Photon336
  • Photon336
@rushwr
Chemistry
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SOLVED
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chestercat
  • chestercat
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Rushwr
  • Rushwr
yp?
Photon336
  • Photon336
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
Photon336
  • Photon336
@Rushwr here it is

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Rushwr
  • Rushwr
partial pressure = mole fraction * total pressure
Rushwr
  • Rushwr
So C?
Photon336
  • Photon336
\[total pressure = P_{t} \frac{ n_{GasA} }{ n_{total} } + P_{t}\frac{ n_{GasB} }{ n_{total} }.....n_{th,Gas}\]
Photon336
  • Photon336
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.
Rushwr
  • Rushwr
Awesome !!!!!!! yu[p

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