A) Isotopes are atoms of the same element that have the same number of protons (atomic number) but different numbers of neutrons (mass number). From the given data, the atomic number for element

$$\text{Relative Atomic Mass of A} = 43.471837 + 20.069123 = 63.54096$$

Rounding to two decimal places:

$$\text{Relative Atomic Mass of A} = \boxed{63.54}$$

9. Step 1: Calculate the mass of carbon in 4.2 g of carbon (IV) oxide ($\text{CO}_2$). Molar mass of C = 12.0 g/mol Molar mass of $\text{CO}_2 = 12.0 + (2 \times 16.0) = 44.0 \text{ g/mol}$

$$\text{Mass of C} = \frac{\text{Mass of C in CO}_2}{\text{Molar mass of CO}_2} \times \text{Mass of CO}_2$$ $$\text{Mass of C} = \frac{12.0}{44.0} \times 4.2 \text{ g} = 0.2727 \times 4.2 \text{ g} \approx 1.1454 \text{ g}$$

Molar mass of H = 1.0 g/mol
Molar mass of $\text{H}_2\text{O} = (2 \times 1.0) + 16.0 = 18.0 \text{ g/mol}$

$$\text{Mass of H} = \frac{\text{Mass of 2H in H}_2\text{O}}{\text{Molar mass of H}_2\text{O}} \times \text{Mass of H}_2\text{O}$$ $$\text{Mass of H} = \frac{2 \times 1.0}{18.0} \times 1.71 \text{ g} = \frac{2.0}{18.0} \times 1.71 \text{ g} = 0.1111 \times 1.71 \text{ g} \approx 0.1900 \text{ g}$$

$$\text{Moles of C} = \frac{\text{Mass of C}}{\text{Molar mass of C}} = \frac{1.1454 \text{ g}}{12.0 \text{ g/mol}} \approx 0.09545 \text{ mol}$$ $$\text{Moles of H} = \frac{\text{Mass of H}}{\text{Molar mass of H}} = \frac{0.1900 \text{ g}}{1.0 \text{ g/mol}} = 0.1900 \text{ mol}$$

Divide by the smallest number of moles (0.09545 mol).
For C: $\frac{0.09545}{0.09545} = 1$
For H: $\frac{0.1900}{0.09545} \approx 1.99 \approx 2$
The ratio of C:H is approximately 1:2.
The empirical formula of the hydrocarbon is $\boxed{\text{CH}_2}$.

10. i) To obtain pure water from a mixture of water and common salt, the best method is distillation.