Suppose there are two known compounds containing the generic elements x and y. you have a 1.00-g sample of each compound. one sample contains 0.25 g of x and the other contains 0.33 g of x. identify plausible sets of formulas for these two compounds. check all that apply.

I think the correct answers are X2Y and X3Y, X2Y5 and X3Y5, and X4Y2 and X3Y, for the following reason: 

If you look at the combining masses of X and Y in each of the two compounds, 

The first compound contains 0.25g of X combined with 0.75g of Y 
so the ratio (by mass) of X to Y = 1 : 3 

The second compound contains 0.33 g of X combined with 0.67 g of Y 
so the ratio (by mass) of X to Y = 1 : 2 

Now, you suppose to prepare each of these two compounds, starting with the same fixed mass of element Y ( I will choose 12g of Y for an easy calculation!) 

The first compound will then contain 4g of X and 12g of Y 
The second compound will then contain 6g of X and 12g of Y 

The ratio which combined the masses of X and the fixed mass (12g) of Y
= 4 : 6 
or 2 : 3 

So, the ratio of MOLES of X which combined with the fixed amount of Y in the two compounds is also = 2 : 3 

The two compounds given with the plausible formula must therefore contain the same ratio.

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.