A sample of seawater contains 78.0 grams of NaCl in 2,025 mL of solution. If the molar mass of NaCl is 58.443 g/mol, what is the molarity of this sample? (3 points)
0 2.08 M NaCl
O 0.659 M NaCl
O 1.84 M NaCl
0.823 M Naci

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.

Alily leaf is a plant whereas a cat is an animala lily leaf cell is a plant cell and a skin cell is an animal cell

The answer is the third option extremely gradual