HESI A2
HESI A2 Chemistry
1. 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.
2. Which compound contains a polar covalent bond?
- A. O
- B. F
- C. Br
- D. H₂O
Correct answer: D
Rationale: The compound 'H₂O' (water) contains a polar covalent bond. In a water molecule, the oxygen atom is more electronegative than the hydrogen atoms. As a result, the electrons in the O-H bonds are unevenly shared, leading to a partial negative charge on the oxygen atom and partial positive charges on the hydrogen atoms. This unequal sharing of electrons creates a polar covalent bond in water. Choices A, B, and C are incorrect because they represent individual elements, not compounds, and do not involve the concept of polar covalent bonds.
3. How many protons does Potassium have?
- A. 18
- B. 19
- C. 20
- D. 21
Correct answer: B
Rationale: Potassium, with the atomic symbol K, has 19 protons, which is equal to its atomic number. The number of protons determines the element's identity, and in the case of Potassium, it is 19. Choice A (18) is incorrect as it does not correspond to Potassium's proton number. Choice C (20) and Choice D (21) are also incorrect as they do not match the actual number of protons in Potassium.
4. 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.
5. If fifty-six kilograms of a radioactive substance has a half-life of 12 days, how many days will it take the substance to decay naturally to only 7 kilograms?
- A. 8
- B. 12
- C. 36
- D. 48
Correct answer: C
Rationale: To decay from 56 kg to 7 kg, the substance needs to go through 3 half-lives (56 kg ÷ 2 ÷ 2 ÷ 2 = 7 kg). Since each half-life is 12 days, the total time required is 12 days per half-life x 3 half-lives = 36 days. Choice A is incorrect because it does not consider the concept of half-lives. Choice B is incorrect because it represents the duration of a single half-life, not the total time required for the decay. Choice D is incorrect as it does not account for the multiple half-lives needed for the substance to decay from 56 kg to 7 kg.
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