what volume of concentrated hcl should be used to prepare 500 ml of a 00 m hcl solution
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HESI A2

HESI A2 Chemistry

1. How much concentrated HCl should be used to prepare 500 mL of a 0.100 M HCl solution?

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. To the nearest whole number, what is the mass of one mole of hydrogen iodide?

Correct answer: C

Rationale: The molar mass of hydrogen iodide (HI) is the sum of the atomic masses of its constituent elements. Hydrogen (H) has a molar mass of approximately 1 g/mol, and iodine (I) has a molar mass of about 127 g/mol. Thus, the molar mass of hydrogen iodide (HI) is approximately 1 + 127 = 128 g/mol. Rounding to the nearest whole number, the molar mass of hydrogen iodide is 128 g/mol, which is closest to choice C. Choice A (2 g/mol) is too low and does not reflect the correct molar mass of hydrogen iodide. Choice B (58 g/mol) is significantly lower than the actual molar mass. Choice D (128 g/mol) matches the calculated molar mass but is not the nearest whole number as requested.

3. What type of starch is glycogen?

Correct answer: B

Rationale: Glycogen is classified as animal starch, not plant starch. It is the storage form of glucose in animals, primarily found in the liver and muscles. Choice A (Plant starch) is incorrect because glycogen is not derived from plants. Choice C (Glucose) is incorrect as glucose is a monosaccharide and not a type of starch. Choice D (Cellulose) is incorrect as cellulose is a structural polysaccharide found in plant cell walls, not the same as glycogen.

4. Cobalt-60 has a half-life of 5 years. If you start with 20 g of cobalt-60, how much is left after 10 years?

Correct answer: C

Rationale: Cobalt-60's half-life of 5 years means that after 5 years, half of the initial amount remains. Therefore, after 10 years, a quarter (half of a half) of the initial amount will remain. Starting with 20 g, after 10 years, 5 g of cobalt-60 will be left. Choice A (15 g) is incorrect because it assumes a linear decrease, not considering the exponential decay characteristic of radioactive substances. Choice B (10 g) is incorrect as it overlooks that after 10 years, more decay has occurred. Choice D (2.5 g) is incorrect as it represents only an eighth of the initial amount after 10 years, not a quarter.

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?

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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