It can be found that 337.5 g of AgCl formed from 100 g of silver nitrate and 258.4 g of AgCl from 100 g of CaCl₂.
<u>Explanation:</u>
2AgNO₃ + CaCl₂ → 2 AgCl + Ca(NO₃)₂
We have to find the amount of AgCl formed from 100 g of Silver nitrate by writing the expression.
![100 g \text { of } A g N O_{3} \times \frac{2 \text { mol } A g N O_{3}}{169.87 g A g N O_{3}} \times \frac{2 \text { mol } A g C l}{1 \text { mol } A g N O_{3}} \times \frac{143.32 g A g C l}{1 \text { mol } A g C l}](https://tex.z-dn.net/?f=100%20g%20%5Ctext%20%7B%20of%20%7D%20A%20g%20N%20O_%7B3%7D%20%5Ctimes%20%5Cfrac%7B2%20%5Ctext%20%7B%20mol%20%7D%20A%20g%20N%20O_%7B3%7D%7D%7B169.87%20g%20A%20g%20N%20O_%7B3%7D%7D%20%5Ctimes%20%5Cfrac%7B2%20%5Ctext%20%7B%20mol%20%7D%20A%20g%20C%20l%7D%7B1%20%5Ctext%20%7B%20mol%20%7D%20A%20g%20N%20O_%7B3%7D%7D%20%5Ctimes%20%5Cfrac%7B143.32%20g%20A%20g%20C%20l%7D%7B1%20%5Ctext%20%7B%20mol%20%7D%20A%20g%20C%20l%7D)
= 337.5 g AgCl
In the same way, we can find the amount of silver chloride produced from 100 g of Calcium chloride.
It can be found as 258.4 g of AgCl produced from 100 g of Calcium chloride.
Answer : The correct expression for equilibrium constant will be:
![K_c=\frac{[C]^8}{[A]^4[B]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5E8%7D%7B%5BA%5D%5E4%5BB%5D%5E2%7D)
Explanation :
Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.
The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.
As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.
The given equilibrium reaction is,
![4A+2B\rightleftharpoons 8C](https://tex.z-dn.net/?f=4A%2B2B%5Crightleftharpoons%208C)
The expression of
will be,
![K_c=\frac{[C]^8}{[A]^4[B]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5E8%7D%7B%5BA%5D%5E4%5BB%5D%5E2%7D)
Therefore, the correct expression for equilibrium constant will be, ![K_c=\frac{[C]^8}{[A]^4[B]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5E8%7D%7B%5BA%5D%5E4%5BB%5D%5E2%7D)
Opposite pairs form ionic bonds, due to this the answer is D - Li and Br; They have unlike charges.
With high cholesterol, you can develop fatty deposits in your blood vessels. Eventually, these deposits grow, making it difficult for enough blood to flow through your arteries. Sometimes, those deposits can break suddenly and form a clot that causes a heart attack or stroke.
The cell notation is:
![Mg(s)|Mg^{+2}(aq)||Ag^{+}(aq)|Ag(s)](https://tex.z-dn.net/?f=Mg%28s%29%7CMg%5E%7B%2B2%7D%28aq%29%7C%7CAg%5E%7B%2B%7D%28aq%29%7CAg%28s%29)
here in cell notation the left side represent the anodic half cell where right side represents the cathodic half cell
in anodic half cell : oxidation takes place [loss of electrons]
in cathodic half cell: reduction takes place [gain of electrons]
1) this is a galvanic cell
2) the standard potential of cell will be obtained by subtracting the standard reduction potential of anode from cathode
![E^{0}_{Mg}=-2.38V](https://tex.z-dn.net/?f=E%5E%7B0%7D_%7BMg%7D%3D-2.38V)
![E^{0}_{Ag}=+0.80V](https://tex.z-dn.net/?f=E%5E%7B0%7D_%7BAg%7D%3D%2B0.80V)
Therefore
![E^{0}_{cell}=0.80-(-2.38)=+3.18V](https://tex.z-dn.net/?f=E%5E%7B0%7D_%7Bcell%7D%3D0.80-%28-2.38%29%3D%2B3.18V)
3) as the value of emf is positive the reaction will be spontaneous as the free energy change of reaction will be negative
Δ![G^{0}=-nFE^{0}](https://tex.z-dn.net/?f=G%5E%7B0%7D%3D-nFE%5E%7B0%7D)
As reaction is spontaneous and there will be conversion of chemical energy to electrical energy it is a galvanic cell.