Chem/Phys Kaplan FL 7
Q: Citric acid is a weak triprotic acid. Because it has multiple acidic protons, a solution of citric acid is: A: An effective buffer over several pH ranges.
Explanation: A buffer is a mixture of a weak acid and its conjugate base (or vice versa), and resists changes in pH. Buffers work best when the pH is near the pKa of the acid. For example, if the pKa of an acid used to make a buffer is 5, the range the buffer will work best at is pH 4-6. The pH varies least from the pKa of the acid when there are equal concentrations of the weak acid and its conjugate base. Because citric acid is triprotic, it has 3 pKa values and 3 different weak acid states (H3A, H2A-, HA2-). H3A, H2A-, HA2- are each able to donate a proton, meaning that each species, along with its conjugate base, can act as a buffer. This gives citric acid the ability to act as a buffer across three different pH ranges.
Q: During the synthesis and purification of the thalidomide derivatives, what is the purpose of washing the organic layer with distilled water? A: To remove any aqueous impurities.
Explanation: A common procedure used to remove impurities after a reaction is to wash the products with the opposite phate (organic or aqueous). Since the thalidomide derivatives are in the organic layer, an aqueous wash should remove any water-soluble impurities.
Q: From the data presented in figure 2, which GOx variant is most stable? A: T132S/T56V
Explanation: A higher melting temperature indicates a more stable protein, because more energy is required to unfold it. According to figure 2, T132Y/T56V has a melting temperature of approximately 44 degrees celsius, which is the temperature at which the protein is 50% unfolded. The melting point is higher for TT132Y/T56V than for any other variant, including wild-type.
Q: Which amino acids are found in the active site of GatD? A: Cys and His
Explanation: According to figure 1, three active site residues of GatD are arginine-128, histidine-189, and cysteine-94. The three-letter codes for these amino acids are Arg, His, and Cys.
Q: Which statement accurately describes how the presence of submerged vegetation can interfere in the determination of stream depth using LIDAR? A: Incoming green light can be reflected by submerged vegetation before it reaches the streambed.
Explanation: According to paragraph 1, "NIR" photons are absorbed or reflected from the water surface, while green light penetrates the water and reflects off submerged vegetation of the bottom." Thus, it is green light that indicates the distance from the laser to the bottom of the stream. However, if that light reflects off submerged vegetation in the water column instead of the streambed, and is picked up at the detector, an artificially shallow measurement of the water's depth can result in incoming green light being reflected by submerged vegetation before it reaches the streambed.
Q: What is the work done by the microwave during the reaction? A: 90kJ
Explanation: According to paragraph 3, the sample was heated for 10 minutes, or 600 seconds, at 150W, which is a measurement of power, or change in energy per unit time. To get work, these two values must be multiplied: power= work/time work= (150w)(600s)= 90,000 kJ
Q: St STP, the volume of N2(g) produced by the complete reduction of 3 moles of NO2 via reaction 3 would be closest to which of the following? A: 78L
Explanation: According to reaction 3: 6NO2+8NH3 --> 7N2+12H2. Thus, when 6 moles of NO2 react ,7 moles of N2 are produced. In this question, only 3 moles of NO2 are reacted, so only 3.5 moles of N2 will be produced. At STP, 1 mole of a gas occupies 22.4L, so the volume occupied by 3.5 moles of N2 is 22.4L/mol * 3.5 mol= 78L
Q: What reaction is catalyzed by MuT? A: Amide group formation
Explanation: According to the passage and reaction 2, MurT uses ammonia and ATP to form D-iso-Gln from D-iso-Glu. The end of the Glu side chain contains a carboxylic acid, while the end of the Cln side chain contains an amide group. This suggests MurT replaces the hydroxyl group on Glu with NH2 from ammonia in an ATP-dependent reaction, leading to the formation of Gln.
Q: Luminol produces stronger and longer-lasting CL with dried blood than with fresh blood. Based on the passage, which of the following is the most plausible explanation for this? A: Iron ions are more available in dried blood than fresh blood, where organic ligands are more likely to interfere with their catalysis of the CL reaction.
Explanation: According to the passage, Fe2+ is a common catalyst for reaction 2, so it is logical to hypothesize that iron ions are catalyzing the CL reaction used to detect blood. The iron ion in blood is complexed with a porphyrin ring in the heme group of proteins like hemoglobin, and since dried blood probably has fewer intact heme groups than fresh blood, iron ions in dried blood should be more available for catalysis. Such interference with CL catalysis by organic molecules is also specifically described in the passage. The lessening of this interference in dried blood, therefore, plausibly explains why luminescence from dried blood is more vigorous.
Q: How do the waves involved in collection of ground penetrating radar (GRP) data compare with the waves involved in collection of LIDAR data? A: GRP waves are the same type of wave, but have longer wavelengths.
Explanation: According to the passage, both GPR and LIDAR use electromagnetic radiation, which consists of transverse waves. GRP uses radio waves, however, which have much lower frequencies and longer wavelengths than the infrared and visible light waves used in LIDAR.
Q: What is the role of the Cu/ZrO2 material in SCR? A: It decreases the activation energy of the reactions.
Explanation: According to the passage, the ZrO2 support material loaded with Cu functions as the catalyst for SCR. Catalysts do not alter the equilibrium position, or the change in free energy, enthalpy or entropy of a reaction. Instead, they speed up a reaction by decreasing its activation energy.
Q: Intrinsically disordered regions (IDRs) are stretches of amino acids within a polypeptide that lack any secondary structure and interact freely with the aqueous environment. Two proteins are analyzed for their tendency to contain IDRs, with scores closer to 1.0 indicating an increased tendency toward IDR formation. Compared with Protein B, protein A likely: A: Contains more hydrophobic amino acids
Explanation: According to the question, IDRs lack secondary structure and interact freely with the aqueous environment. Protein A has significantly more areas that score near 1 on the scale than does protein B, indicating an increased tendency toward IDR formation. Since these IDRs "interact freely with the aqueous environment," they are likely to be composed primarily of hydrophilic amino acids, as hydrophobic amino acids tend to be buried within protein structures.
Q: When an electron falls from n=3 to n=2 in a hydrogen atom, what is the value of the energy released, given that A is the energy needed to move an electron from the ground state of a hydrogen atom to an infinite distance from the atom? A: 0.14 A
Explanation: According to the question, the energy released by an electron falling from infinity to the ground state is A. As electrons fall from a higher to a lower energy state within the atom, energy is also released. The higher the principal number (n), the greater this energy release. Therefore, since A represents the maximum amount of energy an electron can release, in going from n= infinity to n=1, but the electron in the question is only falling from n=3 to n=2, the energy released will be smaller than A. The only choice with a non-zero value smaller than A.
Q: Based on the results of the experiment, which of the following conclusions can be drawn about the reaction in figure 1? A: PAH increases the reaction rate.
Explanation: According to the table, the rates of the reaction depicted in figure 1 are higher in trial 1 than in trial 2 (trial 2's numbers are in pM, not nM), hence the presence of PAH increases reaction rate by several orders of magnitude, and the statement in the correct answer is true. On the other hand, enzymes only affect the kinetics of a reaction, not its thermodynamics; since delta H (enthalpy change) is a thermodynamic measure, unaffected by PAH, the second roman numeral is incorrect. Finally, in general, reaction rates increase with temperature, but there is no reason to believe- nor data in table 1 to indicate- that the reaction rate doesn't continue to increase at temperatures higher than 30 degrees celsius. The conclusion in roman numeral 3 can't be drawn.
Q: Which of the following changes made to the experiment would create a variant of Compound 8 that contains an isotopic substitution in the position marked by the asterisk? A: Use 15N labeled ammonium chloride
Explanation: An isotopic substitution requires a radioactive 15N. According to Scheme 1, while the nitrogen in the left-most ring is derived from compound 2, the nitrogen in the right-most ring is introduced with the addition of ammonium chloride. -Radioactive isotopic substitution requires 15N not 14N.
Q: In the reaction above, what is the change in the oxidation state of nitrogen? A: 0
Explanation: Because NH3 is a neutral molecule, the oxidation numbers of its atoms must add up to zero; each H is assigned an oxidation number of +1, yielding an oxidation number of -3 for N. Since NH4+ has a charge of +1, the oxidation numbers of its atoms must add up to +1; each H has an oxidation number of +1, again yielding an oxidation number of -3 for N. Since the oxidation number of nitrogen does not change in this reaction the answer is 0.
Q: If the blood pressure of the 182 cm tall individual discussed in the passage is measured at the level of the head instead of the heart, what will most likely be the approximate mean arterial pressure? A: 39 mmHg
Explanation: Change in gravitational potential energy density is equal and opposite to change in hydrostatic pressure, where h is the height of blood above a reference point. Since the height term is larger at the head than at the heart, subtracting a larger pgh term in equation 1 will yield a lower arterial blood pressure at the head than at the heart- where pressure is 83mmHg. Arterial pressure must still be greater than 0mmHg, however, or blood would not continue circulating.
Q: The action of phenylalanine hydroxylase on phenylalanine results in: A: Conservation of stereochemistry, since the primary chiral center is not altered.
Explanation: Chiral centers are sp3 hybridized, and attached to four different substituents. The carbon that is attached to the hydroxyl group in tyrosine is sp2 hybridized in both phenylalanine and tyrosine, so it is not a chiral center. The chiral center in both molecules is the alpha carbon (relative to the carboxyl group). Since this carbon is not changed in the reaction, its stereochemistry is conserved.
Q: Which of the following correctly represents the products when silver (I) diamine anion, Ag(nH3)2+, dissociates in water? A: Ag+ + 2NH2
Explanation: Complex ions are composed of a central metal ion and surrounding ligands. The ligands are typically compounds that can donate an electron pair, making them Lewis bases. The bonds formed between the metal and the ligands are weak coordinate covalent bonds, in which both electrons in each bond come from the ligand. In aqueous solution, these bonds can be broken. In particular, the silver diamine anion Ag(NH3)2+ dissociates into Ag+ + 2NH3. Although a small amount of NH3 will go on to react with H2O forming OH- and NH4+, none of the covalent bonds between nitrogen and hydrogen will be disrupted by the mere presence of water.
Q: Based on figure 1, an individual for whom the maximum increase in aortic circumference is 15% would most likely be: A: >80 years old with high arterial elastance.
Explanation: Figure 1 graphs circumferential tension with respect to the increase in aortic circumference. An individual with a maximum aortic circumference of increase of 15% (the line to the extreme left on the graph) has high arterial
Q: Which of the following atoms has the largest first ionization energy? A: Li
Explanation: First ionization energy is the amount of energy required to remove one electron from a neutral atom. First ionization energy decreases moving down a group because the valence electrons in larger atoms are located farther away from the nucleus, and therefore experience weaker attraction to the protons in the nucleus. All the answer choices are Group 1 alkali metals, so they each have a single valence electron, as well as one more proton in the nucleus. All the answer choices are group 1 alkali metals, so they each have a single valence electron, as well as one more proton in the nucleus than the number of inner, shielding electrons. Thus, Li, the alkali metal in the uppermost row of the periodic table, with its valence electron closest to the nucleus, will have the highest first ionization energy.
Q: Strong, soluble bases such as alkaline earth metal hydroxides are not used as antacids, but are often used in laboratory titrations. What volume of an aqueous 1.5 M Ba(OH)2 solution would be required to neutralize 10mL of stomach acid? A: 0.5 mL
Explanation: For neutralization to occur, the moles of hydroxide ions, OH-, from the base, must be the same as moles of H+ from the acid. 0.15N HCl corresponds to 0.15 M HCl since HCl is a monoprotic acid. The number of moles of H+ contained in 10mL of stomach acid is therefore: 0.15 mol/L * 0.01l= 1.5*10^-3 mol. - Therefore, 1.5*10^-3 moles of OH- are needed to neutralize the acid. Because each mole of Ba(OH)2 can give up two moles of OH- upon dissociation, only 7.5*10^-4 moles of Ba(OH)2 are needed for neutralization. Since the concentration of Ba(OH)2 is 1.5M, the volume needed to get 7.5*10^-4 moles of Ba(OH)2 is: 1.5M= 7.5*10^-4/xL xL= 7.5*10^-4/1.5 m= 5*10^-4 L - 5*10^-4 L is equal to 5*10^-1mL, or 0.5mL
Q: Which of the following best explains why 1-chloropentane and 1-pentanol can be separated by fractional distillation? A: Pentanol is capable of hydrogen bonding, and chloropentane is not.
Explanation: Fractional distillation is used to separate a mixture of liquids by heating them until they evaporate at different temperatures based on their differing boiling points. While both 1-chloropentane and 1-pentanol have similar molecular weights, 1-pentanol possessed an alcohol group that allows for hydrogen bonding between molecules, and 1-chloropentane does not. Because of this, 1-pentanol boils at a higher temperature (138 degrees celsius) than 1-chloropentane (108 degrees celsius).
Q: Which of the following amino acid sequences is most likely to be found on the interior of a globular protein? A: AFWIL
Explanation: Globular proteins are primarily water soluble, which means that for the most part, amino acids with hydrophilic side chains are found on their exterior. During folding, hydrophobic amino acids are more likely to fold inward and therefore will be found on the interior of the protein, away from water. the hydrophobic amino acids have nonpolar side chains and include glycine, proline, phenylalanine, alanine, leucine, isoleucine, and valine.
Q: Which is the structure of gluconolactone? A: (See kaplan for answer)
Explanation: Glucose is oxidized by glucose oxidase, so the correct answer should be a version of glucose that is oxidized. One way to monitor oxidation by looking for carbons that have more bonds to oxygen. A lactone is a cyclic carboxylic ester. However, it is not necessary to know this to deduce the correct answer, which should simply be a more oxidized version of the glucose molecule.
Q: Which is the identity of Compound 2? A: Glutamic acid
Explanation: Glutamic acid is an amino acid characterized by the backbone amine and carboxylic acid, with the three-carbon carboxylic acid side chain attached to the alpha carbon. The structure of compound 2 matches glutamic acid.
Q: A scientist uses thin-layer chromatography on silica plates to separate a mixture of phenylalanine and tyrosine. Does phenylalanine has a higher or lower Rf than tyrosine? A: Higher, because tyrosine is more polar than phenylalanine
Explanation: In TLC, the retention factor (Rf) of a compound is equal to the distance traveled by the compound divided by the distance traveled by the solvent front. Silica plates are very polar. Thus, a more polar molecule, with higher affinity for the polar silica stationary phase, will have a lower Rf. Since tyrosine is more polar (due to an additional hydroxyl group) than phenylalanine, it has a greater preference for the stationary phase, and a lower Rf.
Q: In which segment of the figure is the magnitude of acceleration decreasing? A: 3
Explanation: In a velocity-time graph, acceleration is represented by the slope, as acceleration is defined as the change in velocity over time. In segments 1 and 2, the graph slopes upwards, indicating velocity is increasing with time, and thus acceleration is also increasing. When the curve begins to level off in segment 3, the slope of the line becomes less steep, becoming nearly horizontal towards the end of the segment, meaning that acceleration is decreasing to nearly zero. In segment 4, there is no change in velocity, and therefore no acceleration, but in segment 5, the slope of the line becomes steep again, meaning the magnitude of acceleration is increasing again, but in the opposite direction. It remains relatively constant in segment 6, before decreasing again in segment 7, as velocity once again levels off. Hence, the magnitude of acceleration is decreasing in segments 3 and 7 only.
Q: Lung compliance is the volume change in the lung per unit pressure change and accounts for 65% of the work done by intercostal muscles. Increased lung elastance would most likely be counteracted by: A: Increased work by intercostal muscles.
Explanation: Increased lung elastance means increased resistance to stretch and decreased compliance. To counter this decreased compliance, work done by the muscles to expand the lungs must increase.
Q: Ozone has a high oxidation potential, and will readily decompose urea: (NH2)2CO+O3 --> N2 + CO2 + 2H2O. In this reaction, what are the oxidation numbers of carbon as reactant and product, respectively? A: +4 and +4
Explanation: It is simpler to figure out the oxidation number of carbon in CO2: Oxygen has an oxidation number of -2, and there are two oxygens, for a total of -4. To balance this, the oxidation number of carbon must be +4. -Next, determine the oxidation number of carbon in urea by finding the oxidation numbers for all the other atoms. Nitrogen has an oxidation number of -3, and there are two nitrogens, for a total of -6. Hydrogen has an oxidation number of +1, and there are 4 hydrogens, for a total of +4. Finally, oxygen has an oxidation number of -2. The oxidation numbers of nitrogen, hydrogen, and oxygen can be totaled: -6+4-2= -4. Since urea is neutral, carbon will have an oxidation number of +4. Since urea is neutral, carbon will have an oxidation number of +4.
Q: If HCl is the only substance present in the stomach, what is the pH? A: -log(0.15)
Explanation: Paragraph 1 states that the human stomach contains about 0.15N HCl. Normality, N, in the context of acids, is equal to equivalents of H+ per liter of solution. Because HCl is monoprotic, and completely dissociates as a strong acid, one mole of Hcl will release one mole of protons in solution. Therefore, 0.15N HCl is equal to 0.15 moles of H+ per liter, meaning that [H+]= 0.15M. The pH of the stomach can then be calculated as pH= -log([H+])= -log(0.15).
Q: Based on the way NH3 interacts with the copper, it can be concluded that NH3: A: Functions as a Lewis base.
Explanation: Paragraph 2 states that copper acts as a Lewis binding site for NH2. A lewis base acts as an electron pair donor, and a Lewis acid acts as an electron pair acceptor. NH3 has a lone pair of electrons, allowing it to act as a lewis base and share its electrons with copper.
Q: A measurement determined that elastic energy density makes up the greatest proportion of the total fluid energy per unit volume? This measurement was most likely made: A: During systole
Explanation: Paragraph 2 states that elastic energy density is stored in the arterial wall during systole. Thus, elastic energy density would contribute the greatest to the total fluid energy per unit volume during systole.
Q: Why is it important that oxygen be included in the gas that is streamed through the catalyst bed? A: To provide one of the reactants for SCR.
Explanation: Paragraph 2 states that the gas streamed through the catalyst bed included 10% 02 by volume. The reason for this is found in reaction 1 of SCR. In other words, oxygen is a reactant.
Q: From the information provided in the passage, which of the following is necessarily true? A: Keq > 1
Explanation: Paragraph 3 states that the reaction proceeds spontaneously to the right. This means that the Gibbs free energy change (delta G) must be negative, and the equilibrium constant (Keq) must be greater than one. * Keg>1 implies that delta G<0
Q: Which of the following CANNOT be the mechanism for the rate-determining step of this reaction? A: Two molecules of A collide with a molecules of B
Explanation: Paragraph 3 states that the slower, first step of the reaction mechanism follows second-order kinetics. The slowest step of a reaction mechanism is the rate-determining step, and a step that involves second-order kinetics involves two reactant molecules. The kinetics of an overall reaction cannot be derived from its stoichiometry, but when a complex reaction has been broken down into a series of elementary reactions, the rate law is dependent upon the kinetics of the slowest of these elementary reactions. In this case, the rate-determining step is second-order, and thus must consist of an interaction between two molecules. This could be two molecules of A, two of B, or one of each of A and B. Because the reaction step depicted in the correct answer involves interaction of three molecules, it would exhibit third-order kinetics, and CANNOT be the mechanism of this reaction.
Q: What conclusion can be drawn from the experimental results described in the passage? A: Residues 94, 128, and 189 are all essential for substrate sequestration.
Explanation: Paragraph 4 states that the concentration of glutamine remained constant throughout the reaction for 94A, 189A, and 128A. This confirms that GatD mutations in 94, 189, and 128 residues completely impair the glutaminase activity of the MurT-GatD complex.
Q: Hydroquinone can react relatively easily and reversibly to form benzoquinone, as shown below in reaction 1. Coenzyme Q10, a naturally occuring hydroquinone derivative and component of the electron transport chain, carriers high-energy electrons: A: In its reduced form.
Explanation: Reduction is gain of electrons, while oxidation is the loss of electrons. By definition, when coenzyme Q10 gains electrons to carry, it is being reduced.
Q: For many applications, it is necessary for physiologists to approximate the resistance of the human body. In order to measure this, a physiologist might ask a willing subject to: A: Hold the positive terminal of a known voltage source with one hand and the negative terminal with the other hand, while the physiologist measures the current.
Explanation: Resistance can be calculated using Ohm's law, V=IR, if voltage and current are known therefore, if a known voltage is applied, and the current running through the person is measured, resistance can be calculated.
Q: Assuming that the standard electromechanical potential for the reduction of O2 to H2O is 0.81V, and for the reduction of gluconolactone to glucose is -0.58V, the E cell of the enzyme-based fuel cell in figure 1 is: A: +1.39V
Explanation: The E cell of an electrochemical cell can be calculated in two main ways. One method is to calculate the difference between the reduction half-cell potentials of the cathode and the anode. The reduction half-reaction of O2 occurs at the cathode and the oxidation half-reaction of glucose occurs at the anode. E cell= E reduction potential at cathode= E reduction potential at anode. E cell= 0.81V- (-0.58 V)- 1.39V
Q: Chemiluminescence of which of the following colors corresponds to radiation with the highest frequency? A: Violet
Explanation: The colors of visible light, in the order of increasing energy and frequency, are ROYGBIV, with wavelengths running from about 400-700nm.
Q: In a person with normal hearing, 60% of the acoustic energy reaching the tympanic membrane is transmitted into the cochlear fluid. Which of the following would most nearly restore normal hearing to a patient for whom only 0.6% of this energy is being transmitted? A: Amplification of incoming sounds by 20dB.
Explanation: The difference between the loudness of two sounds, measured in decibels, where I1 and I2 are the intensities of the two sounds in W/m^2, is dB= 10(log| I2/ I1 | -The ability to perceive (hear) a sound is of course dependent on how much of that sound's energy is actually transmitted to the cochlear fluid. Normally, according to the question, about 60% of the acoustic energy reaching the tympanic membrane is transmitted to the inner ear, but in this patient, only 0.6% of that energy is being transmitted. Plugging these values into the above equation: 10log |60%/0.6%|= 10log|100|= 10*2= 20dB -Amplifying incoming sounds by 20dB, therefore, should create sound intensities in the patient's inner ear approximately equal to those in a person with normal hearing.
Q: A student is standardizing NaOH using potassium hydrogen phthalate (KHP), an acid with a molecular weight of 204g/mol. The student dissolves 204.22g of KHP in 50mL of water and then titrates with the NaOH of unknown concentration. Which of the following would NOT cause faulty results? A: Using an unknown amount of water that was more than 50mL
Explanation: The equivalence point will be reached when the equivalents of acid (KHP) originally present are equal to the equivalents of added base (NaOH). In this case, the amount of acid is known precisely (204.22g). Changing the amount of water used to dissolve this acid will not change the amount of acid itself. Because it would not change the outcome of the titration.
Q: Accurately measuring the level of NO in a sample of air using CL requires first knowing how much O3 is present in the sample. This is most likely because: A: The rate of NO2 production in reaction 1 is governed by second order kinetics.
Explanation: The intensity of CL in reaction 1 is proportional to the rate of formation of NO2+, which is the slower of the two steps. The two species on the left side of reaction 1 are NO and O3. If CL from the reaction can be used to measure the NO content in air, the rate of formation of NO2* must depend on the concentration of NO. If the O3 concentration must also depend upon the concentration of O3. The most straightforward and common type of chemical reaction fitting these requirements is a concerted reaction following second order kinetics, described by the equation: rate= k[NO][O3].
Q: Which of the following reaction mechanisms is included in the formation of compound 3? A: Nucleophilic attack by compound 2 on a carbonyl carbon of compound 1.
Explanation: The nitrogen in compound 2 is the nucleophile that attacks the carbonyl carbon of compound 1, as seen in the reaction mechanism pictured below. That same nitrogen initiates a second nucleophilic attack on the second carbonyl carbon to close the ring, forming the cyclic imide in compound 3.
Q: The osmium copolymer reacting with glucose oxidase in figure 1 acts as: A: An oxidizing agent
Explanation: The osmium in the copolymer (Os3+) oxidizes the reduced form of glucose as shown in figure 1. In this case, the Os3+ acts as an oxidizing agent, because it oxidizes another species (the glucose oxidase in its reduced form) and it, in turn, is reduced to Os2+.
Q: An overabundance of which of the following amino acids is most likely to result in symptoms similar to those seen in PKU? A: Tryptophan
Explanation: The passage indicates that many of the symptoms of pKU occur because of alterations of amino acid transport into the brain. In individuals with PKU, an overabundance of phenylalanine swamps the transport system and prevents other aromatic amino acids from reaching the brain, also increasing blood osmolarity, and causing symptoms such as those described in paragraph 2. An overabundance of tryptophan, which is also aromatic, could well cause similar problems.
Q: While the active ingredients in most antacids are soluble in the acidic environment of the stomach, their solubility is significantly reduced in pure water. If a very finely ground powder of one such compound (CaCO3, for example), is mixed in a 1:5 ratio by mass with deionized water, which of the following is most likely true? A: A colloidal mixture will be formed.
Explanation: The question indicates that compounds such as CaCO3 are not very soluble in pure water, so only a small proportion of it will dissociate into its component ions, Ca2+ and CO2-/3. Most of the powder will remain undissolved, but because it is "very finely ground," the particles will become suspended. Such a heterogenous mixture is known as a colloid. -Based on this, the correct answer is confirmed.
Q: DD-transpeptidase is a bacterial enzyme that cross-links peptidoglycan chains to form rigid cell walls. Threonine is often found in these interbridges. Which of the following is also a likely component? A: Serine
Explanation: The question is essentially asking for the amino acid most similar to threonine. Of all the amino acids, serine is the most similar to threonine, because both have polar, uncharged side chains that contain a hydroxyl group.
Q: Which of the following compounds has the most stable conjugate base? A: HOOC-CH2F
Explanation: The stronger an acid is, or the more that acid dissociates, the most stable its conjugate base will be. Therefore, the answer choice that represents the strongest acid will be the correct choice. Acid strength is increased by electron withdrawing groups, as they are able to stabilize the negative charge once a proton has been lost. Conversely, the most electron donating groups a molecule has, the less acidic it will be. Because fluorine is the strongest electron withdrawing group in the four answer choices, the compound in HOOC-CH2F, will be the most acidic, and have the most stable conjugate base.
Q: Upon oxidation to the 3-aminophthalate anion, the number of sp2 hybridized carbon atoms in luminol: A: Remains unchanged
Explanation: There are 8 carbon atoms in luminol: 6 in the benzene ring, and 2 carbonyl carbons. Every one of them is involved with one double bond and two single bonds; all of them are therefore sp2 hybridized. In the 3-aminophthalate anion, no carbons have been removed or added, and all of the carbons remain sp2 hybridized. The only change is that the two amide functionalities have been converted into carboxylates. The two relevant carbon atoms, however, are still carbonyl carbons and are still sp2 hybridized.
Q: PG mostly consists of D-amino acids, which contribute to bacterial resistance to proteolysis. Which of the following hypotheses best explains why this is most likely true? A: Most proteases are stereospecific
Explanation: This question requires the incorporation of some basic biological knowledge with knowledge of stereochemistry. Most proteases are stereospecific, and only cleave L-amino acids. Thus, D-peptides are more likely to display resistance to proteolysis, which is advantageous to bacteria because this resistance prevents their degradation and enhances pathogenicity.
Q: If it takes time t for an NIR laser pulse traveling at the speed of light c to be detected by the sensor in airborne LIDAR, which formula can be used to calculate the approximate distance between the source and the water's surface? A: (c*t)/2
Explanation: Total distance traveled equals speed multiplied by time, or in this case, d=c*t. In figure 1, it can be seen that the NIR laser light reflects off the water's surface; thus to reach the detector, it must travel down to the surface and then back again. This distance to the surface, therefore, is half the total distance traveled.
Q: Which of the following figures shows a possible path for the light returning to the detector after reflection from the stream's bottom? A: (See Kaplan for answer)
Explanation: When light travels between two media with different refractive indices, it changes speed; the higher the index of refraction, the slower it travels. This change in speed makes light rays traveling at an angle bend. in this case, the underwater light returning to the detector encounters the interface with air, which has a lower index of refraction than water, so the light will either bend away from a normal drawn perpendicular to the interface, or (if it strikes the interface at too extreme an angle) experience total internal reflection. Since the question specifically asks about light that is "returning to the detector," this light must have been refracted away from the normal rather than internally reflected.
Q: Which of the following is the empirical formula of a compound containing 23g of sodium, 16g of sulfur, and 32g of oxygen? A: Na2SO4
Explanation: While a molecular formula shows how many atoms of each element are in a compound, the empirical formula of a compound shows the most reduced whole-number ratio of the elements in the compound. -In order to determine the empirical formula of a compound from the mass of each element, the first step is to convert grams into moles. The molar masses of sodium, sulfur, and oxygen are approximately 23, 32, and 16g/mol respectively. Sodium: 23g/ (23g/mol)= 1 mole Sulfur: 16g/ (32g/mol)= 0.5 moles Oxygen: 32g/(16g/mol)= 2 moles -This provides the ratio of the elements to one another. However, non-whole numbers, such as 0.5, are not used when writing empirical formulas. Therefore, in the second step of the calculation, everything must be multiplied by the smallest value that yields a whole number of moles for all three elements. In this case, multiplying by two gives 2 moles of sodium, 1 mole of sulfur, and 4 moles of oxygen. The empirical formula of the compound can then be written as Na2SO4.
Q: Reaction 2 was attempted in the presence of peroxidase at various concentrations. Which of the following interpretations of the results shown in the chart below is correct? A: Three times as many CL photons per second were emitted at 30nM than at 10nM
Explanation: intensity of light is measured in terms of energy per unit time per unit area (W/m^2). In measuring CL and similar phenomena, the distance at which the measurement is taken and the cross-sectional area of the light collector are standardized within any given experiment; hence, the intensity of the measured light is directly proportional to the power (J/s) of emitted light. Since the light emitted in Reaction 2 all has the same wavelength (425nm), all its photons have identical energies; this implies that the intensity of measured CL is directly proportional to the number of photons being emitted per unit time. In the experiment, according to the chart, the intensity of CL at 30nM peroxidase was triple the intensity of light at 10nM of peroxidase.
Q: Which of the following is likely the structure of the anhydride that was used to produce compound 7? A: (See kaplan for answer)
Explanation: Compound 1 in scheme 1 shows the generic structure of the anhydride used in the synthesis. The wavy lines in the diagram indicate the sites on the anhydride that can vary. Compound 7 is one of thalidomide analogues in figure 1; it has a single five-membered ring on the left, not a double ring system like the other analogues. This ring originated from the anhydride. The only anhydride that matches this structure is depicted in the correct answer on kaplan.
Q: Which energy values in figure 1 could be altered by the introduction of a catalyst? A: 1 and 2 only
Explanation: In figure 1, the energy differences labeled 1 and 2 are the activation energies of the reaction (2 can be thought of as the activation energy of the reverse reaction), 3 & 4 are the potential energies of the reactants and products, and 5 is the potential energy difference of the reaction as a whole. -Catalyst: A substance that accelerates a reaction by lowering one or more of the activation energies in the reaction; it does not change the delta H of the reaction, which is a state function. Since a catalyst can lower the activation energies of a reaction, only energies 1 and 2 could be affected. Their values could be decreased by lowering the energy of the transition states (the hilltops).
Q: How many grams of AlCl3 are produced when 3 grams of Al(OH)3 react completely with excess HCl? A: 5.05g
Explanation: Since the question stem states that Al(OH)3 is reacted with excess Hcl, Al(OH)3 is the limiting reagent, and all 3g of it will be consumed. According to the balanced equation in Reaction 1, each mole of Al(OH)3 yields one mole of AlCl3. Moles of AlCl3 produced can be calculated as follows: 3g Al(OH)3* ((1mol Al(OH)3/(27+3(16+1)gAl(OH)3)= 3g Al(OH)3 * ((1mol Al(OH)3)/ (78g Al(OH)3)) 3/78 Mole Al(OH)3 -The theoretical yield of AlCl3 is therefore 1/26 mol. The number of grams of AlCl3 can be found by multiplying this value by its molecular weight: 1/26 mole AlCl3 * (27+3(35.5)g)/1mole AlCl= 133.5/26g
