What is the direct function of the energy released from the nuclear chain reaction in a nuclear power plant? turning the blades
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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:
⇄
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: There are 0.499 moles present in molecules of Boron trichloride.
Explanation:
According to mole concept, there are molecules present in 1 mole of a substance.
Hence, number of moles present in molecules are as follows.
Thus, we can conclude that there are 0.499 moles present in molecules of Boron trichloride.