the three important allotrophic forms of phosphorus are red white and
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HESI A2

HESI A2 Chemistry

1. The three important allotropic forms of phosphorus are red, white, and ___________.

Correct answer: C

Rationale: The three important allotropic forms of phosphorus are red, white, and black. These forms indicate the different physical properties and reactivity of phosphorus under various conditions. Red phosphorus is more stable and less reactive than white phosphorus, while black phosphorus is the least reactive form. Choice C, 'black,' is the correct answer as it completes the sequence of allotropic forms of phosphorus. Choices A, 'green,' B, 'gray,' and D, 'silver,' are incorrect as they do not represent recognized forms of phosphorus.

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

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.

3. Here are the solubilities of four substances at 0°C, in grams of solute per 100 mL of water. If the temperature increases to 20°C, what would you expect to happen to the solubility figures?

Correct answer: C

Rationale: Solubility generally tends to increase with temperature for most solid solutes in liquid solvents due to higher kinetic energy leading to better solute-solvent interactions. As the temperature increases from 0°C to 20°C, all four solubility figures are expected to increase. Choice A is incorrect because solubility tends to increase with temperature. Choice B is incorrect as well for the same reason. Choice D is incorrect because the solubility of solid solutes typically increases with temperature.

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?

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. What charge do Group VIA elements typically have?

Correct answer: B

Rationale: Group VIA elements, also known as Group 16 elements, typically have a charge of -2. This is because they have 6 valence electrons and tend to gain 2 electrons to achieve a stable octet configuration, resulting in a -2 charge. Choice A (-1) is incorrect as Group VIA elements need to gain 2 electrons for stability, not just 1. Choice C (-3) is incorrect because Group VIA elements do not need to gain 3 electrons to achieve stability. Choice D (0) is incorrect as Group VIA elements need to gain electrons to reach a stable configuration, resulting in a negative charge.

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