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1 May, 20:00

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A container of what volume would be required to store 60.0 moles of a gas at a pressure of 20.0 atm and a temperature of 50.0 Celsius?

79.6 L

12.3 L

1250 L

8060 L

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  1. 1 May, 20:20
    0
    N: 60 mol

    p: 20 atm = 20*1013hPa = 20260 hPa

    T: 50°C = 323 K

    R: 83,14 hPa·dm³/mol·K

    V:?

    ...

    pV = nRT

    V = nRT/p

    V = (60*83,14*323) / 20260

    V = 79,529 dm³ ≈ 79,6 dm³

    79,6 dm³ = 79,6 L

    :•)
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