Answer: The concentrations of
at equilibrium is 0.023 M
Explanation:
Moles of
= 
Volume of solution = 1 L
Initial concentration of
= 
The given balanced equilibrium reaction is,

Initial conc. 0.14 M 0 M 0M
At eqm. conc. (0.14-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO%5D%5Ctimes%20%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Now put all the given values in this expression, we get :

By solving the term 'x', we get :
x = 0.023 M
Thus, the concentrations of
at equilibrium is 0.023 M
An exchange reaction consists of both synthesis and decomposition reactions.
Here’s a complex example: AB + CD → AC + BD.
Another example might be: AB + CD → AD + BC.
Answer: The particles in a solid are packed very close to each other.
Answer:
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Explanation:
SORRY