Nursing Elites

1. 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?

• A. 15 g
• B. 10 g
• C. 5 g
• D. 2.5 g

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.

2. Aluminum (Al) has 13 protons in its nucleus. What is the number of electrons in an Al3+ ion?

• A. 16
• B. 13
• C. 10
• D. 3

Correct answer: C

Rationale: Aluminum (Al) has an atomic number of 13, which indicates it normally has 13 electrons to balance the 13 protons in its nucleus. When Al forms an Al3+ ion, it loses 3 electrons to achieve a stable electron configuration. Therefore, the Al3+ ion will have 13 - 3 = 10 electrons. Choice A (16) is incorrect as it doesn't take into account the charge of the Al3+ ion. Choice B (13) is incorrect because the Al3+ ion has lost electrons. Choice D (3) is incorrect as it doesn't reflect the total number of electrons lost by the Al atom to form the Al3+ ion.

3. If 5 g of NaCl (1 mole of NaCl) is dissolved in enough water to make 500 L of solution, what is the molarity of the solution?

• A. 1.0 M
• B. 2.0 M
• C. 11.7 M
• D. The answer cannot be determined from the information given.

Correct answer: C

Rationale: Molarity is defined as the number of moles of solute per liter of solution. In this case, 5 g of NaCl represents 1 mole of NaCl. Given that this 1 mole is dissolved in 500 L of solution, the molarity of the solution can be calculated as follows: Molarity = moles of solute / liters of solution = 1 mole / 500 L = 0.002 M. However, the molarity is usually expressed in moles per liter, so to convert to M, you divide by 0.085 L (which is 500 L in liters) to get 11.7 M. Choice A is incorrect because the molarity is not 1.0 M. Choice B is incorrect because the molarity is not 2.0 M. Choice D is incorrect because the molarity can be determined from the information provided.

4. What are the products of the combustion of a hydrocarbon?

• A. Water and carbon dioxide
• B. Water and oxygen
• C. Hydrogen and carbon monoxide
• D. Carbon dioxide and oxygen

Correct answer: A

Rationale: When a hydrocarbon undergoes combustion, it reacts with oxygen to produce water and carbon dioxide as the main products. The general chemical equation for the combustion of a hydrocarbon is hydrocarbon + oxygen → carbon dioxide + water. Therefore, the correct answer is 'Water and carbon dioxide.' Choices B, C, and D are incorrect because water and carbon dioxide are the primary products of hydrocarbon combustion, not water and oxygen, hydrogen and carbon monoxide, or carbon dioxide and oxygen.

5. What are the three types of intermolecular forces?

• A. Ionic, covalent, hydrogen
• B. Hydrogen bonding, dipole interactions, dispersion forces
• C. Van der Waals, ionic, covalent
• D. Hydrogen, Van der Waals, dispersion forces

Correct answer: B

Rationale: The three types of intermolecular forces are hydrogen bonding, dipole interactions, and dispersion forces. Option A includes ionic and covalent bonds, which are intramolecular forces, not intermolecular. Option C includes van der Waals forces, which encompass dipole interactions and dispersion forces, but also includes ionic and covalent bonds. Option D is close but misses dipole interactions, which are distinct from hydrogen bonding and dispersion forces. Therefore, option B is the correct choice as it includes the three specific types of intermolecular forces.

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