well, I will try to do dimensional analysis because I have no idea.
constant * cross sectional area
inversely proportional to length
1 hr = 3600 s
4.1 cm^2* 1m^2/10^4 cm^2 = 4.1*10^-4 m^2
Q = k * A/L * delta c * time
try
Q = 1.24*10^-9 m^2/s * 4.1*10^-4 m^2 /.0106 m * 1.37 kg/m^3 * 3600 s
gives kilograms in an hour
The diffusion constant for the alcohol ethanol in water is 1.24x10^-9 m2/s. A cylinder has a cross-sectional area of 4.10 cm^2 and a length of 2.06 cm. A difference in ethanol concentration of 1.37 kg/m^3 is maintained between the ends of the cylinder.
In one hour, what mass of ethanol diffuses through the cylinder?
3 answers
Damon was close.
m= (DAΔC)t/L = 1.24e-9*4.10e-4*1.37*3600/2.06e-2
= 1.217e-7
m= (DAΔC)t/L = 1.24e-9*4.10e-4*1.37*3600/2.06e-2
= 1.217e-7
THANK YOU!!!!