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Photon336

  • one year ago

@rushwr

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  1. Rushwr
    • one year ago
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    yp?

  2. Photon336
    • one year ago
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    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

  3. Photon336
    • one year ago
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    @Rushwr here it is

  4. Rushwr
    • one year ago
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    partial pressure = mole fraction * total pressure

  5. Rushwr
    • one year ago
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    So C?

  6. Photon336
    • one year ago
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    \[total pressure = P_{t} \frac{ n_{GasA} }{ n_{total} } + P_{t}\frac{ n_{GasB} }{ n_{total} }.....n_{th,Gas}\]

  7. Photon336
    • one year ago
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    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.

  8. Rushwr
    • one year ago
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    Awesome !!!!!!! yu[p

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