The total number of neutrons in the nucleus of a k-37 atom is 18 neutrons.
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Answer:
The answer is 0.01365.
Explanation:
C(t) = 0.0225te^?0.0467t
We can derivate,
C'(t) = d/dt ( 0.0225 te^-0.0467t)
after solving,
C'(t) = 0.0225 (e^-0.0467t - (0.0467e^-0.0467t . t)
Plug in t =5,
Answer is C'(t) = 0.01365.
Answer:
0.174 M
Explanation:
The same can be solve by using Nernst's equation
The Nernst's equation is:
![Ecell=E^{0}_{cell}-\frac{0.0592}{n}log\frac{[anodic]}{[cathodic]}](https://tex.z-dn.net/?f=Ecell%3DE%5E%7B0%7D_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7Dlog%5Cfrac%7B%5Banodic%5D%7D%7B%5Bcathodic%5D%7D)
For silver cell
n= 1
As both the compartments have silver nitrate solution,t the standard emf of cell will be zero.
Given
Ecell = 0.045
[Anodic compartment]= 1 M
Putting values


Taking antilog and solving
[AgNO3]=0.174 M