Answer:
a. in pure water Solubility (x) = 1.26 x 10⁻⁴M
b. in 0.202M M⁺² Solubility (x) = 9.963 x 10⁻¹²M
The large drop in solubility is consistent with the common ion effect.
Explanation:
a. Solubility in pure water
Given: M(OH)₂ ⇄ M⁺² + 2OH⁻
I --- 0 0
C --- x 2x
E --- x 2x
Ksp = [M⁺²][OH⁻]² = (x)(2x)² = 4x³ => x = CubeRt(Ksp/4)
solubility in pure water = x = CubeRt(8.05 x 10⁻¹²/4) = 1.26 x 10⁻⁴M
b. Solubility in presence of 0.202M M⁺² as common ion.
Given: M(OH)₂ ⇄ M⁺² + 2OH⁻
I --- 0.202M 0
C --- +x +2x
E --- 0.202M + x 2x
≈ 0.202M
Ksp = [M⁺²][2x]² = (0.202)(2x)² = (0.202)(4x²) = 8.05 x 10⁻¹²
=> x = (8.05 x 10⁻¹²)/(0.202)(4) = 9.963 x 10⁻¹²M
Answer:
true
because with the both states we increase the surface of reaction
Answer:
1 Oxygen atom has a mass of 16 grams, thus diatomic oxygen weights twice as much- 32 grams. Hence 32 grams is needed.
Answer:
The answer to your question is 33.4 ml
Explanation:
Data
volume 1 = V1 = 42 ml
temperature 1 = T1 = 20°C
temperature 2 = T2 = -60°C
Volume 2 = V2 = x
Process
1.- Convert celsius to kelvin
T1 = 20 + 273 = 293°K
T2 = -60 + 273 = 233°K
2.- Use the Charles' law to solve this problem
Solve for V2
V2 = 
3.- Substitution
V2 = 
4.- Simplification
V2 = 
5.- Result
V2 = 33.4ml
