<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:
Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
When Ni(OH)₂ starts precipitate :
Ksp of Ni(OH)₂ = [ Ni²⁺ ] [ OH²⁻ ]
5.5x10⁻¹⁶ = [ 0.18 ] [ OH²⁻ ]
[ OH²⁻ ] = 5.5x10⁻¹⁶ / 0.18
[ OH⁻ ] = 5.5 × 10⁻⁸ M
pOH = 7.2
therefore , pH = 14 - 7.2
pH = 6.8
Thus, Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
To learn more about pH here
brainly.com/question/15289741
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I’m pretty sure the answer would be natural
They are atoms. a molecule is made up of atoms.
