Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
8 g of Mg and 80 g of O2 react according to 2Mg + O2 → 2MgO. What mass m of MgO can form?
In the reaction 2H2 + O2 → 2H2O, 30 g of H2 reacts completely (the other reactants are in excess) and 201.1 g of H2O is collected. Find the percent yield Y.
You need 250 mL of 0.25 mol/L solution and have a 0.5 mol/L stock solution. What volume V1 of stock solution do you need?
300 mL of a 0.25 mol/L solution is mixed with 200 mL of a 1.2 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
What volume V does 490 g of Ar occupy at 350 kPa and 90 °C? Treat it as an ideal gas.
A fixed amount of CO₂ goes from 300 kPa, 2 L, 15 °C to 100 kPa and 240 °C. Find the new volume V2.
A 250 g block of lead (c = 0.129 J/(g·°C)) at 150 °C is dropped into 200 g of water (c = 4.18 J/(g·°C)) at 19 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find the heat q released when 25 g of CH3OH (M = 32.04 g/mol) burns completely: 2CH3OH + 3O₂ → 2CO₂ + 4H₂O(l). Give q as a positive number.
0.1 mol/L of H₂ and 0.1 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 9. Find the equilibrium concentration [HI].
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.5 at the reaction temperature. A container starts with 0.4 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
Find the pH pH of a 0.25 mol/L solution of formic acid (Ka = 0.00018).
A buffer contains 0.1 mol/L benzoic acid and 0.5 mol/L of its sodium salt. Ka = 6.3 × 10⁻⁵. Find the pH pH of the buffer.
A reaction with rate law −rA = k·CA² has k = 0.25 L/(mol·min) and starts at 1.5 mol/L. Find the concentration CA after 2 min.
A reaction has a rate constant of 0.02 min⁻¹ at 35 °C and an activation energy of 55 kJ/mol. Find the rate constant k2 at 55 °C.
Answer: m = 13.26 g
Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.
Answer: Y = 75 %
Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.
Answer: V1 = 125 mL
Molarity: concentration c = n/V in mol/L, with n = m/M. Diluting adds solvent but not solute, so the number of moles stays the same: c₁·V₁ = c₂·V₂. When two solutions of the same solute are mixed, the moles add up and so do the volumes.
Answer: c = 0.63 mol/L
Molarity: concentration c = n/V in mol/L, with n = m/M. Diluting adds solvent but not solute, so the number of moles stays the same: c₁·V₁ = c₂·V₂. When two solutions of the same solute are mixed, the moles add up and so do the volumes.
Answer: V = 105.8 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 10.69 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Tf = 23.87 °C
Heat and enthalpy: heating a substance takes q = m·c·ΔT. In a calorimeter the heat lost by the hot object equals the heat gained by the cold one. The enthalpy change of a reaction follows from enthalpies of formation: ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants), each multiplied by its coefficient.
Answer: q = 566.8 kJ
Heat and enthalpy: heating a substance takes q = m·c·ΔT. In a calorimeter the heat lost by the hot object equals the heat gained by the cold one. The enthalpy change of a reaction follows from enthalpies of formation: ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants), each multiplied by its coefficient.
Answer: [HI] = 0.12 mol/L
Chemical equilibrium: for aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]c[D]d/([A]a[B]b), using concentrations at equilibrium. An ICE table (Initial, Change, Equilibrium) links the starting amounts to the equilibrium amounts through one unknown x, scaled by the coefficients.
Answer: [PCl5] = 0.1377 mol/L
Chemical equilibrium: for aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]c[D]d/([A]a[B]b), using concentrations at equilibrium. An ICE table (Initial, Change, Equilibrium) links the starting amounts to the equilibrium amounts through one unknown x, scaled by the coefficients.
Answer: pH = 2.179
pH: pH = −log₁₀[H₃O⁺] and pOH = −log₁₀[OH⁻], with pH + pOH = 14 at 25 °C. Strong acids and bases ionise completely. A weak acid only partly: Ka = x²/(c − x), where x = [H₃O⁺]. A buffer of a weak acid and its conjugate base follows Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]).
HA ⇌ H⁺ + A⁻: x mol/L of acid ionises.
Answer: pH = 4.9
pH: pH = −log₁₀[H₃O⁺] and pOH = −log₁₀[OH⁻], with pH + pOH = 14 at 25 °C. Strong acids and bases ionise completely. A weak acid only partly: Ka = x²/(c − x), where x = [H₃O⁺]. A buffer of a weak acid and its conjugate base follows Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]).
Answer: CA = 0.8571 mol/L
Reaction rates: for a first-order reaction −dC/dt = kC, so C = C₀·e−kt and the half-life ln 2/k does not depend on concentration. For a second-order reaction 1/C = 1/C₀ + kt. The rate constant rises with temperature following Arrhenius: k = A·e−Ea/(RT).
Answer: k2 = 0.07401 min⁻¹
Reaction rates: for a first-order reaction −dC/dt = kC, so C = C₀·e−kt and the half-life ln 2/k does not depend on concentration. For a second-order reaction 1/C = 1/C₀ + kt. The rate constant rises with temperature following Arrhenius: k = A·e−Ea/(RT).
ExamProblems · examproblems.com/courses/general-chemistry