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 H2
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.