This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here's how to calculate the temperature change: Question 7: Copper (II) Sulphate reacts with Barium Chloride according to the equation: CuSO_4(aq) + BaCl_2(aq) → CuCl_2(aq) + BaSO_4(s) H = -17.7 kJ/mol Volume of 1 M CuSO_4 = 900 cm^3 Volume of 1 M BaCl_2 = 600 cm^3 Heat capacity of solution c = 4.2 J/g^ Density of solution = 1 g/cm^3 Step 1: Calculate the moles of each reactant. Convert volumes from cm^3 to L: 900 cm^3 = 0.900 L 600 cm^3 = 0.600 L Moles of CuSO_4: Moles = Concentration × Volume Moles of CuSO_4 = 1 mol/L × 0.900 L = 0.900 mol Moles of BaCl_2: Moles of BaCl_2 = 1 mol/L × 0.600 L = 0.600 mol Step 2: Determine the limiting reactant. From the balanced equation, CuSO_4 and BaCl_2 react in a 1:1 molar ratio. Since 0.600 mol of BaCl_2 is less than 0.900 mol of CuSO_4, BaCl_2 is the limiting reactant. Therefore, 0.600 mol of reaction occurs. Step 3: Calculate the total heat released (q). The enthalpy change for the reaction is H = -17.7 kJ/mol. q = Moles reacted × | H| q = 0.600 mol × 17.7 kJ/mol q = 10.62 kJ Convert to Joules: q = 10.62 × 1000 J = 10620 J Step 4: Calculate the total mass of the solution. Total volume of solution: Total volume = 900 cm^3 + 600 cm^3 = 1500 cm^3 Mass of solution: Mass = Total volume × Density m = 1500 cm^3 × 1 g/cm^3 = 1500 g Step 5: Calculate the temperature change ( T). Use the formula q = mc T: T = (q)/(mc) T = 10620 J1500 g × 4.2 J/g^ T = (10620)/(6300) ^ T = 1.6857... ^ Rounding to three significant figures: T = 1.69 ^ The temperature change is 1.69 ^. Last free one today — make it count tomorrow, or type /upgrade for unlimited.