Provide three specific examples of how this cloning is used.
1 answer:
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Correct answer: Option D, <span>K = 5.04 × 10^52
</span>
Reason:
We know that,
Ecell =
,
where n = number of electrons = 2 (in present case)
K = equilibrium constant.
Also, Ecell = <span>+1.56 v
Therefore, 1.56 = </span>
Therefore, log (K) = 52.703
Therefore, K = 5.04 X 10^52
</span>
Reason:
We know that,
Ecell =
where n = number of electrons = 2 (in present case)
K = equilibrium constant.
Also, Ecell = <span>+1.56 v
Therefore, 1.56 = </span>
Therefore, log (K) = 52.703
Therefore, K = 5.04 X 10^52