Nursing Elites

1. What can stop the penetration of alpha particles?

• A. Aluminum foil
• B. Glass
• C. Piece of paper
• D. Plastic

Correct answer: C

Rationale: Alpha particles can be stopped by a piece of paper due to their low penetration power. The paper acts as a shield, effectively blocking the alpha particles from passing through. In contrast, materials like aluminum foil, glass, and plastic are not as effective as a simple piece of paper in stopping alpha particles. Aluminum foil is more effective against beta particles, gamma rays, and x-rays due to its higher density. Glass and plastic also provide some protection against beta particles and gamma rays, but they are less effective than a piece of paper against alpha particles.

2. What charge do Group IA elements have?

• A. +1
• B. +2
• C. +3
• D. 0

Correct answer: A

Rationale: Group IA elements, also known as alkali metals, have a +1 charge. They readily lose one electron to achieve a stable electron configuration, forming ions with a single positive charge. This makes +1 the correct choice. Choices B, C, and D are incorrect because alkali metals in Group IA typically lose one electron, so they do not have a +2, +3, or 0 charge.

3. 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.

4. Which of these intermolecular forces would result in the lowest boiling point?

• A. Dipole-dipole interaction
• B. London dispersion force
• C. Keesom interaction
• D. Hydrogen bonding

Correct answer: B

Rationale: The London dispersion force is the weakest intermolecular force among the options provided. These forces are present in all molecules and are caused by temporary fluctuations in electron density, resulting in temporary dipoles. Since London dispersion forces are generally weaker than dipole-dipole interactions, Keesom interactions, and hydrogen bonding, a substance with London dispersion forces as the primary intermolecular force would have the lowest boiling point due to the weaker intermolecular forces holding the molecules together. Dipole-dipole interactions, Keesom interactions, and hydrogen bonding are stronger intermolecular forces compared to London dispersion forces, resulting in higher boiling points for substances that exhibit these interactions.

5. What are the products of the combustion of a hydrocarbon?

• A. Water and carbon dioxide
• B. Water and oxygen
• C. Hydrogen and carbon monoxide
• D. Carbon dioxide and oxygen

Correct answer: A

Rationale: When a hydrocarbon undergoes combustion, it reacts with oxygen to produce water and carbon dioxide as the main products. The general chemical equation for the combustion of a hydrocarbon is hydrocarbon + oxygen → carbon dioxide + water. Therefore, the correct answer is 'Water and carbon dioxide.' Choices B, C, and D are incorrect because water and carbon dioxide are the primary products of hydrocarbon combustion, not water and oxygen, hydrogen and carbon monoxide, or carbon dioxide and oxygen.

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