HESI A2
HESI A2 Chemistry Practice Questions
1. Which of these types of intermolecular force is weakest?
- A. Dipole-dipole interaction
- B. London dispersion force
- C. Hydrogen bonding
- D. Ionic bonding
Correct answer: B
Rationale: The correct answer is B, London dispersion force. London dispersion forces are the weakest type of intermolecular force among the options provided. These forces arise from temporary fluctuations in electron distribution within molecules, leading to temporary dipoles. London dispersion forces are present in all molecules and are generally weaker than dipole-dipole interactions, hydrogen bonding, and ionic bonding. Dipole-dipole interactions are stronger than London dispersion forces as they involve permanent dipoles in molecules. Hydrogen bonding is stronger than both London dispersion and dipole-dipole interactions as it is a special type of dipole-dipole interaction that occurs when hydrogen is bonded to highly electronegative atoms like oxygen or nitrogen. Ionic bonding is the strongest type of intermolecular force among the options, but it is not the correct answer for the weakest type of force.
2. Balance this equation: Fe + Cl2 → FeCl3
- A. 2Fe + 2Cl2 → 2FeCl3
- B. 2Fe + 3Cl2 → 2FeCl3
- C. 3Fe + 2Cl2 → 3FeCl3
- D. 3Fe + 3Cl2 → 6FeCl3
Correct answer: B
Rationale: In the given equation, Fe combines with Cl to form FeCl3. To balance the equation, we need to have the same number of each element on both sides. Since Cl is represented as Cl2 in the equation, we need 3 Cl2 molecules to balance Fe, resulting in 2Fe + 3Cl2 → 2FeCl3. Choice A is incorrect because it only balances Fe but not Cl2. Choice C is incorrect as it balances Fe but not Cl2. Choice D is incorrect as it balances Fe but overbalances Cl2.
3. How many moles of potassium bromide are in 25 mL of a 4 M KBr solution?
- A. 0.035 mol
- B. 0.1 mol
- C. 0.18 mol
- D. 1.6 mol
Correct answer: B
Rationale: To find the moles of potassium bromide in 25 mL of a 4 M KBr solution, we first need to convert the volume from milliliters to liters. 25 mL is equal to 0.025 L. Then, we use the formula moles = molarity x volume in liters. Substituting the values, moles = 4 M x 0.025 L = 0.1 mol. Therefore, there are 0.1 moles of KBr in 25 mL of a 4 M solution. Choice A, 0.035 mol, is incorrect as it does not properly calculate the moles. Choice C, 0.18 mol, and choice D, 1.6 mol, are also incorrect as they are not the result of the correct calculation based on the given molarity and volume.
4. To the nearest whole number, what is the mass of one mole of hydrogen iodide?
- A. 2 g/mol
- B. 58 g/mol
- C. 87 g/mol
- D. 128 g/mol
Correct answer: C
Rationale: The molar mass of hydrogen iodide (HI) is the sum of the atomic masses of its constituent elements. Hydrogen (H) has a molar mass of approximately 1 g/mol, and iodine (I) has a molar mass of about 127 g/mol. Thus, the molar mass of hydrogen iodide (HI) is approximately 1 + 127 = 128 g/mol. Rounding to the nearest whole number, the molar mass of hydrogen iodide is 128 g/mol, which is closest to choice C. Choice A (2 g/mol) is too low and does not reflect the correct molar mass of hydrogen iodide. Choice B (58 g/mol) is significantly lower than the actual molar mass. Choice D (128 g/mol) matches the calculated molar mass but is not the nearest whole number as requested.
5. What is the correct electron configuration for nitrogen?
- A. 1s² 2s²
- B. 1s² 2s² 2p²
- C. 1s² 2s² 2p³
- D. 1s² 2s² 2pâ´
Correct answer: C
Rationale: The electron configuration of nitrogen is determined by its atomic number, which is 7. Nitrogen has 7 electrons. Following the order of filling orbitals, the electron configuration for nitrogen is 1s² 2s² 2p³. This means the first energy level is filled with 2 electrons in the 1s orbital, the second energy level is filled with 2 electrons in the 2s orbital, and 3 electrons in the 2p orbital. Each orbital can hold a specific number of electrons, and nitrogen, with its 7 electrons, fits this configuration. Choice A is incorrect because it does not account for all the electrons in the nitrogen atom. Choice B is incorrect as it only represents 6 electrons, not the 7 electrons in nitrogen. Choice D is incorrect as it represents 8 electrons, which is not the correct electron configuration for nitrogen.
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