A mixture consisting of 0.140 mol N2, 0.370 mol O2, 0.104 mol CH4, and 0.220 mol CO2 occupies a volume of 8.48 L at 27°C. What is the total pressure of the mixture (atm)? R = 0.08206 atm L mol^-1 K^-1.
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Given: - Moles of N2 = 0.140 mol - Moles of O2 = 0.370 mol - Moles of CH4 = 0.104 mol - Moles of CO2 = 0.220 mol Total moles = 0.140 mol + 0.370 mol + 0.104 mol + 0.220 mol Total moles = 0.834 mol Show more…
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