Nursing Elites

1. What is the number of protons in the atomic nucleus of an alkali metal?

• A. 9
• B. 10
• C. 11
• D. 12

Correct answer: C

Rationale: The number of protons in the atomic nucleus of an alkali metal is 11. Alkali metals, belonging to group 1 of the periodic table, have 1 electron in their outer shell, which corresponds to 1 proton in their nucleus. Therefore, the correct answer is option C: 11. Choice A (9) is incorrect because it does not match the number of protons in an alkali metal. Choice B (10) is incorrect as it is also not the correct number of protons for an alkali metal. Choice D (12) is incorrect as it is not the typical number of protons found in the nucleus of an alkali metal.

2. What can stop the penetration of gamma radiation?

• A. Aluminum foil
• B. Glass
• C. Several feet of concrete
• D. Piece of paper

Correct answer: C

Rationale: Gamma radiation is highly penetrative and requires dense materials to block it effectively. While aluminum foil and glass are not sufficient to stop gamma radiation, several feet of concrete is needed due to its high density and ability to absorb gamma radiation effectively. A piece of paper is too thin and lacks the density required to block gamma radiation, making it an ineffective shield.

3. 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.

4. 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.

5. 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.

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