Complete the questions in the photo

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: $IO_3^-(aq) + I^-(aq) + H^+(aq) \rightarrow I_2(aq) + H_2O(l)$

• 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: $I^- \rightarrow I_2$ Balance I: $2I^- \rightarrow I_2$ Balance charge: $2I^- \rightarrow I_2 + 2e^-$
  • Reduction: $IO_3^- \rightarrow I_2$ Balance I: $2IO_3^- \rightarrow I_2$ Balance O with $H_2O$: $2IO_3^- \rightarrow I_2 + 6H_2O$ Balance H with $H^+$: $2IO_3^- + 12H^+ \rightarrow I_2 + 6H_2O$ Balance charge: $2IO_3^- + 12H^+ + 10e^- \rightarrow I_2 + 6H_2O$
• Step 4: Equalize the number of electrons transferred. Multiply the oxidation half-reaction by 5. $5 \times (2I^- \rightarrow I_2 + 2e^-) \implies 10I^- \rightarrow 5I_2 + 10e^-$

• Step 5: Add the balanced half-reactions and cancel electrons. $10I^- \rightarrow 5I_2 + 10e^-$ $2IO_3^- + 12H^+ + 10e^- \rightarrow I_2 + 6H_2O$

  $10I^- + 2IO_3^- + 12H^+ \rightarrow 6I_2 + 6H_2O$
• Step 6: Simplify the equation by dividing by 2. $5I^- + IO_3^- + 6H^+ \rightarrow 3I_2 + 3H_2O$ 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 $KI = \frac{\text{mass}}{\text{molar mass}} = \frac{8.3 \text{ g}}{166 \text{ g/mol}} = 0.05 \text{ mol}$. 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: $5I^- + IO_3^- \rightarrow 3I_2 + 3H_2O$. The mole ratio of $I^-$ to $IO_3^-$ is 5:1. Moles of $IO_3^- = \frac{1}{5} \times \text{moles of } I^- = \frac{1}{5} \times 0.05 \text{ mol} = 0.01 \text{ mol}$.
• 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 $KIO_3 = \text{moles} \times \text{molar mass} = 0.01 \text{ mol} \times 214 \text{ g/mol} = 2.14$ 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: $CH_3CH_2OH + H_2O \rightarrow CH_3COOH + 4H^+ + 4e^-$ Reduction half-reaction: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
• Step 3: Balance the electrons. Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2. $3(CH_3CH_2OH + H_2O \rightarrow CH_3COOH + 4H^+ + 4e^-) \implies 3CH_3CH_2OH + 3H_2O \rightarrow 3CH_3COOH + 12H^+ + 12e^-$ $2(Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O) \implies 2Cr_2O_7^{2-} + 28H^+ + 12e^- \rightarrow 4Cr^{3+} + 14H_2O$
• Step 4: Add the balanced half-reactions and cancel terms. $3CH_3CH_2OH + 3H_2O + 2Cr_2O_7^{2-} + 28H^+ \rightarrow 3CH_3COOH + 12H^+ + 4Cr^{3+} + 14H_2O$
• Step 5: Simplify by canceling common terms ($H^+$ and $H_2O$). $3CH_3CH_2OH + 2Cr_2O_7^{2-} + 16H^+ \rightarrow 3CH_3COOH + 4Cr^{3+} + 11H_2O$ 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