Answer:
a.
![Keq=([HCO_3^-][OH^-])/([CO_3^(2-)])](https://img.qammunity.org/2022/formulas/chemistry/college/td80tctyfoylw07jbcz79qbsedxuuq8ely.png)
b.
![Keq=[O_2]^3](https://img.qammunity.org/2022/formulas/chemistry/college/kmtio2c3hxyy8nevwjjays5ow10qzawawu.png)
c.
![Keq=([H_3O^+][F^-])/([HF])](https://img.qammunity.org/2022/formulas/chemistry/college/s07oy3jud0vyihpixni3504mbfwvk6ui13.png)
d.
![Keq=([NH_4^+][OH^-])/([NH_3])](https://img.qammunity.org/2022/formulas/chemistry/college/fro3qy49zqxyelg5u61batgf7stwadqqgy.png)
Step-by-step 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=([HCO_3^-][OH^-])/([CO_3^(2-)])](https://img.qammunity.org/2022/formulas/chemistry/college/td80tctyfoylw07jbcz79qbsedxuuq8ely.png)
b.
![Keq=[O_2]^3](https://img.qammunity.org/2022/formulas/chemistry/college/kmtio2c3hxyy8nevwjjays5ow10qzawawu.png)
c.
![Keq=([H_3O^+][F^-])/([HF])](https://img.qammunity.org/2022/formulas/chemistry/college/s07oy3jud0vyihpixni3504mbfwvk6ui13.png)
d.
![Keq=([NH_4^+][OH^-])/([NH_3])](https://img.qammunity.org/2022/formulas/chemistry/college/fro3qy49zqxyelg5u61batgf7stwadqqgy.png)
Regards!