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14 July, 13:28

You have a partially filled party balloon with 2.00 g of helium gas. you then add 2.74 g of hydrogen gas to the balloon. assuming constant temperature and pressure, how many times bigger is the party balloon - comparing before and after the hydrogen gas has been added?

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  1. 14 July, 13:42
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    First, we need to get moles of He:

    moles of He = mass/molar mass of He

    when the molar mass of He = 4 g/mol

    and when the mass = 2 g

    by substitution:

    moles of He = 2 g / 4 g/mol

    = 0.5 moles

    and when V = nRT/P and n is the number of moles

    so, V1 = 0.5RT/P

    then, we need moles of H2 = mass / molar mass

    = 2.74g / 2g/mol

    = 1.37 moles

    ∴ moles of He + moles of H2 = 0.5 moles + 1.37 moles

    = 1.87 moles

    so, V2 = 1.87RT/P

    from the V1 and V2 formula:

    ∴V2/V1 = 1.87 / 0.5

    = 3.74

    ∴ the party balloon is 3.74 times bigger
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