NS exam 3 Chemistry

B is correct. This is a multi-step question, though each step is relatively straightforward. The gravitational potential energy at the runner's height is: PE = (60 kg)(10 m/s2)(0.5 m) = 300 J Most of this energy is conserved as the runner hits the ground and his muscles capture the energy as spring potential energy, so the question only asks about the lost energy, amounting to 10%, or 30 J. Don't worry about the mechanics of which foot lands first and how much of the kinetic energy each one absorbs. The question stem doesn't give that kind of information, so there's nothing for it anyway. So, an additional 30 J is needed per stride, and a stride occurs every 0.8 s. Thus: P = (30 J)/(0.8 s) = 40 W This is between answer choices B and C. So is it really 40 W? Well, no, 8 goes into 30 less than 4 times, so the answer should be lower. Also, g is closer to 9.8 than 10, so that also caused the value of energy, used in the power calculation, to be overestimated. 37 W, then, must be the match. A: This answer implies that the runner does not need to produce any power output to account for the energy loss. Since the runner is losing energy that must be recouped with every step, this choice is incorrect. C: This answer is slightly too large. (30 J)/(0.8 s) is less than 40 W. (We can deduce this easily by recognizing that 32/0.8 = 40, and 30 J is smaller than 32. Thus, our answer should be less than 40 W, and choice C is incorrect. D: This answer is far too large to be reasonable.

A 60-kg runner raises his center of mass approximately 0.5 m with each step. Although his leg muscles act as a spring, recapturing the energy each time his feet touch down, there's an average 10% loss with each compression. What must the runner's additional power output be to account for just this loss, if he averages 0.8 s per stride? A. 0 W B. 37 W C. 46 W D. 331 W

C is correct. Charles' law states that there is a direct relationship between the volume of an ideal gas and its temperature, when pressure is constant. Note that the temperature must be in Kelvin! We can approximate the initial temperature as 300 K and the final temperature as 320 K. (3.0 L)/(300 K) = V2/(320 K) [(3.0)(320)] / (300) = (3.0)(1.1) = 3.3 L A, B: Since volume and temperature of a gas are directly proportional, we would not expect the volume of the balloon to decrease. D: This answer would be obtained by incorrectly using Celsius temperature instead of converting to Kelvin.

A balloon has a volume of 3.0 L at 25°C. What is the approximate volume of the balloon at 50°C? A. 1.5 L B. 2.0 L C. 3.3 L D. 6.0 L

A large Ksp suggests that a substance is more soluble than a substance with a lower Ksp.

A large Ksp suggests that a substance is ?

D is correct. Since the crate is initially at rest, the Vi = 0 m/s, the acceleration a = 2.0 m/s2, and the displacement = 6 m. The appropriate kinematic equation is vf2 = vi2 + 2ad. Substituting the variables gives: vf2 = 02 + 2(2)(6) vf2 = 24 vf = ~5 m/s A, B, C: These answers are the result of miscalculation.

A person pushes horizontally on a 50-kg crate, causing it to accelerate from rest and slide across the surface. If the push causes the crate to accelerate at 2.0 m/s2, what is the velocity of the crate after the person has pushed the crate a distance of 6 meters? A. 1 m/s B. 2 m/s C. 3 m/s D. 5 m/s

B is correct. The other product of Reaction 1 has a formula of C2H4O2. Ethanoic acid, a carboxylic acid, has this formula. The reaction between acetic anhydride will involve the acylation of the alcohol functional group of the salicylic acid, with the by-product being acetic acid, also known as ethanoic acid. A: Ethanol is an alcohol and has a formula of C2H6O, meaning that it cannot be the product that we're looking for. C: Ethanal is an aldehyde and has a formula of C2H4O; again, this is the incorrect formula. D: Methyl formate is an ester and does have the correct formula of C2H4O2. However, as mentioned above, Reaction 1 yields a by-product of ethanoic acid, which preserves the bond between the methyl group and the carbonyl carbon that was present in the acetic anhydride. For methyl formate to form instead, the methyl group would have to rearrange somehow (magic?) and bond to an oxygen atom to form the ester shown below:

A. Ethanol B. Ethanoic acid C. Ethanal D. Methyl formate molecular forms

B is correct. We first need to determine if there is a limiting reagent and then determine the theoretical yield. The moles of acetic anhydride can be determined by using the volume and density information in step 2 of the procedure and the molecular weight given in the passage. 7 mL x 1.08 g/mL x 1 mol/102 g ~ 7 x 10-2 mol We can also calculate the moles of salicylic acid from the number of grams used in step 1 and the molecular weight given in the passage. 5.0 g x 1mol/138 g ~ (5/1.4) x 10-2 ~ 3.5 x 10-2 mol Since the stoichiometry is 1:1 from Reaction 1, the acetic anhydride is in excess and the limiting reagent is the salicylic acid. The stoichiometric ratio between salicylic acid and the acetyl salicylic acid is also 1:1; therefore, the theoretical yield can be calculated using the molecular weight from the passage. 3.5 x 10-2 mol x 180 g/mol ~ 3.5 x 2 x 10-2 x 102 ~ 7 g Since the question indicates that the actual yield was 3.9 g, which is about 4 g, the percent yield is (4/7) x 100 ~ 60%. A, C, D: These answers are the result of miscalculation

Acetylsalicylic acid, commonly known as aspirin, is primarily used for analgesic effect. But the drug is also capable of reducing both fever and inflammation. In low doses, the drug is shown to reduce the risk of heart attack and stroke. The side effects of aspirin include gastrointestinal discomfort, stomach ulcers and bleeding. Subsequently, drugs have been developed to reduce these effects. Aspirin (180 g/mol; pKa = 3.5) is easily prepared by reacting salicylic acid (138 g/mol; pKa = 2.97) with acetic anhydride (102 g/mol) as shown in Reaction 1. 2. Using a graduated cylinder, add 7 ml of acetic anhydride (d = 1.08 g/cm3) to the Erlenmeyer flask. A student determined that her yield of aspirin was 3.9 g. What was her percent yield? A. 45% B. 60% C. 78% D. 92%

D is correct. When the leg "absorbs" kinetic energy, it is converted to elastic potential energy. While this process involves the loss of some energy as heat, we can assume that it is a perfectly elastic process here for the sake of simplicity. Thus, the leg needs to hold up to 125 J of elastic PE. The formula for potential energy contained in a spring is PE = (1/2) kx2. Since PE and KE are interconverted as the individual runs, we can write this formula as KE = (1/2) kx2, which can be rearranged to yield k = 2 KE/x2. k = (2)(125 J) / (0.10 m)2 k = (250 J) / (.01 m2) k = 25,000 N/m A: This answer may result from neglecting to divide by x2. B, C: These answers result from miscalculation.

An artificial leg designed for use by runners is spring-based, to mimic the compression required of a muscle during hard running. For safety reasons, it was determined that the leg should be able to absorb as much as 125 J of kinetic energy without compressing more than 10 cm, or the runner would be likely to stumble. What should the spring constant be? A. 250 N/m B. 2,500 N/m C. 12,500 N/m D. 25,000 N/m

A is correct. An arrhythmia is an irregular time period between heartbeats, not necessarily a rate that is too slow or too fast. Since the question asks about atrial fibrillation, and the P wave is attributed to contraction of the atria, the correct answer should show the P wave being affected. This does not mean that the atria are not contracting at all (eliminate choice B), but it indicates that blood is not efficiently being pumped from the atria to the ventricles. B, D: These choices incorrectly associate arrhythmia with a regular heart rate. C: Absence of an R wave would be indicative of ventricular problems, and is more dangerous problem than AFib. The R wave is a strong, sharp electrical signal coming from the ventricles. Complete absence of an R wave suggests ventricular contraction is likely not occurring or fibrillating at best, which is life-threatening and is associated with myocardial infarction, commonly known as a "heart attack."

Atrial fibrillation is the most common heart arrhythmia, causing palpitations, fainting and chest pain. According to the information in the passage, what is most likely to be observed in the ECG during atrial fibrillation? A. Missing P waves and an irregular rate B. Complete absence of P waves and a regular rate C. Complete absence of R waves and an irregular rate D. Missing R waves and a regular rate

C is correct. The passage states that the cell is meant to act like a galvanic cell, meaning that it proceeds in a spontaneous fashion. Since galvanic cells always have cell potentials that are greater than 0, we can eliminate any negative options (choices A and B) immediately. To decide between C and D, we need more information, namely the reduction potentials of the species involved. Paragraph 1 gives E°reduction of O2 as 0.816 V and E°reduction of AQDS as -0.10 V. Since both species cannot reduce at once, we must reverse the sign of one of the reduction potentials to find the oxidation potential of the relevant reaction. For galvanic cells, reverse the less positive reduction potential, which is -0.10 V here. The associated oxidation potential, then, is +0.10 V. Finally, since we now have a reduction and an oxidation, we can simply add the values directly to yield 0.816 V + 0.10 V = 0.916 V. A, B: Galvanic (spontaneous) cells do not have negative standard cell potentials, so these choices can be eliminated. We know the starch battery is meant to act like a galvanic cell because it is described as "generat[ing] sufficient power," meaning that it must run spontaneously. D: This choice is attained by using the reduction potentials of O2 and NAD+. However, for this question, we need to use the potentials for the two species directly involved in the battery (according to Figure 1). The figure shows AQDS at the anode of the battery and O2 at the cathode, so these are the species we must use.

Biostarch batteries work to convert the chemical energy stored in starch bonds into electrical energy. The process involves the breakdown of starch into glucose 6-phosphate (MW = 260 g/mol) followed by its oxidation to 6-phosphogluconate along with the reduction of NAD+ (E°red = -0.32 V) to NADH. The high energy electrons in NADH are then transferred in a series of redox reactions, first to the electron carrier AQDS (E°red = -0.10 V), and then the through electrodes and ultimately to the reduction of O2 (E°red = 0.816 V) to H2O as shown in Figure 1. What is the standard cell potential for the starch battery created? A. -1.136 V B. -0.916 V C. 0.916 V D. 1.136 V

Boyle's Law indirect relationship between pressure and volume Boyle's Formula P1 x V1 = P2 x V2

Boyle's Law

Charles's Law direct relationship between volume and temperature Charles's Formula V1/T1 = V2/T2

Charles's Law

A is correct. A process known as "hydration" or "solvation" occurs when the attractive force of an ion molecule causes a thin shell of water molecules to surround it. In the case of hydronium (H3O+), each of the H atoms attracts the O atom in an H2O molecule due to hydrogen bonding. These H2O molecules cause a "shell" of water molecules to surround the hydronium. B: This is misleading. Regular H+ ions almost never exist in H2O solutions. Instead, they combine with an H2O molecule to form H3O+. Bases are more than capable of neutralizing H3O+ ions. C: Although this describes an actual biological process in muscle cells, it has no connection to the hydration shell around H3O+ molecules. D: The effects of hydration shells have nothing to do with enhancing the mobility of hydronium.

During strenuous exercise, lactic acid buildup in cells causes the creation of a hydronium complex known as the Eigen cation (H9O4+). If water molecules then experience hydrogen bond attractions to the Eigen cation, this attractive force: A. results in a semi-stable shell of water molecules around the hydronium. B. results in an inability of hydronium to be neutralized by bases the way normal H+ ions would be. C. results in the ability of muscle cells to reverse both the hydronium creation process and the lactic acid creation process once sufficient oxygen is once again made available. D. results in mobility of hydronium within the environment surpassing the mobility of regular water molecules.

B is correct. In an acidic environment, a base such as ammonia (NH3) will dissolve into its conjugate acid, ammonium (NH4+), to a greater extent than would have been the case in a neutral or a basic environment. An environment with lower pH is an environment that is more acidic. A: The pH of an environment that becomes more acidic will be lower, not higher. C, D: An acidic environment will result in more ammonia reacting to form ammonium, not less.

Efforts to treat lactic acid buildup in muscles were attempted using dissected muscle specimens in the laboratory. One of these experiments involved ammonium formation from dissolved ammonia. Under conditions of excessive lactic acid: A. the final concentration of ammonium will be higher than otherwise due to higher pH. B. the final concentration of ammonium will be higher than otherwise due to lower pH. C. the final concentration of ammonium will be lower than otherwise due to higher pH. D. the final concentration of ammonium will be lower than otherwise due to lower pH.

The thermodynamic concept of entropy is critical for understanding how certain biological structures are formed. It is particularly relevant for understanding how protein folding takes place, and why hydrophobic amino acid residues generally wind up clustered together on the inside of globular proteins in aqueous solution. This principle also explains how phospholipids form a bilayer membrane with the hydrophobic alkyl tails in the interior of the membrane and the hydrophilic phosphate heads facing out. To understand this principle, let us carefully compare what would happen if hydrophobic residues within a protein face an aqueous solution versus what would happen if hydrophilic residues face an aqueous solution, keeping in mind the polarity and excellent hydrogen-bonding properties of water. If hydrophilic—that is, polar—residues are facing the aqueous solution, then water will be able to hydrogen-bond freely with those residues, meaning that it will have relatively high entropy, which is energetically favorable. In contrast, water molecules will not be able to hydrogen-bond effectively with nonpolar residues, and as a result will form a highly-ordered solvation shell to minimize interactions with those residues. This highly-ordered shell represents a decrease in entropy, which is energetically unfavorable.

Entropy and the Solvation Layer

C is correct. The electron configuration for beryllium is 1s22s2. The azimuthal quantum number for an s-type orbital is l = 0, which defines the shape of the electron cloud (a spherically-shaped electron wave, with no nodes). For boron, since the s-type orbital in the valence shell cannot hold more than two electrons, the third valence electron must be placed into a p-type orbital, giving an electron configuration of 1s22s22p1. The azimuthal quantum number for a p-type orbital is l = 1 (a dumbbell-shaped electron wave with one node at the nucleus). The 2p orbital is slightly higher in energy than the 2s orbital, making it slightly easier to ionize this electron. Choice C is the only answer option that correctly describes this situation. The shapes of the s-type and p-type orbitals are shown below. (Note that multiple possible shapes exist for p-type orbitals.) A: Boron has three valence electrons, while beryllium has two valence electrons. B: The 2s orbital for beryllium can only hold two electrons, and they have opposite spin. The only option for the third electron for boron is to place it into one of the three 2p orbitals, which do not yet contain any electrons for beryllium. D: This choice incorrectly describes the possible value of the magnetic quantum number for a ground state valence electron of beryllium. The valence electrons of beryllium are in s-type orbitals, which have an l value of 0, requiring that the only possible value of m be 0.

Figure 1 shows a drop in the first ionization energy in going from beryllium to boron. Which of the following best explains the source of this drop? A. Boron contains one valence electron, while beryllium contains none. B. After putting two electrons in the valence s-type subshell for beryllium, the third valence electron for boron must enter an orbital that already contains an electron of opposite spin. C. The two valence electrons for ground state beryllium are in an orbital with an azimuthal quantum number of l = 0, whereas the third valence electron in ground state boron must be in an orbital with an azimuthal quantum number of l = 1. D. The two valence electrons for ground state beryllium are in an orbital with a magnetic quantum number of m = +1, whereas the third valence electron in ground state boron must be in an orbital with a magnetic quantum number of m = -1.

D is correct. The question stem states that free radicals can cause cancer, which is a result of poorly regulated cell growth and division. Cancer is generally a product of mutations in DNA that disrupt these processes. Therefore, we must choose the answer that targets DNA, which is choice D. A, B, C: Damage to these structures/molecules would be unlikely to lead to uncontrolled cell growth (cancer)

Free radicals from ionizing radiation are highly unstable and have carcinogenic effects. These effects are most likely result from damage to: A. proteins. B. phospholipids. C. the Golgi apparatus. D. nucleic acids.

Gay Lussac's Law the direct relationship between pressure and temperature Gay Lussac's formula P1/T1 = P2/T2

Gay Lussac's Law

C is correct. The balanced chemical reaction is: CaCO3 (s) + 2 HCl (aq) → CO2 (g) + H2O (l) + CaCl2 (aq) From the periodic table, the formula weight of calcium carbonate is 40 + 12 + 3(16) = 100 g/mol. 100 g of calcium carbonate therefore represents one mole, and based on the reaction, this will produce one mole of carbon dioxide gas. Remember, one mole of any gas at STP (standard temperature and pressure) has a volume of 22.4 L! Thus, the volume of gas produced will be 22.4 L at STP. 100 g CaCO3 x 1 mol/100 g x 1 CO2/1 CaCO3 x 22.4 L/mol = 22.4 L CO2 A: Since CO2 is produced in this reaction, this answer does not make sense. B, D: These options may result from making a mistake when writing out the balanced chemical reaction.

How many liters of carbon dioxide at STP are produced by reacting 100 g of calcium carbonate with an excess of hydrochloric acid? A. 0.0 L B. 11.2 L C. 22.4 L D. 44.8 L

B is correct. First, note that methane has a molecular formula of CH4. Thus, its molecular weight is approximately 12 + 4(1) = 16 g/mol. 30 g CH4 x (1 mol/16 g) x (890 kJ/mol) = 1.8 x 103 kJ (if we round 890 to 900 kJ/mol) Note that the answer options have units of joules (J), not kilojoules (kJ), and choice A should be eliminated. Converting the units gives: 1.8 x 103 kJ x (103 J /1 kJ) = 1.8 x 106 J A: This is the answer in kilojoules (kJ), but it includes units of joules (J), making it incorrect. C, D: These answers are the result of miscalculation

How much heat is produced from the complete combustion of 30.0 g of methane, if the enthalpy of reaction is -890 kJ/mol? A. 1.7 x 103 J B. 1.7 x 106 J C. 4.7 x 106 J D. 4.7 x 109 J

C is correct. This is a tricky question, because the passage states that nature has no perfectly elastic springs, which means energy cannot be completely conserved. This does not, however, mean that all the energy is lost, nor does it make sense for the potential energy stored in a spring to be negative (ruling out answer choices A and B). Then, to distinguish between choices C and D, we need to obtain an estimate of what the value would be if energy were completely conserved, since that could serve as a baseline. The potential energy of the rabbit at the peak of its height can be estimated as PE = (3)(10)(0.5) = 15 J. Since that value was calculated using the estimate of 10 m/s2 for g, instead of 9.81 m/s2, we should recognize that it is an overestimate. That is, the actual value that would be obtained if no energy was lost would be slightly less than 15 — therefore, answer choice D, 14.7 J, seems about right for how much we need to correct for our estimation of g. That is, choice D corresponds to the value that we would calculate if no energy was lost. Finally, we are left with choice C, 10 J, which corresponds to a non-trivial loss of energy due to inefficiency, which is what the question stem suggests that we should expect (because of the wording "imperfect spring").

If a 3-kg rabbit's leg muscles act as imperfectly elastic springs, how much energy will they hold if the rabbit lands from a height of 0.5 m and its legs are compressed by 0.2 m? A. -0.6 J B. 0 J C. 10 J D. 14.7 J

D is correct. Cell 2 can only accelerate while it is being pushed/in contact with cell 1, so statement I is false. Since the cells experience drag from the medium, cell 2 will decelerate after the collision, making statement II correct and statement III incorrect. The drag will cause cell 2 to continuously decelerate until it comes to rest.

If a flat Petri dish containing a single layer of cells suspended in viscous culture medium is tapped, some cells will collide into each other. If cell 1 collides into cell 2 on such a plate, which of the following describes what happens to cell 2 after the collision, assuming an elastic collision and non-zero drag from the medium? I. Cell 2 continuously accelerates. II. Cell 2 moves with decreasing speed. III. Cell 2 moves with constant speed. IV. Cell 2 decelerates until it reaches a velocity of 0 m/s. A. I only B. II only C. III only D. II and IV only

A is correct. Capacitance (C, measured in farads) is the amount of charge stored per volt, expressed with the equation C = q/V. This means that charge (q) = VC. (Do you remember the home shopping network, QVC?) The capacitance is given as 20 x 10-6 F. Substituting the voltage (12 V) and capacitance into the equation gives us the charge (in coulombs). (20 x 10-6 F)(1.2 x 101 V) = 24 x 10-5 C = 2.4 x 10-4 C = 0.24 x 10-3 C = 0.24 mC Remember when making the exponent larger, we must make the coefficient smaller by the same factor. B, C: These answers constitute order-of-magnitude errors. They may result from failing to realize that 20 μF = 20 x 10-6 F, or from confusing the relationship between coulombs (C) and millicoulombs (mC). D: This answer results from miscalculation.

If a modern portable defibrillator uses as 12 V battery and a 20 μF capacitor, what is the total charge stored on the plates of the capacitor? A. 0.24 mC B. 24 mC C. 24 C D. 60 C

C is correct. SDS-PAGE is an electrophoretic technique which involves the binding of the anionic detergent SDS to a polypeptide chain. SDS binding denatures and imparts an even distribution of charge per unit mass to the protein, resulting in fractionation by approximate size alone during electrophoresis. A: Affinity chromatography is used to separate molecules, including proteins, based upon their specific interaction with the stationary phase. During protein purification, this is often an antigen-antibody, receptor-ligand, or enzyme-substrate interaction. B: Ion-exchange chromatography separates charged particles based on their affinity for charged elements of the ion exchange column. D: NMR spectroscopy is not a separatory technique.

If a scientist chooses not to employ a SEC column, but wishes to fractionate a protein sample on the basis of only molecular weight, what technique would be most appropriate? A. Affinity chromatography B. Ion-exchange chromatography C. SDS-PAGE D. NMR spectroscopy

A is correct. By rearranging Equation 1, ΔG = Gl - Gg, an expression for the Gibbs free energy of the gel phase may be found: Gg = Gl - ΔG. The Gibbs free energy change for the phase transition of raffinose is given by ΔG = ΔH - TΔS. Substituting thermodynamic values from Table 1: ΔG = (16.0 kcal/mol) - [(3.15 x 102 K)(5.05 x 10-3 kcal/K•mol)] ΔG = 16.0 kcal/mol - 1.60 kcal/mol = 14.4 kcal/mol Substituting into the rearranged form of Equation 1: Gg = Gl - ΔG = 17.9 kcal/mol - 14.4 kcal/mol = 3.5 kcal/mol, choice A. B: This value is Gl + ΔG, not Gl - ΔG. C, D: These answers are the result of miscalculation.

If the Gibbs free energy of the ΔG= Glc - Gg liquid-crystalline phase for an aqueous dispersion of DPCC and raffinose is 17.9 kcal/mol, which of the following values is nearest the Gibbs free energy of the associated gel phase? A. 3.5 kcal/mol B. 32.3 kcal/mol C. 566.1 kcal/mol D. 601.9 kcal/mol

Another high-yield fundamental concept is half-life (t1/2), which is the time required for one-half of the parent isotopes in a sample to decay into daughter (radiogenic) isotopes. If we know the half-life of a material, then we can determine how much of a sample is lost (1- ½n) or remains (½n) at any given time, as expressed in the number of half-lives that have passed (n).

If we know the half-life of a material, then we can determine how much of a sample is lost

D is correct. As with other forms of chromatography, increasing the column length will enhance the resolution of the column, leading to more completion fraction by SEC. This is because the material of the matrix provides the physical means of separating the proteins. If the proteins come in contact with a longer length of matrix, the differences in retarding forces experienced by the proteins will have a greater cumulative influence on the migration of the proteins, lengthening the differences in their retention times. A: There is no passage information to suggest a relationship between solvent (buffer) polarity and the function of the SEC column. B: Applying a vacuum will increase the flow rate of buffer through the column. Increasing the flow rate will only increase the rate at which eluent is collected; it will not improve separation. C: Increasing the concentration of protein in the sample will likely decrease the resolution of the column, by increasing the chance of overlapping within the fractions of collected eluent.

More complete fractionation of proteins using an SEC column could be achieved by using a: A. nonpolar solvent. B. vacuum across the column. C. larger protein sample concentration. D. longer column.

C is correct. The anion in phosphorous acid is phosphite, PO33-. When comparing the -ous acids and -ic acids, the -ous acids will have 1 fewer oxygen atoms than their -ic counterparts. A: This is perphosphoric acid. B: This is phosphoric acid. D: This is hypophosphorous acid.

Phosphorous acid, a common ingredient used for potable water treatment, has a molecular formula of: A. H3PO5 B. H3PO4 C. H3PO3 D. H3PO2

C is correct. According to paragraph 4, the stationary phase exerts a retarding effect only against those proteins with molecular weight between the inclusion and exclusion limits of the column. Because differences in the rate of migration through the column are a function of size exclusion arising from that retarding effect, it would be expected that two proteins with a molecular weight exceeding the exclusion limit of the column would elute simultaneously. A: If two molecules of unequal weight eluted simultaneously, it could be explained by the influence of shape on the protein's migratory rates; however, shape having no influence on migration rate could not explain how two proteins of unequal molecular weights elute simultaneously. B: If true, the smaller protein would get trapped by the porous matrix and the elution volumes/times for the proteins would be different. D: This could not account for differences in their elution profiles (particularly when separated on the basis of size).

Students fractionating two proteins of unequal molecular weight noted that the proteins eluted from the SEC column simultaneously. Which of the following statements best explains this observation? A. The shape of a molecule is unrelated to its rate of migration. B. One of the proteins has a molecular weight beneath the exclusion limit. C. The molecular weight of both proteins exceeds the exclusion limit. D. Both proteins possess equal charge at the buffered pH.

The change in entropy (∆S) of a system is commonly understood as the change in the degree of randomness or disorder the system contains. Entropy, like enthalpy, increases with temperature and as a material changes phase from solid to liquid to gas.

The change in entropy (∆S) of a system is commonly understood as

A is correct. We are given R, V, and time, and the question asked us to determine current (charge/time, or C/s). We do not need to use the time since we should know Ohm's law, V = IR. If I = V/R and the question asks for the maximum current, we should use the smallest available resistance. I = V/R = (500 V) / (20 Ω) = (50) / (2) = 25 amps = 2.5 × 101 A B, C: These answers result from miscalculation. D: This is the answer obtained if we used the maximum resistance (100 Ohms) of the skin.

The electrical resistance of dry skin is 100 kΩ, but can be lowered to 20 Ω if electrode contact area is large and conducting gel is used on the skin. If Vmax of the defibrillatror 500 V and lasts 0.01 s, what is the maximum possible current delivered to the heart during defibrillation? A. 2.5 × 101 A B. 2.5 × 10-2 A C. 5.0 × 10-2 A D. 5 A

D is correct. The second paragraph tells us that the radiation is strong enough to cause molecular electronic transitions by exciting electrons to higher energy levels in molecular orbitals. This indicates that that the radiation can either excite or eject electrons (depending on its energy) and can create free radicals (atoms with unpaired valence electrons). These properties correspond to Roman numerals I, II and III. A: RN II and RN III are also correct. B: RN III is also correct. C: RN I is also correct.

The high-energy radiation produced by the γ rays has sufficient energy to: The radiation is strong enough to cause molecular electronic transitions by exciting electrons to higher energy levels in molecular orbitals. I. generate free radicals. II. excite electrons to higher energy levels. III. eject electrons from molecular orbitals. A. I only B. I and II only C. II and III only D. I, II, and III

B is correct. For the interaction to be spontaneous, the free energy change of the reaction must be negative. The question states that the enthalpy and free energy changes for the interaction have opposite signs. Thus, ΔG is negative and ΔH is positive, making Roman numeral I true. For a reaction to be both endothermic (positive ΔH) and spontaneous (negative ΔG), the reaction must involve an increase in entropy (ΔS is positive). This is according to the equation: ΔG = ΔH - TΔS. Thus, II is also true because ΔS is positive. III: The quantity TΔS, not ΔS, must be greater than ΔH. For this reason, RN III does not have to be true. A: RN II is also correct. C: RN I is also correct, while RN III is incorrect. D: RN III is incorrect.

The results of a separate study of the thermodynamic parameters for the interactions of proteins with cyclohexanol and quartenary ammonium salts indicate that the hydrophobic solute-solute interaction is spontaneous, and that ΔH and ΔG have opposite signs. Which of the following must be true for this interaction when temperature is a positive value? I. ΔH > ΔG II. 0 < ΔS III. ΔH < ΔS A. I only B. I and II only C. II and III only D. I, II, and III

C is correct. During inspiration, contraction of the diaphragm and external intercostal muscles leads to expansion of the thoracic cavity and a decrease in intrapleural 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. A: Choice A indicates the surface tension would increase, which would increase the respiratory effort required to inflate the lungs. The efficiency of gas exchange across the alveolar membrane is primarily a function of the differences in partial pressures of gases in inspired air versus the partial pressure of those gases in the pulmonary capillaries. B: The absence of pulmonary surfactant increases surface tension, increasing the minimum radial size of alveoli required to overcome the collapsing force of surfactant at a given pressure of inspired air. When surface tension exceeds the average alveolar radii, collapse of the small airways occurs, eliminating choice B. D: Differences in efficiency of gas exchange wouldn't account for the increased lung-expansion difficulty IRDS patients experience, eliminating D.

Underproduction of pulmonary surfactant in IRDS leads to decreased compliance of alveolar tissue. Based upon this information, which of the following must be true regarding pulmonary surfactant? A. Its adsorption to the water-alveolar interface increases surface tension, preventing alveolar collapse due to intra-thoracic pressure. B. Its absence decreases the minimum radial size of alveoli able to avoid collapse at a given pressure of inspired air. C. Its adsorption to the water-alveolar interface decreases surface tension, decreasing the pressure difference required to inflate the airway. D. Its presence increases the efficiency of gas exchange across the alveolar membrane by decreasing the surface area of the alveolus at a given pressure of inspired air.

D is correct. This question is asking us to remember what factors would contribute to the net charge on a phenylalanine molecule. Being an amino acid, phenylalanine has an acidic carboxy group that will be protonated at a pH of 1 (remember, such a pH is highly acidic). Additionally, it has a basic amino group that will also be protonated. Finally, it contains a neutral toluene side chain. In total, the charge will be (0 from the carboxy group) + (+1 from the amino group) + (0 from the side chain) = +1. The image below depicts phenylalanine at this pH. A: This would be the charge on a phenylalanine molecule at a very basic pH (>9), not a pH as acidic as 1. B: This would be the charge on a phenylalanine molecule at or around physiological pH (7.4). C: This answer implies that half of the phenylalanine molecules have protonated amino groups, while the other half do not. This is not true at pH = 1.

What is the net charge on a phenylalanine molecule at pH 1? A. -1 B. 0 C. +0.5 D. +1

B is correct. Acetylsalicylic acid is a weak acid, with a pKa of 3.5. Therefore, the pH of this solution must be less than the pKa, because the compound is primarily in its acid form and a pH of 3.5 would mean that the concentration of weak acid and conjugate base were equal (a buffer). Choices C and D can be eliminated. A pH of 1.0 for an acid whose concentration is 0.10 M would require complete dissociation (as in a strong acid), eliminating choice A. Alternatively, the pH can be determined from the equilibrium expression: Ka = [H+][A-]/[HA] 10-3.5 = x2/(0.1-x) Since x will be small, we can approximate 0.1-x ~ 0.1, giving: 10-3.5 = x2/(0.1) 10-3.5(0.1) = x 2 10-4.5 = x2 [10-2.25][10-2.25] = x2 10-2.25 = x Since x equals the hydrogen ion concentration, taking the -log(10-2.25) = 2.25. A: This would be correct if acetylsalicyclic acid were a strong acid, but it is weak. C, D: These answers are the result of miscalculation.

What is the pH of a 0.10 M aqueous Aspirin (180 g/mol; pKa = 3.5) solution of acetylsalicylic acid? A. 1.0 B. 2.3 C. 3.5 D. 4.1

C is correct. Both the vertical displacement of the runner's steps and the angle of her body from the vertical increase the energy required to realign the ground reaction forces. Recall that tanθ = sinθ/cosθ, so when θ is close to zero, so is tanθ. When θ is close to 90°, tanθ becomes arbitrarily large. The more the runner leans into the run, the greater (or closer to 90°) tanθ is, and the greater the work expended by the runner must be, according to the work equation given in the passage (Wx = Fz tan(θ) Δz). Also, in this equation, the passage states that Δz represents the vertical displacement of a single step. Thus, the higher the vertical displacement ("high, bouncing strides..."), the greater the energy expenditure. A, B: When the runner keeps her spine fairly vertical, θ is close to 0°, meaning that tanθ is close to zero as well. According to the work equation from the passage, the smaller tanθ is, the less work is expended. This question is asking when the runner expends the most energy (in other words, when she must do the most work), rather than the opposite. D: According to the equation Wx = Fz tan(θ) Δz, a large Δz corresponds to a large Wx (work) value. This answer choice describes the opposite, as "long, low strides" indicates a small Δz. A small Δz corresponds to a small work value, indicating that the runner is expending less energy when taking these low strides. This question asks for the opposite — when the runner expends the most energy.

When does a runner output the most additional energy to keep the ground reaction forces most nearly vertical and through her body's center of mass? A. When she takes high, bouncing strides and keeps her spine fairly vertical B. When she takes long, low strides and keeps her spine fairly vertical C. When she takes high, bouncing strides and leans her upper half into her run D. When she takes long, low strides and leans her upper half into her run

D is correct. The compound mentioned in choice D is alternatively termed tert-butyl chloride (below). An SN1 reaction proceeds by way of a mechanism in which the leaving group (in this case, a chloride) dissociates in a kinetically slow step, producing a planar carbocation which is then rapidly attacked by a nucleophile. Steric congestion tends to promote SN1-type mechanisms, so a tertiary alkyl halide, like tert-butyl chloride, would favor pushing the leaving group off to form the carbocation. A, B: Chloromethane and chloroethane (below) are primary alkyl halides and are not very sterically congested. Therefore, they would probably favor an SN2-type substitution mechanism. C: While the leaving group in 2-chloropropane (below) is on a secondary carbon, which is somewhat sterically congested, it is not as crowded as in the tertiary carbon of 2-chloro-2-methylpropane.

Which of the following alkyl chlorides is most likely to undergo an SN1-type reaction? A. Chloromethane B. Chloroethane C. 2-chloropropane D. 2-chloro-2-methylpropane

B is correct. This question is asking us to determine why arginine is more basic than lysine. The reason must be related to how arginine is better able to handle being protonated, as this is the essence of being a base. Since, in its protonated form, arginine has electron-donating groups via resonance with other nitrogens, it is a more stable conjugate acid. The resonance structures of arginine at this position are shown below. Note that the backbone amino and carboxylic acid groups are deprotonated, meaning that this is the structure of arginine at relatively high pH (albeit not high enough to deprotonate the side chain). A: The stability of arginine's conjugate base is not in question. C: A lack of electron-donating groups does not stabilize a conjugate acid. D: The stability of lysine's conjugate base is not in question.

Which of the following best explains why arginine is more basic than lysine? A. The electron-donating groups around the basic nitrogen on arginine make its conjugate base less stable. B. The electron-donating groups around the basic nitrogen on arginine make its conjugate acid more stable. C. The lack of electron-donating groups on lysine make its conjugate acid more stable. D. The lack of electron-withdrawing groups on lysine make its conjugate base more stable.

D is correct. The visible spectrum contains electromagnetic signals with wavelengths ranging from 400 nm to 700 nm. The wavelength of light emitted during a particular electronic transition is determined by the energy difference (ΔE) between the final and initial energy levels. The energy of each level can be determined using Equation 1 and the principal quantum number in question. For light to be emitted at all, an electron must travel from a higher to a lower level, a process that releases energy. This automatically eliminates options A and B; however, we must actually perform calculations to choose between C and D

Which of the following electronic transitions for a hydrogen atom would result in the emission of a photon that would be visible to the human eye? A. n = 1 to n = 3 B. n = 2 to n = 4 C. n = 2 to n = 1 D. n = 4 to n = 2

C is correct. In order to emit a photon carrying energy, an electron must go from a higher energy level to a lower energy level. Only choice C shows such a transition. A, B, D: These transitions represent a move from a low to a higher energy level.

Which of the following electronic transitions for hydrogen would result in the emission of a quantized amount of energy? A. n = 1 → n = 2 B. n = 2 → n = 3 C. n = 5 → n = 4 D. n = 4 → n = 6

C is correct. Only Roman numerals (RNs) I and III are supported by the passage and supporting science. For RN I, the information about Rutherford is located in paragraph 2. Alpha particles are the nuclei of helium atoms, 4He2+. Since the vast majority of alpha particles pass straight on through the gold foil without deflection or rebounding, they must not have interacted significantly with anything having significant mass or charge, making RN I correct. Regarding RN III, as mentioned in paragraph 2, the deflections were caused by electrostatic repulsion. In order for the alpha particles to rebound or be deflected, the nuclei of the gold atoms must have the same sign of electrical charge as the alpha particle, causing a repulsive electrical force (according to Coulomb's law). Therefore, RN III is correct. The general structure of the atom is depicted below. Note the positive charge of the nucleus, which contains protons (positive) and neutrons (neutral). In the specific case of gold, this nucleus contains 79 protons. II: This is not consistent with the experiment because the alpha particles would not easily pass through the atom if most of its volume was occupied by matter having significant mass

Which of the following explain(s) the observations made by Rutherford? I. Most of the volume of the gold atom is empty space. II. The mass of an atom is evenly distributed throughout the entire volume of the atom. III. The alpha particles and the gold nuclei are both positively charged. A. I only B. I and II only C. I and III only D. II and III only

A is correct. All but one of the hydrohalic acids completely dissociate. The one exception is hydrofluoric acid, which is a weak acid in aqueous solution due to its unusually high degree of covalent bonding between the fluorine and the hydrogen. B, C, D: These are all classic strong acids

Which of the following is a weak acid in aqueous solution? A. HF B. HCl C. HBr D. HI

D is correct. A gamma particle is a photon of electromagnetic energy, which does not have mass. A: An alpha particle consists of two protons and two neutrons, having a mass of 4 amu. B: A beta particle is the nuclear equivalent of an electron, which has a mass of approximately 1/1800 of a proton. C: A positron is the antiparticle of an electron and has its same mass.

Which of the following particles is expected to have the LEAST mass? A. An alpha particle B. A beta particle C. A positron D. A gamma particle

A is correct. Lithium (atomic symbol Li) is in group 1, which is sometimes called the alkali metal group. This group has the largest atomic radii of the periodic table. Thus, lithium has the largest (not the second-largest) atomic radius in period 2, and choice A is the correct answer to this NOT question. Group 2 elements, such as magnesium and calcium, are known as the alkaline earth elements and would have the second-largest atomic radii. Be sure to avoid confusing periods (rows) of the table with groups (columns)! Period 2, where lithium is found, is highlighted below. B, C, D: From the periodic table, we can see that iodine (atomic symbol I) is in group 7 and period 5. Group 7 is the halogen group, which is marked by high electronegativity; iodine, in particular, has the highest electronegativity in period 5. (Note that fluorine is the most electronegative atom overall, but it is not located in period 5!) Iron (atomic symbol Fe) is a d-block element, or transition metal. Neodymium (atomic symbol Nd) is an f-block element, or lanthanide element, which are sometimes referred to as the rare earth elements. All of these characterizations are correct.

Which of the following statements is NOT true concerning the element that it describes? A. Lithium has the second-largest atomic radius in period 2. B. Iodine has the highest electronegativity in period 5. C. Iron is a transition element. D. Neodymium is a rare earth element

A is correct. This question asks us about the wavelength of EMR that ejects an electron from an atom. Shorter-wavelength EMR (such as γ rays) carries much more energy than longer-wavelength EMR (such as radio waves). Therefore, we must look for the answer choice that involves the highest-energy EMR. The closer an electron is to the nucleus, the harder it is to eject. Because sp-hybridized orbitals have the most s character of all of the answer choices, they contain the electrons that are hardest to eject. B: sp2-hyrbridized orbitals have 33% s character, while sp-hybridized orbitals have 50% s character. C: sp3-hyrbridized orbitals have 25% s character, while sp-hybridized orbitals have 50% s character. D: Radiation that excites an electron but does not eject it is of lower energy than radiation that is able to eject an electron.

Which of the following types of electromagnetic radiation would have the shortest wavelength? A. Radiation that ejects an electron from an sp orbital B. Radiation that ejects an electron from an sp2 orbital C. Radiation that ejects an electron from an sp3 orbital D. Radiation that excites but does not eject an electron from an sp3 orbital

D is correct. If SEC separates on the basis of differences in molecular weight, the technique's resolution should increase if the pair of proteins differ in their molecular weights to a greater extent. A, C: The numerical values of the molecular weights do not matter nearly as much as the difference between these weights. B: Resolution between separable proteins will be greatest if their molecular weights differ by the largest amount possible, so long as the molecular weights of both proteins are within the size selectivity range of the column. As such, choice B is the opposite of the correct answer.

Which of the following types of proteins would be most efficiently fractionated by an SEC column? A. Two proteins with very high molecular weights B. Two proteins with similar molecular weights C. Two proteins with very low molecular weights D. Two proteins with substantially different molecular weights

There are a number of ways that this can happen, and when it does, the atom is forever changed. There is no going back - the process is irreversible. There are four primary types of decay: alpha decay, beta decay (β+ and β−), gamma decay, and electron capture. In alpha decay, an alpha particle, containing two protons, two neutrons, and a +2 charge, is emitted. In beta-minus decay, a neutron is converted into a proton in the nucleus, and a β− particle (an electron) is ejected to maintain charge balance. In beta-plus decay, a proton is converted into a neutron, and a β+ particle (a positron) is emitted to preserve charge. Gamma decay involves the emission of a gamma ray, which is a high-energy photon, from an excited nucleus. Finally, in electron capture, a nucleus "grabs" an electron, which changes a proton into a neutron.

four primary types of decay: alpha decay, beta decay (β+ and β−), gamma decay, and electron capture

Redox reactions can be carried out in special devices known as electrochemical cells. These cells must have two electrodes, which are where the redox half-reactions occur. The electrode where oxidation happens is known as the anode, while the electrode where reduction happens is known as the cathode. Therefore, a surplus of electrons is generated at the anode (because electrons are lost during oxidation), and they travel to the cathode. In a galvanic (or voltaic) cell, a spontaneous redox reaction is used to generate a positive potential difference that can drive current. The total standard potential generated by a cell, Ecell, can be calculated from the standard reduction potentials of the half-reactions. The simplest way of defining Ecell is presented below: Ecell = E°cathode − E°anode In contrast to a galvanic cell, an electrolytic cell uses a connected power source to conduct a nonspontaneous redox reaction. While galvanic cells have positive Ecell values (indicating spontaneity), electrolytic cells are characterized by negative Ecell values.

galvanic cell vs electrolytic cell

1000 kDa "log" means log base 10. Some log values to be familiar with are log(0.01) = -2, log(0.1) = -1, log(1) = 0, log(3) ≈ 0.5, log(10) = 1, and log(100) = 2, from which you can predict log values for larger powers of 10.

log (MW) = 6 DA

Power is defined as work divided by time (P = W/t). Its units are watts (W), and 1 W = 1 J/s. The idea of power is essentially that a given amount of work could be expended either quickly or slowly, and that a system capable of doing so quickly is more "powerful" than a system in which the energy represented by a given amount of work is dissipated slowly. Rearranging the units of J/s indicates that power can also be expressed as a constant force multiplied by a constant velocity (P = Fv). Another important definition of power occurs in the context of circuits, where power is defined as current multiplied by voltage (P = IV). This is also derived from the general concept of power being work divided by time, but it is best memorized separately for the MCAT.

power equals

On Test Day, you will need to distinguish between the behaviors and properties of two important classes of electrolytes: strong and weak acids and bases. A strong acid or base dissociates completely in water, while a weak acid or base dissociates incompletely. For example, HBr and HF both dissociate in water to produce H+ ions, but at equilibrium, a solution of HBr will produce more ions in solution than HF, because HBr is a strong acid while HF is a weak acid. The table below shows the strong acids and bases that are commonly tested on the MCAT. Because strong acids and bases completely ionize, the [H+] and pH (or [OH−] and pOH) are easily calculated, since the concentration of protons or hydroxyl groups is equal to the molarity of the solution. Thus, a 0.4 M solution of KOH has a [OH−] of 0.4, and a 0.007 M solution of HI has a [H+] of 0.007. For solutions so dilute that the H+ or OH− normally present in pure water (1 × 10−7 M) is comparable or larger to the amount produced by acid/base ionization, the ions present due to water must be considered.

problem 31 acids to know

springs for the MCAT is that the potential energy stored in a spring can be expressed as PEelastic = ½kx2

springs

Of the biochemical lab techniques tested on the MCAT, sodium dodecyl sulfate-polyacrylamide gel electrophoresis (SDS-PAGE) is quite possibly the most important to understand. The purpose of this technique is simple: it allows proteins to be separated by their mass alone. You may recall that gel electrophoresis accomplishes a similar goal for DNA and RNA: it uses an electric field to separate these molecules by length alone. So why can't gel electrophoresis also function to separate proteins? The answer is twofold. First, proteins vary widely in their structure, or folding patterns. If we simply tried to run a mixture of proteins through a gel electrophoresis apparatus, these variations would impact the proteins' migration. Additionally, simple gel electrophoresis relies on the fact that DNA and RNA molecules contain a uniform negative charge, causing them to travel toward the positive pole of the apparatus. In contrast, proteins can have positive, negative, or neutral charges, and these charges may not be uniformly distributed throughout the molecule. SDS-PAGE addresses both of these issues. First, to eliminate the effects of differences in shape, SDS-PAGE uses a strong anionic detergent: sodium dodecyl sulfate. The "sulfate" portion of this name denotes the negatively-charged head of the molecule, while the "dodecyl" refers to a long hydrocarbon chain that forms the molecule's tail. The SDS molecule denatures native proteins into their unfolded polypeptide states, which prevents protein shape from impacting the separation. To address charge differences and ensure that the proteins actually travel down the gel, SDS coats the proteins with an even distribution of charge per unit mass. Specifically, when the highly anionic SDS associates with the polypeptide backbone, the intrinsic charge of the polypeptide becomes negligible in comparison to the negative charges due to SDS. Since the protein is now highly negative, it will travel toward the positive end of the gel apparatus. Otherwise, SDS-PAGE functions similarly to standard gel electrophoresis. The larger the protein, the more hindered it is as it moves down the gel, and the shorter the distance it travels toward the positive end. In contrast, smaller proteins can travel through the pores of the gel more easily, so they migrate farther toward the positive pole.

summary of SDS-PAGE

Let's begin with SN1 reactions, which - contrary to what you may expect - take place in two distinct steps. In the first step, the leaving group dissociates from the target substrate molecule, leaving behind an unstable carbocation. Since the product is so unstable, this step is slow and thus rate-determining. (This is where the "1" in SN1 comes from; it denotes the fact that only one molecule - the substrate - is involved in this first step.) In the second step, the nucleophile attacks the carbocation to form a more stable product molecule. Since the carbocation intermediate is planar, or lacks stereochemistry, SN1 reactions with chiral substrates produce an even mix of enantiomers termed a racemic mixture. Carbocation stability is the primary factor in SN1 reaction rate. For this reason, SN1 reactions favor tertiary substrates (as tertiary is the most stable carbocation structure) and occur to a negligible extent with primary substrates. SN2 reactions are even simpler, as they take place in a single step. In this mechanism, the strong nucleophile displaces the leaving group by attacking from the rear, often described as a "backside attack." This reverse attack inverts the relative stereochemistry of the molecule. (The absolute stereochemistry - namely R vs. S configuration - is inverted as well, provided that the priority of the nucleophile matches the priority of the leaving group, which is nearly always the case.) For a brief instant before the leaving group fully dissociates, the central carbon of the substrate is at least partially bound to five substituents. This high-energy state is known as a pentavalent transition state. With everything happening at once, steric hindrance is a major limiting factor of SN2 reactions, with primary substrates reacting the most rapidly. Solvents have a notable effect on the rates of these reactions. Polar protic solvents (such as water and ethanol) tend to stabilize ions in solution. Since they can stabilize the carbocation, these protic solvents are best used for SN1 reactions. However, SN2 reactions rely not on a carbocation, but on a strong nucleophile displacing the leaving group. Protic solvents tend to stabilize (weaken) this nucleophile, so they should not be used for SN2 procedures. Instead, polar aprotic solvents, such as acetone, are a better choice.

summary sn1 sn2

Surfactant Surfactants are amphipathic molecules (containing both hydrophilic and hydrophobic regions) that reduce the surface tension of a liquid. For the MCAT (and in your body), the most important example is pulmonary surfactant, which reduces the surface tension in the alveoli, allowing them to remain inflated when the lung is compressed during respiration. The general category of surfactants also includes detergents, the amphipathic structure of which allows them to denature proteins (an example being sodium dodecyl sulfate, or SDS, which is used in an electrophoretic technique known as SDS-PAGE) or solubilize lipids.

what are Surfactant?

C is correct. Let's begin with RN I. As stated in the passage, the P wave is associated with the depolarization and contraction of the atria. A rest potential is created when sodium ions are pumped to the outside surface of the membrane of a cell. When ion channels open, the sodium ions spontaneously flow back into the cell, resulting in depolarization and an action potential. Roman numeral (RN) I is thus correct. Regarding RNs II and III, the passage states that the Q and R waves are associated with the contraction of the ventricles. Like atrial contraction, ventricular contractions result from depolarization, which would involve the inward movement of cations like sodium ions. Thus, RN II and III are accurate as well. IV: The passage states that the T and U waves occur due to repolarization, which takes place when positive ions are pumped back out of the cell to return to rest potential prior to the next heartbeat. This is the opposite of what the question asks for. A: RN II and RN III are also correct. B: RN I is also correct. D: RN IV is incorrect.

what are most likely to be explained by the movement of sodium ions into a cardiac neuron? I. P II. Q III. R IV. T A. I only B. II and III only C. I, II, and III only D. I, II, III, and IV

The third law of thermodynamics allows us to calculate absolute entropy, but not absolute enthalpy. enthalpy does not approach 0 as temperature approaches 0 K

what law allows us to calculate absolute entropy?

Reactions that increase the number of moles of substances in the system (or produce more gas particles) typically increase the entropy of the system. Entropy generally increases when a solid or liquid is dissolved in a solvent. Entropy increases when the solubility of a gas decreases and it escapes from a solvent. Entropy generally increases as molecular complexity increases

when does entropy increase in a literal sense?

From a problem-solving point of view, it is useful to remember that work can be expressed in multiple ways. A common definition is W = F⋅d⋅cos(θ), but other ways to calculate work exist. For instance, the work-energy theorem states that the work performed on or by an object is equal to the change in its kinetic energy: W = KEfinal - KEinitial. It is therefore important to be alert to which parameters are given in specific MCAT problems in order to approach power effectively.

work =