Posted by Kristy on Wednesday, November 10, 2010 at 5:10pm.
0.1 L x 0.300 M NaOH = 0.0300 moles NaOH.
0.1 L x 0.300 M HNO3 = 0.0300 moles HNO3.
Mass water (200 mL) x specific heat water x (Tfinal-Tinitial) = q in joules. This is delta H.
delta H/mol is delta H from above divided by 0.03 mole.
Then you need to put it in kJ/mol for the question.
Ahh I had this right but calculated it wrong. Thanks a lot :)
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