Tuesday
September 16, 2014

Homework Help: Chemistry

Posted by Clay on Wednesday, September 26, 2007 at 1:21pm.

I'm trying to find the dry gas pressure of H2 in mmHg. The barometric pressure is 765.3 mmHg, the vp of H20 is 23.76 mmHg and the gas collected was 25.6 mL.

this is what i did.
765.3 mmHg - 23.76 mmHg - (25.6 Ml/13.6 mmHg) = 739.66 = 740 mmHg 93 sig figs).

Then I have to use the combined gas law to convert this volume of the H2 gas collected to the new volume at STP.

So i used the formula v2 = P1 V1 T2/ T1 P1

I converted the volume of H2 into atm and got .974 atm (740 mmHg/760). Then I multiplied .974 atm by .0244 L (the leftover liquid) and 298 K (the temp.) then I divided by 273 K and 1 atm.

(.974 atm x .0244 L x 298 K)/(273 K x 1 atm) = .0295 L

Is that correct?

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