The solubility of Lead(II)Fluoride is 2.17 × 10⁻³ g/L in water at 25°C.
At a specific solution temperature, a solid salt compound can entirely dissolve in pure water up to a predetermined molar solubility limit. The dissociation stoichiometry ensures that the molarities of the constituent ions are proportionate to one another. The saturable nature of the solution causes them to also coexist in a solubility equilibrium with the solid component. At this temperature, a solubility product constant Ksp is calculated using the solubility product of their molarity values.
Lead (II) fluoride has the following solubility equilibrium for its saturated solution:
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![K_s_p = [Pb^2^+][F^-]^2](https://tex.z-dn.net/?f=K_s_p%20%3D%20%5BPb%5E2%5E%2B%5D%5BF%5E-%5D%5E2)
This compound dissociates in a 1:2 ratio of ions. For the compound dissolved in pure water, the Ksp is expressed in terms of the molar solubility "x" as:


Here,
× 
4.1 × 10⁻⁸ = 4 x³
x³ = 1.025 × 10⁻⁸
x³ = 10.25 × 10⁻⁹
x = 2.17 × 10⁻³ g/L
Therefore, the solubility of Lead(II)Fluoride is 2.17 × 10⁻³ g/L.
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Answer:
the covalent molecule has OA.A
To calculate the amount of the compound in the units of grams, we need to first obtain for the molar mass of the compound. We calculate as follows:
MgCl2 = <span>24.31 + (2 x 35.45) = 95.21 g/mol
</span><span>340 g MgCl2 ( 1 mol / 95.21 g ) = 3.57 mol MgCl2
Hope this answers the question. Have a nice day.</span>
Club soda so A, you just gotta match the color that it says for it to the chart and it’ll show the identity the ph which is four I believe It said
Answer:
it's a measure of how far down light can penetrate through the water column. ... Because water clarity is closely related to light penetration, it has important implications for the diversity and productivity of aquatic life that a system can support