The answer is b since on a it says H2 but on the right side there is no H idk if you forgot to put H there so im guessing b. Have a good day
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
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Answer:
m= 29.645 g
Explanation:
Density:
Density is equal to the mass of substance divided by its volume.
Units:
SI unit of density is Kg/m3.
Other units are given below,
g/cm3, g/mL , kg/L
Formula:
D=m/v
D= density
m=mass
V=volume
Symbol:
The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.
Given data:
density of wood = 0.77 g/cm³
volume= 38.5 cm³
mass= ?
Solution:
d= m/v
m= d × v
m= 0.77 g/cm³× 38.5 cm³
m= 29.645 g
Answer:
[ S2- ] = 4.0 E-47 M
Explanation:
- PbS(s) → Pb2+ + S2-
- HgS(s) → Hg2+ + S2-
∴ Ksp PbS = 3.4 E-28 = [Pb2+]*[S2-]
∴ [Pb2+] = 0.181 M
∴ Ksp HgS = 4.0 E-53 = [Hg2+]*[S2-]
∴ [Hg2+] = 0.174 M
∴ Ksp PbS > Ksp HgS ⇒ precipitate first Hg2+:
∴ [ Hg2+ ] = 1.0 E-6 M
⇒ [S2-] = 4.0 E-53 / 1.0 E-6 = 4.0 E-47 M