Right Ronneydomnick, let's go. 20. a) Which salt is more soluble in water at 70°C? Step 1: Locate 70°C on the x-axis. Step 2: Read the solubility for Salt A and Salt B at 70°C from the graph. 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. Step 3: Compare the solubilities. Salt B has a higher solubility at 70°C. The more soluble salt is: $$\boxed{\text{Salt B}}$$ 20. b) What is the amount of salt in 50g of saturated solution of salt A at 120°C? Step 1: Determine the solubility of Salt A at 120°C from the graph. At 120°C, the solubility of Salt A is approximately 75 g per 100 g of H₂O. Step 2: Calculate the mass of the saturated solution formed by 100 g of water. Mass of saturated solution = Mass of Salt A + Mass of water Mass of saturated solution = $75 \text{ g} + 100 \text{ g} = 175 \text{ g}$ Step 3: Calculate the amount of salt in 50 g of saturated solution. $$\text{Mass of Salt A} = \frac{\text{Mass of Salt A in 175 g solution}}{\text{Mass of 175 g solution}} \times \text{Desired mass of solution}$$ $$\text{Mass of Salt A} = \frac{75 \text{ g}}{175 \text{ g}} \times 50 \text{ g}$$ $$\text{Mass of Salt A} = \frac{3}{7} \times 50 \text{ g}$$ $$\text{Mass of Salt A} = \frac{150}{7} \text{ g}$$ $$\text{Mass of Salt A} \approx 21.43 \text{ g}$$ The amount of salt in 50g of saturated solution of salt A at 120°C is: $$\boxed{21.43 \text{ g}}$$ 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. $$\boxed{\text{Iron(III) chloride}}$$ 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: $$\boxed{2\text{Fe(s)} + 3\text{Cl}_2\text{(g)} \rightarrow 2\text{FeCl}_3\text{(s)}}$$ 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. $$\boxed{\text{Iron(II) chloride and hydrogen gas. Chlorine is a stronger oxidizing agent than hydrogen chloride.}}$$ 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. $$\boxed{\text{When gases react, they do so in volumes that bear a simple whole-number ratio to one another and to the volumes of the gaseous products, provided that all volumes are measured at the same temperature and pressure.}}$$ 22. b) 50cm³ of Sulphur (IV) Oxide is exploded in 50cm³ of Oxygen gas. Determine the volume of the residual gas. Step 1: Write the balanced chemical equation for the reaction. Sulphur (IV) Oxide is $\text{SO}_2$. Oxygen gas is $\text{O}_2$. The product is Sulphur (VI) Oxide, $\text{SO}_3$. $$2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{SO}_3\text{(g)}$$ Step 2: Determine the volume ratios from the balanced equation. The volume ratio of $\text{SO}_2 : \text{O}_2 : \text{SO}_3$ is $2 : 1 : 2$. Step 3: Identify the limiting reactant. 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. Step 4: Calculate the volumes of reactants consumed and products formed. 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 $2:2$) Step 5: Calculate the volume of residual gas. 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: $$\boxed{75 \text{ cm}^3}$$ Send me the next one 📸