Of the objects below, ________ is the most dense. Of the objects below, ________ is the most dense.
1 answer:
You might be interested in
The solubility product, ksp of PbBr₂ is 2.102 × 10⁻⁶
<h3>What is solubility?</h3>The solubility of a solute is defined as the maximum amount of that solute that can be dissolved in a known quantity of solvent at a given temperature.
<h3>What is a solubility product? </h3>Some salts are sparingly soluble in a solvent. For them, we calculate the solubility product.
It is an equilibrium constant that defines the relationship between a solid and its respective ions in an aqueous solution in equilibrium.
The greater the solubility product, the greater the solubility and vice versa.
Here, the solubility of PbBr₂ = 2.96 g/l
Molar solubility of PbBr₂ = = 2.96/367 = 8.07 × 10⁻³
At equilibrium,
1 mole of PbBr2 dissociates into 1 mole of Pb²⁺ ions and 2 moles of Br⁻
Let the molar concentration of Pb²⁺ be x, then the molar concentration of Br⁻ is 2x
Ksp = x.(2x)²
= 4x³
Substitute, x = 8.07 × 10⁻³
Ksp = 4 (8.07 × 10⁻³)³
= 2.102 × 10⁻⁶
Thus, The ksp of PbBr₂ is 2.102 × 10⁻⁶
Learn more about solubility product:
#SPJ4
Solubility of a compound in water can be referred to as the amount of the compound that can be dissolved in 1 L of the solvent (water) at any given temperature. Solubility of a compound can be expressed in the units of g/L or mg/L.
Given that the solubility of calcium carbonate in water = 14 mg/L
We have to calculate the volume of water that can dissolve 11 g of calcium carbonate.
Converting 11 g calcium carbonate to mg:
Volume of water that would dissolve 11000 mg calcium carbonate
=
=785.7 L
Rounding the volume 785.7 L to two significant figures, we get 790 L water.
Therefore, we would need 790 L water to completely dissolve 11 g of calcium carbonate.