Answer:
The molecules in the water become more separated due to heat.
Answer is: the equilibrium concentrations fluorine anion are 0.004 M and lead cation are 0.002 M.<span>
Chemical reaction: PbF</span>₂(aq) → Pb²⁺(aq) + 2F⁻(aq).<span>
Ksp = 3,2·10</span>⁻⁸.
[Pb²⁺] = x.
[F⁻] = 2[Pb²⁺] = 2x<span>
Ksp = [Pb²</span>⁺] ·
[F⁻]².
Ksp = x · 4x².
3,2·10⁻⁸ = 4x³.
x = ∛3,2·10⁻⁸ ÷ 4.
x = [Pb²⁺] = 0,002M = 2·10⁻³ M.
[F⁻] = 2 · 0,002M = 0,004 M = 4·10⁻³ M.
Answer:
Y = 92.5 %
Explanation:
Hello there!
In this case, since the reaction between lead (II) nitrate and potassium bromide is:

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

And the resulting percent yield:

Regards!