Locate 70°C on the x-axis.

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: 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 g + 100 g = 175 g Step 3: Calculate the amount of salt in 50 g of saturated solution. Mass of Salt A = Mass of Salt A in 175 g solutionMass of 175 g solution × Desired mass of solution Mass of Salt A = 75 g175 g × 50 g Mass of Salt A = (3)/(7) × 50 g Mass of Salt A = (150)/(7) g Mass of Salt A ≈ 21.43 g The amount of salt in 50g of saturated solution of salt A at 120°C is: 21.43 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. 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: 2Fe(s) + 3Cl_2(g) → 2FeCl_3(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 (FeCl_2) and hydrogen gas (H_2). Reason: Chlorine gas (Cl_2) is a stronger oxidizing agent than hydrogen chloride gas (HCl). Chlorine oxidizes iron to its +3 oxidation state, while hydrogen chloride oxidizes iron to its +2 oxidation state. 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. 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 SO_2. Oxygen gas is O_2. The product is Sulphur (VI) Oxide, SO_3. 2SO_2(g) + O_2(g) → 2SO_3(g) Step 2: Determine the volume ratios from the balanced equation. The volume ratio of SO_2 : O_2 : SO_3 is 2 : 1 : 2. Step 3: Identify the limiting reactant. Initial volumes: SO_2 = 50 cm^3, O_2 = 50 cm^3. According to the ratio, 50 cm^3 of SO_2 would require (1)/(2) × 50 cm^3 = 25 cm^3 of O_2. Since 50 cm^3 of O_2 is available, O_2 is in excess, and SO_2 is the limiting reactant. Step 4: Calculate the volumes of reactants consumed and products formed. Volume of SO_2 reacted = 50 cm^3 Volume of O_2 reacted = 25 cm^3 Volume of SO_3 produced = 50 cm^3 (since the ratio SO_2 : SO_3 is 2:2) Step 5: Calculate the volume of residual gas. Volume of unreacted O_2 = Initial volume of O_2 - Volume of O_2 reacted Volume of unreacted O_2 = 50 cm^3 - 25 cm^3 = 25 cm^3 Residual gas consists of unreacted O_2 and produced SO_3. Total volume of residual gas = Volume of unreacted O_2 + Volume of SO_3 produced Total volume of residual gas = 25 cm^3 + 50 cm^3 = 75 cm^3 The volume of the residual gas is: 75 cm^3 Send me the next one 📸