Questions 9

HESI A2

HESI A2 Test Bank

Chemistry Hesi A2 Questions

Question 1 of 5

A chemist takes 100 mL of a 40 g NaCl solution and dilutes it to 1L. What is the concentration (molarity) of the new solution?

Correct Answer: C

Rationale: Initially, the chemist has 40 g of NaCl in 100 mL of solution. To find the initial molarity, we need to calculate the number of moles of NaCl using the molar mass of NaCl (58.44 g/mol). After dilution to 1 L, the molarity of the new solution can be calculated by dividing the moles of NaCl by the total volume in liters. Therefore, the concentration (molarity) of the new solution is 0.40 M NaCl. Choice A (0.04 M NaCl) is incorrect because it doesn't consider the correct molar concentration after dilution. Choice B (0.25 M NaCl) is incorrect as it also doesn't account for the correct molar concentration post-dilution. Choice D (2.5 M NaCl) is incorrect as it is too concentrated given the initial amount of NaCl and the dilution factor.

Question 2 of 5

What is the correct name of MgO?

Correct Answer: B

Rationale: The correct name of MgO is Magnesium oxide. Mg represents the chemical symbol for magnesium, and O represents the chemical symbol for oxygen. When these elements combine, they form magnesium oxide. Option A, Manganese oxide, is incorrect as it refers to a compound of manganese and oxygen, not magnesium. Option C, Magnesium oxate, is not a valid chemical compound name. Option D, Magnesium hydroxide, refers to a different compound consisting of magnesium, oxygen, and hydrogen.

Question 3 of 5

Which substance shows a decrease in solubility in water with an increase in temperature?

Correct Answer: C

Rationale: Potassium iodide (KI) shows a decrease in solubility in water with an increase in temperature. This is due to the dissolution of KI in water being an endothermic process. When the temperature rises, the equilibrium shifts toward the solid state, leading to a decrease in solubility. Therefore, as the temperature increases, KI becomes less soluble in water. Choice A (NaCl) and Choice D (CaCl) do not exhibit a decrease in solubility with an increase in temperature. NaCl and CaCl are generally more soluble in water at higher temperatures. Choice B (Oxygen) is a gas and not typically considered in solubility discussions involving solids or liquids dissolving in water.

Question 4 of 5

A chemist takes 100 mL of a 40 g NaCl solution and dilutes it to 1L. What is the concentration (molarity) of the new solution?

Correct Answer: C

Rationale: Initially, the chemist has 40 g of NaCl in 100 mL of solution. To find the initial molarity, we need to calculate the number of moles of NaCl using the molar mass of NaCl (58.44 g/mol). After dilution to 1 L, the molarity of the new solution can be calculated by dividing the moles of NaCl by the total volume in liters. Therefore, the concentration (molarity) of the new solution is 0.40 M NaCl. Choice A (0.04 M NaCl) is incorrect because it doesn't consider the correct molar concentration after dilution. Choice B (0.25 M NaCl) is incorrect as it also doesn't account for the correct molar concentration post-dilution. Choice D (2.5 M NaCl) is incorrect as it is too concentrated given the initial amount of NaCl and the dilution factor.

Question 5 of 5

What is the oxidation state of the nitrogen atom in the compound NH3?

Correct Answer: B

Rationale: In the compound NH3, nitrogen is bonded to three hydrogen atoms. Hydrogen is always assigned an oxidation state of +1. Since the overall charge of NH3 is zero, the oxidation state of nitrogen must be -1 to balance out the hydrogen's +1 oxidation state. Therefore, the correct oxidation state of the nitrogen atom in NH3 is -1. Choice A (-3) is incorrect because it does not account for the electronegativity of hydrogen. Choice C (+1) and Choice D (+3) are incorrect as the nitrogen atom in NH3 needs to balance the +1 oxidation state of each hydrogen atom, resulting in a total of -3 to maintain the compound's charge neutrality.

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