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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the symbol for the elements that contains six rlectrons in each of its neutrual atoms are number of 6
That is True XD
Hope that helps :D
Hybrid Orbitals: are used to describe the orbitals in covalently bonded atoms sp,sp2,sp3
You can find the hybridization by adding the number of bonded atoms and the number of lone pairs.
For example in BF3. The central atom (B) is bonded to three atoms. So the hybridization is sp2
In NH3, the central atom (N) is bonded to three atoms and has one lone pair. The hybridization is sp3
