Air at 100 c flows across a thin, flat plate of subliming naphthalene that is 3.4 m long. the reynolds number at the trailing edge of the plate is at the upper limit for a laminar boundary layer (re 5e5). estimate: (a) the average rate of sublimation in kmol/s*m2; and (b) the local rate of sublimation 0.5 m from the leading edge. vapor pressure of naphthalene = 10 torr viscosity of air 0.0215 cp; molar density of air 0.0490 kmol/m3; and diffusivity of naphthalene in air = 0.94x10-5 m2/s

Ithink that’s it’s b

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.

Ibelieve it is b. proton.hope my answer has you : )