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23 August, 01:43

93.0 mL of O2 gas is collected over water at 0.930 atm and 10.0 C. what would be the volume of this dry gas at STP

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  1. 23 August, 03:29
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    Let us assume that the oxygen behaves as an ideal gas such that we can use the ideal gas equation to solve for the number of moles of O2.

    PV = nRT; n = PV/RT

    Substituting the known values,

    n = (0.930 atm) (93/1000 L) / (0.0821 L. atm/mol. K) (10 + 273.15K)

    n = 3.72 x 10^-3 mols

    At STP, the volume of each mol of gas is equal to 22.4 L.

    volume = (3.73 x 10^-3 mols) x (22.4 L/1 mol)

    volume = 0.0833 L or 83.34 mL
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