What are the answers to the following multiple-choice questions about titration and stoichiometry?

Here are the answers to your questions:

1. c) Proceed to the closest washbasin and thoroughly wash with cold water.

2. a) Not permitted at any time.

3. c) Open shoes such as slippers.

4. b) Volumetric flask.

5. d) Burette.

6. b) Weighing balance.

7. d) Determining the concentration of an unknown solution using a standard solution.

8. b) A solution whose concentration is precisely known.

9. b) Anhydrous sodium carbonate ($\text{Na}_2\text{CO}_3$).

10. a) 12 g of C-12 isotope.

11. b) Mass of solute (g) / Volume ($\text{dm}^3$).

12. d) A solution that changes color at the equivalence point without an external indicator (e.g., $\text{KMnO}_4$).

13. Step 1: Write the balanced half-reactions for the redox reaction between $\text{MnO}_4^{-}$ and $\text{C}_2\text{O}_4^{2-}$ in acidic medium. Reduction: $\text{MnO}_4^{-} + 8\text{H}^{+} + 5\text{e}^{-} \to \text{Mn}^{2+} + 4\text{H}_2\text{O}$ Oxidation: $\text{C}_2\text{O}_4^{2-} \to 2\text{CO}_2 + 2\text{e}^{-}$

    Step 2: Balance the electrons transferred by multiplying the reduction half-reaction by 2 and the oxidation half-reaction by 5.
    $2\text{MnO}_4^{-} + 16\text{H}^{+} + 10\text{e}^{-} \to 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$
    $5\text{C}_2\text{O}_4^{2-} \to 10\text{CO}_2 + 10\text{e}^{-}$

    Step 3: Add the balanced half-reactions to get the overall balanced equation.
    $2\text{MnO}_4^{-} + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^{+} \to 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}$

    The balanced mole ratio of $\text{MnO}_4^{-}$ to $\text{C}_2\text{O}_4^{2-}$ is $\boxed{\text{2:5}}$.

14. c) 60 °C.

15. b) $\text{C}_2\text{H}_2\text{O}_4 \cdot 2\text{H}_2\text{O}$.

16. d) To analyze a mixture of NaOH and $\text{Na}_2\text{CO}_3$.

17. d) Two.

18. Step 1: Write the chemical equation for the half neutralization of $\text{Na}_2\text{CO}_3$ to $\text{NaHCO}_3$ with HCl.

    $$\text{Na}_2\text{CO}_3 + \text{HCl} \to \text{NaHCO}_3 + \text{NaCl}$$

    Step 2: Determine the mole ratio from the balanced equation.
    From the equation, 1 mole of $\text{Na}_2\text{CO}_3$ reacts with 1 mole of HCl.

    The mole ratio of HCl to $\text{Na}_2\text{CO}_3$ for half neutralization is $\boxed{\text{1:1}}$.

19. Step 1: Identify the given concentrations and total concentration. Concentration of NaOH = 4 g/dm³ Concentration of $\text{Na}_2\text{CO}_3$ = 10.6 g/dm³ Total concentration = 14.6 g/dm³

    Step 2: Calculate the percentage composition of NaOH.

    $$\text{Percentage of NaOH} = \frac{\text{Mass of NaOH}}{\text{Total mass}} \times 100\%$$ $$\text{Percentage of NaOH} = \frac{4 \text{ g/dm}^3}{14.6 \text{ g/dm}^3} \times 100\%$$ $$\text{Percentage of NaOH} \approx 27.4\%$$

    Step 3: Calculate the percentage composition of $\text{Na}_2\text{CO}_3$.

    $$\text{Percentage of Na}_2\text{CO}_3 = \frac{\text{Mass of Na}_2\text{CO}_3}{\text{Total mass}} \times 100\%$$ $$$$