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Chemistry 20 Online
OpenStudy (anonymous):

HELP!!! A gas mixture contains 1.35g N2 and 0.82g O2 in a 1.48-L container at 21∘C. Calculate the partial pressure of O2 and N2

OpenStudy (anonymous):

Are you given the total pressure?

OpenStudy (anonymous):

i have the mole fraction n2=.65 02=.35

OpenStudy (anonymous):

And they are not said to be in equilibrium?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

Okay. I will try making my own assumptions and then if you manage confirm if are correct in case you have the marking scheme.

OpenStudy (anonymous):

Total pressure = nRT/V= 1*8.31*(21+273)/1.48*10^-3=1.65*10^6 Pa. Partial pressure of x= mole of x*total pressure/total number of moles. For n2= 0.65*1.65*10^6/1=962000Pa For O2= 0.35*1.65*10^6/1= 518000Pa.

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