This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
My calculations for Exercise 5 are arithmetically correct based on the values provided in the problem statement for Exercise 5. For Exercise 5, the problem explicitly states: "The average volume of A used was 22.50 cm^3." I used this value for V_A in my calculations. While the raw data in Exercise 4, when averaged (even with the identified errors in concordancy), might lead to a slightly different average titre (e.g., 22.55 cm^3 if you average 22.75 cm^3, 22.50 cm^3, and 22.40 cm^3), for Exercise 5, we must use the specific value given in its problem statement, which is 22.50 cm^3. Therefore, my calculations for Exercise 5 are as follows: a) (i) Concentration of A in mol/dm^3 Molar mass of NaHSO_4 = 23 + 1 + 32 + (4 × 16) = 120 g/mol C_A = 10.6 g/dm^3120 g/mol = 0.08833 mol/dm^3 The concentration of A is 0.0883 mol/dm^3. a) (ii) Concentration of B in mol/dm^3 The balanced equation is NaHSO_4(aq) + NaOH(aq) Na_2SO_4(aq) + H_2O(l), so the mole ratio n_A:n_B is 1:1. Using C_A = 0.08833 mol/dm^3, V_A = 22.50 cm^3, and V_B = 25.0 cm^3: C_B = C_AV_An_BV_Bn_A = (0.08833 mol/dm^3) × (22.50 cm^3) × 1(25.0 cm^3) × 1 C_B = 0.079497 mol/dm^3 The concentration of B is 0.0795 mol/dm^3. 3 done, 2 left today. You're making progress.