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Yesterday, 20:38

A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank.

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  1. Yesterday, 21:45
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    Mole fraction of N₂O = 0.330

    Mole fraction of SF₄ = 0.669

    Pressure (N₂O) = 39.12 kPa

    Pressure (SF₄) = 79.12 Pa

    Total Pressure = 118.25 kPa

    Explanation:

    Given

    Given mass of dinitrogen difluoride N₂0 = 5.53 g

    Given mass of sulphur hexafluoride = 17.3 g

    volume of tank V = 8 L = 0.008 ml

    R ideal gas constant = 8.31 J / mol·K

    Temperature = 26.9 C = 299.9 k

    Pressure = ?

    Number of moles = ?

    Number of moles of N₂O in tank = given mass / molar mass = 5.53 / (14.0 x 2 + 16.0) = 0.1256 mol

    Number of moles SF₄ in tank = given mass / molar mass 17.3 / (32.1 + 19.0 x 4) = 0.254 mol

    Mole fraction of N₂O = Number of moles of N₂0/total number of moles

    = 0.1256 / (0.1256+0.254) = 0.330

    Mole fraction of SF₄ = 0.254 / (0.1256+0.254 = 0.669

    For Pressure

    PV=nRT

    Pressure (N₂O) = (0.1256) (8.31) (299.9) / (0.008) = 187.22 = 39127.05 Pa = 39.12 kPa

    Pressure (SF₄) = (0.254) (8.31) (299.9) / (0.008) = 79126.36 Pa = 79.12 Pa

    Total Pressure = 39.12 + 79.12 = 118.25 kPa
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