Answer:
![5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=5.31%2A10%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Explanation:
For a chemical reaction, equilibrium is a state at which the rate of the forward reaction equals that of the reverse reaction. The equilibrium constant Keq is a parameter characteristic of this state which is expressed as a ratio of the concentration of the products to that of the reactants.
For a hypothetical reaction:
xA + yB ⇄ zC
The equilibrium constant is :
![Keq = \frac{[A]^{x}[B]^{y}}{[C]^{z} }](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BA%5D%5E%7Bx%7D%5BB%5D%5E%7By%7D%7D%7B%5BC%5D%5E%7Bz%7D%20%7D)
The given reaction involves the decomposition of H2O into H2 and O2

The equilibrium constant is expressed as :
![Keq = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Since Keq = 5.31*10^-10
![5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=5.31%2A10%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Your answer to your question is: 1s² 2s² 2p⁶
<h2>

</h2>
Explanation:
1. Water decomposition
- Decomposition reactions are represented by-
The general equation: AB → A + B.
- Various methods used in the decomposition of water are -
- Electrolysis
- Photoelectrochemical water splitting
- Thermal decomposition of water
- Photocatalytic water splitting
- Water decomposition is the chemical reaction in which water is broken down giving oxygen and hydrogen.
- The chemical equation will be -

Hence, balancing the equation we need to add a coefficient of 2 in front of
on right-hand-side of the equation and 2 in front of
on left-hand-side of the equation.
∴The balanced equation is -
→ 
2. Formation of ammonia
- The formation of ammonia is by reacting nitrogen gas and hydrogen gas.
→ 
Hence, for balancing equation we need to add a coefficient of 3 in front of hydrogen and 2 in front of ammonia.
∴The balanced chemical equation for the formation of ammonia gas is as follows -
→
.
- When 6 moles of
react with 6 moles of
4 moles of ammonia are produced.
The answer is (b) for example lemon juice is an acid and it tastes sour.
Hope this helps :).