posted by Kristy .
A 100.0 ml sample of 0.300 M NaOH is mixed with a 100.0 ml sample of 0.300 M HNO3 in a coffee cup calorimeter. Both solutions were initially at 35.0 degrees celcius; the temperature of the resulting solution was recorded at 37.0 degrees celcius. Determine the delta H (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HNO3. Assume no heat is lost to the calorimeter or the surroundings; and the density and heat capacity of the resulting solution are the same as water.
What is delta H?
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 :)