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: Calculate the molar mass of ethane (C_2H_6). The molar mass of carbon (C) is approximately 12.01 g/mol. The molar mass of hydrogen (H) is approximately 1.008 g/mol. M_C_2H_6 = (2 × 12.01 g/mol) + (6 × 1.008 g/mol) M_C_2H_6 = 24.02 g/mol + 6.048 g/mol M_C_2H_6 = 30.068 g/mol Step 2: Apply Graham's Law of Diffusion. Graham's Law states that the ratio of the rates of diffusion of two gases is inversely proportional to the square root of the ratio of their molar masses: Rate_1Rate_2 = sqrt((M_2)/(M_1)) Let gas 1 be ethane (C_2H_6) and gas 2 be the unknown gas. Given: Rate_C_2H_6 = 3.6 × 10^-6 mol/h Rate_unknown = 1.3 × 10^-6 mol/h M_C_2H_6 = 30.068 g/mol Substitute the values into Graham's Law: 3.6 × 10^-6 mol/h1.3 × 10^-6 mol/h = sqrt(M_unknown)30.068 g/mol 2.7692 = sqrt(M_unknown)30.068 g/mol Step 3: Solve for the molar mass of the unknown gas (M_unknown). Square both sides of the equation: (2.7692)^2 = M_unknown30.068 g/mol 7.6685 = M_unknown30.068 g/mol M_unknown = 7.6685 × 30.068 g/mol M_unknown = 230.56 g/mol Rounding to two significant figures (based on the given rates): M_unknown = 230 g/mol The molar mass of the unknown gas is 230 g/mol. --- a) Kr and O_2 Step 1: Determine the molar masses of Kr and O_2. Molar mass of Krypton (Kr) = 83.80 g/mol Molar mass of Oxygen (O_2) = 2 × 16.00 g/mol = 32.00 g/mol Step 2: Identify which gas diffuses more rapidly. The gas with the lower molar mass diffuses more rapidly. Since M_O_2 (32.00 g/mol) is less than M_Kr (83.80 g/mol), oxygen (O_2) diffuses more rapidly. Step 3: Calculate the relative rates of diffusion. Using Graham's Law: Rate_O_2Rate_Kr = sqrt(M_Kr)M_O_2 Rate_O_2Rate_Kr = sqrt(83.80 g/mol)32.00 g/mol Rate_O_2Rate_Kr = sqrt(2.61875) Rate_O_2Rate_Kr = 1.618 Oxygen (O_2) diffuses more rapidly. The relative rate is 1.618. b) N_2 and acetylene (C_2H_2) Step 1: Determine the molar masses of N_2 and C_2H_2. Molar mass of Nitrogen (N_2) = 2 × 14.01 g/mol = 28.02 g/mol Molar mass of Acetylene (C_2H_2) = (2 × 12.01 g/mol) + (2 × 1.008 g/mol) M_C_2H_2 = 24.02 g/mol + 2.016 g/mol M_C_2H_2 = 26.036 g/mol Step 2: Identify which gas diffuses more rapidly. The gas with the lower molar mass diffuses more rapidly. Since M_C_2H_2 (26.036 g/mol) is less than M_N_2 (28.02 g/mol), acetylene (C_2H_2) diffuses more rapidly. Step 3: Calculate the relative rates of diffusion. Using Graham's Law: Rate_C_2H_2Rate_N_2 = sqrt(M_N_2)M_C_2H_2 Rate_C_2H_2Rate_N_2 = sqrt(28.02 g/mol)26.036 g/mol Rate_C_2H_2Rate_N_2 = sqrt(1.076125) Rate_C_2H_2Rate_N_2 = 1.037 Acetylene (C_2H_2) diffuses more rapidly. The relative rate is 1.037.