[heathermm]

When I have 2.0g NaOH dissolved in 149.0g water in a calorimeter at 24.0 degrees celcius the temp went up to 29.0 degrees celsius.  If the calorimeter was constant at 11 cal/degrees celcius how would I find the delta H for the solution of one mole of NaOH and water???
I tried it 3 times and got crazy numbers
i used the equation (g)(J/degree C)(Ti-Tf)=(g)(J/degree C)(Ti-Tf)?
but i am not sure this is the right equation?

[Demotivator]

The total heat evolved gets absorbed by the water plus the calorimeter, so the heat capacities of both are invoved. The constant for water is Cw = 1 cal/g degree.
For the calorimeter: Cc= 11 cal/degree
Also, you're doing the experiment on 2 g NaOH. The problem asks for heat of 1 mole, so everything gets multiplied by molecularweight/grams NaOH = 40/2.
So deltaH is:
deltaH = (40/2)[(grams water)(Cw)(Ti-Tf) + (Cc)(Ti-Tf)] calories per mole

[pizza1512]

I know that..

exothermic reactions give out heat...

edothermic reaction need heat...

but what is an athermic reaction?...

Gan anyone give an example of one?...