HESI A2
HESI A2 Chemistry Questions
1. Which of the following substances is a base?
- A. Water
- B. Sodium chloride
- C. Ammonia
- D. Salt
Correct answer: C
Rationale: The correct answer is 'Ammonia' (Choice C) as it is a common example of a base. Bases are substances that release hydroxide ions (OH-) in aqueous solutions, helping to increase the pH level. Ammonia is a weak base that can accept a proton (H+) to form ammonium hydroxide. In contrast, water (Choice A), sodium chloride (Choice B), and salt (Choice D) are not bases; water is neutral, while sodium chloride and salt are neutral compounds composed of a cation and an anion.
2. 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.
3. 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.
4. 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.
5. What can stop the penetration of beta radiation particles?
- A. Plastic
- B. Glass
- C. Aluminum foil
- D. Concrete
Correct answer: C
Rationale: Beta radiation particles are high-energy, fast-moving electrons or positrons. Aluminum foil is effective in stopping beta radiation due to its ability to absorb and block these particles. When beta particles interact with the aluminum foil, they lose energy and are absorbed, preventing their penetration. Plastic and glass are not as effective as aluminum foil in stopping beta radiation. While concrete provides some shielding against beta particles, aluminum foil is a more suitable material for this purpose as it offers better absorption and blocking capabilities.
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