During an experiment, 17.5 mL of oxygen gas were collected over water at room temperature (25 C) and 100.2 kPa of atmospheric pressure. The vapor pressure of water at this temperature is 2.6 kPa.
a) What is the pressure of the "dry" oxygen gas?
b) How many moles of oxygen gas were produced?
c) What is the molar volume of oxygen gas at the conditions in the laboratory?

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  1. 1. Ptotal = pO2 + pH2O
    100.2 = pO2 + 2.6
    Solve for pO2.

    2. Use PV = nRT and solve for n.

    3. Convert mols you have to what volume would be 1 mol.

    Post your work if you get stuck.

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