you contain two odorous gases in vials with porous plugs gas a has twice the mass of gas b which observation is most likely

HESI A2

HESI A2 Chemistry

1. You contain two odorous gases in vials with porous plugs. Gas A has twice the mass of Gas B. Which observation is most likely?

A. You will smell Gas A before you smell Gas B.
B. You will smell Gas B before you smell Gas A.
C. You will smell Gas A but not Gas B.
D. You will smell Gas B but not Gas A.

Correct answer: A

Rationale: The rate of effusion of a gas is inversely proportional to the square root of its molar mass. Since Gas A has twice the mass of Gas B, Gas A will effuse more slowly than Gas B. Therefore, you will likely smell Gas A before you smell Gas B as Gas A will escape and diffuse through the porous plug at a slower rate compared to Gas B. Choice A is correct because Gas A, with its higher molar mass, will take longer to effuse through the porous plug, causing you to smell it first. Choices B, C, and D are incorrect as they do not consider the relationship between molar mass and effusion rate.

2. What is the name of the compound CH₃-CH₂-CH₂-CH₃?

A. Cyclobutane
B. Butane
C. Butene
D. Butyne

Correct answer: B

Rationale: The compound CH₃-CH₂-CH₂-CH₃ is named butane. Butane is a straight-chain alkane comprising four carbon atoms connected by single bonds. The prefix 'but-' denotes the presence of four carbon atoms, while the suffix '-ane' indicates it is an alkane with single bonds between the carbon atoms. Choice A, Cyclobutane, is incorrect as it refers to a cyclic hydrocarbon with four carbon atoms in a ring structure. Choice C, Butene, is incorrect because it is an alkene with a double bond between two carbon atoms, not a saturated hydrocarbon like butane. Choice D, Butyne, is also incorrect as it is an alkyne with a triple bond between two carbon atoms, unlike the single bonds in butane.

3. The three important allotropic forms of phosphorus are red, white, and ___________.

A. green
B. gray
C. black
D. silver

Correct answer: C

Rationale: The three important allotropic forms of phosphorus are red, white, and black. These forms indicate the different physical properties and reactivity of phosphorus under various conditions. Red phosphorus is more stable and less reactive than white phosphorus, while black phosphorus is the least reactive form. Choice C, 'black,' is the correct answer as it completes the sequence of allotropic forms of phosphorus. Choices A, 'green,' B, 'gray,' and D, 'silver,' are incorrect as they do not represent recognized forms of phosphorus.

4. 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.

5. Which of the following is the weakest intermolecular force?

A. Dipole interactions
B. Hydrogen bonding
C. Van der Waals forces
D. Dispersion forces

Correct answer: D

Rationale: Dispersion forces, also known as London dispersion forces, are the weakest intermolecular forces. They are temporary attractive forces that occur due to momentary shifts in electron distribution within molecules. While dipole interactions, hydrogen bonding, and Van der Waals forces are stronger intermolecular forces, dispersion forces are the weakest because they arise from short-lived fluctuations in electron density. Dipole interactions involve permanent dipoles in molecules, making them stronger than dispersion forces. Hydrogen bonding is stronger than dipole interactions and involves hydrogen atoms bonded to highly electronegative atoms. Van der Waals forces encompass dipole-dipole interactions and dispersion forces, making them stronger than dispersion forces alone.