Answer:
The 
of a substrate will be "10 μM".
Explanation:
The given values are:
![[Substract] = 40 \ \mu M](https://tex.z-dn.net/?f=%5BSubstract%5D%20%3D%2040%20%5C%20%5Cmu%20M)
Reaction velocity, 
As we know,
⇒ ![Vo=\frac{K_{cat}[E_{t}][S]}{K_{m}+[S]}](https://tex.z-dn.net/?f=Vo%3D%5Cfrac%7BK_%7Bcat%7D%5BE_%7Bt%7D%5D%5BS%5D%7D%7BK_%7Bm%7D%2B%5BS%5D%7D)
On putting the estimated values, we get
⇒ 
⇒ 
⇒ 
On subtracting "40" from both sides, we get
⇒ 
⇒ 
Answer:
C. Valence electrons
Explanation:
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Answer:
The answer is B. I took the test and made a 100
Explanation:
Answer:
0.51
Explanation:
Given the Nernst equation;
E= E° - 0.0592/n logQ
E= 355 mV or 0.355 V
E° = 0.34 - 0= 0.34 V
n= 2(two electrons were transferred in the process)
Equation of the reaction;
H2(g) + Cu^2+(aq) -----> 2H^+(aq) + Cu(s)
Substituting values;
0.355 = 0.34 - 0.0592/2 log([H^+]/1)
0.355 - 0.34 = - 0.0296 log [H^+]
0.015/-0.0296 = log [H^+]
Antilog (-0.5068) = [H^+]
[H^+] = 0.311 M
pH = -log[H^+]
pH= - log(0.311 M)
pH = 0.51
