Hey i need help with this 1 question. I've looked through my AP Chem book but it does not really help much with this type of problem:
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Juan performed a calorimetry experiment to determine ∆H for the following reaction: KOH (s) --> K(aq) + OH(aq)
His calorimeter contained 64.0 g of water at an initial temperature of 20.7°C. When he added 1.65 g of KOH (s) to the calorimeter, the temperature rose to 26.9°C. What value of ∆H can Juan obtain fromhis data?
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I really dont know where to start and the only formula i know for this type of problem is q=ms∆t but that does not really help finding ∆H. The only ∆H problem i have worked with are ones concerning Hess's Law.
Thanks in advanced.
Your problem is that you're getting confused by all the diffrent symbols used to represent stuff in Chem. What does ∆H stand for? The change in enthalpy, or the change in heat. Now, what does q (from the equation you said you've used, q=ms∆t) stand for? Heat, specifically the change in heat occuring because of a change in mass or temperature. So the ∆H you need to find is the same as q. All you really have to do is plug the data into the equation q=ms∆t and chug out an answer! Have fun....
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I disagree!
you can find the energy
Q=mcat
Q = 64 * 4.18 * 5.8 = 1552
If all they were looking for was the Q then they would have not given you the mass of KOH
when you add 1.65 g of KOH, that means you are adding 1.65/(39 + 16 + 1) = 0.0295 moles. So...if 0.0295 moles produces 1552 Joules of energy, that means that the delta H of dissociation for KOH = 1552/0.0295 = 52.7 kJ/mol
You know you're an AP student if...
you think studying is fun.
you constantly find yourself saying "we had homework?"
everything you know about sex, you learned in english class.
If you try to fail, and succeed, which have you done?
Oops. I guess I shoulda read the question more clearly. Thanks forcatching that, chessmaster. Sorry 'bout that, Panman. That teaches me a lesson...
The hardest thing about riding horses is the ground
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