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OpenStudy (brzybro):

What is the volume of 19.87 mol of ammonium chloride (NH4Cl) at STP?

likeabossssssss (likeabossssssss):

this goes in math may u close this and put it there

OpenStudy (chpatterson):

PV = nRT V = nRT / P P = 1 atm n = 19.87 mol R = 0.0821 L * atm / mol * k T = 0 + 273K = 273K Plug and solve: V = (19.87 mol) (0.0821)(273K) / (1 atm) V = 445.088

OpenStudy (chpatterson):

There you go now you can close this question.

sam (.sam.):

You've posted on OpenStudy Feedback, please ask your future Chemistry questions here http://openstudy.com/study#/groups/Chemistry Thanks :)

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