Asked by grissy
How many moles of O2 contained in 5.25L of a gas at 26 degrees C and 1.2 atm?
Answers
Answered by
Steve
Since PV=nRT, we have
PV/nT = R is constant.
So, use STP and your given numbers:
(1*1)/(1*273) = (1.2*5.25)/((273+26)n)
Just solve for n.
PV/nT = R is constant.
So, use STP and your given numbers:
(1*1)/(1*273) = (1.2*5.25)/((273+26)n)
Just solve for n.
Answered by
DrBob222
I think there is an extra 273 (on the left side) in this solution that doesn't belong there.
PV = nRT
n = PV/RT
n = 1.2*5.25/(0.08206*299)
n = 0.26
PV = nRT
n = PV/RT
n = 1.2*5.25/(0.08206*299)
n = 0.26
Answered by
Steve
Oh, yeah: 22.4L at STP, not 1L.
(1*22.4)/(273*1) = (1.2*5.25)/(299n)
n = 0.256
My 273 is included in the .08026 factor you inserted.
(1*22.4)/(273*1) = (1.2*5.25)/(299n)
n = 0.256
My 273 is included in the .08026 factor you inserted.
Answered by
DrBob222
right. and I tried to get 0.08206 out of 1/273 but no matter how hard I tried it wouldn't work. But 22.4/273 = 0.08205 or
22.404/273 = 0.08206 which is the number usually quoted. I expect a mathematician will tell me that 22.404 is a lot of s.f. that probably don't belong there.
22.404/273 = 0.08206 which is the number usually quoted. I expect a mathematician will tell me that 22.404 is a lot of s.f. that probably don't belong there.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.