The small part in the center of the atom is called a nucleus
Answer: 
Explanation:
The balanced chemical equation will be:

Here Ag undergoes oxidation by loss of electrons, thus act as anode. Nickel undergoes reduction by gain of electrons and thus act as cathode.

Where both
are standard reduction potentials.
![E^0_{[Ag^{+}/Mg]}=+0.80V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FMg%5D%7D%3D%2B0.80V)
![E^0_{[Ni^{2+}/Ni]}=-0.25V](https://tex.z-dn.net/?f=E%5E0_%7B%5BNi%5E%7B2%2B%7D%2FNi%5D%7D%3D-0.25V)
![E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Ag^{+}/Ag]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BNi%5E%7B2%2B%7D%2FNi%5D%7D-%20E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D)

The standard emf of a cell is related to Gibbs free energy by following relation:

= gibbs free energy
n= no of electrons gained or lost =?
F= faraday's constant
= standard emf

The Gibbs free energy is related to equilibrium constant by following relation:

R = gas constant = 8.314 J/Kmol
T = temperature in kelvin =
K = equilibrium constant



Thus the value of the equilibrium constant at
is 
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Answer:
3.65 g / ml correct to 3 sig. fig.
Explanation:
The computation of the concentration required is shown below:
As we know that
[A] = mass of solute ÷ volume of solution
Before that first find the mass of solute
Given that
Initial weight = 5.55g
And,
Final weight = 92.7 g
So,
Mass of KCl is
= 92.7 - 5.55
= 87.15 g ~ 87.2 g
Now the KCi is fully dissolved, so the volume is 23.9 ml
So, concentration is
= 87.2 g ÷ 23.9 ml
= 3.65 g / ml correct to 3 sig. fig.
Answer:
D
Explanation:
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