Acetylene burns in air according to the following equation
Acetylene burns in air according to the following equation. C2H2(g) + 5/2 O2(g) 2 CO2(g) + H2O(g) H0rxn = -1255.8 kJ Given that H0f of CO2(g) = -393.5 kJ/mol and H0f of H2O(g) = -241.8 kJ/mol, what is H0f of C2H2(g)?
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Acetylene burns in air according to the following equation. C2H2(g) + 5/2 O2(g) 2 CO2(g) + H2O(g) H0rxn = -1255.8 kJ Given that H0f of CO2(g) = -393.5 kJ/mol and H0f of H2O(g) = -241.8 kJ/mol, what is H0f of C2H2(g)
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