Nursing Elites

1. Balance this equation: Zn + HCl → ZnCl + H2.

• A. Zn + 2HCl → ZnCl + H2
• B. Zn + HCl → 2ZnCl + H2
• C. 2Zn + 2HCl → 2ZnCl + H2
• D. Zn + 4HCl → ZnCl + H2

Correct answer: C

Rationale: The given unbalanced equation is Zn + HCl → ZnCl + H2. To balance it, we need to have equal atoms on both sides of the equation. The balanced equation is 2Zn + 2HCl → 2ZnCl + H2. This balanced equation shows that two atoms of Zn combine with two molecules of HCl to form two molecules of ZnCl and one molecule of H2. Choice A is incorrect because it does not balance the equation. Choice B is incorrect as it does not have the same number of atoms on both sides. Choice D is incorrect because it does not balance the equation properly, resulting in an unequal number of atoms on both sides.

2. The molar mass of glucose is 180 g/mol. If an IV solution contains 5 g of glucose in 100 g of water, what is the molarity of the solution?

• A. 0.28M
• B. 1.8M
• C. 2.8M
• D. 18M

Correct answer: C

Rationale: To calculate the molarity of the solution, we first need to determine the moles of solute (glucose) and solvent (water) separately. The molar mass of glucose is 180 g/mol. First, calculate the moles of glucose: 5 g / 180 g/mol = 0.02778 mol of glucose. Next, calculate the moles of water: 100 g / 18 g/mol = 5.56 mol of water. Now, calculate the total moles in the solution: 0.02778 mol glucose + 5.56 mol water = 5.5878 mol. Finally, calculate the molarity: Molarity = moles of solute / liters of solution. Since the total mass of the solution is 100 g + 5 g = 105 g = 0.105 kg, which is equal to 0.105 L, the molarity is 5.5878 mol / 0.105 L = 53.22 M, which rounds to 2.8M. Therefore, the correct answer is 2.8M. Choices A, B, and D are incorrect because they do not reflect the accurate molarity calculation based on the moles of solute and volume of the solution.

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. How many times more acidic is a substance with a pH of 3 compared to a substance with a pH of 5?

• A. 8
• B. 2
• C. 100
• D. 1,000

Correct answer: D

Rationale: The pH scale is logarithmic, indicating that each pH unit change reflects a 10-fold difference in acidity level. Going from pH 5 to pH 3 involves a difference of 2 units, which translates to a 100-fold increase in acidity level (10^2 = 100 for each unit). Therefore, a substance with a pH of 3 is 1,000 times more acidic than a substance with a pH of 5 (100 * 10 = 1,000). Choice A (8) is incorrect as it does not consider the logarithmic nature of the pH scale. Choice B (2) is incorrect because it represents the difference in pH units, not the increase in acidity level. Choice C (100) is incorrect as it miscalculates the increase in acidity level, which is 1,000 times and not 100 times.

5. How many moles of potassium bromide are in 25 mL of a 4 M KBr solution?

• A. 0.035 mol
• B. 0.1 mol
• C. 0.18 mol
• D. 1.6 mol

Correct answer: B

Rationale: To find the moles of potassium bromide in 25 mL of a 4 M KBr solution, we first need to convert the volume from milliliters to liters. 25 mL is equal to 0.025 L. Then, we use the formula moles = molarity x volume in liters. Substituting the values, moles = 4 M x 0.025 L = 0.1 mol. Therefore, there are 0.1 moles of KBr in 25 mL of a 4 M solution. Choice A, 0.035 mol, is incorrect as it does not properly calculate the moles. Choice C, 0.18 mol, and choice D, 1.6 mol, are also incorrect as they are not the result of the correct calculation based on the given molarity and volume.

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