Answer:
t = 125.3 seconds
Explanation:
Molar mass of CO2 = 12+2(16) = 66
Molar mass of N2 = 2(14)= 28
rate of diffusion of N2 = volume/ time = 280cm³/70s
= 4cm³/s
let rate of CO2 = rate of diffusion of CO2 = volume/time
= 400/t
Using Graham's law of diffusion,
rN2/rCO2 = √M(CO2)/M(N2)
4/400/t =√44/28 = 4t/400= √11/7
t/100 = 1.253 , t= (100)(1.253)
t = 125.3 seconds
hence it takes CO2 125.3 seconds to diffuse through the membrane
25.13(5512)
4568%of the time is 56789
5678
4333
a. Moles of gold = 35.12/197 = 0.18 moles. b. number of gold molecules = moles x 6.02 x 10^23 = 0.18 x 6.02 x 10^23= 1.08 x 10^23
