Answer: -
Concentration of PbI₂ = 1.5 x 10⁻³ M
PbI₂ dissociates in water as
PbI₂ ⇄ Pb²⁺ + 2 I⁻
So PbI₂ releases two times the amount of I⁻ as it's own concentration when saturated.
Thus the molar concentration of iodide ion in a saturated PbI₂ solution = [ I⁻] =
= 1.5 x 10⁻³ x 2 M
= 3 x 10⁻³ M
PbI₂ releases the same amount of Pb²⁺ as it's own concentration when saturated.
[Pb²⁺] = 1.5 x 10⁻³ M
So solubility product for PbI₂
Ksp = [Pb²⁺] x [ I⁻]²
=1.5 x 10⁻³ x (3 x 10⁻³)²
= 4.5 x 10⁻⁹
The percent composition of Cesium and Fluorine in CsF are 87.493 % and 12.507 % respectively.
<h3>What is a percentage composition?</h3>
Percent composition is very simple. Percent composition tells you by mass what percent of each element is present in a compound.
Molecular mass of Cs = 132.9054519
Molecular mass of F = 18.9984032
Molecular mass of CsF = 132.9054519 + 18.9984032 = 151.9038551
CsF: Percent composition by element:
Cesium (Cs) = = 87.493 %
Fluorine (F) = = 12.507 %
Hence, the percent composition of Cesium and Fluorine in CsF are 87.493 % and 12.507 % respectively.
Learn more about the percentage composition here:
brainly.com/question/20614202
SPJ1
Answer:
10 g/ml
Explanation:
divide mass by volume means divide 1000 by 100 and your answer will be 10