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July 24, 2014

Homework Help: Chemistry (please check)

Posted by Mary on Sunday, June 30, 2013 at 8:31pm.

2) Calculate the Molar Enthalpy of Neutralization (ΔHn) in kJ/mol of the reaction between a monoprotic acid and a monoprotic base, given the following information:

The temperature change equals 5.06C,
50.0 mL of 1.00 M concentration of Acid
50.0 mL of 1.00 M concentration of Base
Heat capacity of the calorimeter is 6.50 J/C.
The specific heat of water is 4.180 J/gC

This is how I calculated
5*5*5.06*6.50= 822.25/4.180= 196.7

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