General Chemistry — Practice Questions flashcards
6 of 50 questions free to study here · 24 topics · the whole deck is built into Willys
Multiple-choice practice questions, one best answer each, on stoichiometry, gases, thermochemistry, bonding, kinetics, equilibrium, pH and buffers, solubility and electrochemistry.
Practice deck · part of General Chemistry I/II · written against: Standard two-semester sequence. Updated 2026-10-10.
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The 6 sample questions as a list
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Written as (n, l, , ), which set of quantum numbers could describe an electron in a 4d orbital?
A) (4, 1, 0, −½)
B) (4, 2, −1, +½)
C) (3, 2, +2, +½)
D) (4, 3, −1, +½)B) (4, 2, −1, +½)Atomic Structure & Quantum Numbers
A d subshell has l = 2, and n = 4 puts it in the fourth shell; = −1 lies within −2 to +2. l = 3 describes a 4f orbital, and n = 3 with l = 2 is a 3d orbital. -
What is the molecular geometry of BrF₅?
A) Octahedral
B) Seesaw
C) Trigonal bipyramidal
D) Square pyramidalD) Square pyramidalVSEPR · Hybridization & Polarity
Bromine has five bonding pairs and one lone pair — six electron domains, an octahedral arrangement. With one position taken by the lone pair, the atoms form a square pyramid; octahedral would need six bonded atoms. -
What volume does 8.00 g of O₂ (32.0 g/mol) occupy at 27 °C and 1.00 atm? (R = 0.0821 L·atm/(mol·K))
A) 6.16 L
B) 12.3 L
C) 5.60 L
D) 0.554 LA) 6.16 LGas Laws
n = 0.250 mol and T = 300 K, so V = nRT/P = 0.250 × 0.0821 × 300 / 1.00 = 6.16 L. Using 27 instead of 300 K gives 0.554 L; 5.60 L applies the 22.4 L/mol value, which holds only at 0 °C. -
1.00 mol of N₂O₄ is placed in an empty 1.00 L flask and N₂O₄(g) ⇌ 2 NO₂(g) is allowed to reach equilibrium. If [NO₂] = 0.40 M at equilibrium, what is ?
A) 0.27
B) 0.50
C) 0.20
D) 0.16C) 0.20Equilibrium Constants & Q
Forming 0.40 M NO₂ uses 0.20 M N₂O₄, leaving 0.80 M, so = (0.40)² / 0.80 = 0.20. Subtracting the full 0.40 from N₂O₄ gives 0.27, and not squaring [NO₂] gives 0.50. -
A 1.0 L buffer contains 0.20 mol of acetic acid and 0.20 mol of sodium acetate ( = 4.74). What is the pH after 0.050 mol of HCl is added, with no volume change?
A) 4.74
B) 1.30
C) 4.52
D) 4.96C) 4.52Buffers & Titrations
HCl converts acetate to acetic acid: acid 0.25 mol, base 0.15 mol, so pH = 4.74 + log(0.15/0.25) = 4.52. Adding the 0.050 mol to the base instead gives 4.96 — added acid must lower the pH. -
In Cr₂O₇²⁻ + 6 Fe²⁺ + 14 H⁺ → 2 Cr³⁺ + 6 Fe³⁺ + 7 H₂O, how does the oxidation number of chromium change, and which species is the oxidizing agent?
A) +3 to +6; Cr₂O₇²⁻ is the reducing agent
B) +6 to +3; Fe²⁺ is the oxidizing agent
C) +6 to +3; Cr₂O₇²⁻ is the oxidizing agent
D) +7 to +3; Cr₂O₇²⁻ is the oxidizing agentC) +6 to +3; Cr₂O₇²⁻ is the oxidizing agentOxidation Numbers & Redox Balancing
In Cr₂O₇²⁻, 2x − 14 = −2 gives Cr = +6, which falls to +3, so dichromate is reduced and is the oxidizing agent. Fe²⁺ is oxidized to Fe³⁺ and is therefore the reducing agent.
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About these cards
Review status: Second-reader reviewed, with corrections applied.
Checked against:
- OpenStax Chemistry 2e, consulted as a reference; no text copied.
- Checked against Zumdahl, Zumdahl & DeCoste, Chemistry, 10th ed. (Cengage, 2018) (reference only; no text copied).
- Original questions written for this deck; no standardized-exam, textbook or question-bank items were copied. Numerical values are hypothetical or rounded, and most questions state the constants and molar masses they use.
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