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4. Use Hess' Law to calculate the enthalpy change (∆H°f) for this reaction:
6C(s) + 6H2(g) + 3O2(g)C6H12O6(s)

using the following equations:
A. C(s) + O2(g)CO2(g)
B. H2(g) + 1/2O2(g)H2O(g)
C. C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l)
D. H2O(l)H2O(g)

∆H°= -393.5 kJ
∆H°= -241.8kJ
∆H°= -2803.0 kJ
∆H°= +40.7 kJ

Why is this change in enthalpy given the term ∆H°f?
4. Use Hess Law to calculate the enthalpy change (AH®t) for this reaction: (12 pts.) 6C(s) + 6H2(g) + 302(0)→ CH12O6/s) usin
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