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3 January, 13:36

What volume of air contains 10.0g of oxygen gas at 273 k and 1.00 atm?

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  1. 3 January, 15:24
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    12.3g 02times1 mol 02/32.0 g 02=mol 02

    .38mol/x=20.95/100

    x=1.8mol air ...

    PV=nRT

    101.3kPa times V = 1.8 mol times 8.31[kPa * L]/{k*mol}*273k

    V=40.3Lair
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