The enthalpy changes H1, H2, and H3 cannot be identified without a reaction diagram or further context.

(a) (i) The enthalpy changes H_1, H_2, and H_3 cannot be identified without a reaction diagram or further context. (a) (ii) Assuming a Hess's Law cycle where H_3 is the sum of H_1 and H_2: Step 1: State the assumed relationship. H_3 = H_1 + H_2 Step 2: Substitute the given values. H_3 = 16 kJmol^-1 + (-770 kJmol^-1) Step 3: Calculate the result. H_3 = 16 - 770 kJmol^-1 H_3 = -754 kJmol^-1 -754 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 (H_2SO_4) is a strong acid and is assumed to fully dissociate both protons in dilute solutions for pH calculations. H_2SO_4 (aq) 2H^+(aq) + SO_4^2-(aq) Step 2: Determine the concentration of hydrogen ions. Since 1 mole of H_2SO_4 produces 2 moles of H^+ ions: [H^+] = 2 × [H_2SO_4] [H^+] = 2 × 0.1 M [H^+] = 0.2 M Step 3: Calculate the pH. pH = -_10[H^+] pH = -_10(0.2) pH ≈ 0.69897 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 (Kc) for the given reaction. The reaction is: 2NO(g) + O_2(g) 2NO_2(g) Kc = [NO_2]^2[NO]^2[O_2] Step 2: Substitute the given equilibrium concentrations. [NO] = 0.05 moldm^-3 [O_2] = 0.13 moldm^-3 [NO_2] = 15.50 moldm^-3 Kc = ((15.50)^2)/((0.05)^2(0.13)) Step 3: Calculate the value of Kc. Kc = (240.25)/((0.0025)(0.13)) Kc = (240.25)/(0.000325) Kc = 739230.769 7.39 × 10^5 dm^3mol^-1 (e) (ii) Step 1: Determine the number of moles of gaseous reactants and products. Reactants: 2NO(g) + O_2(g) 2 + 1 = 3 moles of gas. Products: 2NO_2(g) 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 (NO and O_2). What's next?