This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
Step 1: Calculate the concentration of acid after adding water for each row. The initial concentration of sulphuric acid is 2 M. The total volume of the solution after adding water is always 50 cm^3. We use the dilution formula M_1V_1 = M_2V_2, where M_1 = 2 M and V_2 = 50 cm^3. For Volume of 2M H₂SO₄ = 30 cm³: M_2 = (M_1V_1)/(V_2) = 2 M × 30 cm^350 cm^3 = (60)/(50) = 1.2 mol dm^-3 For Volume of 2M H₂SO₄ = 20 cm³: M_2 = (M_1V_1)/(V_2) = 2 M × 20 cm^350 cm^3 = (40)/(50) = 0.8 mol dm^-3 For Volume of 2M H₂SO₄ = 10 cm³: M_2 = (M_1V_1)/(V_2) = 2 M × 10 cm^350 cm^3 = (20)/(50) = 0.4 mol dm^-3 Step 2: Calculate (1)/(Time) × 1000 (s⁻¹) for each row. For Time taken = 10.0 s: (1)/(10.0 s) × 1000 = 100.0 s^-1 For Time taken = 15.0 s: (1)/(15.0 s) × 1000 = 66.66... ≈ 66.7 s^-1 For Time taken = 25.0 s: (1)/(25.0 s) × 1000 = 40.0 s^-1 Step 3: Complete the table with the calculated values. The completed table is: |c|c|c|c|c| Volume of & Volume of water & Concentration of acid & Time taken & (1)/(Time) × 1000 \\ 2M H_2SO_4 & added (cm^3) & after adding water (mol dm^-3) & (s) & (s^-1) \\ 50 & 0 & 2.0 & 6.0 & 166.7 \\ 40 & 10 & 1.6 & 7.5 & 133.3 \\ 30 & 20 & 1.2 & 10.0 & 100.0 \\ 20 & 30 & 0.8 & 15.0 & 66.7 \\ 10 & 40 & 0.4 & 25.0 & 40.0 \\ Send me the next one 📸