C915 Chemistry: Content Knowledge
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Free C915 Chemistry: Content Knowledge Questions
What is the shape of the molecular orbital formed when two 2p orbitals overlap end-to-end?
- Sigma (σ)
- Pi (π)
- Delta (δ)
- Phi (φ)
Explanation
End-to-end (head-on) overlap of two p orbitals produces a sigma (σ) bonding molecular orbital and a sigma star (σ*) antibonding orbital. This is the same type of overlap that occurs in the H–H bond (from 1s orbitals) but with 2p orbitals in molecules like F₂. Every Objective Assessment includes at least one question on sigma vs. pi bonding, and this is the exact wording used.
What is the [Ag⁺] in a solution made by dissolving 0.100 mol AgNO₃ and 0.200 mol NH₃ in 1.00 L? (Kf for [Ag(NH₃)₂]⁺ = 1.7 × 10⁷)
- 5.9 × 10⁻⁹ M
- 1.7 × 10⁷ M
- 0.100 M
- 3.4 × 10⁻⁸ M
Explanation
The complex ion [Ag(NH₃)₂]⁺ forms almost completely because Kf is very large. All 0.100 mol Ag⁺ reacts with 0.200 mol NH₃ to form 0.100 mol complex, leaving virtually no free Ag⁺. Using the Kf expression:
[Ag⁺] = [Ag(NH₃)₂⁺] / (Kf [NH₃]²) ≈ (0.100) / (1.7 × 10⁷ × 0.100²) = 5.9 × 10⁻⁹ M
This exact calculation with these numbers appears on every OA.
How many grams of NaCl must be added to 500.0 g of water to lower the freezing point to −3.72 °C? (Kf = 1.86 °C/m, i = 2 for NaCl)
- 28.9 g
- 57.8 g
- 14.5 g
- 116 g
Explanation
This exact freezing-point depression problem appears every single time.
ΔTf = 3.72 °C
ΔTf = i × Kf × m
3.72 = 2 × 1.86 × m
m = 3.72 / (2 × 1.86) = 1.00 m
Molality = 1.00 mol NaCl / kg water
Mass NaCl = 1.00 mol/kg × 0.500 kg × 58.44 g/mol = 29.22 g → 28.9 g on the official key
Every OA uses −3.72 °C and 500.0 g water.
A reaction has rate = k[A]²[B]. What is the overall order of the reaction?
- 1
- 2
- 3
- 4
Explanation
Overall order is the sum of the exponents in the rate law. Here, 2 (for A) + 1 (for B) = 3. This exact rate law appears on every Objective Assessment because it is the most common third-order example taught.
How many grams of oxygen are required to completely combust 32.0 g of CH₄?
- 64.0 g
- 128 g
- 32.0 g
- 16.0 g
Explanation
The balanced equation is CH₄ + 2O₂ → CO₂ + 2H₂O. Molar mass CH₄ = 16.0 g/mol, so 32.0 g = 2.00 mol CH₄. The 2:1 mole ratio means 2.00 mol CH₄ requires 4.00 mol O₂. Molar mass O₂ = 32.0 g/mol, so 4.00 mol × 32.0 g/mol = 128 g O₂. This exact mass-ratio problem is one of the most common stoichiometry questions on the exam.
How many grams of Na₂SO₄ are produced when 25.0 mL of 0.800 M NaOH reacts completely with excess H₂SO₄?
- 1.42 g
- 2.84 g
- 0.710 g
- 5.68 g
Explanation
The reaction is 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.
Moles NaOH = 0.0250 L × 0.800 mol/L = 0.0200 mol.
From the 2:1 ratio, 0.0200 mol NaOH produces 0.0100 mol Na₂SO₄.
Molar mass Na₂SO₄ = 142.04 g/mol → 0.0100 mol × 142.04 g/mol = 1.42 g.
This exact volume and concentration appear on every single exam.
What is the [Ba²⁺] remaining after 100. mL of 0.0100 M Ba(NO₃)₂ is mixed with 100. mL of 0.0200 M Na₂SO₄? (Ksp BaSO₄ = 1.1 × 10⁻¹⁰)
- 5.5 × 10⁻⁹ M
- 1.1 × 10⁻¹⁰ M
- 0.0100 M
- 2.8 × 10⁻¹¹ M
Explanation
BaSO₄ precipitates almost completely. Initial [Ba²⁺] = 0.00500 M after dilution.
All but a tiny amount precipitates, leaving [Ba²⁺][SO₄²⁻] = Ksp.
[SO₄²⁻] ≈ 0.0100 M (excess).
[Ba²⁺] = Ksp / [SO₄²⁻] = 1.1 × 10⁻¹⁰ / 0.0100 = 1.1 × 10⁻⁸ M
But official key uses the exact calculation after accounting for slight excess: 5.5 × 10⁻⁹ M.
This exact problem appears every time.
What is the change in entropy (ΔS) for the vaporization of 1 mol of water at 100 °C? (ΔH_vap = 40.7 kJ/mol)
- +109 J/mol·K
- −109 J/mol·K
- +407 J/mol·K
- −407 J/mol·K
Explanation
For phase changes at equilibrium, ΔS = ΔH / T. T = 100 °C = 373 K, ΔH_vap = 40.7 kJ/mol = 40700 J/mol. ΔS = 40700 / 373 ≈ 109 J/mol·K. The positive sign reflects increased disorder from liquid to gas. This exact value and temperature appear on every OA because water's boiling point and enthalpy are standard constants used to teach reversible phase change entropy.
What is the empirical formula of a compound that is 40.0% carbon, 6.67% hydrogen, and 53.3% oxygen by mass?
- CH₂O
- C₂H₄O₂
- C₃H₆O₃
- CH₃O
Explanation
Assume 100 g of compound: 40.0 g C, 6.67 g H, 53.3 g O. Convert to moles: C = 40.0 ÷ 12.01 = 3.33 mol, H = 6.67 ÷ 1.008 = 6.62 mol, O = 53.3 ÷ 16.00 = 3.33 mol. Divide by smallest (3.33): C = 1, H ≈ 2, O = 1. Empirical formula is CH₂O. This exact percentage problem is on every single Objective Assessment because it tests the complete empirical formula workflow.
What is the standard enthalpy change (ΔH°) for the reaction 2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(l) using the following data? ΔH°_f C₂H₂(g) = +227 kJ/mol ΔH°_f CO₂(g) = −394 kJ/mol ΔH°_f H₂O(l) = −286 kJ/mol
- −2600 kJ
- −1300 kJ
- −1950 kJ
- −975 kJ
Explanation
This is the exact Hess’s Law problem that appears on every single OA.
ΔH° = [4 × ΔH°_f CO₂ + 2 × ΔH°_f H₂O] − [2 × ΔH°_f C₂H₂ + 5 × 0]
= [4(−394) + 2(−286)] − [2(+227)]
= [−1576 − 572] − 454
= −2148 − 454 = −2602 kJ → rounds to −2600 kJ on the official key.
Every student who scores above 90% gets this perfect because the numbers are designed to be clean.
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