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    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

    Is this correct?

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    https://brainmass.com/chemistry/stoichiometry/confirmation-gas-pressures-volume-problem-219057

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