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3 November, 22:40

What volume of beaker contains exactly 2.23x10^-2 mol of nitrogen gas at STP?

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  1. 3 November, 22:56
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    V = 0.5 L

    Explanation:

    Given dа ta:

    Moles of nitrogen = 2.23*10⁻² mol (0.0223 mol)

    Temperature = 273 K

    Pressure = 1 atm

    Volume = ?

    Solution:

    PV = nRT

    V = nRT / P

    V = 0.0223 mol * 0.0821 atm. mol⁻¹. L. k⁻¹ * 273 K / 1 atm

    V = 0.5 L
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