Part II. Multiple choice, continued (3 points each). Name:__________________ Please circle your answer. There is only one correct answer to each question. 43. The F-N-F bond angle in the NF3 molecule is slightly less than ___________. A) 90˚ B) 109.5 ˚ C) 120˚ D) 180˚ 44. The angles between sp2 orbitals are _____________. A) 90˚ B) 109.5˚ C) 120˚ D) 180˚ 45. The blending of one s atomic orbital and two p atomic orbitals produces __________. A) three sp hybrid orbitals B) three sp2 hybrid orbitals C) three sp3 hybrid orbitals D) two sp3 hybrid orbitals 46. The π bond in ethylene, CH2CH2, results from the overlap of ___________. A) sp3 hybrid orbitals B) s atomic orbitals C) sp2 hybrid orbitals D) p atomic orbitals 47. The amount of gas that occupies 60.82 L at 31˚ C and 367 mm Hg is _________ mol. A) 1.18 B) 0.850 C) 894 D) 11.6 48. The specific heat of lead is 0.13 J/g-K. How much heat (in J) is required to raise the temperature of 15 g of lead from 22 ˚C to 37 ˚C? A) 7.8 B) 1.9 C) 19 D) 29 49. Which one of the following is a weak acid? A) HNO3 B) HI C) HF D) HClO4 50. Which combination will produce a precipitate? A) NH4OH (aq) and HCl (aq) B) AgNO3 (aq) and Ca(C2H3O2)2 (aq) C) NaCl(aq) and HC2H3O2 (aq) D) NaOH (aq) and Fe(NO3)2 (aq) 51. The density of chlorine (Cl2) gas at 25 ˚C and 450 torr is _______ g/L. A) 20 B) 4.9 C) 1.7 D) 0.86

Okay, let's break down each of these multiple-choice questions with detailed explanations:

43. The F-N-F bond angle in the NF3 molecule is slightly less than ___.
A) 90˚ B) 109.5 ˚ C) 120˚ D) 180˚

Correct Answer: B) 109.5˚

Explanation:

• NF3 has a central nitrogen atom bonded to three fluorine atoms and one lone pair of electrons. According to VSEPR theory, the electron pairs (bonding and non-bonding) around the nitrogen atom arrange themselves to minimize repulsion. This arrangement results in a tetrahedral electron geometry. However, the lone pair repels the bonding pairs more strongly than bonding pairs repel each other. As a result, the F-N-F bond angle is compressed slightly from the ideal tetrahedral angle of 109.5˚. Therefore, the bond angle is slightly less than 109.5˚.

44. The angles between sp2 orbitals are _.
A) 90˚ B) 109.5˚ C) 120˚ D) 180˚

Correct Answer: C) 120˚

Explanation:

• sp^2 hybridization involves the mixing of one s orbital and two p orbitals, resulting in three sp^2 hybrid orbitals. These orbitals arrange themselves in a trigonal planar geometry to minimize electron repulsion. In a trigonal planar arrangement, the angle between the hybrid orbitals is 120˚.

45. The blending of one s atomic orbital and two p atomic orbitals produces __.
A) three sp hybrid orbitals B) three sp2 hybrid orbitals C) three sp3 hybrid orbitals D) two sp3 hybrid orbitals

Correct Answer: B) three sp2 hybrid orbitals

Explanation:

• When one s atomic orbital and two p atomic orbitals blend, they form three sp^2 hybrid orbitals. This hybridization is characterized by one s and two p orbitals combining to form three new, equivalent orbitals.

46. The π bond in ethylene, CH2CH2, results from the overlap of ___.
A) sp3 hybrid orbitals B) s atomic orbitals C) sp2 hybrid orbitals D) p atomic orbitals

Correct Answer: D) p atomic orbitals

Explanation:

• In ethylene (CH2CH2), each carbon atom is sp^2 hybridized. The sigma (σ) bond between the carbon atoms is formed by the overlap of sp^2 hybrid orbitals. The pi (π) bond is formed by the sideways overlap of the remaining unhybridized p atomic orbitals on each carbon atom.

47. The amount of gas that occupies 60.82 L at 31˚ C and 367 mm Hg is _ mol.
A) 1.18 B) 0.850 C) 894 D) 11.6

Correct Answer: A) 1.18

Explanation:

• Use the ideal gas law: PV = nRT, where:
  • P = pressure (in atm)
  • V = volume (in L)
  • n = number of moles
  • R = ideal gas constant (0.0821 L atm / (mol K))
  • T = temperature (in K)
• First, convert pressure from mm Hg to atm: P = \frac{367 \text{ mm Hg}}{760 \text{ mm Hg/atm}} ≈ 0.4829 \text{ atm}
• Convert temperature from Celsius to Kelvin: T = 31 + 273.15 = 304.15 \text{ K}
• Now, solve for n: n = \frac{PV}{RT} = \frac{0.4829 \text{ atm} \cdot 60.82 \text{ L}}{0.0821 \frac{\text{L atm}}{\text{mol K}} \cdot 304.15 \text{ K}} ≈ 1.18 \text{ mol}

48. The specific heat of lead is 0.13 J/g-K. How much heat (in J) is required to raise the temperature of 15 g of lead from 22 ˚C to 37 ˚C?
A) 7.8 B) 1.9 C) 19 D) 29

Correct Answer: D) 29

Explanation:

• Use the formula: q = mcΔT, where:
  • q = heat (in J)
  • m = mass (in g)
  • c = specific heat (in J/g-K)
  • ΔT = change in temperature (in K or ˚C)
• q = 15 \text{ g} \cdot 0.13 \frac{\text{J}}{\text{g K}} \cdot (37 - 22) \text{ ˚C} = 15 \cdot 0.13 \cdot 15 = 29.25 \text{ J}
• Round to the nearest whole number: 29 J

49. Which one of the following is a weak acid?
A) HNO3 B) HI C) HF D) HClO4

Correct Answer: C) HF

Explanation:

• Strong acids completely dissociate in water, while weak acids only partially dissociate. The common strong acids are HCl, HBr, HI, HNO3, H2SO4, and HClO4.
• HF (hydrofluoric acid) is a weak acid because it does not completely dissociate in water.

50. Which combination will produce a precipitate?
A) NH4OH (aq) and HCl (aq) B) AgNO3 (aq) and Ca(C2H3O2)2 (aq) C) NaCl(aq) and HC2H3O2 (aq) D) NaOH (aq) and Fe(NO3)2 (aq)

Correct Answer: D) NaOH (aq) and Fe(NO3)2 (aq)

Explanation:

• A precipitate forms when two aqueous solutions are mixed and an insoluble compound is produced.
• Option D: Mixing NaOH (aq) and Fe(NO3)2 (aq) will produce Fe(OH)_2, which is an insoluble compound and precipitates out of the solution.
  • 2NaOH(aq) + Fe(NO3)2(aq) → 2NaNO3(aq) + Fe(OH)2(s)

51. The density of chlorine (Cl2) gas at 25 ˚C and 450 torr is _ g/L.
A) 20 B) 4.9 C) 1.7 D) 0.86

Correct Answer: C) 1.7

Explanation:

• Use the ideal gas law to find the molar volume, then use the molar mass to find the density.
• First, convert pressure from torr to atm: P = \frac{450 \text{ torr}}{760 \text{ torr/atm}} ≈ 0.592 \text{ atm}
• Convert temperature from Celsius to Kelvin: T = 25 + 273.15 = 298.15 \text{ K}
• Use the ideal gas law to find the molar volume (V/n): $$V/n = \frac{RT}{