Answer:
800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .
Explanation:
3000 lb of 13% solution is required .
Total adhesive in weight = 3000 x .13 = 390 lb of adhesive
Available = 500 lb of 10% solution = 50 lb of adhesive
Rest = 390 - 50 = 340 lb required .
rest mass of solution = 3000 - 500 = 2500 lb
mass of adhesive required = 340 lb
Let the mass of 20% required be V
mass of adhesive = .20 V
.20 V = 340
V = 1700
rest of the volume = 2500 - 1700 = 800 lb which will be of pure solvent
So 800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .
Answer :
The equilibrium concentration of CO is, 0.016 M
The equilibrium concentration of Cl₂ is, 0.034 M
The equilibrium concentration of COCl₂ is, 0.139 M
Explanation :
The given chemical reaction is:

Initial conc. 0.1550 0.173 0
At eqm. (0.1550-x) (0.173-x) x
As we are given:

The expression for equilibrium constant is:
![K_c=\frac{[COCl_2]}{[CO][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCOCl_2%5D%7D%7B%5BCO%5D%5BCl_2%5D%7D)
Now put all the given values in this expression, we get:

x = 0.139 and x = 0.193
We are neglecting value of x = 0.193 because equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.139
The equilibrium concentration of CO = (0.1550-x) = (0.1550-0.139) = 0.016 M
The equilibrium concentration of Cl₂ = (0.173-x) = (0.173-0.139) = 0.034 M
The equilibrium concentration of COCl₂ = x = 0.139 M
<span>The particles are far apart from each other.</span>
Answer:
CaCO3
Explanation:
The molecule is formed by the calcium cation Ca+2 and the carbonate anion CO3−2.