Answer: The concentrations of 
at equilibrium is 0.023 M
Explanation:
Moles of = 
Volume of solution = 1 L
Initial concentration of = 
The given balanced equilibrium reaction is,
Initial conc. 0.14 M 0 M 0M
At eqm. conc. (0.14-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO%5D%5Ctimes%20%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Now put all the given values in this expression, we get :
By solving the term 'x', we get :
x = 0.023 M
Thus, the concentrations of at equilibrium is 0.023 M
Potassium 23.5g/39.0983g/mol = 0.601mol
The Ratio of reactants is 2 to 1 so (0.601mol)/2 = 0.3005mol
Therefore 0.3005mol of F2 is needed to find liters use
formula V = nRT/P (V)Volume = 22.41L
(T)Temperature = 273K or 0.0 Celsius
(P)Pressure = 1.0atm
<span>(R)value is always .08206 with atm n = 0.3005moles
(273)(.08206)(0.3005)/1 = V V = 6.7319 Liters</span>
Use the equation q=ncΔT.
q= heat absorbed our released (in this case 1004J)
n= number of moles of sample ( in this case 2.08 mol)
c=molar heat capacity
ΔT=change in temperature (in this case 20°C)
You have to rewrite the equation for c.
c=q/nΔT
c=1004J/(2.08mol x 20°C)
c=24.1 J/mol°C
I hope this helps
it should either be 5 or 10
