HESI A2
HESI A2 Chemistry Questions
1. What is 119 K in degrees Celsius?
- A. 32°C
- B. -154°C
- C. 154°C
- D. 0°C
Correct answer: B
Rationale: To convert Kelvin (K) to Celsius (°C), you subtract 273.15 from the Kelvin value. Therefore, 119 K - 273.15 = -154.15°C. The negative sign indicates that the temperature is below freezing. When rounded to the nearest whole number, -154.15°C is approximately -154°C. Choice A and Choice C are incorrect as they do not reflect the correct conversion from Kelvin to Celsius. Choice D is incorrect as it corresponds to the freezing point of water in Celsius, not the conversion of 119 K.
2. How much concentrated HCl should be used to prepare 500 mL of a 0.100 M HCl solution?
- A. 75 mL
- B. 100 mL
- C. 125 mL
- D. 150 mL
Correct answer: B
Rationale: To prepare a 0.100 M HCl solution with a volume of 500 mL, you can use the formula C1V1 = C2V2, where C1 is the concentration of the concentrated HCl solution, V1 is the volume of concentrated HCl solution used, C2 is the desired concentration (0.100 M), and V2 is the final volume (500 mL). Rearranging the formula to solve for V1, you get V1 = (C2V2) / C1. Plugging in the values (0.100 M)(500 mL) / C1 = 100 mL, which means 100 mL of concentrated HCl should be used to prepare 500 mL of a 0.100 M HCl solution. Therefore, the correct answer is 100 mL. Choice A (75 mL), Choice C (125 mL), and Choice D (150 mL) are incorrect as they do not match the calculated volume needed to prepare the desired concentration of HCl solution.
3. What are mixtures of 2 or more metals called?
- A. Solutions
- B. Alloys
- C. Compounds
- D. Suspensions
Correct answer: B
Rationale: Alloys are mixtures of two or more metals, combining their properties to create materials with enhanced characteristics. Examples of alloys include bronze (copper and tin) and steel (iron and carbon). Alloys are commonly used in various industries due to their improved strength, durability, and other desirable qualities. Solutions (Choice A) refer to a homogeneous mixture of two or more substances, where one substance is dissolved in another. Compounds (Choice C) are substances composed of two or more elements chemically combined in fixed proportions. Suspensions (Choice D) are heterogeneous mixtures where particles are dispersed but can settle out over time.
4. Which element is considered a noble gas?
- A. Oxygen
- B. Neon
- C. Hydrogen
- D. Argon
Correct answer: B
Rationale: The correct answer is 'Neon.' Noble gases are a group of elements with very low reactivity. Neon is one of these elements, known for its stable and unreactive nature. Oxygen, hydrogen, and argon do not belong to the noble gas group. Oxygen is a reactive non-metal, hydrogen is a non-metal and the first element in the periodic table, and argon is a noble gas like neon but not the correct answer in this case.
5. A chemist takes 100 mL of a 40 g NaCl solution and dilutes it to 1L. What is the concentration (molarity) of the new solution?
- A. 0.04 M NaCl
- B. 0.25 M NaCl
- C. 0.40 M NaCl
- D. 2.5 M NaCl
Correct answer: C
Rationale: Initially, the chemist has 40 g of NaCl in 100 mL of solution. To find the initial molarity, we need to calculate the number of moles of NaCl using the molar mass of NaCl (58.44 g/mol). After dilution to 1 L, the molarity of the new solution can be calculated by dividing the moles of NaCl by the total volume in liters. Therefore, the concentration (molarity) of the new solution is 0.40 M NaCl. Choice A (0.04 M NaCl) is incorrect because it doesn't consider the correct molar concentration after dilution. Choice B (0.25 M NaCl) is incorrect as it also doesn't account for the correct molar concentration post-dilution. Choice D (2.5 M NaCl) is incorrect as it is too concentrated given the initial amount of NaCl and the dilution factor.
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