Find k for this redox reaction when e = +1.05 v: ni + 2ag+ → ni2+ + 2ag

Number of electrons that was displaced in this reaction = 2.
E⁰ = (0.0592/n)log K

log K = (E⁰*n)/0.0592 = 1.05*2/0.0592

K = 10^(1.05*2/0.0592)= 2.97 *10³⁵

This is what i found on google “when one phosphate group is removed by breaking a phosphoanhydride bond in a process called hydrolysis, energy is released, and atp is converted to adenosine diphosphate (adp). likewise, energy is also released when a phosphate is removed from adp to form adenosine monophosphate (amp).” i hope this you .