Answer:
Four
Explanation:
AlCl₃(aq) ⟶ Al³⁺(aq) + 3Cl⁻(aq)
One mole of AlCl₃ produces 1 mol of Al³⁺ and 3 mol of Cl⁻.
That's four moles of ions.
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
.
From the balanced equation:
<span>1mol C3H8 requires 5mol O2 for combustion </span>
<span>Molar mass C3H8 = 44g/mol </span>
<span>8.8g C3H8 = 8.8/44 = 0.2mol C3H8 </span>
<span>This will require 5*0.2 = 1.0mol O2 </span>
<span>Molar mass O2 = 32g/mol </span>
<span>Therefore 32g of O2 required.
</span>
The molarity of the diluted solution is 0.33 M
From the question given above, the following data were obtained:
Molarity of stock solution (M₁) = 0. 5 M
Volume of stock solution (V₁) = 100 mL
Volume of diluted solution (V₂) = 100 + 50 = 150 mL
<h3>Molarity of diluted solution (M₂) =? </h3>
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
<h3>M₁V₁ = M₂V₂</h3>
0.5 × 100 = M₂ × 150
50 = M₂ × 150
Divide both side by 150
M₂ = 50 / 150
<h3>M₂ = 0.33 M</h3>
Therefore, the molarity of the diluted solution is 0.33 M
Learn more: brainly.com/question/24625656
B) O^2-
Hope this helped :)