Locate 70°C on the x-axis.

At 70°C, Salt A has a solubility of approximately 48 g per 100 g of H₂O.
At 70°C, Salt B has a solubility of approximately 52 g per 100 g of H₂O.

Salt B has a higher solubility at 70°C.
The more soluble salt is:

20. b) What is the amount of salt in 50g of saturated solution of salt A at 120°C?

At 120°C, the solubility of Salt A is approximately 75 g per 100 g of H₂O.

Mass of saturated solution = Mass of Salt A + Mass of water
Mass of saturated solution = $75 \text{ g} + 100 \text{ g} = 175 \text{ g}$

The amount of salt in 50g of saturated solution of salt A at 120°C is:

21. a) Identify solid P.
Solid P is formed by the reaction of dry chlorine gas with iron wire. This reaction produces iron(III) chloride.

21. b) Write an equation for formation of solid P.
The balanced chemical equation for the formation of iron(III) chloride from iron and chlorine gas is:

21. c) Suppose the gas used in the set up was dry hydrogen chloride gas; what would be the product obtained after the reaction? Give a reason for your answer.
If dry hydrogen chloride gas was used, the product obtained would be Iron(II) chloride ($\text{FeCl}_2$) and hydrogen gas ($\text{H}_2$).
Reason: Chlorine gas ($\text{Cl}_2$) is a stronger oxidizing agent than hydrogen chloride gas ($\text{HCl}$). Chlorine oxidizes iron to its +3 oxidation state, while hydrogen chloride oxidizes iron to its +2 oxidation state.

22. a) State the Gay-Lussac's law.
Gay-Lussac's Law of Combining Volumes states that when gases react, they do so in volumes that bear a simple whole-number ratio to one another and to the volumes of the gaseous products, provided that all volumes are measured at the same temperature and pressure.

22. b) 50cm³ of Sulphur (IV) Oxide is exploded in 50cm³ of Oxygen gas. Determine the volume of the residual gas.

Sulphur (IV) Oxide is $\text{SO}_2$. Oxygen gas is $\text{O}_2$. The product is Sulphur (VI) Oxide, $\text{SO}_3$.

The volume ratio of $\text{SO}_2 : \text{O}_2 : \text{SO}_3$ is $2 : 1 : 2$.

Initial volumes: $\text{SO}_2 = 50 \text{ cm}^3$, $\text{O}_2 = 50 \text{ cm}^3$.
According to the ratio, $50 \text{ cm}^3$ of $\text{SO}_2$ would require $\frac{1}{2} \times 50 \text{ cm}^3 = 25 \text{ cm}^3$ of $\text{O}_2$.
Since $50 \text{ cm}^3$ of $\text{O}_2$ is available, $\text{O}_2$ is in excess, and $\text{SO}_2$ is the limiting reactant.

Volume of $\text{SO}_2$ reacted = $50 \text{ cm}^3$
Volume of $\text{O}_2$ reacted = $25 \text{ cm}^3$
Volume of $\text{SO}_3$ produced = $50 \text{ cm}^3$ (since the ratio $\text{SO}_2 : \text{SO}_3$ is $2:2$)

Volume of unreacted $\text{O}_2 = \text{Initial volume of O}_2 - \text{Volume of O}_2 \text{ reacted}$
Volume of unreacted $\text{O}_2 = 50 \text{ cm}^3 - 25 \text{ cm}^3 = 25 \text{ cm}^3$
Residual gas consists of unreacted $\text{O}_2$ and produced $\text{SO}_3$.
Total volume of residual gas = Volume of unreacted $\text{O}_2$ + Volume of $\text{SO}_3$ produced
Total volume of residual gas = $25 \text{ cm}^3 + 50 \text{ cm}^3 = 75 \text{ cm}^3$
The volume of the residual gas is: