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.
ScanSolve AI Answer
Fresh day Nomvelo, let's solve. 5.3 Investigation III 5.3.1 Write down the independent variable for this investigation. The independent variable is the factor that is intentionally changed by the experimenter to observe its effect. In this investigation, the concentration of H_2SO_4 is varied across the runs. The independent variable is the concentration of H_2SO_4. 5.3.2 Use the collision theory to explain how the average rate of the reaction is affected in this investigation. In this investigation, the concentration of H_2SO_4 is increased from Run 1 to Run 3. According to collision theory, increasing the concentration of a reactant means there are more reactant particles per unit volume. This leads to a higher frequency of collisions between the reacting particles. Since more collisions occur per unit time, the frequency of effective collisions (collisions with sufficient energy and correct orientation) also increases, resulting in a faster reaction rate. 5.3.3 Calculate the percentage purity of the aluminium. Step 1: Calculate the total volume of H_2 gas produced in RUN 3. The reaction time is given as 2.6 minutes. Convert this to seconds. Time = 2.6 min × 60 s/min = 156 s The average rate of H_2 production in RUN 3 is 40 cm^3 · s^-1. Total Volume of H_2 = Average Rate × Time Total Volume of H_2 = 40 cm^3 · s^-1 × 156 s = 6240 cm^3 Step 2: Calculate the moles of H_2 gas produced. The molar gas volume at 60^ C is given as 27000 cm^3 · mol^-1. Moles of H_2 = Total Volume of H_2Molar Gas Volume Moles of H_2 = 6240 cm^327000 cm^3 · mol^-1 = 0.23111... mol Step 3: Calculate the moles of pure Al reacted. From the balanced chemical equation: 2Al(s) + 3H_2SO_4(aq) → Al_2(SO_4)_3(aq) + 3H_2(g) The mole ratio of Al to H_2 is 2:3. Moles of Al = Moles of H_2 × 2 mol Al3 mol H_2 Moles of Al = 0.23111... mol × (2)/(3) = 0.15407... mol Step 4: Calculate the mass of pure Al reacted. The molar mass of Al is approximately 26.98 g · mol^-1. Mass of Al = Moles of Al × Molar Mass of Al Mass of Al = 0.15407... mol × 26.98 g · mol^-1 = 4.1568... g Step 5: Calculate the percentage purity of the aluminium. The initial mass of the impure aluminium powder is 5 g. Percentage Purity = Mass of Pure AlMass of Impure Al Sample × 100\% Percentage Purity = 4.1568... g5 g × 100\% = 83.136... \% Rounding to one decimal place: Percentage Purity = 83.1\% Drop the next question.