<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.
Answer: C) Both are made up of monomers that are linked by covalent bonds
Explanation: Polymers are large molecules which are formed by combination of small repeating units called as monomers.
Natural polymers are found in nature such as cellulose whereas synthetic polymers are synthesized in laboratories such as Nylon 6,6.
All the polymers, whether natural or synthetic are made up of monomers and are joined by covalent bonds.
Answer:
a.
![Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BHCO_3%5E-%5D%5BOH%5E-%5D%7D%7B%5BCO_3%5E%7B2-%7D%5D%7D)
b.
![Keq=[O_2]^3](https://tex.z-dn.net/?f=Keq%3D%5BO_2%5D%5E3)
c.
![Keq=\frac{[H_3O^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
d.
![Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BNH_4%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_3%5D%7D)
Explanation:
Hello there!
In this case, for the attached reactions, it turns out possible for us to write the equilibrium expressions by knowing any liquid or solid would be not-included in the equilibrium expression as shown below, with the general form products/reactants:
a.
![Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BHCO_3%5E-%5D%5BOH%5E-%5D%7D%7B%5BCO_3%5E%7B2-%7D%5D%7D)
b.
![Keq=[O_2]^3](https://tex.z-dn.net/?f=Keq%3D%5BO_2%5D%5E3)
c.
![Keq=\frac{[H_3O^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
d.
![Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BNH_4%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_3%5D%7D)
Regards!