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When 0.689 g of Ca metal is added to 200.0 mL of 0.500 M HCl(aq), a temperature increase of 112C is observed. Assume the soluti...Asked by Libit
When 0.678 g of Ca metal is added to 200.0 mL of 0.500 M HCl(aq), a temperature increase of 11.0C is observed.
Assume the solution's final volume is 200.0 mL, the density is 1.00 g/mL, and the heat capacity is 4.184 J/gC.
(Note: Pay attention to significant figures. Do not round until the final answer.)
The molar heat of reaction, H rxn, for the reaction of
Ca(s) + 2H+(aq) -> Ca2+(aq) + H2(g)
THANKS! :)
Assume the solution's final volume is 200.0 mL, the density is 1.00 g/mL, and the heat capacity is 4.184 J/gC.
(Note: Pay attention to significant figures. Do not round until the final answer.)
The molar heat of reaction, H rxn, for the reaction of
Ca(s) + 2H+(aq) -> Ca2+(aq) + H2(g)
THANKS! :)
Answers
Answered by
DrBob222
q = mass water x specific heat water x delta T.
Then q/0.678 = ?J/g. Convert that to kJ/mole.
Then q/0.678 = ?J/g. Convert that to kJ/mole.
Answered by
DrBob222
OK. I just read your earlier post.
200 x 4.184 x 11 is correct.
And that number divided by 0.678 is correct for J/g.
Just take that number (J/g) and divide by 1000 to convert to kJ, then multiply by molar mass Ca to convert to kJ/mole.
200 x 4.184 x 11 is correct.
And that number divided by 0.678 is correct for J/g.
Just take that number (J/g) and divide by 1000 to convert to kJ, then multiply by molar mass Ca to convert to kJ/mole.
Answered by
Anonymous
It is negative fyi
Answered by
labkid
why do you multiply by 200? how to the units of mL just disappear then? what do you do with 1.00 g/mL?
Answered by
student
just type it in. it works when its negative.
Answered by
DrBob222
You would times it by the momentum currency of memorandum which is stated in Hess's law of physics
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