Answer : The value of equilibrium constant (K) is, 424.3
Explanation : Given,
Concentration of
at equilibrium = 0.067 mol
Concentration of
at equilibrium = 0.021 mol
Concentration of
at equilibrium = 0.040 mol
The given chemical reaction is:

The expression for equilibrium constant is:
![K_c=\frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
Now put all the given values in this expression, we get:


Thus, the value of equilibrium constant (K) is, 424.3
This was because the experiment showed that a substance could emit radiation even while it was not exposed to light.
It would be “To obey the law of conservation of mass”
Answer:

Explanation:
Hello!
In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

Best regards!