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:
They both have the same number of atoms
Explanation:
The number that indicates the amount of particles in a compound is the Avogadro's number (NA).
It does not matter the mass of compound we have, If we have 1 mol we will be sure that we are talking about 6.02×10²³ particles
6.02×10²³ represents the amount of atoms in twelve grams of 12-pure carbon and it is considered a reference to measure the amount of all kinds of substances present in a given system.
Answer:
either first or second if not them try d but I'm pretty sure a also I'm sorry if I getbyou this wrong I dearly apologize
Hello! Let me try to answer this :)
Thanks and please correct if there are any mistakes ^ ^
