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18 December, 16:30

At standard temperature and pressure, a 0.50 mol sample of H2 gas and a separate 1.0 mol sample of O2 gas have the same (A) average molecular knetic energy

(B) average molecular speed

(C) volume

(D) effusion rate

(E) density

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Answers (1)
  1. 18 December, 16:57
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    Option E density

    Explanation:

    This is actually pretty easy to explain. At the beggining it states that at STP which are 1 atm and 273 K, we have two samples of H2 but in different quantities. So using the ideal gas equation, we calculate the volume of each gas:

    PV = nRT (1)

    Where:

    P: pressure in atm

    V: volume in L

    n: moles

    R: gas constant which is 0.082 L atm / K mol

    T: temperature in K

    So from this equation, we solve for V:

    V = nRT/P

    Replacing data for both samples we have:

    V1 = 0.5 * 0.082 * 273 / 1 = 11.19 L

    V2 = 1 * 0.082 * 273 / 1 = 22.38 L

    Now, to verify that is option E, let's write the expression for density:

    d = m/V (2)

    Where:

    d: density

    m: mass

    To calculate the mass, we use the molar weight of hydrogen (2 g/mol) and the moles of the samples so:

    m1 = 0.5 * 2 = 1 g

    m2 = 1 * 2 = 2 g

    Now, replacing in (2):

    d1 = 1 / 11.19 = 0.0893 g/L

    d2 = 2 / 22.38 = 0.0893 g/L

    As d1 = d2 we can conclude that option E is the correct option.
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