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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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What volume is occupied by 20. 0 g of nitrogen (N2) gas at 30 degree C. and 800 mmHg?

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0.004 gm/cm³

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Q: What volume is occupied by 20. 0 g of nitrogen (N2) gas at 30 degree C. and 800 mmHg?
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The gases in a mixture have the following partial pressures at the same volume and temperature 1013000 Pa nitrogen 700 mmHg argon 1.20 ATM of helium 790 torr oxygen and 0.50 ATM of carbon dio?

Convert all the pressures into one unit and then, using Dalton's Law of Partial Pressure, you just add all the pressures together. Using these conversions(101325Pa equals 1 ATM, 1 ATM equals 760 mmHg, 1 mmHg equals 1 torr), calculate each pressure into one unit(I used mmHg) 1013000 Pa nitrogen equals 7598 mmHg of nitrogen(make sure you round using sig figs) 700 mmHg of argon equals 700 mmHg of argon 1.20 ATM of helium equals 912 mmHg of helium 790 torr oxygen equals 790 mmHg of oxygen 0.50 ATM of carbon dioxide equals 380 mmHg of Carbon Dioxide Now add all the pressures and you get: 7598+700+912+790+380= 10380 mmHg If you need to have the answer in a particular unit: 10380 mmHg = 10380 torr = 13.66 ATM = 1384000 Pa = 1,384 kPa


A sample of dry air contains 210 mL of oxygen 780 mL of nitrogen and 10.0 mL of all other gases What is the pressure in mm of Hg due to the nitrogen if the total pressure was 1000.0 mm?

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