis products benzoic acid and t-butyl alcohol revealed that both are colorless substances between 20°C and 30°C. Would paper chromatography without UV light visualization be an effective technique for researchers to use to separate these drug components? A. Yes, because they have different molecular masses and capillary action affects them in different ways. B. No, because there would not be any visible distinctiveness between the two migrating substances. C. It depends on whether the chromatography had a thin-layer component. D. It depends on the pH of the solvent.

B is correct. Without the option of UV visualization, paper chromatography will only be effective if there is a visible difference between the substances being separated. Two colorless substances such as benzoic acid and t-butyl alcohol would likely migrate at different speeds, but it would not be possible to identify their positions.

metalloids

B, Si, Ge, As, Sb, Te, Po, At

properties of alkanes - what will decrease their BP/densities?

BP/density/MP: increase with increasing length branching chains decrease BP/density

beta anomer compared to alpha anomer, and Fischer projection -in a Fischer, where is the beta?

Beta = same direction (both functional groups point upward/downward) alpha = functional groups point in different directions Fischer projection: Beta is on the left more stable than alpha anomer, will be favored 2:1

Place the following compounds in order of increasing boiling point: propane, propanal, propanol, butanol. A. Butanol, propanal, propane, propanol B. Propane, propanal, propanol, butanol C. Butanol, propanol, propanal, propane D. Propane, propanol, butanol, propanal

Boiling point correlates with the strength of intermolecular forces between molecules of a compound. The two alcohols in the list, butanol and propanol, can experience intermolecular hydrogen bonding. As such, they display greater attraction between particles and have correspondingly high boiling points. Between the two, butanol has a higher BP due to its larger molecular mass. Next, while its molecules cannot hydrogen bond, propanal has a strongly polarized carbonyl group that allows it to exert a dipole-dipole force. This gives it a higher boiling point than propane, a hydrocarbon. In conclusion, we have ordered our compounds from highest to lowest BP as butanol > propanol > propanal > propane, an order that we must reverse to answer the question.

When Bs1A is assembled as an octamer, what is most likely to be true regarding L76, L77, and L79? A. They are oriented toward the solvent-exposed exterior of the protein assembly. B. The assembly would incur an entropic penalty if they occupied a solvent-exposed site. C. Unlike in a monomer, they are not situated within the hydrophobic cap. D. Their physiochemical properites are not substantially dependent on their hydrophobicity.

BslA consists of one α helix and 13 β-strands that form two distinct β-sheet faces. A smaller, three-stranded β-sheet (β-strands CAP1, CAP2, and CAP3) straddles the two main β-sheet faces, forming an almost flat surface. This region, known as the hydrophobic cap, displays numerous solvent-exposed hydrophobic amino acids; however, when Bs1A monomers assemble as an octamer, these amino acids are shielded from polar solvent molecules by solvent-exposed β-sheet faces of other Bs1A molecules. Key terms: α helix , β-strands, β-sheet faces, CAP1, CAP2, CAP3, hydrophobic amino acids, hydrophobic cap, octamer Contrast: in monomeric form, the Bs1A hydrophobic cap contains unusual solvent-exposed hydrophobic amino acids; in the octamer, protein packing buries them under solvent-exposed β-sheet faces of other Bs1A molecules B is correct. Entropic penalty refers to the thermodynamically disfavored requirement of forming a cage of polar solvent molecules around surface-exposed hydrophobic portions of a molecule. In the case of Bs1A protein folding, an entropic penalty would be incurred if L76, L77, and L79, which in the octamer are sheltered from solvent molecules, were to become exposed to these molecules, requiring formation of an orderly solvation shell around that hydrophobic portion of the Bs1A cap by solvent molecules and thus incurring an entropic penalty. A: The passage indicates that in the octamer, L76, L77, and L79 are buried within the protein assembly, away from solvent molecules. C: Whether in a monomer or an octamer, the residues remain part of CAP1, one of the three strands comprising the hydrophobic cap. D: The term "physiochemical property" refers to the physical and chemical properties of the leucine residues, both of which are strongly influenced by the hydrophobic side chains.

what makes an acidic buffer?

Buffer solutions, which are intended to resist changes in pH, consist of a weak acid or weak base and its corresponding salt. The passage states that the buffer solution is acidic (pH = 3.8), so we can assume that a base is not present in the solution. Therefore, any Roman numeral (RN) that is a weak acid or a salt formed from a weak acid is a potential component of the buffer mentioned in the passage.

At pH = 7.3, what is the bond order of the shortest bond to oxygen in glycine? A. 0.5 B. 1 C. 1.5 D. 2

C - why?

p-nitrophenol has a pKa of 7.16, whereas ethanol has a pKa of 16. What best explains this difference? A. Anion stability achieved by the electron-donating effects of the NO2 substituent group B. Anion stability achieved by the electron-withdrawing effects of the NO2 substituent group C. Anion stability achieved by delocalized p-orbitals across the phenolic salt D. Anion stability achieved by the longer carbon backbone

C is correct. Acidity is determined by the stability of the conjugate base. The more stable the conjugate, the stronger the original acid. In p-nitrophenol, the benzene backbone as well as the nitrogen in the NO2 substituent are sp2 hybridized, creating a plane of delocalized p-orbitals capable of stabilizing the excess negative charge on the alkoxide.

Place the following elements in order of increasing atomic radius: F, Ne, Rb, Sr. A. Rb < Sr < F < Ne B. Sr < Rb < Ne < F C. Ne < F < Sr < Rb D. F < Ne < Rb < Sr

C is correct. Atomic radius increases as one moves down and to the left within the periodic table. Of the four elements listed, rubidium is the closest to the bottom left of the table, closely followed by strontium. In contrast, neon is closest to the upper right, implying that it is the smallest of the four.

Compared to its analogous acyl chloride, a carboxylic acid is considered: A. more electrophilic due to resonance. B. more electrophilic due to the inductive effect. C. less electrophilic due to resonance. D. less electrophilic due to the inductive effect.

C is correct. Both chlorine and oxygen are technically capable of electronic delocalization. However, the resonance structure in which the oxygen atom forms a double bond with the carbonyl carbon is more significant than the corresponding structure in which Cl forms the same double bond. Thus, although both atoms are very electronegative, the dominant resonance contribution is more significant for the oxygen species. The resulting increase in electron density around the carbonyl carbon confers decreased electrophilicity.

In general, chronic consumption of trans fats are considered to induce more negative cardiovascular outcomes in humans than chronic consumption of comparable cis fats. Which of the following statements provides a reasonable hypothesis as to why this might be the case? A. The hydrocarbon tails of trans fats experience less London dispersion forces as compared to those of comparable cis fats, making them more likely to be solids at body temperature and cause obstruction in blood vessels. B. Trans fats are more easily solubilized by bile acids in the small intestine than comparable cis fats, making them more likely to be absorbed by the circulatory system and cause obstruction in blood vessels. C. The hydrocarbon tails of trans fats experience less electrostatic repulsion when in close proximity to one another, as compared to those of comparable cis fats, making them more likely to be solids at body temperature and cause obstruction in blood vessels. D. Trans fats are more easily solubilized by chylomicrons and absorbed by the microvilli of the small intestine than comparable cis fats, making them more likely to be absorbed by the circulatory system and cause obstruction in blood vessels.

C is correct. Both the question and the answer choices are rather wordy, making process of elimination our best bet. Answer choice (A) seems to make sense on the surface; however, London dispersion forces rely (as far as the MCAT is concerned) far more on molecular mass than molecular structure. The difference in structure is not really significant here, so we'll table this one and search for something better. Answer choice (B) is way off; the solubilization of fats by bile acids is FAR more dependent on size than structure. Answer choice (C), however, makes more sense; trans fats "slot" together more effectively than cis fats would, reducing electrostatic repulsion overall. Answer choice (D), similarly to (B), is pretty far from the mark; chylomicrons, like bile acids, operate mostly on size and are completely structure-blind.

Which of these reagents would best serve to fully reduce all carbonyl groups present on this molecule? A. Sodium borohydride B. Sodium hydride C. Lithium aluminum deuteride D. Potassium cyanide

C is correct. Chemically, deuterium and hydrogen behave identically. Thus, lithium aluminum deuteride may be considered to be equivalent to lithium aluminum hydride (LAH). LAH is a strong reducing agent and is certainly capable of reducing both carboxylic acids and aldehydes. It would therefore be the most appropriate choice. A: Sodium borohydride would be perfectly adequate for selective reduction of the aldehyde, but it lacks the reducing power necessary to reduce the carboxylic acid. B: Sodium hydride is a strong base, but is not a hydride source and thus is not a reducing agent. D: Potassium cyanide is a nucleophile and a weak base. It is not an appropriate choice for any reduction reactions.

Which types of intramolecular bonds are found in an ethanol molecule? I. Nonpolar covalent II. Polar covalent III. Coordinate covalent **what is a coordinate covalent bond? A. I only B. II only C. I and II only D. I, II, and III

C is correct. Covalent bonds are formed when the two atoms involved share electrons. In a nonpolar covalent bond, these electrons are distributed equally, requiring the two atoms to have similar electronegativities. Here, such a bond certainly exists between ethanol's two carbons; additionally, its C-H bonds can be considered nonpolar. In contrast, a polar covalent attachment requires that electrons are shared in an unequal fashion. These bonds connect C-O and O-H. III: To form a coordinate covalent bond, one atom must donate both electrons. While common in Lewis acid-base reactions, this is not seen in the case of ethanol.

A student performs an experiment and determines that the aqueous decomposition of chloromethyl methyl ether at 650°C is first order with a half-life of 118 seconds. Image If the student began the experiment with 5 moles of chloromethyl methyl ether, how much remained after 354 seconds? A. 125 mg B. 625 mg C. 5 x 104 mg D. 1 x 105 mg

C is correct. Don't let all of the extra information (i.e. the reaction schematic) distract you! This question provides a half-life and then requires us to find out how much of the original molecule is left after (354/118) = 3 half-lives. The molar mass of chloromethyl methyl ether (C2H5ClO) is 80.51 g, which we can round to 80 g. A compound loses 50% of its current amount after each half-life, so 3 half-lives means (½)3 = 1/8 will be left. 5 moles of chloromethyl methyl ether x 80 g/mol = 400 g. After 3 half-lives, 400 g (1/8) = 50 g = 50 g • 103 mg/g = 5 x 104 mg.

A monobactam structure is characterized by a β-lactam ring that is not fused to a neighboring cyclic group. The general structure of a monobactam is shown here. Which of the following statements is true regarding the C-N bonds in the lactam ring? A. The sigma bond allows for free rotation. B. The electronegative characteristics of the nitrogen atom allow for significant charge dispersal. C. A substantial amount of ring strain is caused by the hybridization of the C-N bond. D. The ring is very stable due to its amide functionality.

C is correct. Due to the double bond character of the amide group, no rotation can exist around the ring's sigma bonds to facilitate the formation of the most stable potential structure. In addition, four-membered rings are generally subject to heavy strain.

The molecular geometry of carbonate ion is: A. trigonal pyramidal. B. square planar. C. trigonal planar. D. tetrahedral.

C is correct. First, draw the Lewis structure of the carbonate (CO32-) anion. From this, we see that the central carbon forms one C=O and two C-O bonds with surrounding oxygen atoms. Since there are three substituents and zero lone pairs, the geometry of this molecule is trigonal planar. A: This is similar to a tetrahedral geometry, but includes three substituents and one lone pair instead of four substituents. In any case, the carbon relevant to the question is only bound to three groups. B: A square planar geometry is formed when a central atom is connected to four substituents (forming a square shape), as well as two lone pairs. As stated above, the carbon atom in carbonate does not contain lone pairs at all. D: For this carbon to be tetrahedral, it would need to be bound to four substituents.

O2 (g) + H+ → H2O (l) **considering this reaction During the reduction of a mole of oxygen to water shown in Figure 1, how much charge is transferred? A. 1 x 105 C B. 2 x 105 C C. 4 x 105 C D. 6 x 108 C

C is correct. First, we need to balance the reaction: O2 (g) + 4 H+ → 2 H2O (l) This reaction is now balanced in terms of atoms, but as a redox reaction, it must also be balanced with regard to charge. We currently have a +4 charge on the left and a 0 charge on the right. To balance, we must add 4 electrons to the left side of the reaction. O2 (g) + 4 H+ + 4 e- → 2 H2O (l) Now, we can see that 4 moles of electrons are transferred per mole of O2. Faraday's constant tells us that approximately 105 coulombs are present per mole of electrons, so: 4 mol e- x 105 C/mol e- = 4 x 105 C

Given the following elementary reaction, what is the rate law? 3A + B → 2C A. Rate = k[3A][B][2C] B. Rate = k[A][B] C. Rate = k[A]3[B] D. This answer cannot be determined without additional information.

C is correct. For elementary reactions, stoichiometric coefficients can be used to write the rate law. These coefficients become exponents according to the following theoretical example: for the reaction aA + bB → cC, rate = k[A]a[B]b. D: This choice would be correct if we had not been told that the reaction was an elementary, or single-step, process.

Formation of a hemiacetal from a monosaccharide results in a molecule that has the same molecular weight as the original monosaccharide. Formation of an acetal from two monosaccharides results in a molecule that has: A. a molecular weight that can vary, depending upon where the acetal linkage forms. B. a molecular weight that is the same as the combined weight of the two monosaccharides. C. a molecular weight that is 18 amu less than the combined weight of the two monosaccharides. D. a molecular weight that is 18 amu more than the combined weight of the two monosaccharides.

C is correct. Formation of a acetal disaccharide requires loss of a molecule of water, so the molecular weight of a disaccharide will be 18 amu (1 O, 2 H) less than the combined mass of the two monosaccharide components. Opening a hemiacetal ring does not require any addition of water.

Tissue engineers have been developing a synthetic mineral apatite to use in prosthetics that is comprised of repeating molybdenum hexacarbonyl [Mo(CO)6] subunits. This bonding is likely to involve which of the following orbitals? A. sp3 B. sp3d2 C. d2sp3 D. dsp3

C is correct. Here, the central molybdenum atom is bound to six substituents. In such a configuration, the molecule takes the shape of an octahedron. The number of substituents, or bonding regions, around the atom is equal to the number of orbitals that hybridize: here, there are two 4d, one 5s, and three 5p atomic orbitals, for an overall hybridization of d2sp3.

A researcher carries out a column chromatography at physiological pH, using a stationary medium with a net positive charge. If a solution containing the following oligopeptides is poured into the column, which oligopeptide will most likely be found in the first fraction collected? A. DDGE B. EILD C. KRVV D. VEGP

C is correct. If the stationary phase has a net positive charge, then oligopeptides with negative charges will be attracted to the stationary phase and will move more slowly through the column. KRVV is lysine-arginine-valine-valine which would have a charge of + 2. The positive charge would cause it to not bind the stationary phase and allow it to elute quickly. A, B, D: These answers all contain aspartate (D) or glutamate (E) residues, which would be negatively charged at physiological pH. Thus, choice C would likely move the quickest through the column and be found in the first fraction collected.

An experimenter proposed employing a material with an index of refraction strongly dependent upon the wavelength of transmitted light for use in an objective lens. How would the use of such a material impact the performance of a microscope? A. Less incident light would be transmitted through the objective. B. Monochromatic light passing through the objective will not converge to a single point. C. Images formed by the eyepiece from transmitted light containing multiple wavelengths will be unfocused. D. The angle of reflection of incident light from the objective lens will increase.

C is correct. If transmitted light rays refracted differently depending on their wavelengths, then the effective focal length of the lens would differ for each wavelength of light. As a result, for any light containing rays of multiple wavelengths, the rays would focus at multiple points, resulting in an unfocused image. This is known as a chromatic aberration, in which optical instrument fails to converge light rays from a source to a single point.

Nickel-cadmium (NiCd) batteries are rechargeable cells used in products ranging from remote-control toys to airplanes. An NiCd battery in the process of discharging is most analogous to:

C is correct. In essence, rechargeable batteries work via reversible redox processes. When such a cell is discharging, it is releasing stored energy; this is similar to the spontaneous process seen in galvanic cells. A: A half-cell includes only one "half" of a redox reaction - either reduction or oxidation. As no redox process can occur without two half-reactions, this choice does not make sense. B: To recharge an NiCd battery, we would need to use an external power source to move electrons in a way that would be naturally nonspontaneous. This mimics the process undergone in an electrolytic cell. D: A concentration cell is a type of galvanic cell. As such, it would never be nonspontaneous.

A student sought to remove the H2O and CO2 gas emitted by the animals in separate chambers instead of using silica gel and soda lime. 224 mL of each gas was collected at a pressure of 1 atm. What will be the expected ratio of the mass of CO2 to that of H2O if a valve between the two chambers was opened, allowing the gasses to freely mingle with one another? (Note: the temperatures of all chambers were held constant.) A. 0.5:1 B. 1:1 C. 2:1 D. 3:1

C is correct. In this question, as the two chambers become connected, the pressure and volume for each gas must change while the temperature is held constant. For both gases, the new volume is the total volume of both chambers. Given that the volumes of the two gases are equal (224 mL each) and that they have the same initial pressure, they will have the same partial pressure in the new chamber. To check for yourself, you can use the following equations to determine the partial pressures: For CO2: PiVi = PfVf: (1 atm)(224 mL) = Pf (448 mL) Pf = 0.5 atm For H2O: PiVi = PfVf: (1 atm)(224 mL) = Pf (448 mL) Pf = 0.5 atm Next, use the partial pressure formula to find moles of each gas: Ptotal = (mol CO2)/(total mol) x Ptotal Ptotal = (mol H2O)/(total mol) x Ptotal Since both gases have the same partial pressure, and since the total number of moles and total pressure are equal for both gases, both gases must have the same number of moles. Therefore, the ratio of the masses of the two gases (mass of each gas = number of moles x molar mass) is the ratio of their molar masses. (Molar mass of CO2) / (molar mass of H2O) = 44 g / 18 g ~ 2.

All of the following are experiments that could test the predictions of the energy and entropy budget in the passage EXCEPT: A. measuring solar flux using a satellite. B. mapping the earth's emitted radiation using a satellite. C. measuring heat transfer due to ocean currents. D. mapping the temperature change between the ocean and the air above it.

C is correct. Ocean currents move heat within the ocean system via convection. The model done by the researchers does not take into account heat movements within a system. A, B: These experiments could be used to verify that the calculated solar radiation is close to the empiric value. D: This experiment could help verify if the energy flux of the ocean system and the calculated entropy of the air sea interface is correct.

What is the most likely formula for the ingredient shown on the label in Figure 1, called dibasic calcium phosphate? A. CaPO4 B. Ca3(PO4)2 C. CaHPO4 D. Ca(H2PO4)2

C is correct. Phosphoric acid is H3PO4, and all three of its hydrogen atoms are acidic. As a result, there are three associated anions (PO43-, HPO42-, and H2PO4-), which are typically named phosphate, hydrogen phosphate, and dihydrogen phosphate, respectively. Alternatively, we can refer to these anions in terms of their ability to act as Brønsted-Lowry bases. For example, PO43- has the ability to gain three protons, so it can be called tribasic phosphate. Calcium ion is Ca2+, and when combined with the dibasic anion, HPO42-, the resulting formula is that in choice C. A: Since calcium ion has a +2 charge while phosphate carries a -3 charge, this formula does not make sense. B: PO43- is the tribasic, not the dibasic, ion. D: The anion in this choice is the monobasic ion.

How much does the rate of effusion of dichloromethane at 200 K change when the temperature is changed to 800 K? Assume all other conditions are identical. A. It increases by 50%. B. It decreases by 50%. C. It increases by 100%. D. It increases by 200%.

C is correct. Rate of effusion depends on root-mean-square speed, which is calculated using the formula vrms = √(3RT)/M. Here, T is temperature and M is molar mass. Since the identity of the gas is constant, the only factor changing is T. Therefore, for two samples of the same gas with identical pressures but different temperatures, the following relationship holds: rate1/rate2 = √(T1/T2) rate2/rate1 = √(800/200) = 2 This means that the new rate is twice as large, or 100% larger than the original rate. `

In an isolated system, entropy is maximized when: I. the system is at equilibrium. II. the system is far from equilibrium. III. the system is unable to perform work. A. I only B. II only C. I and III only D. II and III only

C is correct. Recall that at equilibrium, ∆G = 0. With no free energy change, the system is unable to perform work (eliminate answer choices A and B). At equilibrium, there are no energy gradients within the isolated system, so energy is maximally dispersed, resulting in maximal entropy (eliminate choice D). II: A system that is far from equilibrium contains large energy gradients which are able to perform work. The formation of large energy gradients requires localization or concentration of energy, which results in decreased entropy.

An ecologist is making several measurements for a sample of methane gas produced by a herd of cattle. The sample contains exactly 1 mole of gas, but when the ecologist calculates its PV/RT ratio, he finds that it is 0.92. Without knowing the conditions under which the gas is held, what is the most likely explanation for these findings? A. Such a ratio is perfectly normal and does not require explanation. B. The relatively high pressure within the flask is causing particle volume to exert a substantial effect. C. The relatively low temperature within the flask is causing intermolecular forces to exert a substantial effect. D. The sample may be held at a high temperature, causing a lower PV/RT ratio than expected.

C is correct. Since the sample contains 1 mole of gas, we can rearrange the ideal gas law to find our expected PV/RT ratio: 1. The given value is lower, implying that either pressure or volume is smaller than expected. For the ideal gas law to work properly, two main assumptions are necessary: that the volume of particles themselves is negligible and that no intermolecular forces are present between particles. However, these assumptions become especially untrue at low temperature and high pressure. Given this information, C is the only sensible answer. A: According to the ideal gas law, we would expect a ratio of 1, making 0.92 somewhat abnormal. B: While high pressure does promote deviation from ideal behavior, the effect of particle volume would cause total volume to be larger than expected. This would produce a PV/RT ratio that is larger, not smaller, than 1. D: Do not fall for the trap of thinking that increasing T will naturally produce a PV/RT ratio that is less than 1! While this seems to make mathematical sense, PV/RT ratio only deviates from 1 when the gas in question is behaving in a non-ideal fashion. Gases deviate from ideal activity at low temperature and high pressure. This choice mentions high temperature instead.

The azimuthal quantum number corresponds with which of the following? A. The potential energy of the electron B. Approximate radial size of an electron cloud C. Approximate geometric shape of the orbital D. Number of valence electrons orbiting a nucleus

C is correct. The azimuthal quantum number for an atomic orbital determines its orbital angular momentum and describes the shape of the orbital. A, B: This value is quantified by the principal quantum number, n. D: This value is not quantified by any quantum number. The number of valence electrons orbiting a nucleus can only be obtained by examining the complete spectroscopic notation of the electron or the location of the atom on the periodic table.

The following table gives standard reduction potentials for seven half-reactions. Which of the following statements is true? A. A cell involving copper and iron can only have an Ecell value of 0.43 or -0.43. B. Al (s) is a better oxidizing agent than Cd (s). C. Cu (s) is a better reducing agent than Fe2+. D. In a cell containing sodium and aluminum, sodium will always oxidize and aluminum will always reduce.

C is correct. The better reducing agent is always the species that is more likely to oxidize. Here, we are given reduction potentials, showing that Fe3+ reduces more spontaneously than Cu2+. To assess oxidation potential, however, we must switch the reactions. We then see that Fe2+ has an oxidation potential of -0.77, while Cu (s) has one of -0.34; solid copper is thus a better reducing agent. A: A copper/iron cell can only have an E°cell value of 0.43 or -0.43. However, a multitude of other values are possible if the cell is constructed under non-standard conditions. B: Oxidizing agents must be able to reduce themselves. Neither Al (s) nor Cd (s) is able to reduce, or gain electrons to become more negative overall. D: This is true for a galvanic cell, but the opposite would occur in an electrolytic (nonspontaneous) cell.

If the majority of the baseball's kinetic energy comes from power generation in the legs and hips, approximately how much energy do the lower extremities produce in the pitch? A. 65 J B. 70 J C. 140 J D. 810 J

C is correct. The kinetic energy of the baseball can be calculated using the equation 1/2 mv2. Paragraph 2 tells us that the average velocity of the ball is 30 m/s and its mass is 150 g, which is equivalent to 0.15 kg. KE = (½)(0.15 kg)(30 m/s)2 KE = (½)(0.15)(900) KE = (0.15)(450) = 67.5 J The ball ends up with 67.5 J of KE, but the question asked for how much energy the lower extremities generated (eliminate choice A). Table 1 shows us the overall efficiency of each energy transfer. Using these, we can calculate how much energy was needed to end up with 67.5 J of KE in the ball. The overall efficiency of the kinetic chain can be calculated by multiplying the efficiencies of each step, and we can round the numbers to make our calculations easier. Efficiency = (0.9)(0.9)(0.7)(0.8) = 9*9*7*8 x 10-4 = (81)(56) x 10-4 ≈ (80)(60) x 10-4 = 4800 x 10-4 = 0.48 ≈ 0.5 Thus, the total energy generated in the lower extremities in order to ensure that 67.5 J makes it to the ball is 67.5 J / 0.5 = 135 J (eliminate choice D). If you are uncomfortable choosing a number that is not exactly what was calculated, you only need to remember that we estimated by rounding up to 0.5. Thus, our denominator in the final calculation, 67.5/0.5, is a bit larger than it should be, making our final value smaller than it would be had we not rounded or used a calculator (exact value of 67.5/ 0.48 = 140 J). Thus, we round up and choose choice C.

The following diagram depicts the general mechanism of an aldol condensation. **which molecule is the nucleophile in aldol condensation? Molecules or ions that act as nucleophiles in this reaction include: A. the carbonyl carbon of the aldehyde. B. the carbonyl carbon of the ketone. C. the enol. D. the acid catalyst.

C is correct. The mechanism shows the enol tautomer of the original aldehyde (a comparatively electron-rich species) attacking the carbonyl carbon of the second aldehyde molecule. As carbonyl carbons are well-known electrophiles due to their partial positive charges, the enol must be our nucleophile.

The actual size of a sample object appearing 5 mm long when viewed under a compound microscope where the objective focal length is 3 mm, the eyepiece focal length is 25 mm, and the tube distance is 150 mm, is most nearly: A. 1 µm. B. 5 µm. C. 10 µm. D. 50 µm.

C is correct. The passage states that MA for a compound microscope equals MoMe. Substituting equations provided in the passage for each variable, along with the values from the question stem, we find that: MA = (d / fo)(25 / fe) = (150 mm / 3 mm)(25 cm / 2.5 cm) = (50)(10) = 500 Therefore, the 5-mm object has been magnified by a factor of 500. The original size of the object must be: (5 x 10-3) / (5 x 102) = 1 x 10-5 = 10 x 10-6 = 10 µm A, B, D: These answer choices result from miscalculation.

Besides directly killing bacteria, how does pasteurization with hot water treat IPB populations? A. By decreasing the solubility of ferrous (II) hydroxide, preventing further bacterial growth B. By decreasing the solubility of ferric (III) oxide, preventing further bacterial growth C. By decreasing the solubility of oxygen, preventing further bacterial growth D. By increasing the solubility of oxygen, preventing further bacterial growth

C is correct. This question asks us to think about other ways in which pasteurization might be an effective tool against IPB besides direct bacterial death. The passage states that IPB oxidize iron compounds and require a certain concentration of oxygen to survive. Oxygen, like all gases, is more soluble in cold water than warm water. Thus, the process of heating water (pasteurization) would lower the solubility of oxygen and reduce the amount available to IPB.

An organic synthesist seeks to identify an efficient stereoselective reagent with which to produce the enantiomer of the biologically active sphingosine molecule pictured in the passage. Which of the following would be the most logical choice to try? A. Tert-butoxide B. DMSO C. Diethyl tartrate D. Propanol

C is correct. This question relies upon prior knowledge that you should master prior to taking the MCAT; the only stereoselective reagent listed is diethyl tartrate.

Which of the following will contribute to the level of FDG signal seen during a PET scan of the colon? I. Blood flow to the tissue II. Availability of oxygen to the tissue III. GLUT4 transport A. I only B. II and III only C. I and II only D. I, II, and III

C is correct. This question requires us to think about whether the three listed factors would affect FDG's location in the scanned area. Statement I is logical, since if there is less blood flow to an area, less FDG will be present in that tissue. Similarly, when less oxygen is available to a tissue, less oxidative respiration occurs, resulting in a smaller need for FDG. This makes statement II correct, as this tissue would have lower amounts of FDG per unit time and would show less activity.

What is the value of the Keq for the following reaction under standard biological conditions? Pyruvate- + NADH + H+ → NAD+ + Lactate- A. 1 x 10-7 B. 5 x 10-5 C. 2 x 104 D. 1 x 107

C is correct. We start by finding Eº for this reaction. The oxidation of NADH generates +0.32 V and the reduction of pyruvate generates -0.19 V. Thus the reaction listed generates +0.13 V. We can use Equation 2 to determine the equilibrium constant, since the E will be zero and Q = Keq. The number of electrons transferred in the overall REDOX reaction is n = 2. E = E° - 0.060/n log Q 0 = 0.13 - 0.06/2 log K 0.13/0.03 = log K 4.2 = log K 104.2 = K The value of the equilibrium constant is a little larger than 1 x 104 and choice C is the best answer.

Suppose that a moderately hard surface such as the hard-packed earth described in the passage is found to decelerate a person making a feet-first landing to a stop in an average of 0.09 seconds. What is the person's deceleration in terms of g (where 1 g = 9.81 m/s2) when falling from a height of two meters? A. 1.00 g B. 2.27 g C. 7.10 g D. 22.22 g

C is correct. Whew, this is a fairly heavy question! It might be one to skip over, mark, and save for later. Note that the numerical answers are fairly far apart, so it's perfectly fine to approximate throughout. Let's start by figuring out exactly what the question is asking. A person is falling from a 2-meter height, then hitting the ground. From the time of impact (where he hits at some unknown velocity) to the end of the collision (where v = 0 m/s), it takes 0.09 s. To find the magnitude of this acceleration, we need to know the velocity at impact. For this, we can use vf2 = vi2 + 2ad. Initial velocity is 0 m/s, distance is 2 m, and acceleration is g, or 9.81 m/s2. Again, since answer choices are far apart, you'll likely experience no problems if you estimate g as 10 m/s2. vf2 = (0 m/s)2 + 2(10 m/s2)(2 m) vf2 = 40 m2/s2 We know that 62 = 36 and 72 = 49, so √40 must fall between 6 and 7. Let's call it 6.4. Now, acceleration = (final velocity - initial velocity) / time. a = (0 m/s - 6.4 m/s) / 0.09 s We get a negative value in the numerator, but all of our answer choices are positive, and the question asks for the person's deceleration. Since the negative sign simply denotes that he is decelerating, we can ignore it going forward. We can also round 0.09 s to 0.1 s. a = 6.4 m/s / 0.1 s a = 64 m/s2 Finally, note that our answer must be written in terms of g. Estimating g as 10 m/s2, 64 m/s2 is equal to 6.4g. However, the value of g is actually lower, so our real answer should be slightly higher than 6.4g. Choice C, 7.10g, seems perfect. A, B, D: These answer choices result from miscalculation.

IR C-H C=C-H C-triple-C-H C=O c=c aromatic?

C-H, 2900 C=C(1650)-H (3100) C-triple-C(2100)-H (3300) C=C (aromatic) = 1600-1450 C=O (1750) a technique that measures molecular vibrations at different frequencies, from which specific bonds can be determined; functional groups can be inferred based on this information

product of all decarboxylations

CO2

Grignard reaction

Carbon-Mg (carbon is nucleophile) and will attack acetone (post charged carbon) In the first step, (acidic or basic) the c next to the Mg will add to the double bond and the group attached to the Mg will be added, then in the second step, water or acid is added to protonate O- and create alcohol with epoxides = secondary alcohol with aldehydes = tertiary(but 2 original groups will remain and R group connected to MgBr will be the third group) with esters = tertiary (but 2 new Rs will add, and one will be the R group from before the ester)

difference between Ecell and E°cell?

Cell isn't standard and cannot assume standard temp, pressure, concentrations, so data is inconclusive

Reaction of 4-methyl-1-pentanol with chromic acid should produce: **what is the strength of chromic acid? A. 4-methyl-1-pentanone. B. 4-methylpentanoic acid. C. 2-methylpentane. D. 4-methyl-1,2-pentanediol.

Chromic acid (H2CrO4), like many chromium-containing reagents, is a strong oxidizing agent. Reaction of a primary alcohol with such a compound will oxidize it as thoroughly as possible. Since a primary alcohol has only one bond to carbon, it possesses the ability to form three bonds to oxygen, creating a carboxylic acid.

Which of the following is NOT a function of the blood-brain barrier? A. Protection of the brain from bodily hormones B. Maintenance of a stable chemical equilibrium for the brain C. Protection of the brain from carbon dioxide poisoning D. Allowing of more glucose to enter the brain than any other tissue

Co2 duh

As increasing amounts of NaCl are added to water: A. boiling point will decrease, melting point will increase, and vapor pressure will decrease. B. boiling point will increase, melting point will decrease, and vapor pressure will increase. C. boiling point will increase, melting point will decrease, and vapor pressure will decrease. D. boiling point will increase, melting point will increase, and vapor pressure will decrease.

Colligative properties include boiling point elevation, freezing (melting) point depression, and vapor pressure depression. As more solute is added, these properties will be seen with an increasing effect.

Why might a critical residue with a high SASA be more likely to be a false negative by protein rigidity analysis than a critical residue with a low SASA? A. A high SASA indicates the residue is located in the protein interior, and therefore is more likely to be critical for overall protein structure. B. A high SASA indicates the residue is located in the protein interior, and therefore is less likely to be critical for overall protein structure. C. A high SASA indicates the residue is located at the surface of the protein, and therefore is more likely to be critical for overall protein structure. D. A high SASA indicates the residue is located at the surface of the protein, and therefore is less likely to be critical for overall protein structure.

D is correct. A high SASA means that the residue in question has a large surface area that is exposed to the solvent, which implies it is likely at the surface of the protein. Surface residues are less likely to be "keystones" for the protein structure, as those sorts of residues tend to be found in the protein's core. A helpful analogy is a modern skyscraper: the outside wall is often made of glass and steel, with a concrete core that serves as the main weight-bearing material. A surface residue that isn't critical for structure may still be critical for the function of the protein and may be important for binding a substrate. A, B: A high SASA means that the residue is easily accessible to the solvent, which means that it is unlikely to be buried deep within the protein core. C: A high SASA does imply that the residue lies on the surface of the protein; however, that positioning means that it is less likely to be structurally important for the protein.

If a leukemia patient has a hematocrit of 60%, with all other factors held constant, what will be the effect on the volume flow rate of her blood? A. It will increase by a factor of 1.5. B. It will increase by a factor of 2. C. It will decrease by a factor of 1.5. D. It will decrease by a factor of 2.

D is correct. At a hematocrit of 60%, the viscosity of blood is about .008 kg∙cm-1s-1. This is twice as much as normal blood. Therefore, using the following equation: we see that the volume flow rate of her blood will decrease by a factor of 2. A, B: These answer choices incorrectly state that increased viscosity will increase the flow rate. C: This answer choice results from miscalculation. Give feedback on this question

Which of the following is most likely to be overexpressed in cancer cells? A. Bak B. Bax C. BH-R D. Bcl2

D is correct. Cancerous cells are cells that are undergoing uncontrolled cell growth. This implies cancer cells adopt mechanisms to bypass apoptotic activation to ensure their survival. This is often executed by overexpression of factors that suppress the functions of apoptosis inducers. Only Bcl2 is an apoptosis inhibitor. A, B: These are apoptosis inducers, meaning their overexpression would lead to increased cell death, not immortality. C: This is described by the passage as the protein substrate for the regulatory Bcl proteins, and is not likely to be significantly over-expressed in cancer.

According to Charles' law, all of the following statements are correct EXCEPT: A. halving the temperature of a gas should cause its volume to double, if it is being held in a flexible container. B. if the volume of one mole of gas is tripled, its temperature must also triple if its pressure is to remain constant. C. quadrupling the volume of a gas will cause its pressure to drop to a quarter of its previous value. D. both A and C.

D is correct. Charles' law states that the volume of a gas is directly proportional to its temperature, assuming that other factors are held constant. Here, we are looking for statements that are either factually incorrect or do not relate to this law. Statement A is untrue, as halving the temperature of a gas should cause its volume to be halved, not doubled. Statement C is true, but it relates to Boyle's, not Charles', law. B: This statement is both accurate and relevant to Charles' law. Relationship between temperature and volume is DIRECT. If temp increases, volume increases too! Pressure is the opposite.

What is a possible set of quantum numbers for one of copper's d-block electrons? A. 2, 0, 0, +1/2 B. 3, 1, 1, -1/2 C. 4, 2, 2, +1/2 D. 3, 2, -1, -1/2

D is correct. Copper is located in the 3d block, giving it a principal quantum number of 3. Since the question stem refers specifically to the d-block electrons, the relevant azimuthal quantum number must be 2. Potential magnetic quantum numbers range from -2 to 2, while the spin number of any electron can only be +1/2 or -1/2.

A magnetic field is measured at point A in Figure 1. Which of the following is true? A. The induced magnetic field points out of the page. B. The induced magnetic field points down the page. C. The induced magnetic field points up the page. D. The induced magnetic field points into the page.

D is correct. Figure 1 shows the microbe-containing electrode in chamber 1, on the left. In paragraph 2, this electrode is described as the source of electrons, which then travel along the wire to chamber 2. In other words, electrons are traveling from left to right, or clockwise. However, remember that current and electrons always flow in opposite directions! Thus, current is moving counterclockwise. Using the right-hand-rule, point your right thumb in the direction of the moving current (to the left), and curl your fingers around the wire. Your fingers should point into the page at point A. The flow of current and the direction of the induced magnetic field are shown below:

The structure of malonic acid, an organic acid with a variety of uses in manufacturing, is shown below. If one mole of sucrose is able to depress the freezing point of a certain volume of water by 3.9°C, what would be the freezing point of the same volume of water when one mole of malonic acid is added? A. 0°C B. -3.9°C C. -7.8°C D. -11.7°C

D is correct. Freezing point depression is a colligative property, meaning that it depends on the number of particles in solution. Sucrose is organic and does not dissociate in water, but carboxylic acids will lose a proton at a pH of 7. Malonic acid, then, dissociates into three particles: its deprotonated form and two free protons. Thus, one mole will depress this volume by three times the amount that sucrose affected it, or 11.7°C. Water typically freezes at 0°C, making our new freezing point -11.7°C. A: This is the freezing point of pure water. B: A freezing point like this would result if malonic acid did not dissociate at all. C: This value would result if malonic acid split into two particles, not three. Give feedback on this question

In the condensation mechanism shown, what is the most likely role of the diethyl succinate in the reaction? A. Enol source B. Electrophile source C. Strong base D. Enolate source

D is correct. In the Stobbe condensation example, diethyl succinate is condensed with benzophenone using tert-butoxide (a strong base) followed by aqueous acid. The diethyl succinate (the molecule on the right in step 1) is the enolate source. This enolate will act as a nucleophile to attack the electrophilic benzophenone. You should remember for test day that enols are good nucleophiles, but their conjugate bases - enolates - are even better.

Which of these molecules is a possible product of acid-catalyzed amide hydrolysis? A. Heptanal B. 2-methylhexanone C. Benzenamine D. Pentanoic acid

D is correct. In the acid-catalyzed hydrolysis of an amide, a proton first attaches to the carbonyl oxygen to add to the electrophilicity of that molecule. Next, water acts as a nucleophile to attack the carbonyl carbon. This allows the amine to function as a leaving group and forms a carboxylic acid. Note that such a product also forms during other hydrolytic reactions with acid derivatives, like esters. A, B: Neither aldehydes nor ketones are produced during the mentioned reaction. C: This molecule is an enamine, which is also not a typical product of acid-catalyzed amide hydrolysis. In fact, it does not even contain any oxygen molecules, which must be present to answer this question.

Electrolysis can be performed on a molten salt, in a process analogous to the electrolysis of water described in the passage. The products in the case of the electrolysis of molten sodium chloride are sodium metal and chlorine gas. Which species described in the passage performs the same role as the sodium metal in the electrolysis of molten sodium chloride? A. Na B. Na+ C. O2 D. H2

D is correct. In this question, we must match up the process described in the passage with a new redox reaction introduced in the question stem. First, let's assess how sodium metal fits into the described process. The relevant equation is 2 NaCl → 2 Na (s) + Cl2 (g). From this, we can see that solid Na is the product of the reduction of Na+. Now, which of the species from the passage is the product of a reduction reaction? Figure 3 shows 2 aqueous H+ ions reducing to H2. Thus, H2 is analogous to Na and is our answer.

17.5 g of ammonium hydroxide (NH4OH) is mixed with 250 mL of an unknown solvent with a density of 1.5 g/mL. The molarity of this ammonium hydroxide solution is: A. 1.3 × 10-3 M. B. 2.0 × 10-3 M. C. 1.3 M. D. 2.0 M.

D is correct. Molarity is measured in moles of solute per liter of solution, so we first must find the moles of ammonium hydroxide present. 17.5 g NH4OH × (1 mol NH4OH / 35 g) = 0.5 mol NH4OH. Now, we can easily find the solution's molarity by taking 0.5 mol NH4OH / 0.25 L solution, or 2.0 M. Note that we can ignore the density of the solvent, as it is irrelevant to these calculations.

With regard to leaving group alone, which of these species is the most unstable? A. Acetyl fluoride B. Acetyl chloride C. Acetyl bromide D. Acetyl iodide

D is correct. Of the four halogens included in these molecules, iodine serves as the best leaving group. To understand this concept, consider the periodic table. As a member of a much lower period than (for example) fluorine, iodine is a very large atom. As such, it can easily delocalize the negative charge gained when it exits as a leaving group. The better the leaving group, the more reactive the compound, and a more reactive molecule is by definition less stable.

Which of the following is most likely to undergo positive beta decay? **when does positron emission occur? A. 14C B. 13C C. 17O D. 22Na

D is correct. Positive beta decay, also known as positron emission, occurs when the proton-to-neutron ratio is too high. Thus, we are looking for the answer choice with the highest proton-to-neutron ratio. Of the options given, 22Na has the highest ratio, as it has 11 protons and 11 neutrons (1:1). It is still not highly likely to undergo positive beta decay, but it is certainly more likely to do so than the other three options.

What is a possible explanation for the high stability of acid anhydrides in comparison to esters? A. The leaving group on an ester is resonance-stabilized, while that of an anhydride is destabilized by a negative charge. B. The leaving group on an acid anhydride is resonance-stabilized, while that of an ester must carry a full negative charge on a single oxygen atom. C. The alkyl substituent of an alkoxide ion is unusually electron-withdrawing, giving esters especially good leaving groups. D. None of the above; acid anhydrides are actually less stable than esters

D is correct. Remember, stability is the opposite of reactivity. Thus, this question is asking why anhydrides are less reactive than esters, which is simply not true. In fact, after acyl halides, anhydrides are the most reactive of the carboxylic acid derivatives. This trend relates to the anhydride's leaving group, which is a carboxylate ion. This species displays resonance between the two oxygen atoms. Since this makes the group stable in solution, it can easily leave and the anhydride can participate in a variety of reactions. A: Esters do not have resonance-stabilized leaving groups. In fact, their LGs are incredibly unstable on their own, making the compound as a whole unlikely to react. B: While this is partially true, it does not explain the question stem, which attempts to trick you into explaining a false statement. C: Alkyl groups are electron-donating, not electron-withdrawing.

Which of the following compounds contains a weak Brønsted-Lowry base? A. Ammonium cation B. Lithium hydroxide C. Sodium amide (NaNH2) D. Potassium acetate

D is correct. The acetate anion (C2H3O2-) is the conjugate base of acetic acid, a weak acid. Since acetate readily gains a proton to form acetic acid in solution, it is able to act as a Brønsted-Lowry base. Note that it is not strongly basic; in contrast, it is stabilized by resonance, making it somewhat weaker. A: Ammonium (NH4+) is acidic. It is ammonia (NH3) that is a classic weak base. B, C: These compounds are strongly basic. LiOH is one of the alkali metal hydroxides, all of which can act as strong bases. NH2- is the conjugate base of ammonia (also a base!), making it extremely unstable and thus highly prone to gaining a proton. It is commonly used as a base in organic chemistry.

If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher than the rest of the baseball field, at what height would the catcher need to hold his glove to catch the pitched ball? (Note: neglect air resistance, estimate the acceleration due to gravity as 10 m/s2, and assume the pitcher is only throwing the ball horizontally.) A. 2.0 m above the ground B. 1.8 m above the ground C. 0.5 m above the ground D. 0.2 m above the ground

D is correct. The passage tells us that pitchers throw the ball at an average of 30 m/s and that the ball travels 18 m horizontally. This means the ball's flight time is: (18 m) / (30 m/s) = (18/30) s = 3/5 s = 0.6 s The ball is released from a position 2 m off the ground (0.2 m from the pitcher's mound and 1.8 m from the pitcher). To calculate the distance the ball falls during 0.6 s, we can use the equation d = v0t + 1/2at2: d = (0 m/s)(0.6 s) + 1/2(10 m/s2)(0.6 s)2 d = 1/2(10)(0.36) = 1/2(3.6) = 1.8 The ball has fallen 1.8 m from an initial height of 2.0 m. Thus, the catcher must hold his glove 0.2 m above the ground to catch the pitch.

While evaluating the timing of pitches, the researchers noticed a couple of high-performing athletes who were able to generate higher velocities by increasing the length of their acceleration phase. How did the power production for these athletes compare to those originally mentioned? A. These high-performing athletes had an increased power output due to the higher velocities. B. These high-performing athletes had a decreased power output due to the increased time spent in the acceleration phase. C. These high-performing athletes had an increased power output due to the increased time spent in the acceleration phase. D. There may have been no change in power output between the two groups.

D is correct. The power output is given by the equation P = W / t, where W is the work done in joules (or energy output) and t is the time period over which the work was performed. If the velocity increased, then the work also increased, but the time spent in the acceleration phase increased as well (and this increased time of acceleration is the only reason given for the increased velocity, meaning that both v and t increased to the same degree). Thus, there should be no significant difference in power output between the two groups.

Presbyopia is diagnosed when the lens of the eye focuses incoming light rays to a position between the retina and the choroid. Which type of lens should be placed in front of the eye to focus light on the retina and correct this condition? A. Flat B. Spherical C. Diverging D. Converging

D is correct. The problem with presbyopia (far-sightedness) is the image is formed behind the retina rather than on the retina. A converging lens will converge light rays before they can pass the retina and focus light closer to the retina. A: A flat lens will not cause the light rays to converge any sooner and will not correct presbyopia. B: A spherical lens can be diverging or converging and may not cause the light rays to converge any sooner; in that case, it would not correct presbyopia. C: A diverging lens would spread out the light before it reaches the retina and will further exacerbate the problem. It will not correct presbyopia

An oncology team wants to treat a patient with a 10 mL tumor with the same protocol that was used for this study. How many moles of porfirmer should they use? A. 4.2 x 10-2 mol B. 4.2 x 10-8 mol C. 5.2 x 10-3 mol D. 5.2 x 10-9 mol

D is correct. The researchers treated their cells with a concentration of 2.5 μg/mL of porfirmer. We're treating a 10 mL volume, so 2.5 μg/mL x 10 mL = 25 μg = 2.5 x 10-5 g. To convert from grams to moles, we need a molar mass. The molar mass of hematoporphyrin is given as 599 g/mol— use 600 g/mol to simplify the calculation. However, the question asks us for moles of porfirmer, NOT hematoporphyrin. The 2nd paragraph tells us that each molecule of porfirmer contains an average of 8 units of hematoporphyrin, so the molar mass of porphyrin is 8 x 600 g/mol = 4800 g/mol. We'll use 5000 g/mol = 5 x 103 g/mol as an approximation. To convert grams to moles, we divide our mass by the molar mass: 2.5 x 10-5 g / 5 x 103 mol = 5 x 10-9 mol. A, C: These answer choices result from miscalculation. B: This answer choice results from solving for the moles of hematoporphyrin, not the moles of porfimer.

The researchers described in the passage used the hydrated form of calcium nitrate, Ca(NO3)2 ● 4 H2O, to prepare their solutions as described in Figure 1. How many grams of calcium nitrate tetrahydrate are required to prepare this solution? 40 mL (give this amount used) x [1 L/1000 mL] x 0.32 mol/L (give this molarity) A. 1.31 g B. 2.10 g C. 2.23 g D. 3.02 g

D is correct. To begin, we need to use the information in step 1 of Figure 1. The number of moles of calcium ion can be found by multiplying the molarity by the volume of solution in liters. Next, the mass of calcium nitrate tetrahydrate can be determined by multiplying this value by the gram formula mass of calcium nitrate tetrahydrate (including the water molecules). 40 mL x [1 L/1000 mL] x 0.32 mol/L x 236.1 g/mol = 3.02 g

Short-chain saturated fatty acids present in some animal products can produce a rancid odor if they are released from a triglyceride. Which of the following methods would prevent this type of rancidity? I. Hydrogenation II. Dehydration III. Sterilization A. I only B. II only C. I and II only D. II and III only

D is correct. To keep things simple, let's begin with Roman numeral I. Hydrogenation will not change a fully-saturated fatty acid, as there is no double bond to be hydrogenated. Eliminate choices A and C. Now we only need to check RN III. Bacteria often have lipases that catalyze the hydrolysis of the ester linkage between the fatty acid and the triglyceride. Removing bacteria through sterilization should slow hydrolytic rancidity, meaning that III should be in our answer and D is correct. For the sake of thoroughness, however, note that hydrolysis is the reaction that releases a free fatty acid from a triglyceride. Without water, hydrolysis can't happen, so dehydration should also be effective in preventing this type of rancidity.

What is the function of a ribonuclease in E. coli? A. It synthesizes DNA. B. It synthesizes RNA. C. It cleaves DNA. D. It cleaves RNA.

D is correct. We can assess the function of this enzyme by analyzing its name. The ending "-ase" is typical for enzymes, and often (though not always) implies that some larger molecule is being broken down. Specifically, nucleases break down nucleic acids. The two major types of nucleic acids are deoxyribonucleic acids (DNA) and ribonucleic acids (RNA). Given the name, we can figure out that a ribonuclease breaks down RNA

An object floats in water with 4/5 of its volume submerged. If the object is then placed in a type of oil with a density half that of water, which of the following is true about the object placed in the oil? A. The object will float just as it did before. B. The object will float lower than it did in the water. C. The object will float higher than it did in the water. D. The object will sink to the bottom.

D is correct. When dealing with the principles of floating, buoyancy, and density (as in the case of a object floating with a certain portion of it submerged), we can use a shortcut equation: Vfluid / Vobj = ρ obj / ρ fluid , where V fluid is the volume of displaced fluid (here, it's water). V fluid / V obj = 4/5, so ρ obj / ρwater = 4/5. The question tells us that ρwater = 2ρoil. This means that 4/5 = ρobj / 2ρoil . Therefore, ρobj / ρoil = 8/5. Since the ratio of densities is larger than 1, we know that the object is more dense than oil, and it will sink to the bottom. The diagram shown below presents the scenario in the question:

Which of these statements accurately describe the behavior of photoelectrons? I. A photoelectron's energy can exceed that of the incident photon. II. A photoelectron's energy cannot equal the energy of the incident photon. III. The threshold frequency and the likelihood of electron ejection are proportional. IV. The intensity of incident light is unrelated to the energy of an ejected photoelectron. A. I and II only B. I and III only C. II and III only D. II and IV only

Due to conservation of energy, the energy of a photoelectron may only be less than that of the incident photon. This concept is due to the existence of the work function, a quantity of energy required for an electron to be released at all. Since some of the incident photon's energy is devoted to overcoming this threshold, not all of it can be converted to kinetic energy for the ejected electron. For choice IV, remember that the intensity of the incident ray determines the number, not the individual energies, of the ejected photoelectrons. III: The threshold frequency is simply the frequency that must be surpassed for a photoelectron to be emitted. It is the energy of the incident photon (and whether it surpasses this threshold) that determines whether ejection occurs.

what does siRNA bind to?

Due to its structure, siRNA is only able to bind to other RNA strands, not to DNA or protein. Therefore, it must interfere with gene expression after transcription has already occurred, but before translation. Specifically, it prevents the translation of mRNA corresponding to the target protein.

photon energy equation using final and initial

E(initial) = hf + E(final)

Which of the following do NOT have proteins with a nuclear localization signal? I. E. coli II. Homo sapiens III. Fungi IV. Archaea

E. coli (a species of bacteria) and archaea do not have nuclei and thus do not have a need for nuclear localization signal on their proteins. Homo sapiens and fungi are eukaryotes with nuclei.

In locations with very low dissolved O2 concentrations in drinking water, the observed Ksp of ferrous (II) hydroxide will be: ***are equilibrium constants affected by concentration?

Equilibrium constants, including Ksp, are not affected by concentration. These constants are only altered by changing temperature, as shown below for a variety of salts: A, C, D: Ksp will not change due to concentration differences, although concentration may affect whether a precipitate forms.

what happens if leucine zippers are created in a hydrophobic environment?

For long questions like this one, begin by summarizing exactly what the question is asking. In short, leucine residues on different parts of a molecule are coming together to form a dimer. Leucine is hydrophobic, since its side chain contains only carbon and hydrogen. If the solvent were also hydrophobic, the face of a leucine zipper could interact just as favorably with the solvent as with the opposite alpha helix. Some leucine residues would likely interact only with the solvent, preventing formation of the dimer entirely. Hexane, with its nonpolar hydrocarbon structure (shown below), is the least polar solvent listed.

The intravenous solutions used to correct fluid imbalances in trauma patients are aqueous solutions of glucose. Approximately how much glucose is needed to make 1.0 L of a 100 mM solution? A. 18 mg B. 34 mg C. 18 g D. 34 g

From the molarity and the volume of the solution, we can calculate the number of moles and then the mass of glucose (C6H12O6, MW = 180 g/mol) needed to prepare the solution. 1.0 L x 0.10 mole/L x 180 g/mole = 18 g glucose

If a pair of gamma rays emitted from the same annihilation event collide with sensors 0.33 nanoseconds apart, what is the minimum distance the annihilation could have occurred from the center of the machine? A. 5 cm B. 10 cm C. 15 cm D. 20 cm

Gamma rays travel at the speed of light, so both of the rays involved here must move at 3 x 108 m/s. Since the two rays arrived 0.33 nanoseconds apart, one ray must have traveled 10 cm farther than the other: (0.33 x 10-9 s)(3 x 108 m/s) = 1 x 10-1 m = 10 cm. According to the passage, the apparatus in question is circular, with the sensors positioned along the outer perimeter of the circle. This may be difficult to visualize, so when in doubt, draw it out! Suppose the radius of the machine is 2 m, as shown below. If the annihilation event occurs 5 cm from the center, and if the axis of gamma ray emission goes directly through the center, then one gamma ray will travel 95 cm to reach the sensors on the perimeter of the circle. The other gamma ray, which is propelled in the opposite direction, will have 105 cm to travel to reach the sensors on the other side. This represents a difference of 10 cm, which would explain the 0.33 ns time disparity outlined in the question. It turns out that this is the closest you can get to the center and still have one ray travel 10 cm farther.

If delta G is negative, what does K have to be?

Given that ΔG° is negative, and using the equation ΔG° = -RTlnK, ln(K) must be positive for ΔG° to be negative. K > 1.

H+ reduced

H2

what happens when Li react with H2o?

H2 gas produced

fingerprint region -where is the region? What are the units for IR?

IR spectrum - units are wavenumber (analog to frequency) region of 1400 to 400, where more complex vibration patterns, caused by the motion of the molecule as a whole, can be seen; it is characteristic of each individual molecule

what happens in gluconeogenesis if the levels of NADH increase?

If NADH levels have already been increased, then NAD+ levels will be unusually low, since the production of NADH naturally involves the depletion of NAD+. With low NAD+ levels, the future production of OAA and pyruvate (two important gluconeogenic substrates) will be decreased, since pyruvate production is coupled with the conversion of NAD+ to NADH, and since OAA is produced from pyruvate. Thus, the rate of glucose production via gluconeogenesis will decrease.

A 50-kg child is riding on a playground merry-go-round. If the radius of the circular path of the merry-go-round is 5.0 m and the frequency is 0.1 hertz, what is the force required to keep the child on the ride?

If the frequency of the merry-go-round is 0.1 Hz, then the period is 10 s. We can calculate the speed of the merry-go-round from V = 2πr/T = 2(3.14)(5.0 m)/(10 s) = 3.14 m/s. Using this in the centripetal force equation: Fc = mv2/r = (50 kg)(3.14 m/s)2/(5.0 m) Approximate the square of the speed at 10 m2/s2. Fc = (50)(10)/(5) = 100 N

Limiting reagent -what is a quick trick?

If the number of moles reactants match the coefficients, then there will be no limiting reagent

what is a primary amine?

Imide, amide, and tertiary amine are all functional groups present in caffeine. All of the functional groups are part of caffeine except for a primary amine, which consists of nitrogen bonded to one R group and two hydrogen atoms. This structure is not part of caffeine.

The surface temperature of the sun (5770 K) is much greater than that of the earth (298 K). As a result, solar photons have a much higher energy than Earth photons. Given that information, how does the frequency and velocity of an Earth photon differ from that of a solar photon? A. The frequencies of the two photons would be identical; the speed of the solar photon would be greater. B. The frequency of the solar photon would be lower; the speed of the photons would be identical. C. The frequency of the Earth photon would be lower; the speed of the solar photon would be greater. D. The frequency of the Earth photon would be lower; the speed of the photons would be identical.

In a question like this, where the stem is in fact two questions for the price of one, it is easiest to tackle the question separately. Recall Planck's law: E = hf. Since Planck's constant is a constant (which you do not need to memorize), what this equation really says is that the greater the energy of a photon, the higher its frequency. Therefore, Earth photons will have a lower frequency than solar photons. The speed of photons is the speed of light, and it does not depend on the energy of the photons. Therefore, the speed of the photons would be identical. A, B: According to Planck's law, the frequency of the solar photon will be higher than the frequency of the Earth photon. C: Photons all travel at the speed of light. Since a photon is massless, it speed is not influenced by its energy.

which compound would have the lower boiling point? -al or -oic acid?

In distillation procedures, the component with the lower boiling point boils off first, leaving the remaining components in the original flask. Thus, this question is basically asking which component has the higher boiling point. The hydrogen bonding created by the carboxylic acid functional group in 2-methylundecanoic acid (shown below) would lead to a vastly higher boiling point than the aldehyde (over 300ºC compared to 170ºC).

Which of the following molecules does NOT have an atom with sp2-type hybridization? A. Carbon dioxide B. Carbonate C. Formaldehyde D. Methanol

In each case, you should draw the Lewis dot structure. In methanol (CH3OH), four atoms form single bonds with the carbon atom, which is sp3 hybridized. The oxygen atom has two single bonds, one to carbon and one to hydrogen, as well as two lone pairs of electrons; it therefore has sp3 hybridization as well. Neither atom has sp2 hybridization. The structure of methanol is shown below. A: In the case of carbon dioxide (below), the carbon is sp hybridized. However, since there are two lone pairs on the oxygen atoms and a double bond between the carbon and each oxygen atom, the oxygen atoms are sp2 hybridized. B: In the case of carbonate (CO32-) the carbon atom is sp2 hybridized, as shown below. C: In the case of formaldehyde (CH2=O) both the carbon and the oxygen atoms are sp2 hybridized. Give feedback on this question

What do we know about reduction potential? Given the values, which way are the electrons going to flow?

In the electron transport chain, electrons are passed from species with less positive reduction potential to those with more positive reduction potential. O2 serves as the final electron acceptor of the electron transport chain and must possess a standard reduction potential that is more positive than any other acceptor in the chain. Of the standard reduction potentials mentioned in the passage, the greatest is that of Fe3+/Fe2+ in cytochrome c, for which E° = +0.22 V. Only choice A exceeds this value. E should be positive if spontaneous. B, C, D: These values are all less than +0.22 V.

Injection of insulin into the bloodstream is LEAST likely to result in which of the following?

Insulin is secreted in response to high blood sugar. If the body detects that there's plenty of blood sugar, then it would want to stop making more sugar (eliminate choice D), to store that sugar as glycogen (eliminate choice A), and to build up fatty acids into fats for storing up energy (eliminate choice C). The thing the body is LEAST likely to do is to stop storing up energy (choice B). In general, think of the function of insulin as causing the body to build up large molecules to store up energy (glycogen, lipids) and to stop the body from breaking down large molecules to provide energy.

According to Figure 3, why is BHT effective at preventing auto-oxidation? A. It increases the activation energy for the reaction between the lipid radical and oxygen. B. It produces a more kinetically favorable intermediate than the auto-oxidation chain reaction. C. It produces a more thermodynamically favorable intermediate than the auto-oxidation chain reaction. D. It competes with free radicals to oxidize the lipid.

It's ok to say it's more thermodynamically stable - if it's an intermediate, just as long as there isn't change in Gibbs free energy overall

calculating equilibrium constants/what is the relationship between the Keq and the Ka?

Ka, like many other K values, is an equilibrium constant. Hence, the association constant for the given equation is exactly the same as the equilibrium constant of the equation as written. Ka = Keq = [EGFR-EGF]/[EGFR][EGF].

prediction of equilibrium (side favored)

Keq = log (-pKeq), pKeq = pKa (left) - pKa (right)

Ka * Kb =

Kw Kw = [H+][OH-] log (Kw) = log (H+) = log (OH-)

what are the units for R (osmotic pressure) **solve for R normally

L*atm/mol*k

Individuals with alleles that cause them to be unable to produce any DNA editing enzymes will: A. have an average risk of developing cancer. B. be at higher risk for having genetic mutations in their gametes. C. be guaranteed to develop cancer. D. be more resistant to UV-radiation instigated mutations.

Lacking the machinery doesn't increase the chance of a mutation occurring, but it does increase the chance of a mutation being passed on to subsequent cells and spreading. This would be the case for all cells in the body, including germline cells, so gametes will be more likely to have genetic mutations. Though the chance of cancer will go up, it's still a matter of chance. Cancer-causing mutations are never impossible nor are they completely certain.

Which of the following best describes the velocity profile of laminar flow from left to right in a stationary tube?

Laminar flow is due to shear forces (friction) between the fluid and the solid surface of the tube. This results in layers having a gradient of velocities, in which the flow is the fastest in the middle of the tube (where friction is low) and slowest near the surface (where friction is high). A: This choice does not demonstrate a gradient of velocities. C: This choice indicates the fastest velocity regions are closest to the tube surface and slowest in the middle, which is the opposite of what would occur with laminar flow. D: This choice indicates that the flow is fastest near the top surface of the tube and slowest near the bottom surface of the tube, which is not what would occur with laminar flow.

what is a lipase? What kind of enzyme is it? Which specific class of enzymes is primarily responsible for the release of free glycerol from stored triglycerides?

Lipases are the enzymes that digest lipids (fats). Most dietary fats originally exist in the form of triglycerides. The diagram below depicts the six major classes of enzymes. Since lipases typically catalyze hydrolysis reactions, they are a subclass of the hydrolases.

Researchers noted that when a lysine residue on a histone is acetylated, its side chain becomes neutral in charge. Combined with the information in the passage, which of the following conclusions will the researchers most likely reach?

Lysine is a basic amino acid and typically has a positively-charged side chain at physiological pH. When lysine is acetylated, this charge becomes neutral, as shown in the figure below. Since DNA is negatively charged due to its phosphate backbone, the charge on lysine allows for tight histone-DNA interactions thanks to electrostatic attraction between the charged atoms on each molecule. Acetylation of lysine makes the residue neutral, lessening these interactions and promoting a looser structure. Loose chromatin structure is typically associated with euchromatin, the less dense, transcriptionally active chromatin structure that appears light under a microscope. In contrast, histone deacetylation will restore the positive charge to the residue, allowing the electrostatic attractions to return. Therefore, deacetylation of lysine residues on histones should lead to a denser chromatin structure and lowered transcription/gene expression (choice A). acetylation = transcription deacetylation = no transcription.

units for reaction rate units for k, 0th 1st 2nd 3rd

M/s 0th: M/s 1st: 1/s 2nd: 1/M*s 3rd: 1/M^2*s

Given: molality + density

MM

solution equation, if given molarity and volumes/or dilution?

MV=MV

what does N correspond to in terms of energy level? what numbers does the sub-shells correspond to (what are these numbers called?)?

N = principal energy level the s, p, d, f sub-shells depend on the principal energy level, 1 has one sub shell, 2 has 2, 3 has 3 (orbitals are within sub shells, s=1, p=3, d=5, f=7)

what is diatomic nitrogen like?

N2 as a very inert gas. It makes up approximately 80% of the air you breathe, yet has no significant chemical reactions with your lungs - or with anything other than nitrogen-fixing plants. This information implies that nitrogen is very inert (unreactive). As such, it would serve as a good artificial atmosphere when working with reagents that might react with oxygen or other gases. The Lewis structure of N2 is shown below.

are solids and liquids included in Keq expressions?

NO

glucose/glycolytic enzymes effect on current? In this experiment, there was 1 NADH (H used for current) per glucose, but if glycolysis is used, then 2 are made/glucose, so then what happens to current?

Next, we must determine the effect of using glucose and the glycolytic enzymes on the current generated by the battery. From Figure 1, we can infer that the amount of current generated should be directly related to the production of NADH. Unlike in the biostarch battery, glycolysis does not require the breakdown of starch. Thus, NADH is generated directly and more quickly via the conversion of glyceraldehyde 3-phosphate (GAP) to 1,3-bisphosphoglycerate (1,3-BPG). The enzymes should produce 1 NADH per GAP molecule, or 2 NADH per glucose molecule (since each unit of glucose breaks down into two GAP molecules). The original experiment only produced 1 NADH per glucose conversion to phosphogluconate, so we can conclude that the second student should observe a higher initial current than the first.

Is adding water across a double bond an oxidation?

No, because 2 protons and 1 O is not a net loss/gain of electrons. This question is asking us to determine which steps in Figure 3 are oxidations. We are told that steps 1 and 3 convert FAD2+ and NAD+ to FADH2 and NADH + H+. This means that the fatty acid must have been oxidized. Step 2 is simply the addition of water across a double bond, which adds two protons and one oxygen to the fatty acid, meaning there is no net oxidation or reduction.

What is the molar solubility of ferrous (II) hydroxide in water at 25°C? The passage states that the Ksp of ferrous (II) hydroxide is 3.2 x 10-14. When Ksp is known, we can determine molar solubility from the dissociation reaction. Fe(OH)2 (aq) + H2O (l) → Fe2+(aq) + 2 OH- (aq) A. 2.1 x 10-8 B. 8.2 x 10-8 C. 2.8 x 10-6 D. 2.1 x 10-5

Note that Fe(OH)2 dissociates into three ions (one Fe2+ and 2 OH-). Given this 2:1 ratio, Ksp = [Fe2+][OH-]2 = [x][2x]2 = 4x3, where x represents the molar solubility. Next, we must divide Ksp by 4, then take its cube root to solve for x. Dividing 3.2 by 4 is more difficult than dividing 32 by 4, so we can manipulate scientific notation and rewrite Ksp in an easier format. Ksp = 4x3 = 3.2 x 10-14 = 32 x 10-15 8 x 10-15 = x3 2 x 10-5 = x

What is the process of binding that miRNA goes through? In miRNA-directed gene silencing, a small RNA binds to an mRNA and directs degradation of the mRNA or prevents translation of the mRNA. Which of the following terms describes the process through which binding occurs?

Only hybridization describes a process of binding through complementary nucleotides.

Describe the energy of the bond between two H atoms

PE of H atoms begins baseline, then due to stability conferred by the bond, it drops reaching a minimum at ideal bond length, but then same charge repulsion occurs when they're too close, and PE goes back up

ideal gas law

PV = nRT

Which of the following 1H-NMR signals would allow for researchers to differentiate between colchicine (amide group) and IA? A. Large peaks at 7 ppm (what gives peaks at 7)? B. Several peaks between 0 and 5 ppm (what gives peaks here?) C. A doublet at 8.5 ppm D. A signal at 3.4 ppm (what gives peaks here)

Peaks around 5-8.5 ppm are indicative of amide (R(C=O)NR2) hydrogens. Only the colchicine has an amide group. The doublet (signal splitting) is due to the hydrogen on the nearby sp3carbon in the cycloheptane of colchicine. A: (large peaks at 7ppm = aryl hydrogens)Several large peaks would indicate the presence of aryl hydrogens: hydrogen atoms that are bonded to a benzene ring sp2 carbon atom. Both IA and colchicine have benzene rings that would provide these signals. B: (Peaks between 0-5 = sp3 hydrogens) Peaks that appear between 0-5 ppm are indicative of sp3 hydrogens. Both compounds have several groups of sp3 hydrogens. D: (Peaks at 3 = hard to determine) This peak could be indicative of several hydrogen types (alcohols, ethers) that would not definitively identify only one of the molecules.

How many resonance structures are exhibited by phenoxide, the conjugate base of phenol? A. 3 B. 4 C. 5 D. 6

Phenoxide, which can also be described as a phenolate ion, has a number of distinct resonance structures. The five resonance forms of this anion are shown below.

The conversion between glucose and pyruvate strongly favors the formation of pyruvate, and yet the gluconeogenic pathway is able to utilize several shared enzymes with glycolysis to create glucose from pyruvate. It is able to do this primarily because/what do these enzymes do?

Production of PEP, glucose, and fructose 6-phosphate by gluconeogenesis-specific enzymes that bypass irreversible steps of glycolysis push the equilibrium of reversible enzymes that function both in glycolysis and gluconeogenesis in the direction of glucose production.

Which of the following enzymes should the researchers add to the cell samples if they want to reverse the general catalytic effects of protein kinase A?

Protein kinase A has an extremely wide variety of specific effects, but this question asks about its general catalytic function. As a kinase, PKA functions to add a phosphate group to its substrate. The opposite of this activity is the removal of phosphate from a substrate, a function which is performed by phosphatase enzymes. We do not need to know the exact reactions catalyzed by each enzyme to answer this question; nomenclature alone shows us that only choice C represents a phosphatase.

D subshell

REMEMBER that 4s will lose electrons before 3d

Al3+ has a reduction potential of -1.66, while Cd2+ has a reduction potential of -0.40. In an electrolytic cell at standard conditions: A. Al3+ will reduce at the cathode and Cd (s) will oxidize at the anode. B. Al (s) will oxidize at the cathode and Cd2+ will reduce at the anode. C. Al (s) will oxidize at the anode and Cd2+ will reduce at the cathode. D. Al3+ will reduce at the anode and Cd (s) will oxidize at the cathode.

Reactions in electrolytic cells occur nonspontaneously. Since Al3+ has a more negative reduction potential than Cd2+, it is Al3+ that will reduce and Cd (s) that will oxidize. Oxidation and reduction always occur according to the mnemonic "REDCAT": reduction happens at the cathode and oxidation takes place at the anode.

IR Spectroscopy: O-H

Retinol differs from retinal in that it contains -OH groups, but does not contain a carbonyl group (C=O). The carbonyl stretching frequency falls in the range of 1700-1750 cm-1, while the O-H stretching frequency is expected to fall in the range of 3200-3500 cm-1. The reaction is thus complete when the signal in the 1700-1750 cm-1 range has completely disappeared.

Fischer projections

S/R counts (do not rotate)

what is the equation for Q, compared to Keq?

Same equation. Q= reaction quotient, compare to Keq, if smaller, reactant side is favored

what is beta minus decay?

Since each answer has multiple parts, we should divide and conquer, eliminating wrong choices as we go. Paragraph 3 mentions that the nuclei emit gamma rays, a form of electromagnetic radiation that consists of high-energy photons. Gamma rays represent ionizing radiation, which makes them biologically hazardous (eliminate C and D). The passage also states that I-131 undergoes β-minus decay. In β-minus decay, a neutron is converted to a proton as an electron is emitted. Therefore, iodine must be converted to an element with one additional proton, which can only be Xe. β-minus decay is depicted below. Note that we can think of this process in two ways: either simply as the emission of an electron, or as the conversion of a neutron into a proton, an electron, and a neutrino (shown in the circle on the bottom right).

Ten moles of the monoprotic, weakly acidic medication aspirin were added to water to make one liter of solution. If the pH of the resulting solution was 5.9, what is the approximate Kb for the non-diffusible form of aspirin? A. 0.1 B. 0.01 C. 0.001 D. 1

Since we are given pH in the question stem, we will not be able to find Kb immediately. Instead, we need to calculate Ka and solve for Kb from that value. The Ka for the dissociation of a generic acid HA can be written as Ka = [H+][A-]/[HA], where all concentrations are measured at equilibrium. In the solution of aspirin described, the initial concentration of drug is 10 M. Since only a small amount of this weak acid will dissociate, this value is a good approximation for our final equilibrium [HA]. Next, we must find the proton concentration. Remember, [H+] = 10-pH. Here, the pH of the solution is 5.9, so [H+] = 10-5.9 M ~ 10-6 M. Since each HA molecule dissociates into equal parts [H+] and [A-], our value for [A-] must be 10-6 M as well. Returning to the Ka expression, Ka = [(10-6 M )(10-6 M)] / (10 - 10-6 M). Remember, we can estimate that [HA] = 10 M, even though its true final value is 10 M - 10-6 M! [(10-6 M )( 10-6 M)] / (10 M) = 10-12 M / 10 M = 10-13 M In water at 25°C, Ka•Kb = 10-14. Given this, Kb = 10-14/Ka = 10 -14/10-13 = 10-1 = 0.1

Which of the following statements is true? I. Acetals cannot form under basic conditions. II. Hemiacetals and hemiketals exist in equilibrium with aldehydes and ketones. III. In multistep reactions, acetals and ketals often serve as protecting groups for vulnerable esters. **what groups to acetals and ketals serve to protect? IV. Ketals contain two -OR groups, while acetals contain only one.

Statement I is true. Under basic conditions, only hemiacetals can be formed. Acetals, which result from the reaction of a hemiacetal with an alcohol, require acidic conditions. Statement II is also correct, as these species rapidly interconvert. III: Acetals and ketals serve as protecting groups for aldehydes and ketones, which are more reactive than esters. As comparatively inert groups, esters do not need protection as much as other functionalities. IV: Both ketals and acetals contain two -OR groups. In contrast, hemiacetals and hemiketals contain only one such group.

CN - weak or strong base?

Strong

GC vs HPLC

Sublimation is the conversion of a substance from its solid state to a gaseous state. During phase changes, temperature will not change since all the energy is being absorbed or emitted to achieve the phase change (in sublimation, energy is absorbed) so we can eliminate choices B and C. Now we must determine whether high pressure liquid chromatography (HPLC) or gas chromatography (GC) would be best. Since the PUFA will be in the gas phase, GC is best.

what properties affect the TLC/retention factor value? **3 things

TLC depends on the differential affinity of a compound for the stationary vs. the mobile phase. Temperature also affects the rate of movement. ("eluting strength"), but it is true as well. Eluting strength depends on how strongly a compound adsorbs onto the adsorbent. Since typical adsorbents are highly polar, eluting strength increases with increasing solvent polarity. Methanol is more polar than pentane and therefore has a greater eluting strength.

how is effective nuclear charge calculated?

Take the proton number and subtract everything that falls before the n-level given Ex. Aresenic, n = 3, Ar = 33 protons - 10 (because n=3 means we keep everything in the n=3 block, but get rid of everything starting with the n=2 block and down)

While the blood is buffered primarily through the equilibrium between carbon dioxide and carbonic acid, coupled with hemoglobin, the blood may also be buffered through other plasma proteins. Which of the following is true?

The amino acids that make up a protein may include many acidic or basic side chain groups. Those side chains can either release or absorb protons, allowing them to help buffer the blood through action as a Brønsted-Lowry acid or base.

common ion effect

The common ion effect occurs when a solution contains a salt that partially dissolves into ions, and an extra quantity of one of these ions is added. This results in the combination of this ion with existing dissolved ions, causing an additional quantity of the salt to precipitate out of solution. In this case, the partially-dissolved salt is Ag3PO4. Adding AgNO3 causes dissolution of extra Ag+ions, which then combine with already-present PO43- ions and precipitate out of solution into the yellow Ag3PO4 precipitate (identified with the mass spectrometer used in Figure 1). Similarly, adding K3PO4 results in dissolution of extra PO43- ions, which then combine with already-present Ag+ ions and precipitate into yellow Ag3PO4. I and II are therefore both correct.

The first step of the formation of the imidazolinone ring of sgBP is most likely accomplished by the: Understand the mechanism and be able to draw the linking of several amino acids, even with rings involved

The cyclic aminal in the Aequorea fluorophore is formed in step 1 by the attack of the nucleophilic amide nitrogen of glycine on the electrophilic carbonyl carbon of serine. In the QYG chromophore of sgBP, serine is replaced by glutamine. According to the first paragraph of the passage "the CP chromophore ... forms an imidazolinone structure through an autocyclization mechanism analogous to that shown for ... Aequorea." Given this, choice A, the attack of the nucleophilic amide nitrogen of glycine 64 on the electrophilic carbonyl carbon of glutamine 62, is most likely to occur during the first step of the formation of the imidazolinone ring of sgBP. (Test 4)

reducing agents for reducing of aldehyde to primary alcohol

The final step in Figure 2 is the reduction of an aldehyde (retinal) to a primary alcohol (retinol). This is a classic case of reduction, which can be conceptualized as the gain of bonds to hydrogen or the loss of bonds to oxygen (or other electron-withdrawing species). LiAlH4 is a reducing agent capable of performing this reaction. **LAH or LiAlH4

state the efficiency equation and then how changes in work, pressure, volume, etc would impact the equation.

The formula for the efficiency of any system is output work/input energy. Therefore, the efficiency of engine 1 is W/QH. Efficiency is increased by increasing W. Since work in Figure 2 is the interior area of the cycle, if W increases, the interior area should increase as well. This makes choice A, which states the opposite, correct for this EXCEPT question.

A child is rolling his toy car (with mass = m) down a ramp. The coefficient of static friction between the car and the ramp is 0.25. When the car is halfway down the ramp, the child pushes down on the car perpendicular to the plane, halting it. What is the minimum force the child must apply to keep the car from starting to roll down the ramp?

The gravitational force pulling the car down the ramp is mg sin θ. (This is the typical value for the gravitational force that acts on an object to drag it down an incline.) To stop the car from sliding down the ramp, we must have an equal and opposite frictional force. Since these forces are equal and opposite, we can set them equal to each other as follows: Ff = mg sin θ Now, remember that frictional force is equal to the product of the appropriate coefficient of friction and the normal force: Ff = μFN Ff = 0.25 x FN = mg sin θ The car itself has a mass m and thus generates a normal force of FN = mg cos θ. Again, this is the standard value for the normal force on an object positioned on an inclined plane. From this information alone, we may be tempted to pick C, which is the difference between the gravitational force and the frictional force. If the child were pushing the car upwards along the plane in a parallel fashion, this choice would be correct. However, the child is actually pushing down on the car, perpendicular to the plane. Thus, the force exerted by the child (Fa) will add to the force created by the mass of the car itself and alter the value for the normal force. Our total FN = mg cos θ + Fa. Substituting, we get: Ff = 0.25 x (mg cos θ + Fa) = mg sin θ mg cos θ + Fa = (mg sin θ) / 0.25 Thus, the force with which the child must push down on the car is Fa = [(mg sin θ) / 0.25] - mg cos θ.

Which of the following represents the pH of the isoelectric point for glycine?

The isoelectric point is the pH at which the net charge of the amino acid is zero, or when the carboxylate is deprotonated (having a negative charge) but the amine group is still protonated (in its ammonium form with a positive charge). In other words, glycine is in its zwitterionic form. This occurs at the first equivalence point. The pH corresponding with the middle of the first sharp rise in pH, at 15 mL of added NaOH solution, is the isoelectric point. Hence, the estimated pH of the isoelectric point for glycine, as seen in Figure 2 and shown below, is 6.0.

secretion of renin -what type of cells?

The juxtaglomerular cells in the kidney are specialized smooth muscle cells mainly found in the walls of the afferent arterioles (with some in the efferent arterioles). The afferent arterioles deliver blood to the glomerulus. The muscles of the small and large intestine are also smooth muscle cells. -involved in controlling the blood volume/pressure -low volume, it will secrete more -renin goes to liver -Angiotensinogen to angiotensin 1 is converted in the lungs, angiotensin 2 is main molecule that causes aldosterone and vasoconstriction

If trifluoroacetic acid is the reagent used in SP synthesis, which of the following would be the most effective for creating an effective support system?

The list of characteristics in the passages states that an ideal support "...must be inert to all reagents and solvents used during SP synthesis." The question tells you that we are using a weak acid in the synthesis, so we want a molecule that will be inert to the actions of an acid. Benzene, the aromatic hydrocarbon shown below, is very stable (inert) in the presence of acids and bases and will not react with the reagents.

properties of furan -how many pi electrons? is it a pyrrole?

The molecule depicted is furan, one of the classic aromatic heterocyclic compounds. Furan is a useful molecule to recognize for the MCAT. II: The molecule has 6 pi electrons, making it aromatic; this makes II incorrect. III: A pyrrole ring is nitrogen-based, not oxygen-based. The structure of pyrrole is shown below. Even if you did not know this structure (or anything about pyrrole), if you could identify the given compound as furan, you would know that it could not also be pyrrole.

if given values for E cell, for reactions involving being oxidized and reduced.

The more positive, the better oxidizing agent; the LESS negative, the better reducer.

A student finishes an experiment involving several bacteria which are highly pathologic in humans. She wishes to dispose of the agar plates and micropipette tips she used. Which of the following procedures should she carry out?

The most effective technique for sterilizing used laboratory materials is using an autoclave and should be the standard procedure followed here. An autoclave brings the materials to a temperature over 120ºC and a pressure over 2 atm, which is enough to kill almost anything.

what happens to Kb as electron donating groups are added to the base?

The pKb is a measure of basicity, where a smaller pKb value corresponds to a stronger base. Methyl-piperazine is a stronger base than piperazine because of the inductive effect generated by the electron-donating N-methyl group of methyl-piperazine. This effect increases the availability of electron charge density available to donate, thereby increasing the strength of N-methyl piperazine to act as a Lewis base. A low Kb means that it's a good base, that it will serve as an electrophile and that it doesn't give away easily. Electron donating groups provide this.

Which of the following compounds may be effectively removed from waste water by a sand filter with IPB? I. 1,4-dichlorobenzene (why is this compound hydrophobic)? II. Methanol III. 1,2-ethanediol

The passage states that filters with IPB produce a polymer that traps hydrophilic compounds, but not hydrophobic ones. Thus, we need to choose the Roman numerals that represent hydrophilic species. Due to their alcohol groups, methanol (RN II) and 1,2-ethanediol (RN III) are both hydrophilic, or water-soluble. I: 1,4-dichlorobenzene is hydrophobic. Note that the two chlorine substituents are positioned opposite each other, causing their dipoles to cancel out overall. The structures of these compounds are shown below:

Which of the following statements is true regarding peptide bonds found in the CK subunits analyzed?

The peptide bond displays partial double bond character because of the delocalization of electron density from the peptide (amide) nitrogen through the peptide carbonyl carbon and onto the peptide carbonyl oxygen (choice A). As such, the peptide bond exhibits resonance stabilization, as shown below. The carbonyl oxygen and amide nitrogen of the peptide linkage are stable and not ionizable at physiological pH (choice B). For steric reasons, the peptide bond is normally found in the trans, not the cis, configuration (choice C). While urea affects higher levels of protein structure, it does not disrupt peptide bonds, which are generally liable only to strongly acidic conditions (choice D).

Cardiac glycosides like ouabain have long been used in the treatment of heart failure. The above reaction of ouabain would most likely result in: A. the formation of 2 ketones. B. the formation of 1 aldehyde and 1 ketone. C. the formation of 1 carboxylic acid and 1 ketone. D. no reaction.

The reaction shown is ozonolysis. Ozonolysis is the cleavage of an alkene or alkyne with ozone (O3) that results in the multiple carbon-carbon bond being replaced by a double bond to oxygen (carbonyl). There is only 1 alkene on ouabain and its cleavage will result in the formation of 1 aldehyde (the alkene C on the right) and 1 ketone (the alkene C on the left).

Figure this out a little better Become familiar with amino acid side chains at various pHs, plus whether they'll H-bond or ionic, ionic is a result of dipoles

The third paragraph of the passage states that a hydrogen bond forms "when both the phenolic side chain of Y63 and the hydroxyl group of Ser157 are protonated" and that "the strength of Y63-S157 bonding is thought to be important in ESPT, which determines the chromophore absorption spectrum." Figure 3 shows that at pH 10, the QYG chromophore tyrosine 63 is protonated and cysteine 157 is deprotonated. The two residues form a "protonation state-dependent ion-dipole" interaction that accounts for the "greater strength of ion-dipole bonds relative to hydrogen bonds." Figure 3B also shows that the side-chain pKa of cysteine is approximately 8.3. At pH 4.5, the side chain will exist in a mainly protonated form. This suggests that the longer maximum absorbance wavelength at pH 4.5 is because the Y63-C157 interaction predominately consists of hydrogen bonding, and that the ion-dipole bonding which predominates at pH 10 promotes the maximum absorption of shorter-wavelength, higher-energy radiation.

Three students in physics lab are given four capacitors, each with a capacitance of 4 μF, and are told to construct a circuit with the maximum possible total capacitance. Student 1 attaches all four capacitors in series. Student 2 attaches all four capacitors in parallel. Student 3 only inserts a single capacitor in his circuit. Which of the following is true?

The total capacitance of capacitors in parallel is simply the sum of their individual capacitances, so student 2 would create a circuit with 16 μF and the highest capacitance. Student 3, by only using a single capacitor, would create a circuit with 4 μF capacitance. When capacitors are connected in series, the reciprocal of the total capacitance is found by adding the reciprocals (like resistors in parallel), so student 1 would only create a circuit with 1 μF capacitance.

Which of the following correctly lists a pair of analogous structures and a pair of homologous structures, respectively?

The wing of a bee and the wing of a bird; the arm of a human and the flipper of a walrus Analogous structures are those structures that evolved independently to carry out the same function. Thus, the wing of a bee and the wing of a bird are analogous structures. Homologous structures are those that have a similar evolutionary history, arising from the same source, even if they now have different functions. The forelimbs of mammals (human arm, walrus flipper, bat wing) would all be homologous despite their different functions. Thus, this choice correctly indicates a pair of analogous structures and then a pair of homologous structures.

Which of the following is most likely NOT a symptom of acylcarnitine translocase deficiency? what happens when fatty acids cannot be broken down by any cells? what will the cells use for energy?

There will be a great abundance of fatty acids in the body and less energy available. It will also mean that glucose will be relied upon much more heavily as a source of ATP. This implies that there will be little glucose in the blood, and that hyperglycemia will not be a symptom of acylcarnitine translocase deficiency.

If a hydrogen NMR spectrum were obtained for a double-stranded amidine dimer, would the amide N-H proton resonances experience any shielding effects? A. Yes, they would exhibit upfield shifts due to shielding effects. B. Yes, they would exhibit downfield shifts due to deshielding effects. C. No, they would not experience any shielding effects. D. Yes, they would exhibit downfield shifts due to shielding effects.

This deshielding effect will lead to increased chemical (downfield) shifts for N-H proton resonances due to the removal of electron density. NMR numbers are the difference between electrons spins as they go from high to low. The more energy in the bond, the greater the magnetic spins. C-H bonds are the lowest energy because the hydrogens are shielded by electric charge. However, as electron withdrawing groups are added, the H's become deshielded and experience a downward shift. Downward = to the left = higher energy

total internal reflection

This is false because total internal reflection can only result when a ray of light begins in a higher-index material and reaches a boundary with a lower-index one (e.g. starting in water and moving towards air). Here, the light ray started in air (n = 1) and moved into water (n ∼ 1.3), making total internal reflection impossible.

Ritonavir is an antiviral drug that mimics the natural substrate for a key viral protease enzyme. As a competitive inhibitor, ritonavir:

This question is asking us for the definition of a competitive inhibitor, which binds to the active site and blocks the substrate from attaching. The mechanism of a generic competitive inhibitor is depicted below.

In the electron transport chain, which of the following is the component that is reduced but never oxidized?

This question is testing your knowledge of the electron transport chain. Almost all elements in the chain are at some point reduced and oxidized. This is the driving force behind the electron transport chain. However, at the end of the chain, oxygen is reduced to form H2O, which is the end fate of the electrons in the chain. The electron transport chain is shown below. Note, again, that O2 is reduced to yield H2O, but O2 is never shown to become oxidized (made to lose electrons).

Buffer choice

To construct the best possible buffer, we should choose the organic acid with the pKa closest to the pH at which the experiment will take place (9.7). This gives us answer choice D. Note that an ideal buffer should have a pKa within 1 pH unit of the expected experimental conditions.

A marine biologist is analyzing light released by several unidentified metals that were found in a deep-sea cavern. The following table lists the type of light ray emitted by each metal sample. Sample 1: infrared Sample 2: ultraviolet Sample 3: visible Electrons in Sample 2 could have made which of the following jumps, as indicated by their principal quantum numbers? A. 4 to 1 B. 5 to 2 C. 5 to 3 D. 6 to 4

To determine the final principal quantum number involved in the transition, we must be familiar with the different series of spectra. The Lyman series (ultraviolet rays) involves any emission in which the ground state of the electron is n = 1. The Balmer series (visible rays) includes emissions in which the final state is n = 2. Finally, the Paschen series (infrared) contains any emission with a final state of n = 3. Here, the initial principal quantum number, or excited state, does not matter when answering this question. B: This transition could have been experienced by Sample 3, as nfinal = 2. This final value generally corresponds to visible light. C: This might refer to Sample 1, which released infrared light - a type that correlates to an nfinal value of 3.

An artificial lung is filled with various gases to reflect those present in the atmosphere on another planet. The lung contains equimolar amounts of nitrogen and oxygen, argon with a mole fraction of 0.20, and carbon dioxide with a mole fraction of 0.46. In this system, the percent composition by mass of argon is: A. 12%. B. 21%. C. 24%. D. 53%.

To find the percent composition of argon in the balloon, the total mass of all other elements present must be calculated. For convenience, we can assume that there are 100 mol of gas in the balloon. Let's begin with argon itself, which has a mole fraction of 0.20. This is equivalent to 20 mol (0.20 × 100 mol), and since Ar has a molecular weight of approximately 40 g/mol, we can conclude that 800 g of Ar are present. For carbon dioxide, which has a molecular weight of 44 g/mol, 46 mol is equivalent to 2024 g. Finally, though we do not know the mole fractions of nitrogen and oxygen, we know that the remaining mole fraction of 0.34 is divided evenly between them, giving us 17 mol (or 544 g) of O2 and 17 mol (or 476 g) of N2. The percent composition of argon is thus (800 g Ar / 3844 g total mass) × 100, or about 21%.

In the passage, the researchers attempted to model chemical bonds and intermolecular forces as springs. Suppose the chemical potential energy in a single disulfide bond was modeled as elastic potential energy stored in a stretched 10 Å spring, with a spring constant of 2 N/m. What would the length of the stretched spring be? (A disulfide bond has a bond dissociation enthalpy of 54 kJ/mol.) A. 3 x 10-13 m B. 3 x 10-10 m C. 1.3 x 10-9 m D. 2.3 x 102 m

To tackle this question, we first need to figure out how much chemical potential energy is contained in a single disulfide bond. To do this, we have to divide the value of 54 kJ/mol by Avogadro's number. We will round Avogadro's number to 6 × 1023 to make the math a little easier: 54 kJ/mol / 6 × 1023 bonds/mol = 9 × 10-23 kJ/bond = 9 × 10-20 J/bond Recall that elastic potential energy has the following equation: PE (spring) = ½ kx2 where k is the spring constant and x is the displacement in the spring. We can rearrange this equation to isolate for the displacement: Now we'll substitute our values into this equation: But don't forget, we have to find the total stretched length of the spring. The passage tells us that 1 Å = 10-10 m, so our spring has a length of 10-9 m. Summing that value and the displacement of the spring gives us a grand total of 1.3 x 10-9 m for the length of the stretched spring.

Troponin isoenzymes are used as an alternative biomarker in the diagnosis of heart attacks. In which of the following muscle types does the troponin complex function in contraction?

Troponin is a complex of three proteins (troponin I, troponin C, and troponin T) required for muscle contraction in skeletal muscle and cardiac muscle, but not smooth muscle. I and III are true; II is not. Choice C is then correct.

How does UV compare to IR spectroscopy?

UV involves induction of electronic excitations = higher energy light

what if no change in UV absorption peak?

UV radiation is higher-energy, higher-frequency EMR than visible light or IR radiation. The passage states that UV absorbance spectroscopy was preformed in order to assess whether the MH preparation process had changed the structure of A. The fact that A's peak UV absorbance remains the same in aqueous and MH solutions suggest that the structure is identical, so choice B is correct. A: The presence of a C=O double bond is detected by IR spectroscopy, not UV spectroscopy. C: Compounds that absorb light in the visible spectrum, not the UV spectrum, are colored.

best way to separate large, aromatic, cyclic

UV spectroscopy conjugated systems, larger = more shifted

how is ultrasound defined in terms of frequency?

Ultrasound is defined as sound with a frequency above the human range of hearing ("ultra" = "above"). To answer this question, then, we need to know what the human range of hearing is. This range is 20 Hz to 20 kHz, so anything greater than 20 kHz qualifies as ultrasound.

Know how to calculate mole fractions -what number should you use when you see mole fractions?

Use 100 moles

An artificial heart valve was tested for its ability to function under extreme conditions, to a maximum flow rate of 4.00 x 10-4 m3/s. What speed would this correspond to for an average red blood cell within a blood vessel of cross-sectional area 5.00 x 10-6 m2? A. 2.00 x 10-9 m/s B. 1.25 x 10-2 m/s C. 4.05 x 10-4 m/s D. 8.00 x 101 m/s

Watch your units! We're given one number in m3/s and another in m2. To get an answer in the units of m/s, just divide the two numbers in the question: Speed = (4 x 10-4 m3/s) / (5 x 10-6 m2) = (4/5) x (10-4/10-6) = 0.8 x 102 m/s = 8 x 101 m/s This can also be solved using the equation flow rate = cross-sectional area x velocity

A researcher analyzes a nucleic acid sample. After looking only at its nucleotide composition, he decisively concludes that this sample represents single-stranded DNA and not another form of nucleic acid. Which of the following compositions most likely represents this sample?

We are told that the researcher can "decisively conclude" that this sample represents single-stranded DNA using only its nucleotide composition. We are therefore looking for the option that could logically represent single-stranded DNA but could not represent another type of nucleic acid. Both options A and D could be the compositions of single-stranded DNA, but choice A has equal proportions of adenine and thymine as well as equal proportions of cytosine and guanine. Choice A, then, could easily represent double-stranded DNA, so the researcher would not be able to conclude whether that sample were single- or double-stranded. Choice D, then, is the only answer remaining.

When NADH is oxidized, what happens to the hybridization of the nitrogen of the nicotinamide ring, and the C4 carbon atom opposite this nitrogen? A. The nitrogen is converted from sp3 to sp2 hybridization, whereas the C4 carbon is converted from sp2 to sp3 hybridization. B. The nitrogen is converted from sp2 to sp3 hybridization, whereas the C4 carbon is converted from sp3 to sp2 hybridization. C. Both the nitrogen and the C4 carbon are converted from sp2 to sp3 hybridization. D. Both the nitrogen and the C4 carbon are converted from sp3 to sp2 hybridization.

We first need to examine the ring nitrogen atom in Figure 1. In the NADH, there are three single bonds shown. You also need to recognize that there is a lone-pair of electrons on the nitrogen (not shown) to complete the octet, and therefore there are four electron domains, corresponding to sp3 hybridization. A: In NADH, there are four single bonds to the C4 carbon atom, which is sp3 hybridized. In NAD+, there are two single bonds and a double bond, which represents three electron domains, and corresponds to sp2 hybridization. B, C: NAD+ contains two single bonds and one double bond to the nitrogen atom, which represent three electron domains and corresponds to sp2 hybridization, with a roughly trigonal-planar geometry.

Using this equation practically: ΔG = ΔH - TΔS.

When determining the conditions under which a reaction is spontaneous, always consider the change in Gibbs free energy. If the ΔG for a reaction is positive, it is nonspontaneous, and if the ΔG for a reaction is negative, it is spontaneous. The ΔG for a reaction can be determined from the change in enthalpy and entropy based on the equation ΔG = ΔH - TΔS. Based on Table 1, the ΔH values for all of the formation reactions (Hf) are negative, which favors spontaneity. However, each reaction starts with the elements in their standard states, which include C (s), O2 (g), H2 (g), and F2 (g). The formation of a more organized compound from these reactants will result in a decrease in entropy (negative ΔS), which will favor a positive ΔG and a nonspontaneous reaction. So, if the magnitude of ΔH is greater than TΔS (which occurs at low temperatures), then the reaction is spontaneous. If the magnitude of ΔH is smaller than TΔS (which occurs at high temperatures), then the reaction is nonspontaneous

Oxidized Zinc

Zn2+

what is a quinone?

a conjugated dicarbonyl compound, to form a quinone, more C to O bonds must be added to phenol = oxidation

HPLC

a form of chromatography in which a small sample is put into a column that can be manipulated with sophisticated solvent gradients to allow very refined separation and characterization; formerly called high-pressure liquid chromatography

acetal

a functional group that contains a carbon atom bonded to 2 -OR groups, an alkyl chain, and a hydrogen atom

what is a concentration cell?

a galvanic cell with 2 1/2 cells, same species in both sides but different concentrations, electrons move because of an imbalance in concentrations

Gabriel (Masonic-Ester) Synthesis

a method of synthesizing amino acids that uses potassium phthalimide and diethyl bromomalonate followed by an alkyl halide; two substitution reactions are followed by hydrolysis and decarboxylation

to be IR active

a molecule must have an indelible dipole

what would you expect to see from a large specific heat on a graph?

a more gradual slope

what happens to the mass number when a particle experiences beta minus decay?

a neutron is turned into a proton, so there is no change in the mass number

triple point

a point on a phase diagram at which a substance exists in equilibrium between all three phases

isothermal process

a process in which temperature (and, therefore, internal energy) remain constant

solubility

a ratio that measures how much solute can dissolve in a solvent at a given temperature

reduction

a reaction in which a species gains electrons

aldol condensation -describe the two types 1. what kind of environment is needed to produce the enol? 2. what kind of environment is needed produced the enolate? 3. what is the extra step of the enol is used rather than the enolate? 4. what is the product? **what is the difference between aldol condensation and aldol addition?

a reaction in which an aldehyde or keytone acts as both the electrophile and nucleophile, resulting in the formation of a carbon-carbon bond in a new molecule called an aldol **condensation takes the final step and eliminates water, forming a double bond on the carbon beta gamma to the carbonyl of the other carbon chain

buffer

a solution containing a weak acid couple with its conjugate salt, acting to precent changes to the solution's pH upon the addition of acidic or basic substances

reducing agent

a species that is oxidized in the process of reducing another species

oxidizing agent

a species that is reduced in the process of oxidizing another species

NMR

a technique that measure sthe alignment of magnetic moments from certain molecular nuclei with an external magnetic field; can be used to determine the connectivity and functional groups in a molecule

UV Spectroscopy -what kind of light does it measure?

a technique that measures absorbance of ultraviolet light of various wavelengths passing through a sample

phosophodiester bond

a type of bond that links the sugar moieties of adjacent nucleotides in DNA 2 per phosphate in DNA, bond between P and O

oxidation states

ability to lose different numbers of electrons and have stability (transition metals)

does a Lewis acid accept or donate electrons?

accepts

best LG on an acetyl?

acetyl iodide

what increases osmotic pressure across a membrane?

added ions, decreases with added water, albumin -pulls water back into capillaries

glucose =?

aldose

an unsaturated fatty acid possesses? -which two main functional groups?

alkene/alkyne and a carboxylic acid

alkyl ester alkyl halide vinyl halide aryl halide aryl alkanoates

alkyl ester: R group of ester is alkyl group alkyl halide: Halide attached to alkane chain vinyl halide: halide attached to alkene aryl halide: halide attached to benzene ring aryl alkanoates: ester group with alkane chain attached to carbonyl and another alkane R group attached to the O of the ester (ethyl propaonate)

reductive amination

always produce an amine from a carboxl

amide

amides are named by dropping the -oic acid and adding -amide

Which of the following molecules is/are most likely to have selective proteins in the BBB to facilitate its passage into the brain? A. Antibodies B. Starch C. Amino acids D. Urea

amino acids duh

weak bases

ammonia

wavenumber

an analog of frequency used for infrared spectra instead of wavelength

chemical shift ó, what are the units?

an arbitrary variable used to plot NMR spectra; measured in parts per million (ppm)

chemical compound

any elements held together by bonds

R/S

assigns priority to atoms surrounding chiral center, is affected by side chain

dimagnetic

at atom or substance that contains no unpaired electrons and is consequently repelled by a magnet

Charles' Law

at constant pressure, the volume of an ideal gas is directly proportional to its temp V/T= V/T

titration is most effective as a buffer

at the 1/2 equivalence point

where does oxidation take place?

at the anode for both galvanic and electrolytic cells

What is the equation for knowing moles titrated? Students then performed a potentiometric titration of captopril in order to determine the captopril content contained in a tablet formulation. Two tablets were ground and homogenized, producing 104.4 (MM=122 g/mol) grams of fine powder. The powder was then dissolved in 100 mL of water and titrated with a solution of 2 x 10-2 M NaOH. The potentiometric titration curve obtained, along with a plot of the rate of change of potential during the titration, is shown in Figure 2.

at the equivalence point, Mbase × Vbase = Macid × Vacid = moles acid.

alpha linkage

atoms are opposed

How much base do you use to react with a compound with 2 possible bonding/acidic hydrogens?

be careful because you want to only react with the Pk1, or you'll have a weak acid mixture. **find the moles of acid and treat with that exact number moles of base

why is arginine more basic?

because the electron donating groups stabilize the hydrogen (weak acid = better base)

Reducing agent

best reducing agents are most likely oxidize themselves. If given reduction potentials, switch signs to find oxidation potential good oxidizers are prone to reduction good reducers are prone to oxidation

UV spectroscopy

big molecule, conjugated systems

How to increase the total Ecell?

can be done by increasing the number of moles of the reactant present

photoelectric energy

cannot be greater than or equal to that of the incident electron

what type of reaction to primary alcohols go through? why?

cannot form carbocation must be SN2

most acidic to most basic

carboxylic acid -phenol-alcohol-amine

Pd/C -are these supplying any atoms? What are the needed for?

catalyze hydrogenation rxns

spingolipids

characterized by at least one long hydrocarbon chain and nitrogen head group (originates from a fatty acid)

electrochemical reaction

chemical reaction that either is driven by or produces electricity

membrane fluidity -what main molecule helps with this? markers for cell signaling -what main molecule does this?

cholesterol glycosphingolipids

stereoisomers -what are the 5 types?

cis-trans, diastereomers, enantiomers, conformational isomers, and meso compounds

NaIO4 -what does this cleave? and what does it not cleave?

cleaves geminal diols, not geminal esters

what does the lactase enzyme do?

cleaves the bond between galactose and glucose (2 components of the disaccharide lactose) 1,4 acetal linkage is cleaved by lactase

what keeps tertiary butanol from degrading?

cold and pyridine

basic procedure for synthesizing a Grignard reagent -what do Grignard reagents act as?

combine solid Mg with an alkyl halide in diethyl ether and reflux all Grignard reagents function as bases =participate in acid-base reactions

what happens when light is reflected?

comes in high but leaves at lower energy, reflection would be a color of light with lower energy state (longer wavelength)

BL definition

common definition of acids as proton donors and bases as proton acceptors

nucleophiles

comparatively e rich and react with partially + species

solution reaction with acids and bases

compare moles acid to moles base and determine which one will be in excess

how is the Ksp calculated?

concentration of each ion raised to the stoichiometric coefficient

relative acidity

consider stability to conj base

urea

contains amide fxn group

what is special about cellulose?

contains beta 1,4 acetal linkages between glucose molecules. Humans dont have the enzymes to cleave these bonds, alpha 1,4 acetal linkages we can cleave

achiral

could contain a plane of symmetry: has a superimposable mirror image

ion-exchange chromatography -could it help to distinguish oxidation state?

could help distinguish oxidation state -yes

what 2 ways are free radicals created? what ways are they not created?

created by UV light and peroxides, not created by cold and water

what is the common ion effect?

creates MORE SALT

entropy

denoted by S

D/L -are these dependent on R/S? which group's placement does it depend on?

depends on placement of amino group, independent of R group

inorganic phosphate

derived from phosphoric acid, the molecule that forms high-energy bonds for energy storage in nucleotide triphosphate like ATP; also used from enzyme regulation

electron-donating

describes groups that push additional electron density toward another atom; stabilizes positive charges and destabilizes negative charges while decreasing acidity

Ionic character

determined by electronegativity difference between bonded elements (look at the distance apart on the periodic table)

what kind of isomers are cis-trans?

diastereomers

what category do epimers fall into? what can we expect from their chemical properties?

diastereomers solubilities/BP/MP similar but not identical

galactose differs from glucose at which carbon? mannose? aldose? what kind of substance is fructose?

differs from glucose C4, mannose at C2 and allies at C3, fructose is a keytose

if a dicarboxyl group is put into water, what will happen?

dissociation which elevates bp

recrystallization

dissolve in solvent soluble at high temp and insoluble at low temp *you don't want contaminates to share the properties of the solution trying to purify

will SDS break covalent bonds during process?

does not break covalent bond

beta unsaturated keytone

double bond between alpha and beta carbons

thermodynamic enolates

double bond is more substituted than kinetic

The half-life of 18F is 110 minutes. If 5 grams of FDG remain after 5 hours and 30 minutes, how much energy was emitted from the patient's body in the form of gamma rays from radioactive decay of the FDG? **what is eV?

eV is bond dissociation energy 511 keV per molecule This question requires us to determine how many half-lives have occurred after 5.5 hours. 5.5 hr x (60 min/1 hr) = 330 minutes, or 3 half-lives. Working backwards from the final amount of 5 g we see that we must have started with 40 g of of the original sample: 5 g x 2 (first half-life) x 2 (second half-life) x 2 (third half-life) = 40 g original sample Thus 35 grams of FDG must have undergone positive beta decay. 35 grams x (1 mol/181 grams) = about 1/6 mol of FDG, which means that 1/3 moles of gamma rays were emitted. (1/3) x (6.022 x 1023) * 511 keV = approximately 1 x 1026 keV.

batteries are an example of what type of cell?

electrolytic

what is the difference between an electrolytic and galvanic cell?

electrolytic, travels against spontaneity, anode (+) while cathode (-)

bohr model

electromagnetic radiation = jumps in a fixed orbit

Pauli exclusion principal

electrons have to have opposite spins

metals

electropositive, low ionization energies, low electron affinities

which elements have larger atomic radii?

elements to the left, atomic radius increases from top to bottom and from right to left

what is the formula for enantiometric excess?

enantiomeric excess = observed optical rotation *100/specific rotation

configurational isomers -what three types of isomers does this term include? **how can they be interconverted?

enantiomers, diastereomers, cis-trans isomers **by breaking bonds

ring strain

energy created in a cyclic molecule by angle strain, torsional strain, and non bonded strain; determines whether a ring is table enough to stay intact

Gibbs free energy -what is the energy available to do?

energy of a system available to do work

what is the ionization energy?

energy required to pull an electron off

ionization energy

energy required to remove an electron from orbit about a gaseous atom into free space. increases from left to right and from bottom to top

classic buffer: what is a way to make a classic buffer if you have strong bases?

equal amounts of weak acid and conj base moles do not have to be equal to each other, strong bases are not good buffers, but if you have much less then the weak acid, it's ok acid should have a pKa closest to desired pH

The table below gives the reduction potentials of two metals. A system is constructed with two electrodes, one made of solid silver and the other composed of solid tin. Current from an external generator is then applied to power a reaction. If the species above are the only ones that react in the solution, the E°cell for this system could be: A. -0.94 V. B. -0.66 V. C. 0.94 V. D. Both A and B are possible values.

ere, the question stem asks for E°cell, or the standard-state cell potential. As a result, we can assume that this apparatus is constructed under standard conditions. Additionally, the stem mentions the use of external current to promote this reaction, implying that it must be nonspontaneous, or electrolytic. Thus, the tin cation would reduce (E° = -0.14 V) while the silver metal would oxidize (E° = -0.80 V). (-0.14) + (-0.80) = -0.94. B, D: While the negative sign here does denote a nonspontaneous reaction, this value could not be obtained with these metals under standard conditions. C: This answer choice corresponds to a galvanic cell, not an electrolytic one. We can easily discern this fact due to the lack of a negative sign. Remember, the question stem implies that our correct answer choice should correspond to an electrolytic cell.

the glycerol backbone of a triacylglycerol is attached to 3 fatty acids via

ester bonds

IUPAC name ethene

ethylene = non IUPAC ethyne = acetylene

HOMO

excited from HOMO to LUMO

do unsubstituted or substituted alkanes have higher boiling points?

exert only London dispersion forces-low BP adding halogen would raise BP because of dipole-dipole interactions

is photo emission endo or exothermic?

exothermic

what would you expect from the Kb of a weak base?

expect it to be closer to 0

what causes a smaller bp elevation?

fewer number of particles

In 250 mL of the MH solution with the most favorable solubility profile (15 mg and 40mg/mol), how many moles of nicotinamide (MW = 122 g/mol) are present?

find how many mL are present, because you already know how many you have for a certain concentration (40mg/mL, means that for every one mL, there are 40 mg. So with the ratio, then you know that there are 15 mg of the substance that you want per mL), and multiply by the number of mL to find grams. Convert grams to moles

empirical formula

find mole ratio

how to make the cell potential more positive?

find reaction that produces the solid cathode (reduction at the cathode) and increase the reaction

what is the formula for flow rate?

flow rate = cross-sectional area x velocity.

what causes spectroscopy, in general?

frequencies of electromagnetic radiation (light) absorbed by the molecules or their response to a magnetic field

threshold frequency

frequency that must be surpassed for a photo electron to be emitted

hemiacetal

functional group that contains a carbon atom bonded to one -OR, one -OH, an alkyl chain, and a hydrogen atom

furan furanose

furganose: sugar molecule (like), alternating -OH groups and CH2-OH on carbons 1 and 4 furan: ring containing 4 carbons and 1 oxygen

Kp

gases, treat like Keq

maltose

glucose + glucose, connected by ALPHA 1,4 linkage

if a current was generated?

glucose was being made

Markovnikov

halogen will bind to more substituted end of double because H bonds to less substituted end in radical run, opposite

What kind of compound is Furan?

heteroaromatic

what is the equation for photon frequency if you're given the principal energy level number?

hf = R (1/n^2(final) - 1/n^2(initial))

what is hf equal to in a photon equation?

hf = the energy that the incident electron has

what is a super critical fluid?

high high temp and pressure

E

highest priority groups on opposite sides of double bond

Z

highest priority groups on same side of double bond

Based on the passage results, how would the mass and stiffness be expected to vary along the length of the membrane? A. They would both be at a maximum value at the base and decrease along the membrane to a minimum at the apex. B. They would both be at a minimum value at the base and increase along the membrane to a maximum value at the apex. C. The stiffness would be highest at the base, while mass would be highest at the apex. D. The stiffness would be lowest at the base, while mass would be lowest at the apex.

his question requires us to examine Figure 2. We need to determine how the mass and stiffness would be expected to vary to produce the results shown in Figure 2, which shows the highest frequency points of resonance at the base and lowest frequency points of resonance at the apex. Paragraph 3 also states that each location can be modeled like an individual spring that will vibrate only at its natural frequency. The equation for the frequency of a spring is f = (1 / 2 π)(√k / m). Thus, in order to have the maximum frequency at the base, the base should have the maximum k and minimum m values while the situation is reversed at the apex. A, B: Frequency and mass are inversely related, so we would expect the maximum of one value to co-occur with the minim of the other. D: Instead, the stiffness is highest at the base, and the mass is lowest at the apex. Give feedback on this question

what does higher boiling point correspond to? -what type of bonds?

hydrogen bonds

ketals -under what conditions are they hydrolyzed?

hydrolyzed by strong acid and H20

ideal gas

hypothetical gas containing particles with zero volume and with no attractive intermolecular forces

Tollens reagent -what does it do? how does it do it?

identifies reducing sugars silver mirror indicates reducing sugar (glycogen has no free, reducing ends)

2nd order

if rate and concentration have an exponential relationship

1st order

if rate and concentration increase by the same factor, raise to the first power

How can NMR be tricky?

if same molecule/same attachment, only count H once and if benzene ring, only count 3

in vitro

if tests are run in vitro and also not, then compare the data for each

esters - what kind of conditions do you need for esters to be leaving groups?

if you wish to create leaving groups, you need acidic conditions

solving for molarity -what to do with density?

ignore the density (use it for molality)

coupling -what happens to the peaks?

in NMR spectroscopy, a phenomenon that occurs when there are protons in such close proximity to each other that their magnetic moments affect each other's appearance in the NMR spectrum by subdividing their peaks into sub peaks; also called splitting

keto stability v enol

in general, keto is more stable, but when enol connected to conjugated system, it's more stable

what is a chalcogen?

in the O2 family

what direction does the current flow?

in the opposite direction as the electrons flow

what does the intensity of the incident ray correspond to?

incident ray determines the number, not the individual energies of the ejected photoelectrons

what do acid catalysts do? what does it do to the carbonyl carbon? what kind of reaction likes an acid catalyst? **which reaction is very efficient in the presence of an acid catalyst?

increase the electrophilicity of the carbonyl carbon -protonation of the carbonyl oxygen makes carbon oxygen bond weaker which makes the C more positive and electrophilic acid catalyst-most efficient way to carry out aldol condensation

torsional strain

increased energy that results when molecules assume eclipsed or gauche staggered conformations

non-bonded strain

increased energy that results when nonadjacent atoms or groups compete for the same space; also called steric strain

what does increasing temperature do to osmotic pressure?

increases it

what causes decreasing entropy?

increasing mass, because this would encourage the solid state which decreases entropy **I is more stable than F

Enantiomers

indistinguishable chemical/physical properties only separated by reaction with chiral carbon

electrolyte

ion in solution that is capable of conducting electricity in that solution

What kind of bond forms between NaH?

ionic

surfactant

is correct. During inspiration, contraction of the diaphragm and internal intercostal muscles leads to expansion of the thoracic cavity and a decrease in intra-pleural pressure. This negative pressure, relative to atmospheric pressure at the entry of the upper airway, generates airflow through the respiratory tree and to its terminal extension—the alveoli. The elastic recoil force of the airway and the surface tension of the water lining the airway oppose expansion of the alveoli due to the influx of atmospheric pressure. Pulmonary surfactant adsorbs to the air-water-alveoli interface, reducing surface tension and the total force resisting expansion. This increases pulmonary compliance—a measure of lung volume change at a given pressure of inspired air—and decreases the work required to expand the lungs at a given atmospheric pressure. This is consistent with choice C. In general, surfactant molecules are amphipathic, meaning that they contain both hydrophobic and hydrophilic regions. The diagram below shows surfactant molecules surrounding a micelle of oil. The hydrophobic tails of the surfactant molecules mix well with the hydrophobic oil, while the hydrophilic heads point away from the oil droplet.

If a given channel in a nerve membrane shows a preference for transporting larger solutes, which of the following would increase the likelihood of a given particle being transported by the channel? A. Removing an electron from a K atom to create a K+ ion B. Subjecting a dipeptide to proteolysis C. Adding an electron to a Cl atom to create a Cl- ion D. Removing two electrons from a Ca atom to create a Ca2+ ion

is correct. The question states that the channel favors larger solutes. Adding electrons to make a negatively charged ion would increase the radius of the particle, making it larger. Thus going from a Cl atom to a Cl- ion would increase the likelihood of it being transported by this channel. A, D: Removing electrons to make positively charged ion would decrease radius. Making the particle in question smaller would make it less likely to be transported by this channel. B: Cleaving a larger molecule (dipeptide) into small pieces (amino acids) would, as with the positive ions in choices A and D decrease the likelihood of this channel transmitting the particle.

Carbon's electron affinity

is higher than N, even though this goes against the trend Due to almost half filled subshell, so Carbon wants one more and N has a half-filled so doesn't want to gain

tautomers

isomers that can interconvert by changing the location of a proton

what condition does SN2 need as far as steric hindrance?

it needs to lack it

what happens if light is emitted?

it's absorbed as photons, electrons then are excited, but then fall and are released, the (re)emission is the same color

how does ethanthiol compare to ethanol?

it's more acidic

What does CrO3 do? what special product does it have?

it's used in oxidation reactions, forms quinones

cyclic ketal

keystone uses H20, makes dial, dial containing 2 016 and keynote containing O15 (from water)

tautomers - what functional groups are tautomers?

keystone, aldehydes, imines, but not imides (anhydride, but with nitrogen in the middle)

sulfur

larger than O, the conj base is better able to delocalize neg charge, increasing stability and thus acidity

free radicals

learn something about the reaction

what absolute configuration does cystine have? R or S?

lone R residue

catalyst

lower Ea but do not impact the delta G (are not thermodynamic)

what do SN1 reactions do to optical activity?

makes a racemic mixture because the molecule is planar when it attacks

sucrose -under what conditions hydrolyzed? what do they make? are they acid sensitive? able to undergo nucleophilic attack?

may be hydrolyzed using HCl acetals are acid sensitive non-reducing sugar, so unable to undergo nucleophilic attack

what does a small Ksp mean?

means that the compound is less soluble (more likely to precipitate out)

what is fusion?

melting

what 2 groups offer great steric protection?

mesylate/tosylate

what needs to be reacted with benzyl alcohol to make an ether?

methyl iodide and sodium hydride

m

molality

constitutional isomers

molecules that have the same molecular formulas but different connectivity; also called structural isomers

larger the Kb

most basic ion

sN2

most rapid in polar aprotic environments

what are 2 important properties of meso compounds?

multiple chiral centers and are not optically active

N (normality) -what do you need to remember to do if asked for molarity in the end?

multiply moles by number of equivalents Remember to divide by Liters if given for solution

E2 -what two requirements? what about for E1?

must involve H anti to the LG strong base = E2 weak base = E1

Balmer series

n = 2

Lyman series

n=1

Paschen

n=3

weak acids

nitrous, hypoclorous, sulfurous, acetic, hydrocyanic

is carbon the only atom that can be a chiral center?

no

what does nonvolatile mean?

no vapor pressure of its own (NaCl)

is a carbonyl a H-bond donator?

no, it's an acceptor, but the donators have to be connected to FON

what solvents are best for mono substitution?

non polar solvents

enantiomers -what can be expected about the chiral center? what chemical/physical properties?

non superimposable stereoisomers that are mirror images of each other. Different in configuration at every chiral center but share the same chemical properties in a nonstereospecific environment. Optical activities are exactly opposites of each other.

Is Keq affected by pressure?

not affected by pressure

Is ubiquinone polycyclic? Can it be reduced? Can it be oxidized? Is it conjugated?

not polycyclic, can be reduced, cannot be oxidized, is conjugated

how is Zeff charge determined? effective nuclear charge

number of protons minus the non valence electrons

what is the formula for specific rotation?

observed optical rotation/path length*concentration

isobaric process

occurs at constant pressure

glycoprotein -what 3 sugars compose the oligosaccharide?

oligosaccharide chain composed of all the following: mannose, glucose, and galactose

anomers

one stereo center differs

Tollens Test/what is a reducing sugar? Is glucose/sucrose?

only aldehydes Tollens' test is intended to identify "reducing sugars," or sugars with the capacity to serve as reducing agents. Specifically, sugars with hemiacetal groups can undergo mutarotation, allowing them to be oxidized by CuO. The process of mutarotation requires ring opening, which occurs at a hemiacetal group. Thus, sugars with hemiacetal groups can be oxidized and can thus function as reducing sugars. We do not need to have the structures of glucose and sucrose memorized to answer this question. Instead, simply note that the question stem tells us that glucose yields a positive Tollens' test. Glucose must therefore contain a hemiacetal group, whereas sucrose (which gives a negative test result) must not. The hemiacetal group on glucose, shown below, marks it as a classic reducing sugar. Remember, a hemiacetal consists of a carbon atom directly attached to one -OR and one -OH group. The same carbon atom is also attached to a hydrogen atom and an R group.

what does Ka change in response to?

only changes in response to temperature

Hunds rule

orbitals will be occupied in order of increasing energy (every orbital must contain one E before they're paired)

oxidation/reduction (oxidation numbers)

oxidation = increase reduction = decrease

anode

oxidation occurs, electrons flow away from the anode

what will CrO3 (chromic acid) do to a reaction? what can it not oxidize?

oxidize all the way to carboxylic acids and keystone, cannot oxidize tertiary OHs

What causes RNA to bind to its polymerase?

oxygen has a negative charge and the amino acid side chains Lys and Arg on RNA polymerase are attracted to it

percent by mass

part/TOTAL

what are the 2 tenets of the ideal gas law?

particles have no negligible volume and no attractive forces between them

what does a hydroquinone look like?

phenol with an extra -OH

What solvents are used during polysubstitution?

polar solvents

acetone

polar, aprotic best for SN2

H2SO4

poor nucleophile, unlikely to cause unwanted side rxns

What is the heisenburg uncertainty principle?

position and momentum cannot both be known

what does Ne do in a Gabriel synthesis?

prevents atmospheric elements from reacting with potassium Thad

Tosylation of benzyl alcohol -what does it accomplish?

prevents it from transferring proton H is transferred to pyridine

what does a salt bridge do?

prevents salt build up by allowing sulfate anions to travel from anode to cathod

what does PCC oxidize best? what into what?

primary alcohol into aldehyde

what is meant by enhancement ratio? **in water **what is being measured?

probably compared to the solubility/effect in water

What happens when an amide is hydrolyzed?

products are carboxylic acid and amine

acetyl chloride + pyridine, what do they do?

protects naked -OH moieties

SN1

protic environments

BL acid

proton donor

nucleon

proton or neutron

Effective nuclear charge

protons - all electrons in principle shell not concerned with **not concerned with oxidation or ions because those deal with outer electrons

electron-withdrawing

pull electron density away from another atom; stabilizes negative charge while destabilizing positive charge, increases acidity

capacitance

q/v

what are the 4 ways to measure/denote concentration?

quantified by mole fraction, molarity, molality, normality

what kind of mixture does it result in, Sn2? will the product rotate polarized light?

racemic mixture will not rotate polarized light

chemical kinetics will raising temp always increase the total amount of products? Does the activation energy forward equal the activation energy reverse? do catalysts change the delta G?

raising temp will not always increase total amount of products activation energy forward does not equal reverse catalysts do not impact delta G

grahams law what is the rate of gas/diffusion proportional to?

rate 1/rate 2 = (mm2/mm1)^1/2 = time1/time2 rate is proportional to 1/(density)^1/2

0th order

rate doesn't change when concentration changes -eliminate from rate law

what is the solution equilibrium? in terms of rate?

rate of dissociation equals the rate of precipitation of the solute, regardless of any additional solute introduced into the mixture

collision theory of chemical kinetics

rate of reaction is directly proportional to the number of collisions that take place between reactants per second

reaction quotient

ratio of the concentrations of the products to the concentrations of the reactants at any point during a reaction, where each reactant and product in the expression is raise to the power of its stoichiometric coefficient. Commonly denoted by Q. Usually compared to the equilibrium content to determine the direction a reaction will proceed to regain equilibrium

Rf

ratio used in thin-layer chromatography to identify a compound; calculated as how far the compound traveled relative to how far the solvent front traveled

oxidation

reaction in which a species loses electrons

E cell greater than 0?

reaction is energetically favorable and maybe spontaneous

Decreased number of alveoli in the lungs leads to respiratory distress because:

reduced surface area in the lungs reduces the rate at which O2 and CO2 can diffuse through the lung epithelium. Many biological processes depend on a very high surface area-to-volume ratio. The large number of tiny alveolar sacs in the lungs, the presence of microvilli in the small intestine, and the folding of the inner mitochondrial membrane are all examples in which an increased surface area allows for biological processes to take place more quickly. When the individual alveoli break down (as, for example, in emphysema) the lungs lose the necessary surface area to allow for effective diffusion of respiratory gases. The alveoli and their larger environment (the human lungs) are shown below.

LAH

reduces aldehyde to primary alcohol

describe mass spectroscopy -how is the sample immobilized in the gas phase? what are the molecules stripped of?

requires immobilization of sample in the gas phase, accomplished via bombardment with ions, molecules passing through mass spectrometer are stripped of surrounding water or lipids (could alter protein) movement through the device is proportional to both mass and charge (cannot be measured directly, only be clocking time of flight-then ratio can be obtained) peaks correlate to mass

how are acetals and ketals in the presence of bases?

resistant

what is an anomeric carbon?

separates 2 oxygens in the structure

fractional distillation

separates compounds with very similar roiling points (less than 25 degrees) -hexanol/pentanol most likely to have similar boiling points

AgCl

silver chloride insoluble

What is Edman degradation? what is CID? what is the end product?

similar to CID (collision induced dissociation = free radicals), sequential cleavage of terminal amino acid resides from polypeptide

nitrate salts

soluble

Ammonium and nitrate

soluble salts

titrant

solution of known concentration added in small volumes to a solution of unknown concentration to reach the equivalence point

nucleophile

species that tends to donate electrons to another atom. With the same attacking atom (OH-, CH3O-) in aprotic solvents, nucleophile strength correlates to basicity. In protic solvents and in situations where the attacking atom is different (OH-,SH-), nucleophile strength correlates to smaller size

common ion effect

states that the molar solubility of one salt is reduced when another salt, having a common ion, is Brought into the same solution

diastereomers -how many chiral centers are shared? how do their physical and chemical properties compare?

stereoisomers that are not mirror images of each other. Diastereomers differ in their configurations at at least one chiral center and share configuration at at least one chiral center. They have similar chemical properties, but different physical properties

conformational isomers

stereoisomers that differ by rotation about one or more single bonds, usually represented using Newman projections

perchloric acid

strong acid

what does E2 need?

strong base

what is a carbonyl's functionality?

strong dipole intermolecular force

lower the Pka

stronger the acid

what will a reaction of tertiary alcohol with h2SO4 and methanol (weak nucleophile) make?

structural and stereoisomers

glucose and fructose

structural isomers

constitutional isomers

structural isomers bonded differently - different construction of bonds

striker synthesis

synthesis of amino acid from aldehyde/keytone, cyanide, and ammonia. Aldehyde + ammonia in acid = imide + CN = + H+ = amino acid condensation followed by hydrolysis

what kind of substrate is ideal for an E1 reaction? what kind of conditions are needed for reaction?

tertiary substrate, needs mild base, heat, polar protic

what kind of geometry does a sulfate ion have?

tetrahedral

In a population of Amish people, the frequency of the recessive autosomal allele for polydactyly is 1.2%. What percent of the population are heterozygotes for the polydactyly allele? A. 0.0144% B. 1.19% C. 2.37% D. 97.6% *what equation do you start with if you are given frequency of the ALLELE? (alleles are short and sweet) *what if you are given the frequencies of the individuals?? (individuals are long and complicated)

that the total number of alleles in the population has to add up to 1: A + a = 1 And the total number of genotypes in the population must also add up to 1: AA + 2Aa + aa = 1 We're told that a = 0.012. (Note that the question gave the frequency of the recessive autosomal allele, not the frequency of individuals in the population that are homozygous recessive!) By the first equation above, A = 0.988. The carriers are the heterozygotes with the genotype Aa. Their frequency is: 2Aa = 2 x 0.988 x 0.012 = 0.988 x 0.024 At this point we've got what looks like a tough calculation to do, so we should probably back up and start estimating. Our calculation is telling us to take 98.8% of 0.024, so our answer is going to be really close to 0.024 (since 100% of any number is just that number itself [e.g. 100% of 56.7 is 56.7]). If we look at the answer choices, the only one that's remotely close is 0.0237, or 2.37%.

decarboxylation

the complete loss of a carboxyl group as carbon dioxide

delta G

the difference between the starting and ending points

cathode

the electrode at which reduction occurs during a cell's oxidation-reduction reaction. Electrons always flow towards the cathode in an electrochemical cell

if the absorbance increased?

the enzyme was involved in debranching linear starch has no debranching so no absorbance seen

absolute configuration

the exact spatial arrangement of atoms or groups in a chiral molecule around a single chiral atom, designated by (R) or (S)

in gas chromatography

the mobile phase is a gas, such as He, liquid is stationary (inert solid supports liquid)

effusion

the movement of gas from one compartment to another through small opening under pressure

how is formal charge determined?

the number of valence electrons minus the number of bonds minus the nonbonding electrons

which elements have the greatest Zeff?

the ones with little shielding. The electrons are close to the nucleus

if asked to mentally rotate at the groups appear to be on the same side (both sticking backward), then what type of absolute configuration does it have?

the opposite of what you think

shielding

the phenomenon of atoms pushing electron density toward surrounding atoms; in NMR spectroscopy, pulls a group further upfield on the spectrum. Deshielding is the opposite-the pulling of electron density by surrounding atoms; in NMR, pulls a group further downfield on the spectrum

steric hindrance

the prevention of a reaction at a particular location in a molecular by substituent groups around the reactive site

transesterification -what type of environment? what type of reaction (sn1 or sn2)?

the process that transforms one ester to another when an alcohol acts as a nucleophile and displaces the alkoxy group on an ester **all hydronium, but alcohol acts as nucleophile (sn1)

NMR - what causes a shift left?

the proximity of hydrogen to electron withdrawing groups suggests it resides in a zone of electron definciency, causes it to resonate at a frequency distinct from other Hs (shift left if more exposed/less shielded) reference point Hs (TMS) are at the extreme right

what is the relationship between Ka and acid strength? what is the relationship between Kb and base strength?

the smaller the Ka, the weaker the acid. A small Ka will correspond to a large pKa. the smaller the Kb, the weaker the base. A small Kb will correspond to a large pKb.

relative configuration

the spatial arrangement of groups in a chiral molecule compared to another chiral molecule

if one-step reaction -what order is a one-step reaction?

then 2 reactants come together making the rate second order

What happens if the ion product is less than the Ksp?

there will be no precipitation

what is tetroxide in relation to tertiary butanol?

they are each others conj acids and bases

when a question asks about a study's significance?

they're asking you to examine the error bars or the p values

what are the properties of Ca and Sr?

they're shiny silver solids at room temperature

carboxylic acid derivatives

thirsters and acyl halides

How can EtBr be used?

to stain singe-stranded DNA, it will bind

enthalpy

total heat content of a system at constant pressure, denoted by H

heat

transfer of energy from a substance with a higher temp to a substance with a lower temp

facilitates the irreversible binding of cephalosporin to the active site of PBP in a manner that is not competitive inhibition.

translation: SUICIDE INHIBITION

what is TLC? what is the stationary phase?

type of chromatography that uses silica gel or alumina on a card as the medium for the stationary phase

boyle's law

under constant temp, the volume of an ideal gas is inversely proportional to its pressure PV = PV

rate laws -can we determine the rate law if it's expressed that the reaction is elementary/single step? do rate laws contain the products? what is the formula?

unless a reaction is specified as an elementary/single step rxn, is rate can only be determined experimentally unlike Keq, should not contain products in equation rate = k (reactant)^coeff * (reactant)^coeff

paramagnetic v dimagnetic

unpaired/paired

what does an SN1 run generate?

unstable intermediate with vacant orbitals

to make an anhydride 1. what are the reactants? 2. what needs to be used for catalyst? 3. what does it need to be combined with after? 4. what is PCl5 commonly used for?

use two carboxylic acids, with Pcl5, extract product, combine with additional acid and heat Use PCL5 if you think acid is not sufficiently reactive = commonly used to convert carbon to acyl halides

HSQC -what is it useful for explaining?

useful for explaining ligand binding interaction which do not involve formation of new covalent bonds (through space, not bonds)

electrochemical cell

uses an external electric source to drive a non spontaneous oxidation-reduction reaction

nomenclature esters methyl acetate

usually -oate methyl acetate = methyl groups attached to both the carbonyl carbon and the oxygen

aqueous solution

water is solvent

what are the pKbs for amine groups?

when contains electron-donating groups that stabilize the + charge lowest pKb = alkyl amines -primary amines then ammonia then secondary amines tertiary aryl amines have low pkas, actually they're more acidic heterocyclic amines, basic, except imidazole and pyridine (proton donators)

Grignard reagent -what will it react with? what will it not react with?

will not react with alkane, needs carbonyl

enols -what will the tautomerize to form?

will tautomerize to form keytones/aldehydes!!

Will a heavier gas diffuse more slowly than a lighter one?

yes

will ethanol evaporate even when VP is less than atmospheric pressure?

yes

does O participate in amide bonding?

yes and it is the most electronegative element