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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let's get these questions solved. 1. A: i) A standard solution? A standard solution is a solution where the concentration is precisely known. It's often prepared using a primary standard, a substance of high purity and stability. 1. A: ii) A disproportionation reaction A disproportionation reaction is a redox reaction where an element in a single oxidation state is simultaneously oxidized and reduced, forming products with both higher and lower oxidation states. 1. B: Volumetric Analysis Problem Given data: K=39, I=127, O=16, H=1. Unbalanced ionic equation: IO_3^-(aq) + I^-(aq) + H^+(aq) → I_2(aq) + H_2O(l) Step 1: Determine the oxidation states of iodine in the reaction. In IO_3^-, oxygen is -2. Let the oxidation state of I be x. x + 3(-2) = -1 → x - 6 = -1 → x = +5. In I^-, the oxidation state of I is -1. In I_2, the oxidation state of I is 0. Step 2: Identify the species being oxidized and reduced. Iodine in I^- (oxidation state -1) is oxidized to I_2 (oxidation state 0). Iodine in IO_3^- (oxidation state +5) is reduced to I_2 (oxidation state 0). This is a disproportionation reaction of iodine species. Step 3: Balance the half-reactions. Oxidation: I^- → I_2 Balance I: 2I^- → I_2 Balance charge: 2I^- → I_2 + 2e^- Reduction: IO_3^- → I_2 Balance I: 2IO_3^- → I_2 Balance O with H_2O: 2IO_3^- → I_2 + 6H_2O Balance H with H^+: 2IO_3^- + 12H^+ → I_2 + 6H_2O Balance charge: 2IO_3^- + 12H^+ + 10e^- → I_2 + 6H_2O Step 4: Equalize the number of electrons transferred. Multiply the oxidation half-reaction by 5. 5 × (2I^- → I_2 + 2e^-) 10I^- → 5I_2 + 10e^- Step 5: Add the balanced half-reactions and cancel electrons. 10I^- → 5I_2 + 10e^- 2IO_3^- + 12H^+ + 10e^- → I_2 + 6H_2O ------------------------------------------------------------------ 10I^- + 2IO_3^- + 12H^+ → 6I_2 + 6H_2O Step 6: Simplify the equation by dividing by 2. 5I^- + IO_3^- + 6H^+ → 3I_2 + 3H_2O This is the balanced ionic equation. Step 7: Calculate the molar mass of KI. Molar mass of KI = 39 + 127 = 166 g/mol. Step 8: Calculate the moles of KI used. Moles of KI = massmolar mass = 8.3 g166 g/mol = 0.05 mol. Since KI dissociates into K^+ and I^-, moles of I^- = 0.05 mol. Step 9: Use the stoichiometry of the balanced equation to find moles of IO_3^-. From the balanced equation: 5I^- + IO_3^- → 3I_2 + 3H_2O. The mole ratio of I^- to IO_3^- is 5:1. Moles of IO_3^- = (1)/(5) × moles of I^- = (1)/(5) × 0.05 mol = 0.01 mol. Step 10: Calculate the mass of KIO_3. Molar mass of KIO_3 = 39 + 127 + 3(16) = 39 + 127 + 48 = 214 g/mol. Mass of KIO_3 = moles × molar mass = 0.01 mol × 214 g/mol = 2.14 g. The mass of KIO_3 required is 2.14 g. 2. A: i) Define a nucleophile. A nucleophile is a chemical species that donates an electron pair to form a chemical bond in reactions. It is attracted to a positively charged center (nucleus). 2. A: ii) State the type of reaction that occurs when an aqueous solution of sodium hydroxide is warmed with bromoethane. The type of reaction is nucleophilic substitution (specifically, hydrolysis). 2. B: Reaction of an alcohol Step 1: Identify the reactants and products. Reactants: Ethanol (CH_3CH_2OH) and acidified potassium dichromate(VI) (K_2Cr_2O_7/H^+). Product: Ethanoic acid (CH_3COOH). Step 2: Write the balanced redox equation. Ethanol is oxidized to ethanoic acid. Dichromate(VI) ion (Cr_2O_7^2-) is reduced to chromium(III) ion (Cr^3+). Oxidation half-reaction: CH_3CH_2OH + H_2O → CH_3COOH + 4H^+ + 4e^- Reduction half-reaction: Cr_2O_7^2- + 14H^+ + 6e^- → 2Cr^3+ + 7H_2O Step 3: Balance the electrons. Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2. 3(CH_3CH_2OH + H_2O → CH_3COOH + 4H^+ + 4e^-) 3CH_3CH_2OH + 3H_2O → 3CH_3COOH + 12H^+ + 12e^- 2(Cr_2O_7^2- + 14H^+ + 6e^- → 2Cr^3+ + 7H_2O) 2Cr_2O_7^2- + 28H^+ + 12e^- → 4Cr^3+ + 14H_2O Step 4: Add the balanced half-reactions and cancel terms. 3CH_3CH_2OH + 3H_2O + 2Cr_2O_7^2- + 28H^+ → 3CH_3COOH + 12H^+ + 4Cr^3+ + 14H_2O Step 5: Simplify by canceling common terms (H^+ and H_2O). 3CH_3CH_2OH + 2Cr_2O_7^2- + 16H^+ → 3CH_3COOH + 4Cr^3+ + 11H_2O This is the balanced equation for the oxidation of ethanol to ethanoic acid. 3. A: i) Define a transition metal. A transition metal is an element that has an incomplete d sub-shell, either in its elemental form or in its common oxidation states. 3. A: ii) State two characteristic properties of transition metals. 1. They form colored compounds. 2. They exhibit variable oxidation states. 3. They form complex ions. 4. They act as catalysts. 3. B: Complex ion formation Step 1: Identify the central ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._