<u>Answer:</u> The atomic radius of iron is 128 pm.
<u>Explanation:</u>
To calculate the radius of the metal having FCC crystal lattice, the relationship between edge length and radius follows:
Where,
a = edge length = 362 pm
r = atomic radius of iron = ?
Plugging values in above equation, we get:
Hence, the atomic radius of iron is 128 pm.
Answer : The concentration of HI (g) at equilibrium is, 0.643 M
Explanation :
The given chemical reaction is:
Initial conc. 0.10 0.10 0.50
At eqm. (0.10-x) (0.10-x) (0.50+2x)
As we are given:
The expression for equilibrium constant is:
Now put all the given values in this expression, we get:
x = 0.0713 and x = 0.134
We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.0713
The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M
Thus, the concentration of HI (g) at equilibrium is, 0.643 M