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HESI A2

Chemistry Hesi A2

1. 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.

2. Which compound has a nonpolar bond in which the electrons are shared equally?

Correct answer: D

Rationale: The compound CH₄, methane, has a nonpolar bond where carbon and hydrogen share electrons equally. This occurs because carbon and hydrogen have similar electronegativities, meaning they have equal abilities to attract shared electrons. Consequently, a nonpolar covalent bond is formed due to the balanced sharing of electrons between these atoms. Choices A, B, and C do not have nonpolar bonds with electrons shared equally. In H₂O (water), there are polar covalent bonds due to the difference in electronegativity between hydrogen and oxygen. In NH₃ (ammonia), the nitrogen-hydrogen bonds are polar because of the electronegativity difference. In Cl₂ (chlorine gas), the Cl-Cl bond is nonpolar, but the question specifies a compound, not an element, and chlorine does not share its electrons equally with another element in a compound.

3. Balance this equation: Zn + HCl → ZnCl + H2.

Correct answer: C

Rationale: The given unbalanced equation is Zn + HCl → ZnCl + H2. To balance it, we need to have equal atoms on both sides of the equation. The balanced equation is 2Zn + 2HCl → 2ZnCl + H2. This balanced equation shows that two atoms of Zn combine with two molecules of HCl to form two molecules of ZnCl and one molecule of H2. Choice A is incorrect because it does not balance the equation. Choice B is incorrect as it does not have the same number of atoms on both sides. Choice D is incorrect because it does not balance the equation properly, resulting in an unequal number of atoms on both sides.

4. 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?

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.

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.

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