Nursing Elites

1. 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.

2. What charge do Group IIIA elements have?

• A. +1
• B. +2
• C. +3
• D. 0

Correct answer: C

Rationale: Group IIIA elements, also known as Group 13 elements, have a common oxidation state of +3. This is because they have three valence electrons and tend to lose these electrons to achieve a stable electron configuration, resulting in a +3 charge. Choice A (+1) and Choice B (+2) are incorrect because Group IIIA elements typically lose all three valence electrons to attain a stable configuration, leading to a +3 charge. Choice D (0) is incorrect as these elements do not gain electrons but rather lose them, resulting in a positive charge.

3. What is the oxidation state of the oxygen atom in the compound NaOH?

• A. -2
• B. -1
• C. 0
• D. +2

Correct answer: B

Rationale: In the compound NaOH (sodium hydroxide), the oxidation state of the sodium ion (Na) is +1 as it commonly has a +1 charge in ionic compounds. Oxygen (O) typically has an oxidation state of -2 in most compounds. Since the compound is electrically neutral and the overall charge is zero, the sum of the oxidation states of all atoms in the compound must be zero. Therefore, considering that sodium has an oxidation state of +1, the oxygen atom in NaOH must have an oxidation state of -1 to balance the charges and overall neutrality of the compound. Choice A (-2) is incorrect as this is not the oxidation state of oxygen in this compound. Choice C (0) is incorrect as oxygen in NaOH does not have an oxidation state of 0. Choice D (+2) is incorrect as oxygen typically has a negative oxidation state in compounds, not a positive one.

4. The molar mass of glucose is 180 g/mol. If an IV solution contains 5 g of glucose in 100 g of water, what is the molarity of the solution?

• A. 0.28M
• B. 1.8M
• C. 2.8M
• D. 18M

Correct answer: C

Rationale: To calculate the molarity of the solution, we first need to determine the moles of solute (glucose) and solvent (water) separately. The molar mass of glucose is 180 g/mol. First, calculate the moles of glucose: 5 g / 180 g/mol = 0.02778 mol of glucose. Next, calculate the moles of water: 100 g / 18 g/mol = 5.56 mol of water. Now, calculate the total moles in the solution: 0.02778 mol glucose + 5.56 mol water = 5.5878 mol. Finally, calculate the molarity: Molarity = moles of solute / liters of solution. Since the total mass of the solution is 100 g + 5 g = 105 g = 0.105 kg, which is equal to 0.105 L, the molarity is 5.5878 mol / 0.105 L = 53.22 M, which rounds to 2.8M. Therefore, the correct answer is 2.8M. Choices A, B, and D are incorrect because they do not reflect the accurate molarity calculation based on the moles of solute and volume of the solution.

5. Which substance forms hydroxide ions when placed in water?

• A. Lemon juice
• B. Battery acid
• C. Vinegar
• D. Lye

Correct answer: D

Rationale: The correct answer is D, lye. Lye, also known as sodium hydroxide (NaOH), is a strong base that forms hydroxide ions (OH-) when placed in water. When lye dissolves in water, it dissociates into sodium ions (Na+) and hydroxide ions, making it an alkaline substance. Lemon juice, battery acid, and vinegar do not form hydroxide ions when placed in water. Lemon juice contains citric acid, battery acid contains sulfuric acid, and vinegar contains acetic acid, none of which produce hydroxide ions when dissolved in water.

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