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What volume does a mixture of 14.2g of helium and 21.6g of hydrogen gas occupy at 28 degrees celcius and .985atm?

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Convert 14.2 g He and 21.6g H2 to moles, add them, and substitute them for n into PV = nRT. Solve for V.

V=359L

My number is 358 when rounded to 3 s.f. Check your figures. I did not round the molar masses.

I almost forgot. Note the correct spelling of celsius.

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