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What is the pressure exerted by a mixture of 2.00 gram of H2and 8.00g of N2at 273 k at 10dm3 vessel?

Updated: 4/28/2022

Wiki User

∙ 7y ago

Best Answer

2.00 g H2 = 1 mole H28.00 g N2 = 8.00g/28 g/mole = 0.285 moles N2

Total moles gas = 1.285 moles

PV = nRT

P = nRT/V = (1.285)(0.08206)(273)/10

P = 2.88 atmospheres

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∙ 7y ago

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Q: What is the pressure exerted by a mixture of 2.00 gram of H2and 8.00g of N2at 273 k at 10dm3 vessel?

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How many moles of oxygen in 10dm3?

No. of moles = mass/relitive molecular mass in this case = 10/16 = 0.625 so that's 0.625 of a mole and a mole of anything contains 6.022 x 1023 atoms = 3.76 x 1023 atoms in 10g of oxygen.

Related questions

What is 10000cm3 in dm3?

1000 cm3 = 1dm3 so 10000/1000 = 10dm3

How many moles of oxygen in 10dm3?

No. of moles = mass/relitive molecular mass in this case = 10/16 = 0.625 so that's 0.625 of a mole and a mole of anything contains 6.022 x 1023 atoms = 3.76 x 1023 atoms in 10g of oxygen.

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