Question

@rushwr

Answer 1

yp?

Answer 2

4. A mixture of three gases has a pressure of 1380 mmHg at at 298 K. The mixture is analyzed and is found to contain 1.27 mol CO2, 3.04 mol CO, and 1.50 mol Ar. What is the partial pressure of Ar?
a. 238 mm Hg
b. 302 mm Hg
c. 356 mm Hg
d. 1753 mm Hg
e. 8018 mm Hg

Answer 3

@Rushwr here it is

Answer 4

partial pressure = mole fraction * total pressure

Answer 5

So C?

Answer 6

\[total pressure = P_{t} \frac{ n_{GasA} }{ n_{total} } + P_{t}\frac{ n_{GasB} }{ n_{total} }.....n_{th,Gas}\]

Answer 7

Yep, partial pressure is equivalent to the total pressure multiplied by the mole fraction.

\[Ar_{pressure} = P_{total}*\frac{ n_{Ar} }{ n_{total} } = \frac{ 1380mmHG }{ 760mmHG } *\frac{ 1.5mol_{Ar} }{ 5.81mol } = 0.468atm\]

\[0.468 atm*(\frac{ 760mmHG }{ atm }) = 0.468*760 = 356 mmHG\]

choice C.

Answer 8

Awesome !!!!!!! yu[p