Answer:

Explanation:
Hello,
In this case, the dissociation reaction is:

For which the equilibrium expression is:
![Ksp=[Pb^{2+}][I^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BPb%5E%7B2%2B%7D%5D%5BI%5E-%5D%5E2)
Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)

In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:
![[Pb^{2+}]=1.39x10^{-3}M](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%3D1.39x10%5E%7B-3%7DM)
![[I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M](https://tex.z-dn.net/?f=%5BI%5E-%5D%3D1.39x10%5E%7B-3%7DM%2A2%3D2.78x10%5E%7B-3%7DM)
Thereby, the solubility product results:

Regards.
Filtration. A filter is a porous barrier. Heterogeneous means it's a visible mixture, i.e large enough particles to see.
Answer: 65.38g of Ca(OH)2 is needed
Explanation:
From The equation of reaction
2 HCl + Ca ( OH ) 2 ⟶ CaCl 2 + 2 H 2 O
NB: Molar mass of HCl= 1+35.5=36.5
Ca(OH)2= 74
From The stoichiometric equation
2mol of HCl(36.5×2=73) require 1mol of Ca(OH)2 (74g)
Hence 64.5g of HCl will require 64.5×74/73= 65.38g of Ca(OH)2