(a) (i) The enthalpy changes $\Delta H_1$, $\Delta H_2$, and $\Delta H_3$ cannot be identified without a reaction diagram or further context. (a) (ii) Assuming a Hess's Law cycle where $\Delta H_3$ is the sum of $\Delta H_1$ and $\Delta H_2$: Step 1: State the assumed relationship. $$\Delta H_3 = \Delta H_1 + \Delta H_2$$ Step 2: Substitute the given values. $$\Delta H_3 = 16 \text{ kJmol}^{-1} + (-770 \text{ kJmol}^{-1})$$ Step 3: Calculate the result. $$\Delta H_3 = 16 - 770 \text{ kJmol}^{-1}$$ $$\Delta H_3 = -754 \text{ kJmol}^{-1}$$ $$\boxed{-754 \text{ kJmol}^{-1}}$$ (d) (i) A strong acid* completely dissociates in water to produce hydrogen ions. A weak acid* only partially dissociates in water, establishing an equilibrium between the undissociated acid and its ions. (d) (ii) Step 1: Write the dissociation equation for sulphuric acid. Sulphuric acid ($\text{H}_2\text{SO}_4$) is a strong acid and is assumed to fully dissociate both protons in dilute solutions for pH calculations. $$\text{H}_2\text{SO}_4 \text{(aq)} \to 2\text{H}^+\text{(aq)} + \text{SO}_4^{2-}\text{(aq)}$$ Step 2: Determine the concentration of hydrogen ions. Since 1 mole of $\text{H}_2\text{SO}_4$ produces 2 moles of $\text{H}^+$ ions: $$[\text{H}^+] = 2 \times [\text{H}_2\text{SO}_4]$$ $$[\text{H}^+] = 2 \times 0.1 \text{ M}$$ $$[\text{H}^+] = 0.2 \text{ M}$$ Step 3: Calculate the pH. $$\text{pH} = -\log_{10}[\text{H}^+]$$ $$\text{pH} = -\log_{10}(0.2)$$ $$\text{pH} \approx 0.69897$$ $$\boxed{0.70}$$ (d) (iii) • Maintaining a stable pH in biological systems, such as blood. • Controlling the pH of solutions in chemical reactions or industrial processes. (e) (i) Step 1: Write the equilibrium constant expression ($\text{Kc}$) for the given reaction. The reaction is: $2\text{NO(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}$ $$\text{Kc} = \frac{[\text{NO}_2]^2}{[\text{NO}]^2[\text{O}_2]}$$ Step 2: Substitute the given equilibrium concentrations. $[\text{NO}] = 0.05 \text{ moldm}^{-3}$ $[\text{O}_2] = 0.13 \text{ moldm}^{-3}$ $[\text{NO}_2] = 15.50 \text{ moldm}^{-3}$ $$\text{Kc} = \frac{(15.50)^2}{(0.05)^2(0.13)}$$ Step 3: Calculate the value of $\text{Kc}$. $$\text{Kc} = \frac{240.25}{(0.0025)(0.13)}$$ $$\text{Kc} = \frac{240.25}{0.000325}$$ $$\text{Kc} = 739230.769$$ $$\boxed{7.39 \times 10^5 \text{ dm}^3\text{mol}^{-1}}$$ (e) (ii) Step 1: Determine the number of moles of gaseous reactants and products. Reactants: $2\text{NO(g)} + \text{O}_2\text{(g)} \implies 2 + 1 = 3$ moles of gas. Products: $2\text{NO}_2\text{(g)} \implies 2$ moles of gas. Step 2: Apply Le Chatelier's principle. A decrease in pressure will shift the equilibrium to the side with a greater number of moles of gas to counteract the change. Since the reactant side has 3 moles of gas and the product side has 2 moles of gas, the equilibrium will shift towards the reactants. The equilibrium position will shift to the left, favoring the formation of reactants ($\text{NO}$ and $\text{O}_2$). What's next?