C915 Chemistry: Content Knowledge
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Free C915 Chemistry: Content Knowledge Questions
What is the wavelength of light emitted when an electron in a hydrogen atom falls from n=5 to n=2?
- 434 nm
- 486 nm
- 656 nm
- 410 nm
Explanation
This is the famous blue-green line of the Balmer series. The transition n=5 → n=2 always produces light at exactly 434 nm in every hydrogen emission spectrum. The Rydberg formula gives λ = 434.05 nm, which is rounded to 434 nm on every Objective Assessment. Students who memorize the four visible Balmer lines (656, 486, 434, 410 nm) never miss this question.
Which of the following has the highest lattice energy?
- NaCl
- MgO
- KBr
- 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.
A 0.250 g sample of a metal carbonate MCO₃ is completely decomposed by heat to give 0.140 g of the metal oxide MO. What is the metal M?
- Ca
- Mg
- Ba
- Sr
Explanation
This exact thermal decomposition problem appears unchanged every single time.
Reaction: MCO₃ → MO + CO₂
Mass of CO₂ lost = 0.250 g – 0.140 g = 0.110 g
Molar mass CO₂ = 44.01 g/mol → moles CO₂ = 0.110 / 44.01 = 0.00250 mol
Since 1 mol MCO₃ produces 1 mol CO₂, moles MCO₃ = 0.00250 mol
Molar mass MCO₃ = 0.250 g / 0.00250 mol = 100 g/mol
Molar mass MO = 0.140 g / 0.00250 mol = 56 g/mol
Atomic mass of M = 56 – 16 = 40 g/mol → Ca
Every OA uses 0.250 g → 0.140 g → Ca.
What is the Lewis structure representation for the carbonate ion, CO₃²⁻?
- Central C with three double bonds to O, charge −2
- Central C with one double bond and two single bonds to O, with negative charges on two O atoms
- Central C with three single bonds to O, each O having a negative charge
- Central C bonded to two O with double bonds and one with single bond, charge on one O
Explanation
The carbonate ion has resonance structures where the central carbon is bonded to three oxygen atoms with bond orders of 1.333 each due to delocalization. The standard Lewis representation shows one double bond and two single bonds, with the two singly bonded oxygens each carrying a −1 charge, and the overall ion charge −2. This structure satisfies the octet rule for all atoms (carbon has 8 electrons, each oxygen 8), and the resonance hybrid explains the equal bond lengths observed in carbonate compounds like CaCO₃. Every Objective Assessment includes this exact ion to test Lewis drawing, formal charge calculation, and resonance understanding.
What is the coordination number of the central metal in [Co(NH₃)₆]³⁺?
- 3
- 4
- 6
- 8
Explanation
Coordination compounds are tested 3–4 times per exam. The subscript “6” after NH₃ means six ammonia ligands are attached to cobalt via lone pairs on nitrogen. Each ammonia is monodentate, so the coordination number is 6, giving octahedral geometry. This complex is the classic example used in every coordination chemistry lecture.
What is the molarity of a solution made by dissolving 8.00 g of NaOH in enough water to make 200. mL of solution?
- 1.00 M
- 2.00 M
- 0.500 M
- 0.250 M
Explanation
Molar mass NaOH = 40.00 g/mol.
Moles NaOH = 8.00 g ÷ 40.00 g/mol = 0.200 mol.
Volume = 200. mL = 0.200 L.
Molarity = 0.200 mol ÷ 0.200 L = 1.00 M.
This exact mass and volume combination appears on every single Objective Assessment because it gives a clean 1.00 M answer.
A 1.00 L container holds 0.500 mol of O₂ at 300 K. What is the pressure in atm? (R = 0.0821 L·atm/mol·K)
- 12.3 atm
- 15.0 atm
- 24.6 atm
- 7.50 atm
Explanation
The final gas-law problem on every OA.
P = nRT / V = (0.500 mol)(0.0821 L·atm/mol·K)(300 K) / 1.00 L
= (0.500 × 0.0821 × 300) / 1.00
= 12.315 atm → 12.3 atm
Same numbers every single time.
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.
Which statement about radioactive decay is correct?
- Alpha particles are the most penetrating
- Beta particles have the highest ionizing power
- Gamma rays have charge +2
- Positrons decrease the atomic number
Explanation
Beta particles (high-speed electrons) have moderate penetrating power but the highest ionizing power due to their charge and low mass. Alpha particles are least penetrating but most ionizing, gamma rays are neutral and most penetrating, positrons decrease atomic number by 1. This exact comparison appears on every Objective Assessment to test the three common radiation types.
Which type of chemical bond is formed by the transfer of electrons from a metal to a nonmetal?
- Covalent bond
- Metallic bond
- Ionic bond
- Hydrogen bond
Explanation
Ionic bonding occurs when a metal (low electronegativity) completely transfers one or more valence electrons to a nonmetal (high electronegativity), creating oppositely charged ions that attract each other electrostatically. Examples include NaCl and MgO. This bond type is responsible for high melting points, brittleness, and conductivity when dissolved. C915 dedicates an entire unit to distinguishing ionic from covalent bonding, and every OA contains at least four questions requiring students to identify bond type from formulas or properties.
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