What is the pressure in atmospheres of 263 grams of N2 gas in a 10.0L container at 25.0oC.
PV = nRT
So: xV = nRT
N2 = N(14.01) * 2 = 28.01 amu
263 / 28.01 = 9.3895037486611924312745448054266 moles of N2 gas
x(10.0L) = 9.38950~(0.08206 * 25.0)
x(10.0L) = 19.262566940378436272759728668331
x = 1.9262566940378436272759728668331 atm
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