Calculate the enthalpy change for the reaction Fe2O3(s) + CO(g) 2FeO(s) + CO2(g) from the following: Fe2O3 (s) + 3CO(g) 2Fe(s) + 3CO2(g) H = _ 23.4kJ , FeO(s) + CO(g) Fe(s) + CO2(g) H = _ 10.9kJ.

Idon’t know what the answer is i wish i could

(no3)⁻n has oxidation number x,o has oxidation number -2.(x*1 -2*3) charge of the nitrate ion, and we know that charge of the nitrate ion =-1.so,(x*1 -2*3) =-1x-6=-1x=+5answer a. +5.

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.