Determine the enthalpy of formation for glucose (C6H1206) given the following enthalpies of formation: CO2(g), -393.5 kJ/mol; H20(), -285.8 kJ/mol and the enthalpy for the reaction shown below. C6H1206(s) + 6O2 (g) → 6CO2(g) + 6H2O() AH = -2801 kJ %D Select one: a. +2122 kJ/mol b. -1275 kJ/mol C. -2122 kJ/mol d. +1275 kJ/mol e. +875 kJ/mol

A scientist measures the standard enthalpy change for the following reaction to be 80.1 kJ : 3Fe,O3(s) + H2(g)2FE3O4(s) + H,O(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of Fe,O3(s) is kJ/mol.

Calculate the standard enthalpy of formation for glucose (C6H12O6) (in kJ/mol), given the following values: ΔH°combustion of glucose = -2800.8 kJ; ΔH°f, CO2 = -393.5 kJ/mol, ΔH°f, H2O = -285.8 kJ/mol.

Tetrachloromethane, CCl4, is formed from methane and chlorine in the following reaction: CH4 (g) + 4 Cl₂ (g) → 4 HCl(g) + CCl4 (g) = Calculate A₁H, the enthalpy change for this reaction, using the enthalpies of formation of CO₂ (g), H₂O(l), and CCl4 (g), (▲ƒ H° = enthalpy changes for the following reactions: CH4 (g) + 2 O2(g) → 2 H₂O(l) + CO₂(g) ΔΗ° :-890.4 kJ/mol-rxn 2 HCl(g) → H₂(g) + Cl₂ (g) ΔΗ° +184.6 kJ/mol-rxn = ΔΗ° = Compound AfH (kJ/mol) H₂O(g) CO₂(g) -285.8 -393.5 kJ/mol-rxn - 95.98 kJ/mol), and the