Kaplan MCAT General Chemistry Review полностью

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 (v_rms) can be calculated by taking the square root of (3RT/M_m). This equation tells us that the v_rms 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 v_rms of tank 2 will therefore be higher. Tank 1, which contains only oxygen, will have a higher molar mass and a lower v_rms 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