Answer : The equilibrium concentration of 
will be, (C) 
Explanation : Given,
Equilibrium constant = 14.5
Concentration of 
at equilibrium = 0.15 M
Concentration of 
at equilibrium = 0.36 M
The balanced equilibrium reaction is,
The expression of equilibrium constant for the reaction will be:
![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 values in this expression, we get:
![14.5=\frac{[CH_3OH]}{(0.15)\times (0.36)^2}](https://tex.z-dn.net/?f=14.5%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%280.15%29%5Ctimes%20%280.36%29%5E2%7D)
![[CH_3OH]=2.82\times 10^{-1}M](https://tex.z-dn.net/?f=%5BCH_3OH%5D%3D2.82%5Ctimes%2010%5E%7B-1%7DM)
Therefore, the equilibrium concentration of will be, (C)
The heat is involved in the production of 5.0 mol of MgO is 180 Kj
calculation
2 Mg (s) +O2 → 2 MgO
From the equation above two moles of MgO is used therefore
72.3 is for 2 moles
that is 72.3 kj = 2moles
? = 5 moles
by cross multiplication
= 72.3 kj × 5/2 = 180 Kj
Answer:
The atomic radius of calcium is approximately 175
Explanation:
Given that the atomic radius of magnesium = 150 pm
The atomic radius of strontium = 200 pm
Therefore, given that calcium comes in between magnesium and strontium in group 2 of the periodic table, the atomic radius should be half way between the length of the atomic radius of magnesium and strontium, given that the atomic radius is not a fixed quantity
Therefore;
The atomic radius of calcium is approximately given as follows;
The approximate atomic radius (200 + 150)/2 = 175 pm.
