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OpenStudy (anonymous):
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 standard conditions?
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OpenStudy (anonymous):
You need to use the formula (PV)/(T) at STP and (PV)/(T) at the experimental conditions. So (1.0atm)(V)/(274 K) = (0.930 atm- the pressure of water at 10 degrees celcius in atm)(0.093L)/(283.15 K) If needed convert the pressure of water to, 1atm = 760 mmHg. Sorry I can't express this it in an easier way ): Hope it helps!
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