Kaplan MCAT General Chemistry Review

(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 H₂ 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