Answer : The value of equilibrium constant (Kc) is, 0.0154
Explanation :
The given chemical reaction is:

Initial conc.
0 0
At eqm.
x x
As we are given:
Concentration of
at equilibrium = 
That means,

The expression for equilibrium constant is:
![K_c=\frac{[SO_2][Cl_2]}{[SO_2Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BSO_2%5D%5BCl_2%5D%7D%7B%5BSO_2Cl_2%5D%7D)
Now put all the given values in this expression, we get:



Thus, the value of equilibrium constant (Kc) is, 0.0154
It requires the allocation of resources.
Fission is similar to mitosis because they both involve splitting.
Answer: <span><span>d. mitosis</span></span>
They certainly can. However, they have other groups that are used to classify a compound.
Answer:
My guess would be 12.5828