if gas a has four times the molar mass of gas b you would expect it to diffuse through a plug

HESI A2

Chemistry HESI A2 Practice Test

1. If gas A has four times the molar mass of gas B, you would expect it to diffuse through a plug ___________.

A. at half the rate of gas B
B. at twice the rate of gas B
C. at a quarter the rate of gas B
D. at four times the rate of gas B

Correct answer: A

Rationale: When comparing the diffusion rates of two gases, according to Graham's law of diffusion, the rate of diffusion is inversely proportional to the square root of the molar mass. If gas A has four times the molar mass of gas B, the square root of the molar masses ratio (4:1) is 2. This means that gas A would diffuse through a plug at half the rate of gas B. Therefore, the correct answer is A, at half the rate of gas B. Choices B, C, and D are incorrect because they do not reflect the correct relationship between the molar masses and the rates of diffusion according to Graham's law.

2. Why does the diffusion rate increase as a substance is heated?

A. The kinetic energy of particles increases.
B. The space between particles increases.
C. The density of particles decreases.
D. The size of particles increases.

Correct answer: A

Rationale: The correct answer is A. When a substance is heated, the kinetic energy of particles increases, causing them to move faster. This increased movement allows the particles to spread out more rapidly, leading to a higher diffusion rate. Choice B is incorrect because heating does not directly affect the space between particles. Choice C is incorrect because heating does not necessarily lead to a decrease in the density of particles. Choice D is incorrect because the size of particles does not necessarily increase when a substance is heated. Therefore, the correct explanation for the increase in diffusion rate is the rise in kinetic energy of particles.

3. Which of the following elements does not exist as a diatomic molecule?

A. boron
B. fluorine
C. oxygen
D. nitrogen

Correct answer: A

Rationale: The correct answer is 'boron.' Diatomic molecules consist of two atoms of the same element bonded together. Boron is an exception and does not exist naturally as a diatomic molecule. On the other hand, fluorine, oxygen, and nitrogen commonly exist as diatomic molecules in their natural states. Fluorine, for example, exists as F2, oxygen exists as O2, and nitrogen exists as N2.

4. What is a benefit of water's ability to make hydrogen bonds?

A. Lack of cohesiveness
B. Low surface tension
C. Use as a nonpolar solvent
D. High specific heat

Correct answer: D

Rationale: The correct answer is D, high specific heat. Water's ability to form hydrogen bonds results in a high specific heat capacity, allowing it to absorb and release a large amount of heat energy with minimal temperature change. This property is essential for moderating temperature changes in organisms and maintaining stable environmental conditions for life processes. Choices A, lack of cohesiveness, and C, use as a nonpolar solvent, are incorrect. Water actually has high cohesiveness due to its ability to form hydrogen bonds, and it is a polar solvent, not nonpolar. Choice B, low surface tension, is also incorrect as water's hydrogen bonding contributes to its relatively high surface tension.

5. What is the correct electron configuration for carbon?

A. 1s²2s²2p¹
B. 1s²2s²2p²
C. 1s²2s²2p³
D. 1s²2s²2p⁶3s¹

Correct answer: B

Rationale: The correct electron configuration for carbon is 1s²2s²2p². This configuration indicates that there are 2 electrons in the first energy level (1s²), 2 electrons in the second energy level (2s²), and 2 electrons in the second energy level (2p²). It adheres to the aufbau principle, which states that electrons fill orbitals starting from the lowest energy level, and the Pauli exclusion principle, which states that each electron in an atom must have a unique set of quantum numbers. Choice A is incorrect because it does not fill the 2p orbital correctly. Choice C is incorrect as it exceeds the number of possible electrons in the 2p orbital. Choice D is incorrect as it includes an electron in the 3s orbital, which is not part of the electron configuration for carbon.