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
Because the water seeped through the bag (? I don’t know the context correct me if I’m wrong)
Answer:
Explanation:
I’m 99.9% sure that it’s an element because it can’t be broken down any more than it already is.
Answer:
100.8 °C
Explanation:
The Clausius-clapeyron equation is:
-Δ
Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'
Isolating for T2 gives:

(sorry for 'deltaHvap' I can not input symbols into equations)
thus T2=100.8 °C
Answer:
it must be testable I think that's the answer