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
Hey hafsatjibril2006, good to see you again. This problem involves Graham's Law of Diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Step 1: Determine the molar masses of oxygen (O_2) and methane (CH_4). • Molar mass of Oxygen (O_2): 2 × 16 g/mol = 32 g/mol • Molar mass of Methane (CH_4): 12 g/mol + (4 × 1 g/mol) = 16 g/mol Step 2: Apply Graham's Law of Diffusion. Graham's Law can be expressed as: Rate_1Rate_2 = sqrt((M_2)/(M_1)) Where Rate is the volume diffused per unit time ((V)/(t)). Let oxygen be gas 1 and methane be gas 2. V_O_2/t_O_2V_CH_4/t_CH_4 = sqrt(M_CH_4)M_O_2 Step 3: Substitute the given values into the equation. Given: • V_O_2 = 200 cm^3 • t_O_2 = 50 s • V_CH_4 = 80 cm^3 • t_CH_4 = ? • M_O_2 = 32 g/mol • M_CH_4 = 16 g/mol 200 cm^3 / 50 s80 cm^3 / t_CH_4 = sqrt(16 g/mol)32 g/mol Step 4: Simplify and solve for t_CH_4. 4 cm^3/s80 cm^3 / t_CH_4 = sqrt((1)/(2)) 4t_CH_480 = (1)/(sqrt(2)) t_CH_420 = (1)/(sqrt(2)) t_CH_4 = (20)/(sqrt(2)) To rationalize the denominator: t_CH_4 = 20sqrt(2)2 t_CH_4 = 10sqrt(2) s t_CH_4 ≈ 10 × 1.4142 t_CH_4 ≈ 14.14 s The time it will take for 80 cm^3 of methane to diffuse is: 14.14 s Drop the next question.