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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Step 1: Define the term rate of reaction. The rate of reaction is the change in concentration of reactants or products per unit time. Step 2: Identify a controlled variable. One controlled variable for this investigation is the mass/surface area of magnesium carbonate, temperature, or pressure. Step 3: Explain why the mass decreases. The mass of the reaction mixture and flask decreases because carbon dioxide gas (CO_2(g)) is produced during the reaction and escapes from the flask. Step 4: Calculate the average rate for curve B. From the graph for curve B: Initial mass of flask + contents at t=0 s is 144.50 g. Final mass of flask + contents at t=120 s is 143.24 g. Mass of CO_2 produced: m(CO_2) = 144.50 g - 143.24 g = 1.26 g Molar mass of CO_2: M(CO_2) = 12.01 g/mol + (2 × 16.00 g/mol) = 44.01 g/mol Moles of CO_2 produced: n(CO_2) = m(CO_2)M(CO_2) = 1.26 g44.01 g/mol ≈ 0.028629857 mol Volume of CO_2 produced (molar gas volume is 24.5 dm^3 · mol^-1): V(CO_2) = n(CO_2) × Molar gas volume = 0.028629857 mol × 24.5 dm^3 · mol^-1 ≈ 0.7014315 dm^3 Average rate of CO_2 production: Rate = V(CO_2) t = 0.7014315 dm^3120 s ≈ 0.005845 dm^3 · s^-1 Rounding to three significant figures: The average rate is 0.00585 dm^3 · s^-1. Step 5: Identify the curve for Experiment 1 and explain using collision theory. Experiment 1 has the lowest HCl concentration (0.1 mol · dm^-3). A lower concentration means a slower reaction rate, which corresponds to the curve with the least steep slope (slowest mass decrease). Therefore, curve A represents Experiment 1. According to collision theory, a lower concentration of HCl means there are fewer HCl particles per unit volume. This leads to a decrease in the frequency of effective collisions between HCl and MgCO_3 particles, resulting in a slower reaction rate. Step 6: Compare the final mass of CO_2 in Experiment 2 and Experiment 3. The final mass of CO_2(g) produced in Experiment 2 will be THE SAME as that in Experiment 3. This is because the problem states that hydrochloric acid is in excess. This means that magnesium carbonate (MgCO_3) is the limiting reactant in both experiments. Since the initial amount of the limiting reactant (MgCO_3) is the same for all experiments, the total amount of product (CO_2) formed will be the same, regardless of the HCl concentration. Send me the next one 📸