Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
10 g of Mg and 80 g of O2 react according to 2Mg + O2 → 2MgO. What mass m of MgO can form?
In the reaction 2Na + Cl2 → 2NaCl, 16 g of Na reacts completely (the other reactants are in excess) and 30.5 g of NaCl is collected. Find the percent yield Y.
You need 500 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?
200 mL of a 0.5 mol/L solution is mixed with 50 mL of a 0.75 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
What volume V does 90 g of CO₂ occupy at 700 kPa and 5 °C? Treat it as an ideal gas.
A fixed amount of He goes from 350 kPa, 20 L, 25 °C to 500 kPa and 210 °C. Find the new volume V2.
A 50 g block of copper (c = 0.385 J/(g·°C)) at 90 °C is dropped into 300 g of water (c = 4.18 J/(g·°C)) at 16 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find ΔH° for the reaction 2C4H10 + 13O₂ → 8CO₂ + 10H₂O(l).
0.25 mol/L of H₂ and 0.25 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 64. Find the equilibrium concentration [HI].
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.02 at the reaction temperature. A container starts with 0.8 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
Find the pH pH of a 0.05 mol/L solution of acetic acid (Ka = 1.8 × 10⁻⁵).
A 1.00 L buffer contains 0.1 mol hypochlorous acid and 0.2 mol of its conjugate base (Ka = 3 × 10⁻⁸). 0.05 mol of NaOH is added. Find the new pH pH.
A second-order reaction (−rA = k·CA²) has k = 0.8 L/(mol·min) and an initial concentration of 0.5 mol/L. Find its first half-life t½.
The rate constant of a reaction is 0.1 min⁻¹ at 40 °C and 0.373 min⁻¹ at 55 °C. Find the activation energy Ea.
Answer: m = 16.58 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 = 74.99 %
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 = 250 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 = 0.55 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 = 6.756 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 22.69 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Tf = 17.12 °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: ΔH° = −5755 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.
Negative: the reaction gives off heat (exothermic).
Answer: [HI] = 0.4 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: [Cl2] = 0.1169 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 = 3.027
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 = 8.222
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: t½ = 2.5 min
Integrated rate laws: zero order: C = C₀ − kt (straight line, t½ = C₀/(2k)); first order: ln(C₀/C) = kt (t½ = ln 2/k); second order: 1/C − 1/C₀ = kt (t½ = 1/(kC₀)). Conversion X means C = C₀(1 − X).
Each later half-life is twice as long, because C₀ has halved.
Answer: Ea = 74.98 kJ/mol
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).
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