You want to make a chain growth polymer, but you only have one initiator available: benzoyl peroxide. for some reason, management has decided to be especially stingy with the budget this year and you only have a small amount in your lab. as such, you need to ration it as much as possible. to complicate matters, you also can't have the reaction taking too long to proceed so you need to balance your usage of initiator. you decide that a good rate is to convert one third of your starting monomer sample (concentration of 2 moles/liter) into polymer in 85 minutes. in the vendor data sheet for benzoyl peroxide you find the following information: "the decomposition of benzoyl peroxide is characterized by a half-life of 50 hours at 60c and an activation energy of 27.3 kcal/mol." trouble is, your reaction vessel needs repair and is currently only capable of off and 70c. what concentration of benzoyl peroxide should you use to get your desired rate? in your lab notebook, you find that you have a ratio of kz/kt = 128 l/mol-hour.

The total number of nucleons

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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.