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21 June, 16:29

Assuming ideal behavior, how many moles of argon would you need to fill a 14.0*12.0*10.0 ft room? assume atmospheric pressure of 1.00 atm, a room temperature of 20.0 ∘c, and 28.2 l/ft3. express your answer with the appropriate units.

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  1. 21 June, 19:52
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    We use the formula:

    PV = nRT

    First let us get the volume V:

    volume = 14 ft * 12 ft * 10 ft = 1,680 ft^3

    Convert this to m^3:

    volume = 1680 ft^3 * (1 m / 3.28 ft) ^3 = 47.61 m^3

    n = PV / RT

    n = (1 atm) (47.61 m^3) / (293.15 K * 8.21x10^-5 m3 atm / mol K)

    n = 1,978.13 mol
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