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.
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Here are the solutions to the questions: 3(j)(i) A buffer solution is a solution that resists changes in pH when small amounts of an acid or a base are added. It typically consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. 3(j)(ii) The pairs of chemicals that would act as buffer solutions are: • H_3PO_4 / NaH_2PO_4 • NH_3 / NH_4Cl 3(b) Step 1: Determine the moles of H_2SO_4 produced from the given moles of O_2. The balanced chemical equation is: 2SO_2(g) + O_2(g) + H_2O_(l) 2H_2SO_4(l) From the stoichiometry, 1 mole of O_2 reacts to produce 2 moles of H_2SO_4. Given 1.5 moles of O_2: Moles of H_2SO_4 = 1.5 mol O_2 × 2 mol H_2SO_41 mol O_2 = 3.0 mol H_2SO_4 Step 2: Calculate the molar mass of H_2SO_4. Given atomic masses: H = 1.0, O = 16.0, S = 32.0 Molar mass of H_2SO_4 = (2 × 1.0) + (1 × 32.0) + (4 × 16.0) = 2.0 + 32.0 + 64.0 = 98.0 g/mol Step 3: Calculate the mass of H_2SO_4 produced. Mass of H_2SO_4 = Moles × Molar mass = 3.0 mol × 98.0 g/mol = 294.0 g The mass of H_2SO_4 that can be prepared is 294.0 g. 3(c)(i) • (α) less than 7: NH_3(aq) with H_2SO_4(aq) • (β) equal to 7: KOH_(aq) with HCl_(aq) • (γ) greater than 7: NaOH_(aq) with H_2CO_3(aq) 3(c)(ii) • For (α) less than 7 (NH_3 with H_2SO_4): This is a titration of a weak base with a strong acid. At the equivalence point, the resulting salt (ammonium sulfate) contains the ammonium ion (NH_4^+), which is the conjugate acid of a weak base. The NH_4^+ ion hydrolyzes in water to produce H^+ ions, making the solution acidic (pH < 7). • For (β) equal to 7 (KOH with HCl): This is a titration of a strong base with a strong acid. At the equivalence point, the resulting salt (potassium chloride) is formed from a strong acid and a strong base. Neither the K^+ nor the Cl^- ions hydrolyze significantly in water, so the solution remains neutral (pH = 7). • For (γ) greater than 7 (NaOH with H_2CO_3): This is a titration of a strong base with a weak acid. At the equivalence point, the resulting salt (sodium carbonate or bicarbonate) contains the carbonate or bicarbonate ion, which is the conjugate base of a weak acid. These ions hydrolyze in water to produce OH^- ions, making the solution basic (pH > 7). 3(d)(i) The factors that determine whether a given ionic solid would dissolve in water include: • Lattice energy: The energy required to break the ionic bonds in the solid. Lower lattice energy generally favors dissolution. • Hydration energy: The energy released when ions are surrounded by water molecules. Higher hydration energy generally favors dissolution. • Temperature: The solubility of most ionic solids increases with increasing temperature. 3(d)(ii) • (α) Na_2CO_3(aq) + ZnCl_2(aq) 2NaCl_(aq) + ZnCO_3(s): This is a precipitation reaction (or double displacement reaction) because two aqueous solutions react to form an insoluble solid (precipitate). • (β) Zn_(s) + CuSO_4(aq) ZnSO_4(aq) + Cu_(s): This is a single displacement reaction (or redox reaction) because a more reactive metal (zinc) displaces a less reactive metal (copper) from its compound. • (γ) CH_4(g) + 2O_2(g) CO_2(g) + 2H_2O_(l): This is a combustion reaction because a hydrocarbon (methane) reacts rapidly with oxygen to produce carbon dioxide and water, releasing energy. Send me the next one 📸