Posted by Lisa on Tuesday, June 28, 2011 at 7:09pm.
I saw your post earlier and didn't understand it; therefore, I passed it by. I assume no one else understood it either since you've received no responses. Perhaps you could rephrase the question because it makes no sense to me as is.
CH3OH (l) +3/2O2 (g) = CO2 (g) + 2H2O(l) H = -726.5 kJ
Part A:
What is the enthalpy change for the reverse reaction?
Express your answer using four significant figures.
Answer I got right: 726.5 kJ
Part B:
Balance the forward reaction with whole-number coefficients:
CH3OH (l) +3/2O2 (g) = CO2 (g) + 2H2O(l) H
Answer I got right: 2,3,2,4
Part C:
What is for the reaction represented by this equation?
Express your answer using four significant figures.
Answer : I am stuck. ?????
Energy, q, w, Enthalpy, and StoichiometrySource(s):
This is the subject that I am on.
Opps I left off what is H
Part C:
What is H for the reaction represented by this equation?
Express your answer using four significant figures.
Answer : I am stuck. ?????
I think the upper post took care of things.
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