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Calculate the actual AG for the reaction in which molecular oxygen is reduced by the electron carrier NADH. 1202 + NADH+H→ H2

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Answer #1

/ O₂ + NADUTH H₂O & NADO E° = 0816-(-0.320) v =+ 1.136 V E=60+ 0.050 log [02] [NADA] s [220] [NADY = 1.136+ 0.059 log (1)12x1

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