MCAT General Chemistry Review

Part II Practice Sections

Answers and Explanations

PRACTICE SECTION 1

ANSWER KEY

1. A

2. C

3. D

4. A

5. B

6. D

7. C

8. C

9. B

10. B

11. D

12. B

13. D

14. A

15. A

16. B

17. C

18. B

19. C

20. C

21. A

22. C

23. D

24. A

25. C

26. D

27. B

28. A

29. B

30. A

31. B

32. C

33. D

34. C

35. A

36. C

37. B

38. A

39. C

40. D

41. C

42. B

43. A

44. A

45. C

46. B

47. B

48. C

49. B

50. A

51. D

52. C

PASSAGE I

1. A

There are two concepts involved here. The first is that both of these species are strong acids, so they will fully ionize in solution; the bivalent species has two H+ ions for every one H+ ion of the monovalent species, so it will produce a more acidic solution. Second is the idea that a more acidic solution will have a lower pKa.

2. C

It is given in the passage that the pH of normal rain is 5.2, so this problem is simply asking you to convert pH = 5.2 into [H+]. Using the Kaplan logarithm estimation procedure on the answers, it is clear that the only concentration that will produce a pH of 5.2 is (C).

3. D

It is important to know that every Arrhenius acid is also a Brønsted-Lowry acid, and that every Brønsted-Lowry acid is also a Lewis acid (the same idea applies for bases). This means that (B) and (C) can be eliminated. Sulfuric acid produces protons, so it is qualified as an Arrhenius acid, and we then know it can be called a Brønsted-Lowry or Lewis acid as well. All of the items are correct.

4. A

We are able to tell that HCO3- is a buffer because it can either donate or accept a proton. Being able to donate its remaining H+ allows it to act as an acid, while its overall negative charge allows it to act as base and accept an H+. (A) is correct because HCO3- will accept a proton from the acid introduced to the blood stream, and act as a buffer. (B) is incorrect because while it will act as a buffer, it will accept an H+ ion, not donate an H+ ion. (C) is incorrect because HCO3- does act as a buffer. (D) is incorrect because there is enough information.

5. B

Write down the steps as you go. First, we need to figure out what the original [H+] is in the rain. Because H2SO4 is bivalent, we double its concentration to find its contribution to the total [H+], which is 4.0 × 10-3 M. We then add this to 3.2 × 10-3 M to get [H+]rain= 7.2 × 10-3 M. If we divide this number by 4 (the factor by which the volume decreased, use V1C1 = V2C2), we get [H+]rain+pure = 1.8 × 10-3 M. Finally, we can use Kaplan’s logarithm estimation strategy to find that the pH is just below 3, so 2.8 is the answer.

6. D

A conjugate base is an acid that loses a proton. (D) shows an acid and a molecule that has lost a molecule of H2O, so it is not an acid/conjugate base pair. The other choices show an acid and conjugate base that has one less proton.

7. C

This question tests your ability to actively read the passage and synthesize the information presented. The student would likely agree with (A) because pollution is a major cause of acid rain, and industrialization over the past 150 years has greatly increased pollution levels. The student would likely agree with (B) because, as shown in the passage, radicals are integral in the formation of acid rain. The student would likely disagree with (C) because more acid content in water would increase the conductive capacity of water. The student would likely agree with (D) because while rain is naturally acidic, acid rain contains dangerous levels of acid.

8. C

The first pKa in this curve can be estimated by eye. It is located between the starting point (when no base had been added yet), and the first equivalence point. This point is at pH of approximately 6.3. The value of the second pKa is notable because it is found in a slightly different way than the first pKa. It is located at the midpoint between the first and second equivalence points. In this curve, that corresponds to pH = 10.3

9. B

Equivalence points are at the midpoint of the quickly escalating slope range. In this titration curve, the value of the first equivalence point is 7.8 and the value of the second is 12.0.

PASSAGE II

10. B

Many MCAT Physical Sciences questions include dimensional analysis calculations. If you do not know the specific heat of water from memory, you’ll want to know this constant on Test Day! (It is reported in paragraph 1.) The correct answer is (B). (A) is tempting, as it has the same magnitude as the specific heat of water in Jg-1K-1. You will need to convert (A) from grams to moles using the molecular mass of pure water, 18 g/mol. (C) and (D) differ from (B) and (D) by factors of 1,000, a frequent source of error in thermochemistry problems (using kJ rather than J).

11. D

This is a discrete question which draws on information from the passage. The passage does not provide an explicit comparison of intensive and extensive physical properties. You can infer from paragraph 1, however, that intensive properties do not depend on the amount of substance present in the measurement, and that extensive properties do. While viscosity is irrelevant to this specific experiment, it is an intensive property of a liquid. Mass, heat, and enthalpy are all extensive properties. (Heat is an extensive property; temperature is an intensive property.) The correct answer is (D).

12. B

Use a calorimetry equation to determine the heat capacity of the calorimeter. This setup requires referring to table 1, which describes how the student calibrated the instrument. The equation must account for all heat inputs and outputs in the closed system. In short, the heat lost by one part of the system must be gained by another part of the system. Here, the cold water heats up, and the hot water cools.

That transition is summarized by the equation mhotCH20(Tf -T i,, hot) = mcold CH20(Tf - Ti, cold) + Kcal (Tf - Ti, cold). Use the density of pure liquid water, 1 g/mL, to convert the volumes of hot and cold water to mass quantities. (B) is the answer. (A) is the specific heat of water, not the heat capacity of the calorimeter. (C) results from reversing the sign of the hot water temperature change, which obtains a negative heat capacity value. (D) is off by a factor of 1,000, using water’s specific heat in kg.

13. D

While it is true that mercury is much more dense than alcohol, the thermometer content should not make a difference in the student’s results within this temperature range [(A) and (B)]. There is no information to suggest the relative precision of the instruments (C). (Had the question asked about a digital thermometer, this would be relevant.) (D) is correct. There should be negligible differences in the imageT obtained by each respective thermometer, though initial and final temperature measurements may vary slightly for each instrument.

14. A

Styrofoam is an excellent, though imperfect, insulator. Calibration accounts for heat loss to the calorimeter. The amount of water is irrelevant in a specific heat measurement, and that the calibration is supposed to account for the properties of the styrofoam (B). Calibration (C) accounts for a heat transfer, not a temperature change. The water temperature equilibrates in order to calibrate the thermometer, but that doesn’t suggest anything about the reaction of other calorimeter contents inside (D).

15. A

The addition of salt ions to pure water disrupts hydrogen bonding between water molecules. When an aqueous solution is more disordered, there are weaker forces between component molecules. Compared to a highly stable hydrogen bond network like water, the salt water solution is disordered enough that its intermolecular bonding is much easier to break, lowering its specific heat.

16. B

Pressure must remain constant; paragraph 3 implies that bomb calorimeters are useful because they maintain constant pressure. Though specific heat is an intensive quantity, it is important to keep track of mass for the overall calorimetry calculation, which can involve different substances with different heat capacities. Heat should not enter or exit the system in a precise measurement. Items I and III are correct so the answer is (B).

17. C

Bomb calorimeters operate at high pressure, so they can accommodate temperature changes in a gas. Coffee cup calorimeters are no longer useful when the water boils. (A), (B), and (D) are distractors. Saltwater (A), as an aqueous ionic solution, is analogous to the fruit punch solution described in the passage. While ethanol (B) boils at a lower temperature than water, its structure is highly similar to that of pure water and it is relatively stable at room temperature. Coffee cup calorimeters can measure the specific heat of a metal like copper (D) if it is placed in water.

QUESTIONS 18–21

18. B

For every 2 mol of HCl, 1 mol of hydrogen gas is produced (assuming excess magnesium). Multiplying both sides of the ratio by 1.5 means that 3 mol HCl under the same conditions should produce 1.5 mol hydrogen gas. At STP, 1 mol of a gas (assuming it to be ideal) occupies 22.4 L so 1.5 mol HCl should occupy 33.6 L.

19. C

Savvy test takers will note that (A) and (D) can be eliminated because the answer is internally inconsistent; favoring a reaction almost always means increasing its rate. To decide between (B) and (C), reason via Le Châtelier’s principle. The stress is heat. An endothermic reaction requires heat and so is more likely to consume the heat (i.e., reduce the stress) than an exothermic reaction, which will add to the stress by producing heat.

20. C

By definition, gamma radiation is a stream of energy that in addition to creating the Hulk has neither mass nor charge. No such thing as delta radiation (D) has been defined. Alpha particles and beta particles have both charge and mass.

21. A

Rust, or corrosion, is the oxidation of a substance when it comes into contact with both water and oxygen. (B) can be eliminated because, if true, this would enhance the reactivity of Al or Zn. (C) can be eliminated because reducing agents are oxidized and so that would make Al or Zn more likely to rust. (D) is incorrect, because Al and Zn still rust, just more slowly than iron. As a metal rusts it oxidizes and thus function as a reducing agent. Self-protective oxides (a common way of finding alkali metals as well) prevent further oxidation by complexing the atom with oxygen.

PASSAGE III

22. C

Compound C is shown to promote the reaction without affecting the overall yield of product. It is not consumed or produced in the net reaction, either. This is enough information to identify compound C as a catalyst, which decreases the activation energy of a reaction by definition. In this case, (C) is more likely than (D). Step 1 of the reaction is the fast step and step 2 is the rate-determining, slow step. Even though the catalyst may have an effect on the fast step, its primary purpose is to speed up the slow step. The bulk of the energy input in the overall reaction is in step 2. The catalyst decreases the activation energy, so it is likely to have its primary effect on the step that requires the greatest energy input.

23. D

To answer this question, you must apply Le Châtelier’s principle, which states that adding a compound to a system will shift the system reaction such that less of that compound is produced. Adding compound A decreases the overall yield of the system. This change pushes the equilibrium in step 1 to the left to produce more of compound AB and less of compound A. Unbound compound B is not involved in any step of the reaction, so adding it will have no effect on any of the steps. Compound C is a catalyst; increasing the concentration of a catalyst may increase the rate of the reaction, but will not affect the equilibrium constant at any given temperature. The correct answer is (D), as compound D is present on the left side of step 2. Adding more of compound D will push the reaction to the right, increasing the amount of product.

24. A

In terms of concentration, Keq is equal to [products]/[reactants]. An increased concentration of products signifies an increased equilibrium constant, so Keq of this reaction will increase along with the equilibrium concentration of BD. The data suggest that this concentration increases with increasing reaction temperature, but the concentration starts to stabilize after the temperature reaches approximately 100° C. Only (A) contains a graph demonstrating a Keq which initially increases with temperature but starts to level off around 100° C. (B) shows a graph which follows the opposite of the correct path. (C) suggests that the equilibrium constant rises faster and faster as temperature increases, which is incorrect. (D) represents an equilibrium constant that does not change with temperature.

25. C

If a reaction is first-order with respect to each of the reactants, the overall rate is directly proportional to the concentration of each reactant. Doubling the concentration of compound AB will double the rate of reaction. In this new experiment, the concentration of compound AB is twice as high as the concentration used in the first experiment, so the rate of formation is also twice as high.

26. D

The data in the two tables shows that the second experiment, which omitted the catalytic compound C, was significantly slower than the first at every temperature except 150° C. Recall that under normal circumstances, a catalyst will speed up a reaction by decreasing its activation energy, thereby allowing the reaction to reach its activation energy more easily. Consequently, increasing the amount of heat will also help the reaction achieve its activation energy. In this particular experiment, the catalyzed experiment was not much faster than the uncatalyzed experiment at 150°C. This suggests that the reaction reached its activation energy without the help of a catalyst at this temperature range.

27. B

Single-replacement reactions involve the direct replacement of one constituent of a molecule with another constituent molecule. In this case, compound C replaced compound A to turn AB into BC. Double replacement reactions, on the other hand, involve the replacement of two different species. An example of such a reaction is AB + CD image AC + BD. Combination reactions require two or more molecules to combine into one molecule, as in step 2. Decomposition reactions require one molecule to break down into two or more molecules, as in step 3.

28. A

Gases are highly compressible, while solids and liquids are not. For this reason, changes in pressure will not have a substantial effect on solids and liquids. An increase in pressure will push the reaction toward the side with less gas molecules, while a decrease in pressure will do the opposite. Because step 1 is the only part of the reaction mechanism that involves gases, it is also the only part that will be affected by a change in pressure.

29. B

This system starts out with two gases (AB and C) and one solution (D) and ends with one gas (A) and two solutions (BD and C). Gases have higher average kinetic energies than solutions do, so they are more disordered (i.e., they have a higher overall entropy, which is a thermochemical measure of the relative disorder in a system). The entropy of the products is lower than the entropy of the reactants, so imageS for the overall reaction is negative. Based on the information given, it is impossible to determine whether imageH is positive or negative. Although heat is added to stimulate the reaction, the passage does not specify how much heat is released at the end. imageG is typically calculated from imageH and imageS. imageH is not known, so imageG is also impossible to determine.

PASSAGE IV

30. A

Newlands’ table suggests that platinum is heavier than gold. According to the modern periodic table, gold is heavier than platinum. (B), (C), and (D) represent pairs of elements that were arranged correctly by Newlands.

31. B

Periods 2 and 3 of the periodic table, which contain only elements in the s-block and the p-block, are the only periods that actually contain exactly eight elements. If most of the elements in the d-block and the f-block were discovered, it would become obvious that elements do not occur in matching octaves. (A) is incorrect because the discovery of all of the s-block and p-block elements would create more matching octaves (although the theory would have to be modified to accommodate the noble gases). (C) is incorrect because the law of octaves does not exclude the possibility of electrons, as long as the electrons are not organized as in Bohr’s theory. (D) is incorrect for the same reason as (C).

32. C

Mendeleev’s table was organized in terms of atomic mass rather than atomic number. Item I did require modification because it was subsequently discovered that an element is characterized by its number of protons (rather than its mass). Item II did not require direct modification of the table and the existing entries were left intact; the only change was the knowledge that future entries would be organized according to their outer orbitals. Item III only strengthened Mendeleev’s original model, since the discovery of gallium fulfilled his prediction that two elements exist between zinc and arsenic. Items II and III did not require modification, so (C) is the answer.

33. D

The passage states that Newlands’s table contains all of the elements that had been discovered at the time. Because his table contains no noble gases, we can assume that they had not been discovered. The table contains uranium, which is an f-block element (A), and several instances of halogens and metalloids [(B) and (C)] are evident throughout the table.

34. C

An element’s first ionization energy, which is defined as the tendency of the element to donate an electron, decreases with increasing atomic number within a period. Because Mendeleev’s predicted element had a higher atomic number than calcium, it is less likely to lose an electron. Atomic radius and ionic radius [(A) and (B)] both decrease with increasing atomic number. Electron affinity increases with increasing atomic number within the same period (D).

35. A

Atomic radius increases as an element gains energy levels and decreases as an element gains protons and electrons; you should know that the largest elements are those that have more energy levels and are found near the bottom of the periodic table. Uranium is the only element on Newlands’ table that has seven energy levels, meaning it has the largest atomic radius.

36. C

If Mendeleev had created his table in response to Newlands’ theory, then Mendeleev would clearly never have made his breakthrough without Newlands’ contribution; if this were the case, it’s reasonable to say that Mendeleev simply modified the law of octaves, which was eventually refined to produce the modern version. (A) is incorrect because it does not suggest that Newlands had any impact on the development of today’s periodic table. (B) also makes it unlikely for Newlands to have made a direct contribution to the evolution of the system, because Mendeleev only learned about Newlands’ work after he had already invented his own table. (D) clarifies the fact that Mendeleev was the first person to publish the modern periodic table and does not suggest that Newlands made any contribution to these findings.

PASSAGE V

37. B

This is a simple ideal gas law question. Using the equation PV = nRT, we are able to substitute: (1 atm)(V) = (1 mole) (R)(318 K), which simplifies to 318R. The answer does not need to be simplified past 318R (you are not allowed to use a calculator during the exam and multiplication by R takes too much time by hand).

38. A

This question tests your knowledge of the ideal gas law. The law shows that volume and temperature have a direct linear relationship (PV = nRT), meaning that as volume increases, so must T (assuming isobaric conditions). (B) shows an indirect linear relationship, (C) an exponential one, and (D) a logarithmic one.

39. C

In Experiment 2, we see that V1 has been halved. So using the knowledge that temperature has been held constant, we know that the pressure must be double what it originally was to maintain the ideal gas law. So the right answer is 2 atm. The other choices all would violate the ideal gas law.

40. D

The a in the van der Waals equation accounts for the attractive forces between gas molecules. The b in the equation accounts for the actual volume that the molecules occupy (B). It isn’t necessary to memorize the van der Waals equation for Test Day, but do know what a and b correct for in the equation.

41. C

The key here is realizing that 64 g O2 is 2 mole O2. Substituting the values given into this law, we get: (2 atm)(3 L) = (2 mole)(R)(T ). This simplifies to 3/R K.

42. B

Dalton’s law of partial pressures says that a gas’s molar fraction multiplied by the total pressure gives the partial pressure supplied by that specific gas. This question asks specifically about 1.5 mole of NO, out of a total of 3 mole, which means that NO has a molar fraction of XNO = 0.5. Solve for Pusing the ideal gas law: (2 L)(P) = (3.0 mol)(R)(300 K), P = 450R atm. This value multiplied by XNO comes to 225R atm.

43. A

This question simply tests your knowledge of the conditions under which the ideal gas law is most relevant. The reference to experiment 1 is included just to mislead you. It is necessary to know only that the gases act closest to ideal when they are at high temperatures and low pressures.

44. A

At first glance, this question looks extremely simple, yet you must realize that if the center divider is receiving different pressures from each side, it will move until pressure from both sides is equal. When the experimenter reduced the molar concentration in V2 by half, the pressure was reduced on the V2 side of the divider, leading to the expansion of V1. It is important to note that (B) and (D) are incorrect because the question explicitly states that the cylinder is allowed to re-equilibrate with the new molar concentrations.

PASSAGE VI

45. C

An oxidation-reduction reaction requires a transfer of electrons from one atom to another. This reaction simply involves the neutralization of a strong acid by a weak base; no electrons are transferred and all of the oxidation numbers stay constant throughout the reaction.

46. B

The conjugate base of a Brønsted-Lowry acid is the product that does NOT include the H+ ion that came from the acid. Even if you did not know this definition, there is a hint to this answer in the passage; because the products in reaction 1 contain a salt and an acid, it is clear that the salt is not the conjugate acid in neutralization reactions. Based on this information, you should be able to determine that the conjugate base is MgCl2. You can determine the percent composition of the cation (Mg2+) in this salt by dividing the molecular weight of the cation (24.3 g/mol) by the molecular weight of the molecule (95.2 g/mol). It should be obvious that 24.3/95.2 is approximately equal to 25/100, which equals 25%.

47. B

The passage states that efficacy is equal to the number of moles of HCl that can be neutralized by one gram of antacid. This means that the most effective antacids are those with the maximum neutralization capacity per gram. Because CO32- can neutralize two H+ ions, Al2(CO3)3 has the capacity to neutralize six HCl molecules. Similarly, each of the other answer choices can neutralize three HCl molecules. (C) and (D) can be eliminated because they have the same neutralization capacity as Al(OH)3, but are significantly heavier molecules. To decide between (A) and (B), consider that Al2(CO3)3 has a molecular weight of 234 g/ mol, while Al(OH)3 has a molecular weight of 78 g/mol. Because Al2(CO3)3 can neutralize only twice as many molecules as Al(OH)3 but has nearly three times the weight, it is clear that Al(OH)3 has the best per-weight efficacy.

48. C

The passage states that NaHCO3 is an antacid, so (A) is incorrect. The reactions suggested in (C) and (D), where H2CO3(aq) decomposes to produce H2O (l) and CO2 (g), is a common reaction that recurs every time H2CO3 is present in aqueous solution. Because nearly all the CO2 is in gas form, it does not significantly affect the pH of the solution; this means that there are no acids left to decrease the pH and, therefore, NaHCO3 is an effective antacid.

49. B

The limiting reagent is the reactant that is completely used up during the reaction while the other reactant still remains. Antacids are alkaline, so if the pH is below 7, there must not have been enough antacid to neutralize all of the HCl. This means that the progress of the reaction was limited by the amount of antacid.

50. A

The passage states that the student initially tested 1 gram of antacid along with 100 mL of 0.1 M HCl. Magnesium hydroxide, Mg(OH)2, has a molecular weight of about 58 g/mol; because the answer choices are all very rough approximations, we can estimate the weight as about 50 g/mol in order to make the calculations easier. At this molecular weight, 1 gram of antacid is approximately equal to (1 g)/(50 g/mol) = 0.02 mol. 100 mL of 0.1 M HCl is equal to (0.1 L) × (0.1 mol/L) = 0.01 mole. Because each molecule of Mg(OH)2 has two OH- ions, only half a mole of Mg(OH)2 is required to neutralize a mole of acid; therefore, only 0.005 moles of antacid is used up. The student started with 0.02 moles of antacid, so he is now left with 0.02 - 0.005 = 0.015 moles. To convert back to grams, we multiply 0.015 moles by 50 g/mol to get 0.75 g—which is equal to 750 mg.

51. D

A stronger antacid can neutralize the same amount of acid while using a smaller quantity of reactant. This means that more reactant will be left over after the neutralization is complete. Often, this leftover reactant will present itself in the precipitate (some antacids, however, will dissolve in the solution; this is why the student’s logic was flawed). (A) is incorrect because the question states that the antacid is present as an excess reagent; increasing the amount of an excess reagent does not increase the amount of product unless more of the limiting reagent becomes available. (B) is incorrect for the same reason.

52. C

Sodium and potassium (and the rest of the alkali metals) form soluble salts, while aluminum, magnesium, and calcium do not; however, you do not need to know this fact in order to answer this question. The increased solubility of the compounds in group B caused the unreacted material to dissolve in solution, so the student was unable to detect any of the starting compound in the precipitate. The compounds in group A are sparingly soluble, so stronger antacids left a larger amount of unreacted solid. The alkalinity of the compound has no bearing on its ability to precipitate (A); also, you should know that strong bases containing sodium and potassium (i.e., NaOH and KOH) are just as alkaline as their counterparts which contain magnesium, calcium, and aluminum. (B) and (D) are accurate statements, but do not explain the student’s results as well.

PRACTICE SECTION 2

ANSWER KEY

1. B

2. D

3. A

4. C

5. B

6. C

7. D

8. B

9. B

10. D

11. A

12. B

13. A

14. B

15. C

16. A

17. C

18. D

19. B

20. A

21. A

22. C

23. A

24. B

25. D

26. C

27. B

28. D

29. A

30. C

31. A

32. A

33. B

34. A

35. B

36. D

37. A

38. C

39. B

40. D

41. B

42. D

43. A

44. C

45. C

46. C

47. A

48. B

49. A

50. D

51. C

52. C

PASSAGE I

1. B

The dissociation reaction tells us that the coefficients for both products are equal to 1. We use this information to write the solubility equation. The solubility equation for the dissociation of CaSO4 is the following:

4.93 × 10-5 = [Ca2+][SO42-]

Because Ca2+ and SO42- have a one-to-one ratio, they can be replaced by the variable x to solve for their concentration. Solving the equation 4.93 × 10-5 = (x)2, we obtain a concentration of 7.02 × 10-3 M for each ion. Because one mole of ion is equivalent to one mole of salt, this is the concentration of CaSO4 needed to equal the Ksp, at which point the solution is saturated. The concentration of CaSO4 ions, 7.02 × 10-3 M, can be multiplied by the volume, 3.75 × 105 L, to obtain a value of 2.63 × 103 moles. Converting moles of CaSO4 to grams is accomplished by multiplying by the molar mass of CaSO4, 136.14 grams/mole, to give a mass of 3.58 × 105 grams needed to reach saturation. (A) is the number of moles, not grams, of CaSO4.

2. D

One way to solve this problem is to calculate the Ca2+ concentration for one mole of each compound using the Ksp and chemical equilibrium equation. Looking at the answer choices, (C) can be ruled out immediately because of the extremely small Ksp, which indicates that very few of the salt ions will dissolve. The other Ksp values are comparable for (A), (B), and (D). Remember to raise the ion to the power of its coefficient when setting up the equilibrium equation. For example, the equilibrium equation for (D) would be Ksp = 6.47 × 10-6 = [Ca2+][IO3-]2. The concentration of calcium for each of these answer choices, respectively, equals 6.93 × 10-5 M, 2.14 × 10-4 M, and 1.17 × 10-2 M. (D) has the highest concentration of Ca 2+ at 1.17 × 10-2 M.

3. A

The concentration of a saturated solution of CaSO4 is equal to approximately 0.015 M, according to the graph. This number was found by determining the corresponding concentration for a conductivity value of 2500 µS/ cm. The concentration of CaSO4 is equal to the concentration of both Ca2+and SO42- ions in solution according to equation 1 because the coefficients are all 1. Therefore, the Ksp = [Ca2+ ][SO42- ] = (.015 M)2, which equals 2.25 × 10-4 M2. The concentration of CaSO4 must be square to obtain the Ksp, so (B) is incorrect. (C) doubles, not squares, the concentration. (D) might be obtained if you had taken the square root, not squared, the concentration.

4. C

As temperature increases in a solution, movement of molecules and ions increase. This will increase the current between the electrodes in the probe and increase the value of conductivity. Temperature’s effect on the solubility product constant cannot be determined without further information. Although many salts have a higher solubility with higher temperatures, not all salts share this property. The only sure way to determine the relationship is by experimentation. An increase in temperature would increase the conductivity, not decrease it, so (B) and (D) are incorrect. (A) is incorrect because it is impossible to determine the relationship between temperature and Ksp.

5. B

The common ion effect is when an ion is already present in the solution and affects the dissociation of a compound that contains the same ion. In this case, Ca(OCl)2 contains calcium and will cause the reaction in equation 1 to shift to the left due to Le Châtelier’s principle. If there is no Ca(OCl)2and therefore no extra Ca2+ present in solution, more CaSO4 will dissociate. (C) is the opposite. (A) is misleading because chlorine gas will not react with SO42-. (D) is incorrect because changing from Ca(OCl)2 to chlorine gas will remove the common ion effect and cause more dissociation of CaSO4.

6. C

An increase in pH means that fewer H+ ions are present in solution. Lowering the concentration of H+ will shift the equilibrium in equation 2 to the right, and fewer HOCl molecules will be in solution. Because the HOCl molecules can oxidize harmful agents, the oxidizing power has been reduced as a result of the pH increase. An increase in pH will not break the HOCl compound (A), though over time it will break down. Although salts do act as buffers (B), it is still possible to change the pH of the pool. As for (D), an increase in pH will decrease [H+] and thus result in fewer H+ ions available to associate with OCl-.

7. D

To make a supersaturated solution you must first heat the solution, which allows additional salt to dissolve. (While not all salts have a higher solubility at higher temperatures, this statement holds true for the majority of salts; an increase in temperature will generally increase the solubility of a collection of salts.) This occurs because at higher temperatures, the Ksp generally increases. The solution can then be cooled and the salt will remain dissolved, creating a supersaturated solution.

8. B

Using the top equation we can write the following:

422

We can manipulate the equation to solve for [H+], where

423

Using the bottom equation we can write the following: Ksp = [Ca2+][CO32-]

We can rearrange this equation to solve for [CO32-], which can then be substituted into the top equation:

424

Substituting into the top equation gives:

425

The Ka equation includes the reactant in the denominator because the reactant is aqueous (in contrast to the Ksp equation which doesn’t include the reactant in the denominator). The Ksp reactant is a salt in its solid form; solids are not included in equilibria equations.

9. B

The color change will occur slightly before the pKa of phenol red is reached. This is because the basic form, which is prevalent above the pKa, has a higher absorptivity than the acidic form. This means that the basic form will absorb light more strongly than the acidic form. At the pKa the basic and acidic forms are equal (definition of pKa), but because of the higher absorptivity of the basic form, the color will begin to change when there still is more acidic than basic molecules of phenol red. (C) and (D) are above the pKa value, after the color change has occurred. (A) is under the pKa but is too low. At pH 4, acidic molecules heavily dominate the solution so the absorptivity difference between acidic and basic forms does not come into play.

PASSAGE II

10. D

Hydrogen is diatomic in its elemental state, i.e., H2 (g). The passage refers to the fact that each gaseous hydrogen atom has a single electron, for two total electrons in H2, but the proper electron configuration of a diatomic substance requires using the bonding-antibonding model from molecular orbital theory. None of the choices are that specific, so they are not correct. (A) is a distortion; substances in their elemental state are presumed to keep their electrons in the ground state. (B) is a distortion as well; hydrogen is diatomic in its elemental state, meaning two electrons.

11. A

(B) is opposite; light is absorbed when the electron moves away from the nucleus. While electrons make the energy transitions, the energy transitions result in light radiation, not the emission of the electron itself, so (C) and (D) are incorrect.

12. B

The Lyman series transitions occur in the UV range (? = 200-400 nm). The Balmer series corresponds to four visible wavelengths (? = 400-700 nm), though the Balmer constant itself is in the UV range. Both the infrared and X-ray ranges of the light spectrum [(C) and (D)] fall far outside the transitions which characterize these spectra. Infrared rays are low-energy and have higher wavelengths, while X-rays are high-energy and have very short wavelengths. It is not necessary to memorize the specific wavelengths in different kinds of radiation as long as you have a general sense of differences in magnitude.

13. A

If you ignore the digression about astronomy, the question is straightforward and does not require referring back to the passage. It simply asks what color of visible light corresponds to 656.3 nm. The visible light spectrum covers 400-700 nm. Using the ROYGBIV mnemonic and prior knowledge that infrared wavelengths are longer than visible wavelengths, you know that red is on the 700 nm end, and that violet is on the 400 nm end.

14. B

This image is an absorption spectrum, in contrast to the emission spectrum presented in the passage. Think of it as the inverse of the emission spectrum. We cannot tell from the passage what a Lyman or Bohr series looks like.

15. C

Deuterium is “heavy” hydrogen, with one neutron, one proton, and one electron. The Balmer series concerns only electron transitions, so there will be no change in the number of peaks. In other words, there will be the same number of peaks, but with more splitting; the nucleus is heavier, which will slow down the transitions. This effect is visible at high resolutions.

16. A

A full explanation is out of the scope of the MCAT Physical Sciences section, but you should know from general physics that Bohr’s model was incomplete because it didn’t incorporate the theory of relativity. We know from the question stem and equation that the Balmer series works only for wavelengths related to the quantized energy levels (B). Though it is not stated explicitly in the passage, Bohr’s quantitative model incorporates atomic number (C) (more detail is unnecessary for now; in short, it affects the atomic radius). Hydrogen atoms have only one electron (D), and we have no indication that other particles are involved.

17. C

According to Bohr’s model, the energy differences between quantized energy levels become progressively smaller the further away the electron moves from the nucleus. Textbooks will sometimes describe this concept as if the energy levels themselves are “narrowing.” (If energy levels are depicted as rings around a nucleus, they will appear to be closer together as you move farther from the nucleus.) Read carefully. (B) describes a drawing like this one, but the graphical representation of this concept does not really explain the difference in energy. We do not have enough information in the passage to determine whether the energy levels themselves are a greater or smaller distance apart. (A) is the opposite of (B), intended to stump readers who might not read through to (C) and (D). Because (C) refers directly to energy differences between energy levels, it is more accurate. (Remember E = hf = h × [speed of light/ wavelength].) (D) is its opposite.

QUESTIONS 18–22

18. D

Only (D) gives a response where both kinds of particles have a mass. Neither neutrons nor gamma rays have mass and so they are unchanged by the actions of a particle accelerator. The dependence on mass arises because a particle accelerator works by means of high-energy ideally elastic collisions. Also, if there is no mass, both kinetic energy and momentum are undefined.

19. B

A solution is available here which doesn’t require an equation. Both the acid and base are monoprotic, thus there are no multiple dissociations. Notice that the acid is twice as concentrated as the base. Thus, double the amount of base will be needed to neutralize the acid or 2 × 15 mL = 30 mL. If you forgot this factor of 2, you might have chosen (A), and if you squared the 2 out of uncertainty (only in physics do you square things when in doubt) you might have chosen (D). C is the total volume of the solution, not the volume just of the base.

20. A

The ideal gas theory assumes that the particles have large interatomic distances and a relative absence of intermolecular forces. This is more true at lower pressures than at higher ones, thus eliminating (C) and (D). (B) implies that high pressure squishes the molecules. According to the kinetic theory of gases, individual atoms are treated as incompressible point properties. The main effect of the pressure is to reduce the interatomic distance, not the intra-atomic distance (i.e., the atomic radius).

21. A

Molecules can only be nonpolar if they have total symmetry around the central atom. Bent molecules lack this symmetry (think of a molecule of water with two lone pairs of electrons and two hydrogen atoms extending from the central oxygen atom) and so a dipole is created which causes polarity in the molecule. A diatomic covalent (B) molecule (such as H2 or O2) is perfectly symmetrical. (C) and (D) are incorrect because, although dipoles can exist in these configurations, symmetrical arrangements are also possible.

22. C

Isotypes of the same element have the same number of protons. Alpha decay results in the loss of two protons and two neutrons, while beta decay (?-decay) causes the gain of one proton. Thus, two beta decays must occur for each alpha decay to ensure that the number of protons in the daughter nucleus is equal to the number of protons in the parent nucleus. The answer is (C), 1:2.

PASSAGE III

23. A

In order to balance the equation, we must first combine the half-reactions. This requires writing all of the products and all of the reactants for both reactions in one equation:

LiCoO2 + xLi+ + xe- + 6C 426Li1-xCoO2 + xLi+ + xe- + LixC6

If a certain compound/particle is on both the left and the right sides of the equation, then it is appearing as both a reactant and a product. Since there is no net change in that specific species, we can omit it from the net reaction; for this reason, we can eliminate the terms “xLi+” and “xe-” from both sides of this equation. This leaves us with (A) as the correct choice.

24. B

You should know that the overall potential of a galvanic cell (E°cell) is equal to the sum of the potentials at the cathode and the anode. The overall potential of a discharging cell is always positive, but not necessarily greater than 1; for this reason, (B) is the only viable option. You can also arrive at this conclusion by realizing that the reaction in a battery is always spontaneous, since the battery must supply energy. Spontaneous reactions always have a negative imageG, which you can plug into the free energy equation (imageG = -nFE°cell). To get a negative imageG from the free energy equation, you must have a positive E°cell.

25. D

The first thing to note here is that lithium acts as an electrolyte in this reaction, meaning that it is not oxidized or reduced; that rules out (A) and (B) immediately. By definition, reduction happens at the cathode and oxidation happens at the anode. However, because this is a reversible reaction (which you should know based on the fact that the reaction is at equilibrium), we cannot use these definitions to distinguish between the species that is oxidized and the species that is reduced. The easiest way to answer this question is by noting that the carbon in the anode is neutral on the left side of the forward reaction and is negatively charged on the right side. This means that carbon is reduced, allowing us to select (D) as the correct answer. You can verify this by checking the cathode side: Because the cathode is made of CoO2-, in which Co has a +3 charge (since oxygen almost always carries a -2 charge, two oxygen atoms add up to a -4 charge; to create an overall -1 charge, Co must be +3), Co3+ must be oxidized in the conversion of LiCoO2 to Li1-xCoO2. The product of this reaction, because it contains less +1 charge from lithium, must contain a more positive charge from cobalt (further confirming that the cobalt is reduced).

26. C

Because of the complexity of this question, the first step should be to eliminate as many choices as possible. There is no indication in the passage that E°cell is positive for the forward reaction (A). The assumption about reaction kinetics in (B) is correct, but it’s likely that the effect of a change in xwill be different for each reaction. One reaction will shift to the left and the other will shift to the right, but the shifts will not be exactly identical to one another; therefore, Keq will change and (B) is incorrect. A galvanic cell gradually discharges while transporting electrons from the anode, where oxidation occurs, to the cathode, where reduction occurs. In the forward reaction, oxidation occurs at the anode and reduction occurs at the cathode. This means that the forward reaction does not represent a discharging cell, so the reverse reaction must be favored when the battery is in use. Because the reverse reaction takes precedence over the forward reaction, Keq is low when the cell is discharging.

27. B

LiCoO2 is usually present in the battery since it is on the left side of the cathode reaction. The presence of the Li(CoO2)2 complex is more important for this question. It is clear that Li(CoO2)2 is a form of the Li(CoO2)1/(1-x) complex, so we can calculate x by writing the equation 1/(1-x) = 2. Simple algebra yields the fact that ½ of the battery’s initial energy is remaining. Because the initial energy was equal to 100 J, the current energy equals 50 J.

28. D

According to the passage, the system contains only lithium as a salt in a solvent. Because it is not part of the cathode or the anode, it always retains its +1 charge. All three items are correct.

29. A

When atoms or molecules are in close proximity for an extended period of time, there is a substantial probability that they will interact with one another. The reaction suggested in (A) would require reduction of Co3+ to Co2+, which is not unlikely to occur in small quantities. Though the passage does not directly indicate that this was the mechanism of production of lithium oxide and cobalt(II) oxide, it is more likely than any of the other answer choices. It would be incorrect to dismiss the possibility of decomposition, but the reaction suggested in (B) (LiCoO2image Li2O + CoO) is stoichiometrically impossible. (C) is incorrect because interaction with environmental oxygen would affect cell 2 just as much as it would affect cell 1. (D) is incorrect because we have no evidence suggesting that any water is present in the system; the passage clarifies that organic solvents are used.

30. C

The scientist believes that battery deterioration is caused primarily by the formation of cobalt(II) oxide and lithium oxide in equal quantities. Therefore, increased amounts of cobalt(II) oxide suggest a decrease in battery capacity. (A) and (B) are incorrect because x is a ratio that is unaffected by the energy storage capacity of a cell. (D) is incorrect because, although a decrease in concentration of LixC6 would suggest a decrease in cell capacity, most of the deterioration is caused by the conversion of LiCoO2 to Li2O and CoO2.

PASSAGE IV

31. A

Ammonia’s heat of vaporization is given in the data and described as low compared with that of water. Thus, its evaporation rate must also be high compared to that of water so life could not have evolved in a liquid ammonia environment as life on earth evolved in a liquid water environment. The phases in (B), (C), and (D) would matter less for life’s evolution from a liquid environment; they are also incorrect applications of the data from the passage.

32. A

Solid water would have been most dense if it were like other substances, and thus sunk to the bottom of any liquid system. Liquid systems would have been frozen from the bottom up, and life could not have evolved in the liquid phase of a watery environment. (B), (C), and (D) do not address this “what-if ” scenario.

33. B

It is the kinetic energy of the molecules that moves them further apart to allow a phase change from solid to gas, as described in the question. The same process would dictate a phase change from solid to liquid, or liquid to gas.

34. A

The electronegativities of the atoms comprising a molecule determine the polarity of that molecule. None of the other choices will make as significant a contribution. (C) is the next most logical answer, but is less correct because forces emanate from the intrinsic nature of electronegativity within the water molecule, not between.

35. B

The crystalline lattice formed for a water-solid uniquely collapses under pressure to become a liquid. (A) is a true statement, though it doesn’t directly address the question.

36. D

Both (A) and (B) are true. Water has a higher heat of vaporization than ammonia and therefore evaporates at a higher temperature. Temperature is a measurement of average kinetic energy, so high temperature means higher average kinetic energy; the molecules are also moving faster.

37. A

Ions dissolved in the lattice break existing intermolecular attractive forces. This process interferes with the formation of a crystal lattice in ice, which explains why ice melts when salt is added.

PASSAGE V

38. C

Asking for the strongest oxidizing agent is equal to asking for the element with the highest electronegativity. Electronegativity increases as we move from the left to the right and from the bottom to the top of the periodic table. Therefore, you are looking for an element that is above and to the right of phosphorus. (C), oxygen, fits this description.

39. B

You’re being asked which elements have a larger atomic radius than phosphorus. Atomic radius increases from right to left and from top to bottom of the periodic table. Therefore, you are looking for elements that are below and to the left of phosphorus. Both K and Pb fit this description, so (B) is the correct answer.

40. D

To answer this question, you must know that phosphorus is a nonmetal. The other answer choices describe properties of nonmetals. (D) describes a property of metals, so it is the correct answer.

41. B

(B) is the correct full electron configuration. (C) places the 3s-electrons in the 3p-orbital. (D) is incorrect because phosphorus does not have a 3d-orbital.

42. D

(A) and (B) are incorrect because alkaline earth elements are generally more dense than alkali metals. (C) is incorrect because alkali metals contain the same number of orbitals as the alkaline earth element in the corresponding row. (D) is a true statement.

43. A

Halogens can naturally exist in the gaseous, liquid, and solid states (iodine).

44. C

(A), (B), (D) are all true properties of transition elements, but they do not greatly contribute to the malleability shown by these elements. Their malleability can be attributed mostly to the loosely held d-electrons.

45. C

(A) is incorrect because phosphorus is not a metalloid. Metalloids do often behave as semiconductors (B), but it is not relevant here. (C) is correct because both arsenic and antimony are in the same group as phosphorus and they have similar properties.

PASSAGE VI

46. C

The reaction presented in the passage, beginning with acetic acid and forming methane and carbon dioxide, is a decomposition reaction. It begins with one reactant and ends with two products, which is typical of a decomposition reaction. A combination reaction (A) would have been the opposite: more reactants than products. A single displacement reaction (B) typically is an oxidation/reduction reaction, which is not demonstrated in this question. A combustion reaction (D) is catalyzed by oxygen and results in carbon dioxide and water as products.

47. A

First, calculate the molecular weight of acetic acid, CH3COOH, which is 60.05 g/mol. Next, recognize that there is a 1:1 molar ratio between this reactant and the product, methane. By beginning with 120.0 grams of acetic acid and dividing by the molecular weight, this yields 1.998 moles of acetic acid and should produce the same number of moles of methane. The molecular weight of methane is 16.04 g/mol. By multiplying this molecular weight by the number of expected moles yield, 1.998, the theoretical yield expected if all of the acetic acid were decomposed would be 32.05 grams of methane product.

48. B

A key issue in harnessing a gaseous product is its phase and transport, so (B) is correct. If compressible, then methane would take up significantly less volume and be easier to transport, making it more likely to be efficient as a common energy source. The passage states that this reaction typically takes place at a temperature well above room temperature. Even if (A) were true, it doesn’t support an argument toward using methane gas as a major energy source. While (C) is true, the production of gas challenges, more than supports, the passage’s argument of making clean energy. Finally, whether or not the reaction is exothermic, the energy source is the product, methane gas, so (D) presents an irrelevant piece of information.

49. A

To build the equation necessary to answer this question, begin with 88.02 grams of CO2 in the numerator. Next, convert this to moles of CO2 by dividing by grams per mole. This eliminates (D), since 88.02 grams is in the denominator and the molecular weight of CO2 is in the numerator. (C) is also out with 88.02 grams in the denominator. Next, recognize that there is a 1:1 ratio of CO2 to CH3COOH, making it unnecessary to convert the number of moles. This eliminates (B), which incorrectly uses a 1:2 ratio. The only choice remaining is (A), which correctly ends by multiplying by the molecular weight of acetic acid; this would yield the number of grams of acetic acid necessary for the proposed reaction.

50. D

The net ionic equation shows all aqueous ions that productively participate in the reaction, eliminating all spectator ions. The only true spectator ion in this equation is sodium, so it should not be present in a net ionic equation, which proves that (A) is incorrect. Next, CH3CH2CH2Cl has a covalent bond between the carbon and chloride atoms, meaning that it is unlikely to ionize in solution; Thus, chloride should not be represented as an ion on the reactant side of the equation, as it is in (A) and (B). Finally, (D) correctly keeps the reactant as one molecule and has the chloride ion written as a product.

51. C

The first step in this question requires you to consider the correctly balanced equation for converting CO2 and H2 to CH4 and H2O (the reactants and products specified in the passage and in the question stem):

CO2 + 4 H2image CH4 + 2 H2O

The correctly balanced equation uses the molar ratio of 1 CO2:4 H2:1 CH4:2 H2O. Next, recall that according to the ideal gas law, one mole of an ideal gas has a volume of 22.4 L at standard temperature and pressure (STP). From this, one can calculate that there are 0.134 moles of hydrogen gas and 0.0893 moles of carbon dioxide available as reactants. Then, calculate the molar equivalents of each, since 4 moles of hydrogen gas are required per mole of CO2. Thus, there are 0.0335 molar equivalents available of hydrogen gas per 0.0893 moles CO2, meaning that the hydrogen gas is the limiting reagent. Finally, because there is one mole of methane produced per four moles of hydrogen gas used, there will be 0.0335 moles of methane produced. (A) does not account for the 4:1 molar equivalency of H2:CH4. (B) uses carbon dioxide as the limiting reagent. (D) reverses the molar ratio of H2:CH4, thus multiplying 0.134 moles by 4 instead of dividing.

52. C

Begin by calculating the theoretical yield of methane, which is the amount (in either grams or moles) of methane expected to be produced if all the hydrogen were fully consumed by the following formula:

CO2 + 4 H2image CH4 + 2 H2O

Recall that at STP, one mole of gas has a volume of 22.4 liters, which allows you to calculate that there are 2.23 moles of hydrogen gas given to react. Next, use the balanced equation for converting CO2 and H2 to CH4 and H2O (the reactants and products specified in the passage and in the question stem). The correctly balanced equation uses the molar ratio of 1 CO2:4 H2:1 CH4:2 H2O. Thus, with 2.23 moles of hydrogen gas, you would expect to form one mole of methane per four moles of hydrogen gas, or 0.558 moles of methane product. Finally, multiply by methane’s molecular weight, 16.04, which results in 8.95 grams of methane; this is the theoretical yield. However, the question stem states that only 8.02 grams of methane were produced. Thus, the percent yield is calculated by dividing the actual yield (8.02 grams) by the theoretical yield (8.95 grams) and multiplying by 100 percent, which results in a percent yield of 89.6 percent.

PRACTICE SECTION 3

ANSWER KEY

1. A

2. B

3. D

4. D

5. D

6. A

7. B

8. C

9. B

10. B

11. C

12. B

13. C

14. A

15. A

16. C

17. B

18. C

19. A

20. A

21. D

22. D

23. A

24. A

25. B

26. A

27. A

28. D

29. C

30. D

31. C

32. D

33. A

34. D

35. D

36. D

37. A

38. C

39. B

40. B

41. D

42. C

43. C

44. B

45. A

46. D

47. D

48. A

49. C

50. D

51. C

52. B

PASSAGE I

1. A

For a molecule to pass from the systemic circulation to the CNS, it must pass between the tightly sealed endothelial cells or through the endothelial cells. Unless a medication has a transporter to cross the cell membrane twice, it is more likely to diffuse between endothelial cells, despite their tightly sealed spaces. Because cell membranes are composed primarily of nonpolar lipid molecules, it will be easiest for a lipophilic, or nonpolar compound to diffuse into the CSF. Although (B) does describe a method for delivering drugs to the CNS, a nonsystemic route would be inefficient for a process as routine as general anesthetic administration; (A) is a better option. (C) is incorrect, because nonpolar or lipophilic molecules will penetrate more effectively than will a polar molecule. (D) is irrelevant; furthermore, a slow-acting general anesthetic would be impractical when physicians are aiming to minimize time under anesthesia and maximize patient comfort.

2. B

The key to understanding the blood-brain barrier is that the endothelial cells are tightly sealed, prohibiting free passage of molecules, unlike the leaky capillary systems of the peripheral circulation. The molecules that are most likely to still move between these cells and enter the CSF are those that are uncharged and not repelled by the hydrophilic cell membranes. Even though charged particles might associate closely with one another (A), this doesn’t affect their passage through the blood-brain barrier. (D) is incorrect because the passage relates no information about the differential solubilities of charged molecules in the bloodstream versus CSF.

3. D

The blood-brain barrier, as described in the passage, provides a tight seal between the systemic circulation and the more sensitive CNS. As a result, this barrier can serve to protect the CNS from potentially damaging substances that can more readily enter the systemic circulation, and then be filtered before entering the CNS. (A) is incorrect because the blood-brain barrier’s adaptive seal is not designed for maximizing nutrient transport; in fact, it limits transport/ transfer between two systems. (B) is incorrect because this barrier necessarily means that many molecules and particles cannot pass between the CNS and systemic compartments, thus making these two micro-environments different. (C) is incorrect because the blood-brain barrier’s main role is not to limit movement of particles or agents from CSF to the systemic circulation, but rather to limit flow in the other direction because the systemic circulation is more readily contaminated.

4. D

According to the passage, molecules most likely to cross the blood-brain barrier are hydrophobic or nonpolar. Thus, ion-dipole and dipole-dipole interactions [(A) and (B)] are unlikely to be the most important forces governing intermolecular interactions among these molecules (because they require charge and/or polarity). Although hydrogen bonding (C) might have a role in these molecules, hydrogen bonds, because they occur between atoms in otherwise polarized bonds (a hydrogen with a partial negative charge and a lone pair or otherwise electronegative atom with a negative charge), would be unlikely to facilitate transport in a hydrophobic environment. Dispersion forces (D) refer to the unequal sharing of electrons that occurs among nonpolar molecules as the result of rapid polarization and counterpolarization; these are likely to be the prevailing intermolecular forces affecting molecules that move easily through the blood-brain barrier.

5. D

All three items are false. A polar molecule can still have a formal charge of zero, because the molecule’s formal charge is the sum of the formal charges of the individual atoms. Each atom could individually still have a positive or negative formal charge, calculated by the formula:

FC = valence electrons - ½ bonding electrons - nonbonding electrons

Thus, a molecule with a formal charge of zero could have multiple polarized bonds and/or multiple atoms with positive/negative formal charges, making it unlikely to permeate through the nonpolar blood-brain barrier. Similarly, having a negative formal charge on the molecule overall would be unfavorable to move through a nonpolar barrier. Finally, two molecules, both with formal charges of zero, could have very different characteristics, making them more or less likely to permeate the blood-brain barrier. For example, one compound could be comprised entirely of nonpolar bonds, with all atoms of formal charge of zero, both of which would make it favorable to pass through the blood-brain barrier. Size also plays a key role, as large molecules do not readily pass through the barrier. A separate molecule with a formal charge of zero could, as described above, contain positive or negative formal charges on different atoms and/or have polar bonds, making it pass through the barrier less readily.

6. A

In cisplatin, a molecule with square planar geometry, two chloride atoms and two ammonia molecules each are bonded directly to the central platinum, without any remaining lone pairs on the central platinum. The NH3 groups bond to the central platinum by donating a lone pair of electrons into an unfilled orbital of the platinum atom. As such, NH3 is acting as a Lewis base, and platinum is acting as a Lewis acid, and they form a coordinate covalent bond. A polar covalent bond (B) is not formed from this type of donation of a lone pair; an example of a polar covalent bond would be the N–H bonds in the NH3 group, with partial negative charge on the nitrogen and partial positive charge on the hydrogen. In (C) and (D), the Pt–N bond is formed by a Lewis acid/ base relationship, which is not the case for a nonpolar covalent bond or ionic bond.

7. B

Although the geometry of this carbon atom is likely to be changed somewhat by the ring strain on the adjacent ring structures, it is still most likely to approximate a tetrahedral geometry. This carbon is bonded to four groups: the two phenyl groups, a nitrogen, and the additional carbon atom in the adjacent carboxyl group. The tetrahedral geometry (B) maximizes the space among these four groups. Octahedral geometry (D) typically refers to a central atom surrounded by six groups, not four.

8. C

Resonance structures serve to spread out formal charge. The most important resonance structures minimize or eliminate the formal charge on individual atoms. As a result, polarity in any single bond might be minimized. (A) is essentially the opposite of this argument. (B) is incorrect; it is possible for molecules with or without resonance to be polar, and this is too broad of a generalization to be true. (D) is incorrect because polar bonds will intrinsically place a partial negative charge on the more electronegative atom.

Important resonance structures would likely further accentuate this inclination, placing extra electrons on more electronegative atoms. Counteracting this effect and essentially “removing” electrons from highly electronegative atoms to counterbalance the natural polarity of the bond would form an extremely high-energy, unfavorable, and thus unimportant, resonance structure.

PASSAGE II

9. B

The order of this reaction can be determined by the equation r = k[A]x [B]y. You must divide r3 by r1 to get: 427428. This simplifies to 4 = 4x, so x = 1. This means that H+ is a first-order reactant.

10. B

The equation for the rate of a reaction is:

Rate = k [reactant1]order1[reactant2]order2

Using the equation for determining order of a reactant detailed in the explanation for the answer to question 9, we find that O2 is second order. Substituting the information into the equation for the rate of a reaction leads to:

Rate = (0.5)[2.0 M]1[2.0 M]2

Simplifying this equation allows us to determine that the rate is 4 M/sec.

11. C

(C) is correct because it describes a catalyst, which is exactly the role Fe2+ - SOD plays. (A) is incorrect because it describes a reactant, and Fe2+ - SOD is not a reactant. (B) is incorrect because it describes a product, and Fe2+ - SOD is not a product. (D) describes a transition state, and is incorrect because transition states are temporary states of highest energy of the conversion of reactant to product.

12. B

(B) is not true (making it the correct answer) because all the colliding particles must have enough kinetic energy to exceed activation energy if a collision is to be effective. (A) is true because particles must collide to react, and thus the rate of reaction is both dependent on and proportional to the number of particles colliding. (C) is the accurate definition of a transition state. (D) is true because “activated complex” is another name for “transition state,” which by definition has greater energy than both reactants and products.

13. C

Section C shows the energy that must be put into the reaction to drive it in a forward direction. Section A in the diagram represents the change in enthalpy; it shows the difference between the starting and the final energy values. Section B does not represent any specific energy value. Section D represents the reverse activation energy. It is much greater than the forward activation energy because the reactant is starting at a much lower level of energy, yet must still reach the same amount of total energy to proceed with the reaction.

14. A

Section A is correct because it shows the difference between the starting and the final energy values.

15. A

The equation to determine the equilibrium constant for a reaction aA + bB = cC + dD is Kc = ([C]c[D]d)/([A]a[B]b). For this reaction, Kc = ([2]1[1]1/([3]2[1]2 = 2/9 = 0.22. When the corresponding values are plugged into the equation, Kc = 2/9, or 0.22.

16. C

According to Le Châtelier’s principle, the addition of a reactant (C) will cause an equilibrium to shift to the right. An increase in volume (A) would have no effect in this particular reaction. An addition of product (B) would cause a shift to the left. A temperature decrease (D) would most likely cause a shift to the left due to fewer numbers of collisions between particles, assuming kinetic energy is proportional to temperature.

QUESTIONS 17–21

17. B

The question says that latent heat flux is caused by evaporation. Therefore, simply identify which value of imageH is related to this phase change. Vaporization is another way of saying evaporation, so (B) is correct. Fusion (A) refers to the change from a solid to a liquid, and sublimation (C) refers to the change from a solid to a gas. Ionization (D) is unrelated to a phase change.

18. C

The correct answer is (C) due to the Henderson-Hasselbach equation, which states that for a weak acid solution, the pH equals the pKa plus the log of the ratio of the concentration of conjugate base to the concentration of acid.

pH = pKa + log([conjugate base]/[acid])

For an acid HX, the conjugate base is the X- ion that is formed when the acid dissociates. Now, if the pKa of an acid is 5, then for its pH to be 6, the log of that concentration ratio must be 1, or in other words, the ratio must be 10. This means that the concentration of conjugate base, and therefore of dissociated acid, must be 10 times the concentration of acid.

19. A

The question requires you to know the solubility rules. Because an electrolyte must dissociate into its component ions in water, look for the substance that will not dissociate. The answer is (A), silver chloride (AgCl), which is insoluble in water. All of the other compounds will readily dissociate into their constituent ions.

20. A

(A) is correct. Because aluminum oxide is alkaline, immersion in an acidic solution would readily strip away the protective coating and allow the acid to oxidize the metal below.

21. D

How can you relate the average velocities of gases at the same temperature? Two gases at the same temperature will have the same average molecular kinetic energy. Because two gases at the same temperature will have the same average molecular kinetic energy, (1/2)m A v A2 = (1/2)m B v B2, which gives mB /m A = (vA/vB)2. Since vA /vB = 2, we have mB /m A = 4. Only (D) lists two gases with a mass ratio around 4:1 (krypton 83.3g/mol and neon 20.2 g/mol).

PASSAGE III

22. D

The solid block becomes smaller in the experiment and so must move from the solid phase to the gaseous phase, sublimation. The other answer choices require a liquid phase, which is not valid because the experimental results demonstrate that no liquid was detected when the solid shrunk in size.

23. A

The solid/liquid equilibrium line has a positive slope and so the phase change from liquid to solid at a constant temperature is the only correct possible phase change. (B) is wrong because the gas phase is not possible as pressure increases; the liquid would change to gas, however, if the temperature were increased while maintaining constant pressure. (C) and (D) are illogical because kinetic energy is equivalent to temperature. Because temperature is constant according to the question stem, there is no kinetic energy change.

24. A

You must deduce from both the experimental results and the chemical equation that the volume does not change. The volume does not change because it is described as a rigid container, and no change in volume is described in the experimental results. As a result, the dry ice must absorb energy from its surroundings (the air in the container) in order to change phase. Because the total energy must remain constant, the air initially present in the container must lose an equivalent amount of kinetic energy.

25. B

The chemical equation shows that the dry ice absorbs heat from its environment to change into a gas, requiring it to overcome the intermolecular forces that organize it in the solid form. (C) and (D) refer to hydrogen bonds, which are not present in carbon dioxide.

26. A

The chemical reaction shows that the conversion of solid carbon dioxide to the gaseous state requires an absorption of heat from the ambient air. This use of heat defines an endothermic reaction, which must therefore characterize the reaction occurring in the container. (C) and (D) refer to an exothermic reaction, which would release heat. (B) is incorrect since an endothermic reaction absorbs/uses heat, which increases the energy of some of the molecules in the reaction; thus, the potential energy should not decrease.

27. A

The air gains dry ice molecules and these exert a pressure on the container, according to the equation PV = nRT, where n represents the number of moles of gas. (B) is the opposite; its volume would increase, according to the same equation, if possible. However this experiment takes place in a rigid container, preventing any change in volume. (C) is incorrect because the question clearly states that the solid changes phase. In this endothermic reaction, air temperature decreases as the solid carbon dioxide absorbs energy in order to undergo sublimation.

28. D

None of the choices shows a substance moving from the solid directly to the gas phase in the process of sublimation. (A) and (B) show fusion and evaporation, respectively, but not sublimation. (C) indicates the heating of a gas; in the experiment, the carbon dioxide solid increases its kinetic energy; the gas is not being heated.

29. C

(C) indicates change from a solid to a gas (sublimation) as described in the passage for dry ice.

QUESTIONS 30-36

30. D

Because imageG = imageH - image can be used to relate free energy to enthalpy and entropy, and T is always positive, a negative imageH (enthalpy) and a positive imageS (entropy) will always give a negative imageG value; that is, the reaction will occur spontaneously. (D) is correct.

31. C

Moving from left to right across the periodic table, atomic radii will decrease. Because calcium is on the left side of the table and gallium is on the right, in the same period, gallium should have a smaller atomic radius. As stated in (C), this is due to the greater number of protons in the nucleus holding the electrons more tightly.

32. D

The major forces that cause gases to deviate from ideal behavior are intermolecular attractions and the volume of the gas molecules. These factors are minimized when gas molecules are far apart and moving quickly, which occurs at low pressures and high temperatures.

33. A

The question says that this gas is ideal, so use PV = nRT. The ideal gas law shows that P is directly proportional to T. So with volume held constant, if pressure is reduced by a factor of 2, temperature will also be reduced by a factor of 2. (A) is correct.

34. D

Though this question begins by telling you about greenhouse gases, it ultimately asks you to identify similarities between CO2 and H2O which N2 and O2 lack. CO2 is not a polar molecule; its linear geometry allows the opposing dipole moments to cancel out, so (A) is incorrect. Because its dipole moments cancel, (B) is incorrect. Furthermore, it lacks hydrogen atoms and is therefore incapable of hydrogen bonding, so (C) is incorrect. However, both CO2 and H2O have polar bonds, while N2 and O2 both have diatomic, nonpolar covalent bonds between two atoms of the same element.

35. D

Fluorine (A) is not a transition metal, so its ionized counterparts will have a valence octet of electrons implying no unpaired electrons. (B), (C), (D) are all transition metals; however (B) and (C), commonly oxidize to the 2 and 3 positive states, respectively, which fill their d subshells. Iron (D) commonly oxidizes to the 2 positive state, which gives one extra s-subshell electron. This unpaired electron helps to give iron its magnetic properties and explains why iron in the body is further oxidized to the 3 positive state (to avoid inductive currents that could damage protein structure).

36. D

Following the Aufbau principle, the order of orbital filling should produce [Ar]4s23d4 as the electronic configuration for chromium. However, there is an overall increase in stability for the molecule if the d-subshell can be half-filled. Because of this, an electron will be moved from the 4s-subshell into the 3d-subshell giving (D) as the correct answer.

PASSAGE IV

37. A

Henry’s law states that the amount of gas dissolved in a liquid is directly proportional to the partial pressure of the gas in equilibrium with the liquid. Therefore, Henry’s law can be used to calculate the concentration of oxygen in water using the partial pressure of oxygen in air. Boyle’s law (B) deals with the relationship between the pressure and volume of gases, but does not address concentration of gases in water. Raoult’s law (C) pertains to the vapor pressure of a mixture of liquids, not a gas dissolved in a liquid. Le Châtelier’s principle (D) addresses changes to an equilibrium state and cannot stand alone to explain the equilibrium between a gas in air and in solution.

38. C

We can use the solubility constant for nitrogen provided in the passage, 8.42 × 10-7 M/torr, to solve this question. Because the units in the constant are in torr, we first convert 0.634 atm to torr by multiplying by 760 torr/1 atm. The partial pressure of nitrogen equals 481.8 torr. Multiplying the pressure by the constant 8.42 × 10-7 M/torr gives us 4.06 × 10-4 M nitrogen. The units for the answer are in g/L, so we multiply by the molar mass of nitrogen, 28 g/mole, to get our answer of 1.14 × 10-2 g/L.

39. B

The solubility constants provided in the passage can be used to determine that helium is the least soluble gas. A soluble gas is not desired because we want to minimize gas bubbles in the body. Moreover, helium is an inert gas, meaning it does not readily react with other gases. Whether helium is diatomic (A) has no bearing on its use in scuba tanks. Many gases are present in trace amounts in the water (D), so this fact alone could not account for the use of helium in scuba tanks.

40. B

This is a classic ideal gas law problem using PV = nRT. You are given the volume, then must convert the temperature to degrees Kelvin to obtain a useable temperature and must convert the mass of oxygen to moles to find nV equals 2.80 L. T equals 273.15 + 13.0 = 286.15 K. The mass, 0.320 kg, equals 320 grams. We divide the mass by the molar mass of oxygen, 32 g/mole, to obtain the moles of oxygen, 10 moles. R equals 0.0821 L atm/(mole K). Plugging in these numbers to the equation, PV = nRT. Solving for P gives a pressure of 83.9 atm.

41. D

Immediate isolation in a hyperbaric chamber is the most effective and common treatment for those suffering from severe decompression sickness. The chamber recreates a high-pressure environment to allow gas bubbles to dissolve back into body fluids and tissues. The chamber can be brought back to normal pressure slowly in order to allow the body to adjust to the decreased pressure. Helium gas administration (A) or gas and air mixture (B) would not rid the body of excess gas bubbles. In fact, it might increase the gases bubbles and make symptoms worse. A hypobaric chamber (C) would certainly make symptoms worse because it decreases the pressure below 1 atm.

42. C

The solubility of gas in liquids decreases with an increase in liquid temperature. The warming of oceans has resulted in less dissolved oxygen and many oxygen-depleted “dead zones.” It is true that carbon dioxide has increased ocean acidity (A), but acidity alone cannot account for decreased oxygen levels. A predator shark may explain why certain fish are dying off (B), but it would not explain the decrease in oxygen. If rainfall did increase water levels in the ocean (D), the oxygen levels would equilibrate (as per Henry’s law) between the ocean and atmosphere to allow more dissolved oxygen in the oceans.

43. C

This is a PV = nRT problem. Find the moles of oxygen by converting 0.38 kg to grams and dividing by the molar mass of oxygen, 32.g/mole. The number of moles equals 11.875. STP indicates a temperature of 273.15 K and a pressure of 1 atm. R equals 0.0821 (L atm/mole K). Plugging these numbers into PV = nRT and solving for V gives a volume of 266 L.

44. B

Multiplying the amount of nitrogen gas in the air by the solubility constant of nitrogen will give the amount of nitrogen that is dissolved in the diver’s blood. The solubility constant for nitrogen can be obtained by dividing the

solubility of nitrogen, 6.2 × 10-4 M, by 1 atm to get 6.2 × 10-4 M/atm. The amount of nitrogen in the air can be obtained by finding the partial pressure of nitrogen. The total pressure, 3 atm, is multiplied by the percentage of nitrogen in the air, 78 percent, to get a partial pressure of nitrogen equal to 2.3 atm. Finally, we multiply 2.3 atm of nitrogen by the solubility constant to obtain a value of 1.4 × 10-3 M for the concentration of nitrogen in the diver’s blood.

45. A

The root-mean-square velocity (vrms) can be calculated by taking the square root of (3RT/Mm). This equation tells us that the vrms increases when molar mass decreases. Tank 2 contains a mixture of helium and oxygen; the helium will lower the average molar mass of the gas molecules because it has a lower molar mass compared to oxygen. The vrms of tank 2 will therefore be higher. Tank 1, which contains only oxygen, will have a higher molar mass and a lower vrms value.

PASSAGE V

46. D

F- is related to less acid production and reduces the risk of dental caries (D). The fluoride ion is related to acid production and reduces the risk of dental caries, according to the passage. Fluoride has not been proven to directly cause or even be related to bacterial death (B, C), nor has it directly been proven to stop bacteria from forming dental caries (A). However, fluoride is related to/correlated with acidity reduction and dental caries reduction. The other choices imply relationships and causations not inferred from the passage.

47. D

The high electronegativity of fluorine means that it is inclined to hold onto or pull electrons. (C) is the opposite of this atomic property because fluorine would not easily give off electrons. While the passage discusses fissures as a cause of caries [(A) and (B)], it does not infer that fluoride is involved directly with its filling or repair.

48. A

You must deduce from the definition of electronegativity that in order for the nucleus to pull on the orbital electrons, it should be closer to the electrons; therefore, a smaller radius is desirable. (B), (C), and (D) all contribute to a decrease in electronegativity because the distractors either favor a larger size of the atom with more electron shells or a smaller positive core of the protons that are responsible for the electronegative attraction in the first place.

49. C

The fluoride ion has the atomic structure of the element fluorine, which would be 1s22s22p5, with an additional electron to make it an anion with a charge of negative one. Using [He] at the beginning of the notation accurately reflects the fact that F- has the same structure as helium, but with the additional electrons as noted. (A) is incorrect because this is the notation for the element fluorine, not the ion. (B) is the structure for oxygen. (D) is incorrect because when using the notation [Ne], one implies that the atom has the structure of that noble gas, with additional shells. [Ne] comes after fluorine in the periodic table, and it would not be correct to add a level 2 shell after completing that shell in [Ne].

50. D

(D) is correct because Heisenberg’s uncertainty principle states that the momentum (m and v) cannot be determined exactly and quantitatively if the location of an electron in an atom is known and conversely, the location cannot be known if the momentum is known. One could get a qualitative measurement of momentum, since the measurement is still possible, but that measurement becomes less accurate as the accuracy of the measurement of position increases. (A), (B) and (C) all violate this principle, as the location is stated to be known in the stem of the question.

51. C

The effective nuclear charge is calculated by the following equation: Zeff = Z - S, where Z is the atomic number (the number of protons in the nucleus), and S is the average number of electrons between the nucleus and the valence electrons (which is the equivalent of saying S is the number of inner or non-valence electrons). Because fluorine’s atomic number is 9, there are also 9 electrons in fluorine (not an ion). Of these 9 electrons, 7 are in the outer shell (valence) and 2 electrons are inner electrons, therefore these electrons would have a Zeff of the following: Zeff = 9 - 2 = 7. Now for the fluoride ion, there are now 10 total electrons. However, there are still only 2 inner electrons, so these 8 outer electrons would have a Zeff of the following: Zeff = 9 - 2 = 7. Without changing the number of inner electrons, the effective nuclear charge will not change. This would affect the radius, however, as fluoride now has 8 valence electrons distributing the same Zeff of 7, whereas fluorine has only 7 electrons sharing the Zeff of 7. This is why anions are larger than their respective unionized species.

52. B

(B) is correct because the highest electron shells, or orbitals, are most loosely held by the nucleons and thus are most available for bonding. 3s is higher than any other level (A), but it is unoccupied in the fluoride ion. (C) and (D) are lower energy levels/orbitals than 2p.