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5 May, 22:42

The reaction: 2 NO2 (g) ↔ N2O4 (g) has an equilibrium constant, Kc, of 170 at 298K. Analysis of this system at 298K reveals that 1.02E-1 mol of NO2 and 8.00E-3 mol of N2O4 are present in a 27.0-L flask. Determine the reaction quotient, Q, for this mixture. Enter your answer in scientific notation.

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  1. 6 May, 00:25
    0
    The reaction quotient Q is 20.7

    Explanation:

    Step 1: Data given

    Number of moles NO2 = 1.02 * 10^-1 = 0.102 moles

    Number of moles N2O4 = 8*10^-3 = 0.008 moles

    Volume = 27.0 L

    Step 2: The balanced equation

    2 NO2 (g) ↔ N2O4 (g)

    Step 3: Calculate concentrations

    Concentration = moles / volume

    [NO2] = moles NO2 / volume

    [NO2] = 0.102 moles / 27.0 L

    [NO2] = 0.00378 M

    [N2O4] = moles N2O4 / volume

    [N2O4] = 0.008 moles / 27.0L

    [N2O4] = 2.96 * 10^-4 M

    Q = [N2O4] / [NO2]²

    Q = (2.96 * 10^-4) / (0.00378²)

    Q = 20.7

    The reaction quotient Q is 20.7

    When Q
  2. 6 May, 01:29
    0
    Answer:do it

    Explanation:

    You are open hrjejhbg scheff
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