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 n. V 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].