HESI A2
Chemistry HESI A2 Practice Test
1. 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.
2. What is the correct electron configuration for lithium?
- A. 1s²2s¹
- B. 1s²2s²
- C. 1s²2s¹2p¹
- D. 1s¹2s¹2p²
Correct answer: A
Rationale: The electron configuration for lithium is 1s²2s¹. Lithium has 3 electrons, and the configuration indicates that the first two electrons fill the 1s orbital, while the third electron fills the 2s orbital. Therefore, the correct electron configuration for lithium is 1s²2s¹. Choice B (1s²2s²) is incorrect as it represents the electron configuration for beryllium, not lithium. Choice C (1s²2s¹2p¹) includes the 2p orbital, which is not involved in lithium's electron configuration. Choice D (1s¹2s¹2p²) is incorrect as it does not accurately represent lithium's electron configuration.
3. Radioactive isotopes are frequently used in medicine. What kind of half-life would a medical isotope probably have?
- A. Seconds-long
- B. Days-long
- C. Years-long
- D. Many years long
Correct answer: B
Rationale: Medical isotopes used in diagnosis and treatment need to have a relatively short half-life to minimize radiation exposure to patients. If the half-life were too long (such as many years) or even years-long, the radiation would persist for too long and could be harmful to the patient. Seconds-long half-lives would not provide enough time for the isotope to be effective. Days-long half-lives strike a balance between providing enough time for the isotope to be used effectively and minimizing radiation exposure.
4. Which of the following factors would not affect rates of reaction?
- A. Temperature
- B. Surface area
- C. Pressure
- D. Time
Correct answer: D
Rationale: Time would not directly affect rates of reaction. The rate of a chemical reaction is determined by factors that affect the frequency of successful collisions between reactant molecules, leading to a reaction. Temperature, surface area, and pressure can influence reaction rates by impacting the kinetic energy of molecules, the exposed surface for collisions, and the concentration of reactants, respectively. However, time, in the context of this question, does not alter the rate of reaction but may affect the extent of the reaction or the amount of product formed over time.
5. 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.
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