Matching total CO2 and initial [H+] determines line shape Steps: - Calculate for P: 10.0 cm³ = 0.010 dm³ of 0.100 mol dm⁻³ H₂SO₄ gives 0.001 mol H₂SO₄, producing 0.001 mol CO₂; initial [H⁺] = 0.200 mol dm⁻³. - Reaction stoichiometry: MgCO₃ + H₂SO₄ → MgSO₄ + CO₂ (1:1) or MgCO₃ + 2HCl → MgCl₂ + CO₂ (2:1 for HCl). - For each choice, compute moles of acid, moles CO₂ (limited by acid), and initial [H⁺] assuming reaction volume ≈ acid volume. - Line Q likely shows 10× higher max volume than P with slower initial rate due to lower [H⁺]. Why D is correct: - 200 cm³ = 0.200 dm³ of 0.100 mol dm⁻³ HCl gives 0.020 mol HCl (0.010 mol CO₂, 10× P's); [H⁺] = 0.100 mol dm⁻³ (half P's), slowing rate per rate law dependence on [H⁺]. Why the others are wrong: - A: 0.050 mol CO₂ (50× P's) with [H⁺] = 1.00 mol dm⁻³ (faster rate, higher max). - B: 0.100 mol CO₂ (100× P's) with [H⁺] = 1.00 mol dm⁻³ (faster rate, much higher max). …