Asked by Joseph
IF CO2 TRAVELS 20.5 CM DOWN A 46.0 CM GLASS TUBE IN 1.3*10^4 AS IN THE GRAHAMS LAW EXPERIMENT, THEN COLLIDES WITH ANOTHER GAS. WHAT IS THE MOLECULAR MASS OF THE SECOND GAS?
Answers
Answered by
bobpursley
So the other gas traveled 46-20.5 in the same time?
Difusion rates are the heart of Graham's law....
now distance=rate=distance/time, so rate and distance are related. So modify Grahm's law...
distance1/distanc2=sqrt(Momlass2/molmass1)
or molmass2= molmass1*(20.5/25.5)^2
Difusion rates are the heart of Graham's law....
now distance=rate=distance/time, so rate and distance are related. So modify Grahm's law...
distance1/distanc2=sqrt(Momlass2/molmass1)
or molmass2= molmass1*(20.5/25.5)^2
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