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What does 1vol. of hydrogen peroxide correspond to in moldm-3? watch

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    The title explains itself

    I need to draw a log(rate) against Log(H2O2) to find the order of reaction.
    The rate equation is rate = k[H2O2]
    T
    Thanks
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    (Original post by Marseille013)
    The title explains itself

    I need to draw a log(rate) against Log(H2O2) to find the order of reaction.
    The rate equation is rate = k[H2O2]
    T
    Thanks
    Imagine that you have 1 litre of 1 volume hydrogen peroxide. This will release 1 litre of oxygen gas (at room temperature), which is equal to approx 1/24 = 0.0417 mol

    The equation for the decomposition is:

    2H2O2 --> 2H2O + O2

    hence moles of hydrogen peroxide = 2 x moles of oxygen.

    Therefore there are 0.0417 x 2 mol of hydrogen peroxide per litre = 0.0833 mol dm-3
 
 
 
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