Agas stream flows at a rate of 14 m3/s at 322 k and 172 kpa. it consists of a mixure of 52 mole % ammonia in air. the gas enters through the bottom of a packed-bed gas absorber, where it flows countercurrent to a stream of pure liquid water that absorb 90% of all of the ammonia, and virtually no air. the absorber vessel is cylindrical with an internal diameter of 2.5 m. (a) neglecting the evaporation of water, calculate the ammonia mol fraction in the gas leaving the absorber. (b) calculate the outlet gas mass velocity (defined as mass flow rate per unit empt

A. volume stays constant. just took took the quiz

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.

It is a ok  i know i have a brain functioning well.                                                                                                                                                           

The answer is a. mars