aluminum al has 13 protons in its nucleus what is the number of electrons in an al3 ion

HESI A2

Chemistry Hesi A2

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. Which one does not name a polar molecule?

A. NH₃
B. H₂S
C. SO₂
D. CO₂

Correct answer: A

Rationale: The correct answer is NH₃. The molecule NH₃ does not represent a polar molecule because nitrogen and hydrogen in this molecule have a small difference in electronegativity that does not result in a significant polar covalent bond. In contrast, molecules H₂S, SO₂, and CO₂ have polar covalent bonds due to larger electronegativity differences, making them polar molecules. Therefore, options B, C, and D are polar molecules, unlike option A.

3. What is the correct name of ZnSO₄?

A. Zinc sulfate
B. Zinc sulfide
C. Zinc sulfur
D. Zinc oxide

Correct answer: A

Rationale: The correct name of ZnSO₄ is zinc sulfate. In this compound, zinc is combined with the polyatomic ion sulfate (SO₄). Sulfate is a common anion formed from sulfur and oxygen atoms. Therefore, the correct name for ZnSO₄ is zinc sulfate. Choice B, Zinc sulfide, is incorrect because sulfide is a different anion (S²⁻) compared to sulfate (SO₄²⁻). Choice C, Zinc sulfur, is incorrect as it does not represent the correct anion in the compound. Choice D, Zinc oxide, is incorrect as it involves an oxygen anion, not sulfate.

4. 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.

5. What type of starch is glycogen?

A. Plant starch
B. Animal starch
C. Glucose
D. Cellulose

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.