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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You're on a roll — Here are the solutions to your questions: 1. a) i) Sodium chloride is obtained from sea water through solar evaporation*. Sea water is channeled into large, shallow ponds (salt pans). The sun's heat causes the water to evaporate, leaving behind concentrated brine. As evaporation continues, sodium chloride crystals precipitate out and are then collected. ii)* I. Equation for the formation of ammonium hydrogen carbonate: NH_3(g) + H_2O(l) + CO_2(g) NH_4HCO_3(aq) II. Equation for the second reaction (formation of sodium bicarbonate): NH_4HCO_3(aq) + NaCl(aq) NaHCO_3(s) + NH_4Cl(aq) iii)* I. Process I is filtration*. This separates the solid sodium hydrogen carbonate from the ammonium chloride solution. II. Process II is heating or thermal decomposition*. This converts sodium hydrogen carbonate into sodium carbonate, water, and carbon dioxide. iv)* I. Calculate the number of moles of sodium carbonate that reacted. The reaction between sodium carbonate and sulphuric acid is: Na_2CO_3(aq) + H_2SO_4(aq) Na_2SO_4(aq) + H_2O(l) + CO_2(g) Step 1: Calculate moles of sulphuric acid. Moles of H_2SO_4 = Molarity × Volume Moles of H_2SO_4 = 0.5 M × (40.0)/(1000) L Moles of H_2SO_4 = 0.02 mol Step 2: Use the mole ratio from the balanced equation. From the equation, 1 mole of Na_2CO_3 reacts with 1 mole of H_2SO_4. Moles of Na_2CO_3 = Moles of H_2SO_4 Moles of Na_2CO_3 = 0.02 mol The number of moles of sodium carbonate that reacted is 0.02 mol. II. Determine the percentage of sodium carbonate in the sample. Step 1: Calculate the molar mass of Na_2CO_3. Molar mass of Na_2CO_3 = (2 × 23.0) + 12.0 + (3 × 16.0) Molar mass of Na_2CO_3 = 46.0 + 12.0 + 48.0 Molar mass of Na_2CO_3 = 106.0 g/mol Step 2: Calculate the mass of pure Na_2CO_3 in the sample. Mass of Na_2CO_3 = Moles × Molar mass Mass of Na_2CO_3 = 0.02 mol × 106.0 g/mol Mass of Na_2CO_3 = 2.12 g Step 3: Calculate the percentage purity. Percentage purity = Mass of pure Na_2CO_3Mass of sample × 100\% Percentage purity = 2.12 g2.15 g × 100\% Percentage purity = 98.6046...\% The percentage of sodium carbonate in the sample is 98.60\%. b)* Two industrial uses of sodium carbonate: Manufacture of glass*. Manufacture of detergents* and soaps. Water softening* (removing hardness from water). 2. a) State the Charles' Law. Charles' Law states that for a fixed mass of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature. b)* Calculate the temperature at which the volume of the gas would be 2.8 × 10^-2 m^3 at 1.0 × 10^5 Pascals. Given: V_1 = 3.5 × 10^-2 m^3 T_1 = 291 K P_1 = 1.0 × 10^5 Pa V_2 = 2.8 × 10^-2 m^3 P_2 = 1.0 × 10^5 Pa We use the combined gas law: (P_1V_1)/(T_1) = (P_2V_2)/(T_2) Since P_1 = P_2, the equation simplifies to Charles' Law: (V_1)/(T_1) = (V_2)/(T_2) Step 1: Substitute the given values. 3.5 × 10^-2 m^3291 K = 2.8 × 10^-2 m^3T_2 Step 2: Solve for T_2. T_2 = (2.8 × 10^-2 m^3) × 291 K3.5 × 10^-2 m^3 T_2 = (2.8 × 291)/(3.5) K T_2 = 0.8 × 291 K T_2 = 232.8 K The temperature at which the volume would be 2.8 × 10^-2 m^3 is 232.8 K. 3. a) State the Graham's law diffusion. Graham's Law of Diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (or density), at constant temperature and pressure. b)* The molar masses of gases W and X are 16.0 and 44.0 respectively. If the rate of diffusion of W through a porous material is 12 cm^3s^-1, calculate the rate of diffusion of X through the same material. According to Graham's Law of Diffusion: Rate_WRate_X = sqrt((M_X)/(M_W)) Given: Rate_W = 12 cm^3s^-1 M_W = 16.0 M_X = 44.0 Step 1: Substitute the given values into the formula. 12 cm^3s^-1Rate_X = sqrt((44.0)/(16.0)) (12)/(Rate)_X = sqrt(2.75) (12)/(Rate)_X ≈ 1.6583 Step 2: Solve for Rate_X. Rate_X = (12)/(1.6583) Rate_X ≈ 7.236 cm^3s^-1 The rate of diffusion of X is approximately 7.24 cm^3s^-1. What's next?