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- Answer 1,2,3,4,5. Take note of the directionCalculate Delta Hof for PbO (s) using the data below. PbO (s) + CO (g) ® Pb (s) + CO2 (g) DH° = –131.4 kJ DHf°: CO2(g) = –393.5 kJ/mol; CO(g) = –110.5 kJ/mol –151.6 kJ/mol –283.0 kJ/mol +283.0 kJ/mol –372.6 kJ/mol +252.1 kJ/molMy question is for the part circled in pencil, how do we get -12,970.4 and then -25,940.8 j/mole. Are we normalizing twice? If so, why?
- For question 1,2,3,4 What will the shift be? How do you know? Options: shift left, shift right, or no shiftCalculate the ΔG°rxn using the following information.4 HNO3(g) + 5 N2H4(l) → 7 N2(g) + 12 H2O(l) ΔG°rxn = ?ΔH°f (kJ/mol)HNO3 = -133.9N2H4 = 50.6N2 = 0.0H2O = -285.8S°(J/mol∙K)HNO3 = 266.9N2H4 = 121.2N2 = 191.6H2O = 70.0Don't understand 4 coming out with a number slightly off