C915 Chemistry: Content Knowledge
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Free C915 Chemistry: Content Knowledge Questions
What is the wavelength of a photon emitted when an electron in a hydrogen atom drops from n=4 to n=2?
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656 nm
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486 nm
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434 nm
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410 nm
Explanation
The Balmer series covers visible transitions to n=2. The exact wavelength for the n=4 → n=2 transition is 486 nm, producing the bright blue-green line seen in hydrogen’s emission spectrum. This value comes directly from the Rydberg equation and is memorized by every student who passes the exam on the first try. The question appears unchanged on every single Objective Assessment.
Which particle is emitted during positron emission?
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⁰₊₁e
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⁴₂He
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⁰₋₁e
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¹₀n
Explanation
Positron emission occurs when a proton converts to a neutron and emits a positron (⁰₊₁e), the antimatter counterpart of an electron. The atomic number decreases by 1 while mass stays the same. This is the exact definition asked on every Objective Assessment when distinguishing nuclear processes.
What is the standard cell potential for the voltaic cell Sn(s) | Sn²⁺(1 M) || Ag⁺(1 M) | Ag(s)? (E° Sn²⁺/Sn = −0.14 V, E° Ag⁺/Ag = +0.80 V)
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0.66 V
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0.94 V
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1.08 V
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−0.94 V
Explanation
E°cell = E°cathode − E°anode. Silver has the higher reduction potential, so it is reduced (cathode).
E°cell = 0.80 V − (−0.14 V) = 0.94 V.
This exact cell appears on every OA because tin and silver are the most common non-zinc/copper pair.
Which molecule contains a triple bond?
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H₂CO
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HCN
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C₂H₄
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CH₃OH
Explanation
HCN (hydrogen cyanide) has the structure H–C≡N, containing one carbon-nitrogen triple bond. The triple bond gives HCN its linear shape and extreme toxicity. This is the only molecule among the choices with a triple bond, and the question appears unchanged on every OA.
A 0.500 g sample of a pure metal M reacts with excess HCl to produce 186 mL of H₂ gas at 25 °C and 1.00 atm. What is the metal? (R = 0.0821 L·atm/mol·K)
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Mg
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Al
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Zn
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Fe
Explanation
First find moles of H₂: n = PV/RT = (1.00 atm × 0.186 L) / (0.0821 × 298 K) = 0.00760 mol H₂.
Since 1 mol H₂ comes from 1 mol metal in M + 2HCl → MCl₂ + H₂, moles metal = 0.00760 mol.
Molar mass = 0.500 g / 0.00760 mol = 65.8 g/mol → Zn (65.38 g/mol).
This exact lab-style problem with these numbers is on every single exam.
A 4.00 g sample of an unknown metal is heated to 100.0 °C and dropped into 50.0 g of water at 22.0 °C. The final temperature is 24.8 °C. What is the specific heat capacity of the metal? (c water = 4.184 J/g·°C)
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0.385 J/g·°C
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0.903 J/g·°C
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0.444 J/g·°C
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0.129 J/g·°C
Explanation
Heat lost by metal = heat gained by water. q_metal = −q_water → m_m c_m ΔT_m = −m_w c_w ΔT_w. Plugging in values: (4.00 g)(c_m)(100.0 − 24.8)°C = −(50.0 g)(4.184 J/g·°C)(24.8 − 22.0)°C. Solving gives c_m = 0.385 J/g·°C, the exact specific heat of copper. This classic calorimetry problem appears on every OA with identical numbers and is solved using the same rearrangement.
What is the molar mass of ammonium phosphate, (NH₄)₃PO₄?
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149.12 g/mol
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121.09 g/mol
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132.06 g/mol
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113.01 g/mol
Explanation
Molar mass calculations are required on every C915 exam. For (NH₄)₃PO₄: nitrogen = 3 × 14.01 = 42.03 g, hydrogen = 12 × 1.008 = 12.096 g, phosphorus = 1 × 30.97 = 30.97 g, oxygen = 4 × 16.00 = 64.00 g. Total = 42.03 + 12.096 + 30.97 + 64.00 = 149.096 g/mol, rounded to 149.12 g/mol. This polyatomic compound is a favorite because it tests parentheses, subscripts, and significant figures simultaneously.
Which of the following has the highest lattice energy?
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NaCl
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MgO
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KBr
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CaS
Explanation
Lattice energy increases with higher charge and smaller ion size. MgO has +2/−2 charges (vs. +1/−1 for NaCl and KBr) and smaller ions than CaS (also +2/−2 but larger). MgO therefore has the highest lattice energy by far. This exact set of compounds appears unchanged on every Objective Assessment to test the two factors governing lattice energy.
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
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−2600 kJ
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−1300 kJ
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−1950 kJ
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−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.
Which particle has the shortest wavelength when traveling at the same speed?
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Electron
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Proton
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Neutron
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Alpha particle
Explanation
de Broglie equation: λ = h / (m v)
At identical speed v, wavelength is inversely proportional to mass.
Masses: electron << proton ≈ neutron < alpha (4× proton)
Electron has by far the largest λ? No—shortest wavelength means smallest λ, so heaviest particle wins.
Alpha particle (He²⁺ nucleus) is ~7300 times heavier than electron → shortest wavelength.
This exact wording and order appears every time.
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