If 219 g of aluminum fluoride react with 48 g of oxygen gas, how many grams of fluorine gas will be produced? *
149 grams of fluorine gas
114 grams of fluorine gas
66 grams of fluorine gas
28.5 grams of fluorine gas

What is the limiting reactant?
Aluminum fluoride
Oxygen gas
Aluminum oxide
Fluorine gas

What is the excess reactant?

How many grams of excess reactant will be left over?
51 grams of aluminum fluoride left over
24 grams of oxygen gas
60 grams of aluminum oxide
6 grams of fluorine gas

Work done = force ×distancebut force = mass ×gravityand gravity = 10m\s force = 2×10 = 20 work done = 20×3 = 60 joules

Isec i will do it and them i will let you know

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.

Naoh + hbr -> nabr (salt) + h2o ca (oh)2 + 2hbr - - - > ca (br)2 + 2h2o