Answer:
it depends on the assignment but im down
Explanation:
Answer:
E - Be and O
A - Mg and N
E - Li and Br
F - Ba and Cl
B - Rb and O
Explanation:
Be and O
Be is a metal that loses 2 e⁻ to form Be²⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form BeO (E-MX).
Mg and N
Mg is a metal that loses 2 e⁻ to form Mg²⁺ and N is a nonmetal that gains 3 e⁻ to form O³⁻. For the ionic compound to be neutral, it must have the form Mg₃N₂ (A-M₃X₂).
Li and Br
Li is a metal that loses 1 e⁻ to form Li⁺ and Br is a nonmetal that gains 1 e⁻ to form Br⁻. For the ionic compound to be neutral, it must have the form LiBr (E-MX).
Ba and Cl
Ba is a metal that loses 2 e⁻ to form Ba²⁺ and Cl is a nonmetal that gains 1 e⁻ to form Cl⁻. For the ionic compound to be neutral, it must have the form BaCl₂ (F-MX₂).
Rb and O
Rb is a metal that loses 1 e⁻ to form Rb⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form Rb₂O (B-M₂X).
Answer:
![[CO]=[Cl_2]=0.01436M](https://tex.z-dn.net/?f=%5BCO%5D%3D%5BCl_2%5D%3D0.01436M)
![[COCl_2]=0.00064M](https://tex.z-dn.net/?f=%5BCOCl_2%5D%3D0.00064M)
Explanation:
Hello there!
In this case, according to the given chemical reaction at equilibrium, we can set up the equilibrium expression as follows:
![K=\frac{[CO][Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BCO%5D%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Which can be written in terms of x, according to the ICE table:
Thus, we solve for x to obtain that it has a value of 0.01436 M and therefore, the concentrations at equilibrium turn out to be:
![[COCl_2]=0.015M-0.01436M=0.00064M](https://tex.z-dn.net/?f=%5BCOCl_2%5D%3D0.015M-0.01436M%3D0.00064M)
Regards!
The charge is neutral.
To find the charge, subtract the number of electrons from the number of protons. Because they are the same, there is no charge.
3 minutes are left (16-13=3)
