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8 October, 17:57

what is the pressure in mm of Hg, of a gas mixture that contains 1g of H2, and 8.0g of Ar in a 3.0 L container at 27 degree C

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  1. 8 October, 18:51
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    4367.72 mmHg

    Explanation:

    Given dа ta:

    Mass of hydrogen = 1 g

    Mass of argon = 8 g

    Volume of container = 3.0 L

    Temperature = 27 °C = 27 + 273 = 300 K

    Pressure of mixture = ?

    Solution:

    Moles of hydrogen:

    Number of moles = mass / molar mass

    Number of moles = 1 g / 2 g/mol

    Number of moles = 0.5 mol

    Moles of argon:

    Number of moles = mass / molar mass

    Number of moles = 8 g / 40 g/mol

    Number of moles = 0.2 mol

    Pressure of hydrogen:

    PV = nRT

    P = nRT/V

    P = 0.5 mol * 0.0821 atm. L. mol⁻¹. K⁻¹ * 300 K / 3.0 L

    P = 12.315 atm / 3.0

    P = 4.105 atm

    Pressure of argon:

    PV = nRT

    P = nRT/V

    P = 0.2 mol * 0.0821 atm. L. mol⁻¹. K⁻¹ * 300 K / 3.0 L

    P = 4.926 atm / 3.0

    P = 1.642 atm

    Total pressure:

    T (total) = P (H₂) + P (Ar)

    T (total) = 4.105 atm + 1.642 atm

    T (total) = 5.747 atm

    atm to mmHg:

    5.747 * 760 = 4367.72 mmHg
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