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    Hi,

    So currently working through some questions and wanted some clarification over why what I did was wrong.

    The question is: calculate the temperature is 10cm^3 of 0.1mol dm^-3 NaOH is reacted with 20cm^3 of 0.1mol dm^-3 HCl. We're told that the enthalpy change of neutralisation is -58kJmol^-1.

    So what I did is:
    n=cv
    n=0.1x30
    n=3mol
    3/1000=0.003
    0.003x-58000=-174
    then rearranged mcΔt to
    Δt=-174/4.18x30 to get -1.39C however that isn't correct. Can anyone tell me why and where I went wrong? The answer in the textbook is 0.46°C.

    Thanks.
 
 
 
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