Astream consisting of 44.6 mole% benzene and 55.4% toluene is fed at a constant rate to a process unit that produces two product streams, one a vapor and the other a liquid. the vapor flow rate is initially zero and asymptotically approaches half of the molar flow rate of the feed stream. throughout this entire period, no material accumulates in the unit. when the vapor flow rate has become constant, the liquid is analyzed and found to be 28.0 mole% benzene.

(a)

sketch a plot of liquid and vapor flow rates versus time from startup to when the flow rates become constant.

(b)

is this process batch or continuous? is it transient or steady-state before the vapor flow rate reaches its asymptotic limit? what about after it becomes constant?

(c)

for a feed rate of 100 mol/min, draw and fully label a flowchart for the process after the vapor flow rate has reached its limiting value, and then use balances to calculate the molar flow rate of the liquid and the composition of the vapor in mole fractions.

The answer is b i think

Tolerance to high levels of salinity

Answer; carbon monoxide (co);

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.