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
Answer:
64.52 mg.
Explanation:
The following data were obtained from the question:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Final amount (N) =.?
Next, we shall determine the rate constant (K).
This is illustrated below:
Half life (t½) = 1590 years
Rate/decay constant (K) =?
K = 0.693 / t½
K = 0.693/1590
K = 4.36×10¯⁴ / year.
Finally, we shall determine the amount that will remain after 1000 years as follow:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Rate constant = 4.36×10¯⁴ / year.
Final amount (N) =.?
Log (N₀/N) = kt/2.3
Log (100/N) = 4.36×10¯⁴ × 1000/2.3
Log (100/N) = 0.436/2.3
Log (100/N) = 0.1896
Take the antilog
100/N = antilog (0.1896)
100/N = 1.55
Cross multiply
N x 1.55 = 100
Divide both side by 1.55
N = 100/1.55
N = 64.52 mg
Therefore, the amount that remained after 1000 years is 64.52 mg
Answer : The current passing between the electrodes is, 
Explanation :
First we have to calculate the charge of sodium ion.

where,
q = charge of sodium ion
n = number of sodium ion = 
e = charge on electron = 
Now put all the given values in the above formula, we get:

Now we have to calculate the charge of chlorine ion.

where,
q' = charge of chlorine ion
n = number of chlorine ion = 
e = charge on electron = 
Now put all the given values in the above formula, we get:

Now we have to calculate the current passing between the electrodes.



Thus, the current passing between the electrodes is, 
Given :
Number of moles , n = 36.25 mol .
Molarity , M = 1.25 M .
To Find :
The volume of water required .
Solution :
Moarity is given by :

So , 
Here , n is number of moles and M is molarity .
Putting all values in above equation , we get :

Therefore , volume of water required is 29 L .
Its air and water. if its multiple choice