what is the pressure exerted by a mixture of 0.250 mol in N2 and 0.400 Mol He 27 °C in a 3.00 L container?

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Question

science chemistry

Answer 1

Answer #1

Volume = V = 3.00 L

Temperature = T = 27 + 273 = 300 K

total moles = 0.250 + 0.400 = 0.650

P V = n R T

P x 3.00 = 0.650 x 0.0821 x 300

P = 5.34 atm

pressure = 5.34 atm

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Answer 2

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