NADH ------------> NAD^+ + H^+ + 2e^- E0 = 0.320v
NO3^- + 2H^+ +2e^- ------> NO2^- + H2O E0 = 0.421v
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NADH + NO3^- + H^+ ------------> NAD^+ + NO2^- + H2O E0cell = 0.741v
n = 2
G0 = -nE0cell*F
= -2*0.741*96500
= -143013J/mole
= -143KJ/mole >>>answer
8 mol of Acetyl-CoA
24 mol of NADH
8 mol of FADH2
Question 4 (1 point) Given the following reaction, what is the AGC for the spontaneous reaction...
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