Iodine, I2, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature in a 83.3-mL

Question

Iodine, I2, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature in a 83.3-mL bulb that contains 0.392 g of I2 vapor at a pressure of 0.562 atm

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Thành Công 3 months 2021-07-30T17:36:27+00:00 1 Answers 10 views 0

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    2021-07-30T17:37:33+00:00

    Answer:

    T = 3.75 K

    Explanation:

    As we know

    PV = nRT

    R =8.3144598 J. mol-1. K-1

    P = 0.562 atm

    V = 83.3 mL

    moles in 0.392 g of I2 = 0.392/mass of I2 = 0.392 grams/253.8089 g/mol = 0.0015 moles

    Substituting the given values, we get

    0.562 atm * 83.3 *10^-3 L = 0.0015 moles * 8.3144598 J. mol-1. K-1 * T

    T = 3.75 K

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