The solubility of Ca(OH)2 in a 0.469M CaCl2 solution at 31°C, at given Ksp of Ca(OH)2 is determined as 1.626 mM.
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Dissociation reaction of Ca(OH)2</h3>
The dissociation reaction of Ca(OH)2 is given as follows;
Ca(OH)₂ ⇄ Ca²⁺ + 2OH⁻¹
x 2x
Concentration of Ca²⁺ = 0.469 M
Ksp = [x][2x]²
ksp = (0.469)(2x²)
ksp = 4(0.469)x²
ksp = 1.876x²
4.96 x 10⁻⁶ = 1.876x²
x² = (4.96 x 10⁻⁶)/(1.876)
x² = 2.643 x 10⁻⁶
x = √(2.643 x 10⁻⁶)
x = 1.626 x 10⁻³ M
x = 1.626 mM
Thus, the solubility of Ca(OH)2 in a 0.469M CaCl2 solution at 31°C, at given Ksp of Ca(OH)2 is determined as 1.626 mM.
Learn more about solubility here: brainly.com/question/23946616
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What you need to do is find 1/8 of 50
you can just divide 50 by 8 to get 6.25
so now you have to find how many days it will take till there are 6.25 grams of iodine left
every 8.1 days its mass is split in half
so start splitting it in half and every time you do, you add 8.1 days
50/2 =25 8.1
25/2 =12.5 + 8.1
12.5/2= 6.25 +8.1
now you have reached 1/8 of the original amount of Iodine-131
so to find how long it took just add 8.1+8.1+8.1
(this is the same as 8.1x3)
which equals 24.3
it will take 24.3 days for Iodine 131 to decay to 1/8 of its original mass.
(good luck on the regent if thats what your studying for :)
Answer:
K remains the same;
Q < K;
The reaction must run in the forward direction to reestablish the equilibrium;
The concentration of 
will decrease.
Explanation:
In this problem, we're adding an excess of a reactant, chlorine gas, to a system that is already at equilibrium. According to the principle of Le Chatelier, when a system at equilibrium is disturbed, the equilibrium shifts toward the side of the equilibrium that minimizes the disturbance.
Since we'll have an excess of chlorine, the system will try to reduce that excess by shifting the equilibrium to the right. Therefore, the reaction must run in the forward direction to reestablish the equilibrium.
The value of K remains the same, as it's only temperature-dependent, while the value of Q will be lower than K, that is, Q < K, as Q < K is the case when reaction proceeds to the right.
As a result, since is also a reactant, its concentration will decrease.
Answer:
The critical temperature of a substance is the temperature at and above which vapour of the substance cannot be liquefied, no matter how much pressure is applied.
