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1 mol N2 = 28g

30C = 303K

At STP, 1mol * 273K is proportional 24L * 760torr. (PV = nRT)

Then we do a little bit of stoichiometry...

10g * 1mol/28g * 24L*760torr/(1mol*273K) * 303K * xL/750torr = 9.6402 L

10.0g of N2 at 30C and 750torr should occupy 9.6402L

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โˆ™ 2010-06-14 23:58:50
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Q: What volume is occupied by 10.0 grams N2 at 30 degrees Celsius and a pressure of 750 torr?
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