Answer:
Explanation:
Hello,
In this case, for the dissociation of calcium fluoride:
The equilibrium expression is:
![Ksp=[Ca^{2+}][F^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BCa%5E%7B2%2B%7D%5D%5BF%5E-%5D%5E2)
In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent 
is computed as follows:
Thus, the molar solubility equals the reaction extent , therefore:
Regards.
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:
For large rivers the problem is not simply a matter of deduction of consumptive use from runoff: it is more complex and the complexity is related to the changes in .
Explanation:
Answer:
Double Displacement (Acid-Base)
Explanation:
