Answer:
a.
![K=([BaO]^2)/([Ba]^2[O_2])](https://img.qammunity.org/2022/formulas/chemistry/high-school/3h2iwcnzw0404dzt8c9o7xfd3rglvhrfh2.png)
b.
![K=([MgO]^2)/([Mg]^2[O_2])](https://img.qammunity.org/2022/formulas/chemistry/high-school/12yzzqvhn19ult8zffy3wh3en1fatmfbg9.png)
c.
![K=([P_4O_(10)]^2)/([P_4][O_2]^5)](https://img.qammunity.org/2022/formulas/chemistry/high-school/33mwfs5ir7mpl7g6aqgygsnrfvb797905y.png)
Step-by-step explanation:
Hello!
In this case, since the equilibrium expression is set up by dividing the products over the reactants and powering to the stoichiometric coefficient, we can proceed as follows:
a. 2 Ba + O2 ⇌ 2 BaO.
![K=([BaO]^2)/([Ba]^2[O_2])](https://img.qammunity.org/2022/formulas/chemistry/high-school/3h2iwcnzw0404dzt8c9o7xfd3rglvhrfh2.png)
b. 2 Mg + O2 ⇌ 2MgO
![K=([MgO]^2)/([Mg]^2[O_2])](https://img.qammunity.org/2022/formulas/chemistry/high-school/12yzzqvhn19ult8zffy3wh3en1fatmfbg9.png)
c. P4 + 5 O2 ⇌ P4O10
![K=([P_4O_(10)]^2)/([P_4][O_2]^5)](https://img.qammunity.org/2022/formulas/chemistry/high-school/33mwfs5ir7mpl7g6aqgygsnrfvb797905y.png)
Best regards.