Answer: (a) The solubility of CuCl in pure water is 
.
(b) The solubility of CuCl in 0.1 M NaCl is 
.
Explanation:
(a) Chemical equation for the given reaction in pure water is as follows.
Initial: 0 0
Change: +x +x
Equilibm: x x
And, equilibrium expression is as follows.
![K_{sp} = [Cu^{+}][Cl^{-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCu%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D)
x =
Hence, the solubility of CuCl in pure water is .
(b) When NaCl is 0.1 M,
,
, 
Net equation: 
= 0.1044
So for,
Initial: 0.1 0
Change: -x +x
Equilibm: 0.1 - x x
Now, the equilibrium expression is as follows.
K' = 
0.1044 = 
x =
Therefore, the solubility of CuCl in 0.1 M NaCl is .
Stoichiomety:
1 moles of C + 1 mol of O2 = 1 mol of CO2
multiply each # of moles times the atomic molar mass of the compund to find the relation is weights
Atomic or molar weights:
C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
CO2 = 12 g/mol + 2* 16 g/mol = 44 g/mol
Stoichiometry:
12 g of C react with 32 g of O2 to produce 44 g of CO2
Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen
And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.
You cannot obtain 72 g of CO2 from 18 g of C.
May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.
The concentrations of a mixture at equilibrium are constant as a function of time because the <span>e forward reaction proceeds at the same rate as the reverse reaction.</span>
Answer:
465mL
Explanation:
Volume of a solution, V =Mass of substance, m/(Molarity of the solution of the substance, M × molar mass of the substance, M.m)
Given in the question,
M=.132M
M.m=23+35.5 = 58.5g/mol
m=3.59g
V= 3.59/(.132×58.5)
V = 0.465L
Volume in mL = volume in L × 1000
= 0.465 × 1000 = 465mL
Therefore, 465mL of a .132M aqueous solution of sodium chloride, NaCl, must be taken to obtain 3.59 grams of the salt
