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A 8.40-L container holds a mixture of two gases at 27 °C. The partial pressures of gas A and gas B,

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A 8.40-L container holds a mixture of two gases at 27 °C. The partial pressures of gas A and gas B, respectively, are 0.162 atm and 0.888 atm. If 0.160 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Mar 19th, 2015

p1=0.162atm

p2=0.888atm

p3=nRT/V= 0.16*(0.0821*273/8.4)

p3=0.427atm

total pressure= p1+p2+p3=0.162+0.888+0.427=1.477atm.........................please best me

Mar 19th, 2015

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