Approximately 1.000 g each of four gasses H2, Ne, Ar, and Kr are placed in a sealed container all under1.5 atm of pressure. Assuming ideal b

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Approximately 1.000 g each of four gasses H2, Ne, Ar, and Kr are placed in a sealed container all under1.5 atm of pressure. Assuming ideal behavior, determine the partial pressure of the H2 and Ne

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Neala 6 months 2021-09-04T11:52:03+00:00 1 Answers 0 views 0

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    2021-09-04T11:53:56+00:00

    Answer:

    The partial pressure of H2 is 0.375 atm

    The partial pressure of Ne is also 0.375 atm

    Explanation:

    Mass of H2 = 1 g

    Mass of Ne = 1 g

    Mass of Ar = 1 g

    Mass of Kr = 1 g

    Total mass of gas mixture = 1 + 1 + 1 + 1 = 4 g

    Pressure of sealed container = 1.5 atm

    Partial pressure of H2 = (mass of H2/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm

    Partial pressure of Ne = (mass of Ne/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm

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