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Chemistry 12 Online
OpenStudy (anonymous):

How many moles of air are in 2.7L at 250K and 93kPa?

OpenStudy (anonymous):

My work: PV = nRT V = 2.7L T = 250K P = 93kPa R = 8.314 *kPa /mol*K 93kPa * 2.7L =n(8.314 *kPa /mol*K)250K --> 251.1 = 2078.5n, n = 0.120808

OpenStudy (anonymous):

@ParthKohli @sauravshakya

OpenStudy (anonymous):

@ghazi @Yahoo! @hba

OpenStudy (anonymous):

working with LaurAK on her problem. Just so you know.

OpenStudy (jfraser):

that's what i got

OpenStudy (anonymous):

Great, thanks!

OpenStudy (anonymous):

Could you check another one please?

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