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Math Help - Solving for heat and deltaH

  1. #1
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    Solving for heat and deltaH

    Solving for heat and deltaH?

    NaOH + KCl --> NaCl + KOH

    Given: 50 ml NaOH in 1.000M,
    50ml KCl in 1.000M
    Specific heat water: 4.18 J/gC
    density of solution: 1.000 g/ml
    change in temp: 6.3000C
    Find J (heat transferred)
    delta H (kJ/mol)


    thanks! i have no idea
    i only got to the part where you multiply density x 1000 to get mass, but im not even sure thats right.
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  2. #2
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    Q = m \cdot c \cdot \Delta T

    You have a 100 mL solution with a density of 1.000g/mL. This means you have: (100 \ \text{mL}) \cdot (1.000 \ \text{g mL}^{-1}) = 100 \ \text{g} of solution.

    So, you have the mass (m), the specific heat capacity of your solution (c, assumed to be that of water), and the change in temperature ( \Delta T). Plug in your values to get the amount of heat released.

    To find \Delta H, notice that there are (50 \ \text{{\color{red}mL}})(0.100 \ \text{mol {\color{blue}L}}^{-1}) = 0.005 \ \text{mol} of NaOH or KCl.

    This means that for every 0.005 mol of NaOH or KCl used, the value you got for Q is the amount of heat released. So how much heat will be released if exactly 1 mol of NaOH or KCl used? You can then figure out \Delta H (Look at the units, they give you big hints as to what you have to do!)
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