Rxn =
Calculate ΔΗο. for the combustion reaction of
CH4 shown below given the following:
AHºf CH4(g) = -74.8 kJ/mol;
AHºf CO2(g) = -393.5 kJ/mol;
AHºf H2O(1) = -285.5 kJ/mol.
CH_(g) + 2 O2(g) > CO2(g) + 2 H2O(1)
=
=
A. -604.2 kJ
B. 889.7 kJ
C. -997.7 kJ
D. -889.7 kJ
E. None of the above

Sorry i dont understand the question can you rephase it?

Ican't see the chart but the density of both nickel and cobalt is 8.90 g/cm3, so it would be one of those.

Given data,temperature, t = 230°c = 503 kconsider the chemical reaction,ch₃oh(l) + 1/2 o₂(g)  → ch₂o(g) + h₂o(l)calculate the enthalpy change for the reaction: δh⁰rxn =  σnδh⁰f(products) -  σmδh⁰f(reactants)substitute the values from the table: δh⁰rxn = { -115.9 + (-285.8)} - {0+(-238.7)}              = -163.0 kjcalculate the entropy change for the reaction.δs⁰rxn = (219 + 70) - (127.2 + 1/2 x 205.2) = 59.2 j/kconsider the expression for free energy: δg⁰rxn =  δh⁰rxn - t  δs⁰rxnsubstitute the values in the above equationδg⁰rxn = -163 x 10³ - (503 x 59.2)              = -192777.6 j              = -192.777 kj