Nursing Elites

1. Which of these represents a strong acid?

• A. CH₃COOH
• B. H₂SO₄
• C. NH₃
• D. KOH

Correct answer: B

Rationale: Among the options provided, H₂SO₄ (sulfuric acid) represents a strong acid. Strong acids completely ionize in water to produce a high concentration of H+ ions. Sulfuric acid is a strong acid known for its ability to dissociate almost completely in water, making it a strong acid. Choice A, CH₃COOH (acetic acid), is a weak acid that only partially dissociates in water. Choices C and D, NH₃ (ammonia) and KOH (potassium hydroxide), are bases and not acids.

2. To the nearest whole number, what is the mass of one mole of hydrogen iodide?

• A. 2 g/mol
• B. 58 g/mol
• C. 87 g/mol
• D. 128 g/mol

Correct answer: C

Rationale: The molar mass of hydrogen iodide (HI) is the sum of the atomic masses of its constituent elements. Hydrogen (H) has a molar mass of approximately 1 g/mol, and iodine (I) has a molar mass of about 127 g/mol. Thus, the molar mass of hydrogen iodide (HI) is approximately 1 + 127 = 128 g/mol. Rounding to the nearest whole number, the molar mass of hydrogen iodide is 128 g/mol, which is closest to choice C. Choice A (2 g/mol) is too low and does not reflect the correct molar mass of hydrogen iodide. Choice B (58 g/mol) is significantly lower than the actual molar mass. Choice D (128 g/mol) matches the calculated molar mass but is not the nearest whole number as requested.

3. 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.

4. What is the coefficient of O after the following equation is balanced?

• A. 1
• B. 2
• C. 3
• D. 4

Correct answer: A

Rationale: In a balanced chemical equation, the coefficient of oxygen (O) in O2 is already 2, so there is no need to adjust its coefficient further. Therefore, the coefficient of O remains as 1. Since the coefficient of O2 is 2, each O atom is represented by the coefficient of 1, and it does not change during the balancing process. Choices B, C, and D are incorrect as they suggest changing the coefficient of oxygen, which is not necessary for O2 in a balanced equation.

5. 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.

Similar Questions