C915 Chemistry: Content Knowledge
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Free C915 Chemistry: Content Knowledge Questions
What is the final safety question you must know before scheduling the OA?
- Never eat or drink in the lab
- Always add acid to water, never water to acid
- Wear open-toed shoes for ventilation
- Store chemicals above eye level
Explanation
The very last question on every single C915 Objective Assessment. The universal lab safety rule for dilution of concentrated acids: always pour acid slowly into water while stirring to prevent violent splashing and boiling. Adding water to acid causes the first drops to create a superheated layer that explodes outward. This rule has been tested verbatim since the course began.
What is the correct IUPAC name for the compound Cr(NO₃)₃?
- Chromium nitrate
- Chromium(III) nitrate
- Chromium trinitrate
- Chromium(II) nitrate
Explanation
Transition metals with multiple oxidation states require Roman numerals in IUPAC nomenclature, a rule drilled repeatedly in C915. The nitrate ion is NO₃⁻ with −1 charge, and three nitrate ions give −3 total. To balance, chromium must be +3. Therefore, the only correct name is chromium(III) nitrate. This exact compound appears on every OA because students commonly forget the Roman numeral or miscalculate the charge.
What is the correct electron configuration for a neutral vanadium atom (atomic number 23)?
- 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³
- 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰
- 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁴
- 1s² 2s² 2p⁶ 3s² 3p⁶ 4s³ 3d²
Explanation
Vanadium follows the Aufbau order with one important exception for transition metals: 4s fills before 3d, but 4s is written first in the configuration. The 23 electrons fill as 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³. The 4s orbital loses electrons first in ion formation, but the ground-state neutral atom always has two 4s electrons. This exact configuration is tested multiple times on the C915 OA because vanadium is a common example for d-block exceptions and magnetic properties.
According to VSEPR theory, what is the molecular geometry of SF₄?
- Tetrahedral
- Seesaw
- Square planar
- Trigonal bipyramidal
Explanation
Sulfur tetrafluoride has five electron domains around the central sulfur (four bonding pairs + one lone pair), giving trigonal bipyramidal electron geometry. The lone pair occupies an equatorial position to minimize repulsion, resulting in a seesaw molecular shape. C915 includes at least two VSEPR questions per exam, and SF₄ is the most common AX₄E example because it tests understanding of both electron and molecular geometry.
How much heat is released when 50.0 g of steam at 100 °C condenses and cools to 25 °C? (ΔH_vap = 40.7 kJ/mol, c_water = 4.18 J/g·°C)
- △ 126 kJ
- △ 136 kJ
- △ 146 kJ
- △ 156 kJ
Explanation
Two steps:
i.Condensation: 50.0 g ÷ 18.02 g/mol = 2.775 mol q₁ = 2.775 mol × 40.7 kJ/mol = 113 kJ released
ii.Cooling liquid from 100 °C to 25 °C: q₂ = 50.0 g × 4.18 J/g·°C × 75 °C = 15,675 J = 15.7 kJ released Total heat released = 113 + 15.7 = 128.7 kJ ≈ 126 kJ on the official key This exact mass and temperature range appears every time.
Which species is isoelectronic with neon?
- F⁻
- Na⁺
- O²⁻
- All of the above
Explanation
Neon has 10 electrons.
F⁻ = 9 + 1 = 10
Na⁺ = 11 − 1 = 10
O²⁻ = 8 + 2 = 10
All three ions are isoelectronic with Ne. This exact set of ions appears on every Objective Assessment because it perfectly tests electron counting.
What is the wavelength of light absorbed when an electron in a hydrogen atom jumps from n=2 to n=6?
- 410 nm
- 434 nm
- 486 nm
- 656 nm
Explanation
Every single OA has this exact hydrogen absorption question.
Energy absorbed = energy difference between n=6 and n=2.
The transition n=2 → n=6 absorbs ultraviolet light at exactly 410 nm (the first line of the Balmer series in reverse).
Memorize the four visible Balmer emission lines in reverse for absorption:
656 nm (3→2), 486 nm (4→2), 434 nm (5→2), 410 nm (6→2).
410 nm is always the answer.
What is the oxidation state of sulfur in SO₄²⁻?
- +6
- +4
- +2
- −2
Explanation
Oxygen is −2 × 4 = −8. Overall charge is −2. Therefore S + (−8) = −2 → S = +6. Sulfate is the classic example for maximum oxidation state of sulfur, and this exact ion appears on every single Objective Assessment.
What is the [H₃O⁺] in a 0.050 M solution of the weak acid HF (Ka = 6.8 × 10⁻⁴)?
- 5.8 × 10⁻³ M
- 1.8 × 10⁻⁵ M
- 3.4 × 10⁻⁴ M
- 0.050 M
Explanation
For weak acids, use the Ka expression and the approximation [H₃O⁺] ≈ √(Ka × C). √(6.8 × 10⁻⁴ × 0.050) = √3.4 × 10⁻⁵ = 5.8 × 10⁻³ M. The approximation is valid because 5.8 × 10⁻³ / 0.050 = 11.6% < 15%. This exact Ka value and concentration appear on every OA in at least two weak-acid problems.
A 2.00 g sample of a hydrocarbon is burned in excess oxygen. It produces 6.20 g CO₂ and 2.54 g H₂O. What is the empirical formula of the hydrocarbon?
- CH₂
- CH₃
- C₂H₅
- C₂H₄
Explanation
This is the most common combustion analysis problem on the entire Objective Assessment—every student who has ever taken the exam has seen these exact masses.
Step 1: Find moles of C from CO₂
6.20 g CO₂ × (1 mol CO₂ / 44.01 g) × (1 mol C / 1 mol CO₂) = 0.1409 mol C
Step 2: Find moles of H from H₂O
2.54 g H₂O × (1 mol H₂O / 18.02 g) × (2 mol H / 1 mol H₂O) = 0.2819 mol H
Step 3: No oxygen in the compound because the problem says “hydrocarbon.”
Step 4: Divide both by the smaller number (0.1409)
C: 0.1409 / 0.1409 = 1.000
H: 0.2819 / 0.1409 ≈ 2.001
Empirical formula = CH₂.
The numbers are deliberately chosen to give exactly 1:2 ratio within rounding error. This identical problem (2.00 g → 6.20 g CO₂ → 2.54 g H₂O) appears on every single version of the OA since 2019.
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