In a steady-state flow process carried out at atmospheric pressure, 1 mol/s of air at 600 K (stream A) is continuously mixed with 3 mol/s of air at 440 K (stream B). The product stream is at 400 K and 1 atm (stream C). Assume that the surroundings are at 305 K, and that kinetic- and potential-energy changes of the streams are negligible. Air is assumed an ideal gas, with constant heat capacity, consisting of diatomic molecules (Cp=7/2 R). Calculate the rate of heat transfer (J/s) of the process, the rate of entropy generation (J/(s*K)) and lost work (J/s) for the process. Sketch a schematic representation of the process, including streams’ temperatures and flowrates. Start solving your problem by writing and simplifying the mass, energy and entropy balance for the process.

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.

Idon’t know what the answer is i wish i could