Asked by kim
the valve between a 3.00-L tank containing O2(g) at 6.00 atm. and a 5.00-L containing N2(g) at 1.20 atm. is opened. Calcualte the ratio of partial pressures (O2:N2) in the container?
Answers
Answered by
DrBob222
At constant T, I would say the volume has changed from 3.00 L to 8.00L; therefore, PO2 is 6.00 x 8/3 = ?? atm
PN2 is 1.2 x 8/5 = ?? atm
Now take the ratio Of O2:N2
PN2 is 1.2 x 8/5 = ?? atm
Now take the ratio Of O2:N2
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