<span>Ka is an equilibrium constant for the partial ionization of "weak" acids in water.</span>
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Complete Question
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Answer:
The solution to this question is shown on the second uploaded image
Explanation:
<u>Answer:</u> The expression for equilibrium constant is ![K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BHOCl%5D%5E2%7D%7B%5BH_2O%5D%5BCl_2%5D%5E2%7D)
<u>Explanation:</u>
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
For the general chemical equation:

The expression for
is given as:
![K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
For the given chemical reaction:

The expression for
is given as:
![K_{eq}=\frac{[HOCl]^2[HgO.HgCl_2]}{[HgO]^2[H_2O][Cl_2]^2}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BHOCl%5D%5E2%5BHgO.HgCl_2%5D%7D%7B%5BHgO%5D%5E2%5BH_2O%5D%5BCl_2%5D%5E2%7D)
The concentration of solid is taken to be 0.
So, the expression for
is given as:
![K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BHOCl%5D%5E2%7D%7B%5BH_2O%5D%5BCl_2%5D%5E2%7D)
6 miles of H2O is equal to 12 Hydrogen molecules and 6 oxygen molecules. Equaling 18 in total.