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9 September, 01:47

The pressure exerted by 0.002 moles of a gas in a 500 mL container at 25oC is:

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  1. 9 September, 04:46
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    0.1715 atm

    Explanation:

    We use the Ideal gas equation to calculate this

    Mathematically;

    PV = nRT

    rearranging, we have

    P = nRT/V

    from the question, we have

    n = 0.002 moles

    R = 0.082

    T = 250 = 250 + 273 = 523 K

    V = 500 ml = 500/1000 = 0.5 L

    Plugging the values we have;

    P = (0.002 * 0.082 * 523) / 0.5

    P = 0.1715 atm
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