Question


8)(20) Use the thermodynamic tables to determine AH for 12H*(aq) + SCH,OH(1) + 4MnO (aq) = 4Mnº(aq) + SCH,COOH(aq) + 11H20(1)



Enthapies of Formation H(aq): 0 kJ/mol; C,H,OH(1): -277.6 kJ/mol; MnO, (aq): -529.9 kJ/mol Mn(aq): -219.4 kJ/mol; CH COOH(aq
0 0
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Answer #1

+479104 (29) 40724 + + 12H+ +5 GH OH (1) Cool +11H2O சரOH (aq) (1) AH = {dHproduct - EoH reactant ) - (12.04 H+ +5AH + 40H GHДН – Гря (Gиви-з)+1х (-26 s. .)+ ця (- 21:)) - [12xo +5х-277 - )+ Их -529-41)] – 2,215 — 3143-8 – 317,6 о - 1388 - 2111 -6 -

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