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: Address Question 11.5. According to the collision theory, an increase in the concentration of reactants affects the reaction rate as follows: • An increase in concentration means there are more reactant particles per unit volume. • This leads to an increase in the frequency of collisions between the reactant particles. • Consequently, there will be a higher number of effective collisions per unit time (collisions with sufficient energy and correct orientation). • A greater number of effective collisions per unit time results in an increased rate of reaction. Step 2: Address Question 11.6. First, write the balanced chemical equation for the reaction: Na_2S_2O_3(aq) + 2HCl(aq) 2NaCl(aq) + SO_2(g) + S(s) + H_2O(l) From the equation, 1 mole of Na_2S_2O_3 produces 1 mole of S. Step 3: Calculate the molar masses. Molar mass of Sulphur (S) = 32.07 g/mol Molar mass of Sodium thiosulphate (Na_2S_2O_3) = (2 × 22.99) + (2 × 32.07) + (3 × 16.00) = 45.98 + 64.14 + 48.00 = 158.12 g/mol Step 4: Calculate the moles of sulphur produced. Given mass of S = 1.62 g Moles of S = Mass of SMolar mass of S Moles of S = 1.62 g32.07 g/mol ≈ 0.0505145 mol Step 5: Calculate the moles of sodium thiosulphate used. From the balanced equation, the mole ratio of Na_2S_2O_3 to S is 1:1. Moles of Na_2S_2O_3 = Moles of S ≈ 0.0505145 mol Step 6: Calculate the mass of sodium thiosulphate used. Mass of Na_2S_2O_3 = Moles of Na_2S_2O_3 × Molar mass of Na_2S_2O_3 Mass of Na_2S_2O_3 = 0.0505145 mol × 158.12 g/mol Mass of Na_2S_2O_3 ≈ 7.987 g Rounding to two decimal places: The mass of the sodium thiosulphate used in experiment 1 is 7.99 g. Last free one today — make it count tomorrow, or type /upgrade for unlimited.