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Veda Glover

Lvl 10
4y ago
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11y ago

i suppose that the 1.00 atm of dry air is actually 1.00m^3 dry air
PV=nRT
=> n=PV/RT

P= 711tor/760torr = 0.936 atm
V= 1.00m^3 = 1000L of dry air

Find mol of dry air
n=(0.936atm)(1000L)/(0.08206)(286.15K)
=39.9 mol of dry air

Mol of CO2
39.9* 0.000375 = 0.0150 mol of CO2 * 44.01 g/mol = 0.660 g CO2

Answer:
0.660g CO2

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Q: What mass of carbon dioxide is present in 1.00 m3 of dry air at a temperature of 25 ∘C and a pressure of 725 torr ?
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