12) what is the mass percent of phosphorus in mgkpo4 6h2o? (15 pts)

11.62%

Explanation:

We are given;

A compound MgKPO₄.6H₂O

We are required to calculate the mass percent of phosphorus in the above compound.

Mass % of an element =(Mass of the element in the compound ÷ Formula mass of the compound) × 100

In this case;

Atomic mass of Mg = 24.305 g

Atomic mass of K = 39.098 g

Atomic mass of P = 30.974 g

Atomic mass of O = 16.0 g

Molar mass of water is 18.02 g/mol

Therefore; the Formula mass of MgKPO₄.6H₂O  will be;

= (1 × 24.305 g) + (1 × 39.098 g) + (1 × 30.974 g) + (4 × 16.0 g) + (6 ×18.02 g)

= 266.497 g

But, mass of P in the compound is 30.974 g

Therefore;

Mass % of P = (30.974 g ÷ 266.497 g) × 100

                    = 11.62%

Thus, the mass percent of phosphorus in the compound is 11.62%

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.