Step 1: Calculate the concentration of acid after adding water for each row. The initial concentration of sulphuric acid is $2 \text{ M}$. The total volume of the solution after adding water is always $50 \text{ cm}^3$. We use the dilution formula $M_1V_1 = M_2V_2$, where $M_1 = 2 \text{ M}$ and $V_2 = 50 \text{ cm}^3$. For Volume of 2M H₂SO₄ = 30 cm³: $$M_2 = \frac{M_1V_1}{V_2} = \frac{2 \text{ M} \times 30 \text{ cm}^3}{50 \text{ cm}^3} = \frac{60}{50} = 1.2 \text{ mol dm}^{-3}$$ For Volume of 2M H₂SO₄ = 20 cm³: $$M_2 = \frac{M_1V_1}{V_2} = \frac{2 \text{ M} \times 20 \text{ cm}^3}{50 \text{ cm}^3} = \frac{40}{50} = 0.8 \text{ mol dm}^{-3}$$ For Volume of 2M H₂SO₄ = 10 cm³: $$M_2 = \frac{M_1V_1}{V_2} = \frac{2 \text{ M} \times 10 \text{ cm}^3}{50 \text{ cm}^3} = \frac{20}{50} = 0.4 \text{ mol dm}^{-3}$$ Step 2: Calculate $\frac{1}{\text{Time}} \times 1000$ (s⁻¹) for each row. For Time taken = 10.0 s: $$\frac{1}{10.0 \text{ s}} \times 1000 = 100.0 \text{ s}^{-1}$$ For Time taken = 15.0 s: $$\frac{1}{15.0 \text{ s}} \times 1000 = 66.66... \approx 66.7 \text{ s}^{-1}$$ For Time taken = 25.0 s: $$\frac{1}{25.0 \text{ s}} \times 1000 = 40.0 \text{ s}^{-1}$$ Step 3: Complete the table with the calculated values. The completed table is: $$\boxed{ \begin{array}{|c|c|c|c|c|} \hline \text{Volume of} & \text{Volume of water} & \text{Concentration of acid} & \text{Time taken} & \frac{1}{\text{Time}} \times 1000 \\ 2\text{M H}_2\text{SO}_4 & \text{added (cm}^3) & \text{after adding water (mol dm}^{-3}) & \text{(s)} & \text{(s}^{-1}) \\ \hline 50 & 0 & 2.0 & 6.0 & 166.7 \\ 40 & 10 & 1.6 & 7.5 & 133.3 \\ 30 & 20 & \textbf{1.2} & 10.0 & \textbf{100.0} \\ 20 & 30 & \textbf{0.8} & 15.0 & \textbf{66.7} \\ 10 & 40 & \textbf{0.4} & 25.0 & \textbf{40.0} \\ \hline \end{array} }$$ Send me the next one 📸