What is the formula mass of the fictitious compound: AB2?

For a molecule AB2, 3.5g of A represents one molar fraction, and 8.00g B represents 2 molar fractions (or 4.0+4.0). Therefore, a direct ratio can be given as 3.5:4.0, or 1:1.14. This means a molecule AnBm will give a mass ratio for A:B of n:1.14xm

For a molecule AB, for every 1g of A, you will have 1.14g of B.

For a molecule AB2, for every 1g of A, you will have 2.28g of B.

For a molecule A2B3, for every 1g of A, you will have (1.14x3/2) 1.71g of B.

A1 a 2 a 3 ooh lol i think that's it

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.