The breakdown of a certain pollutant X in sunlight is known to follow first-order kinetics. An atmospheric scientist studying the process fills a 20.0Lreaction flask with a sample of urban air and finds that the partial pressure of X in the flask decreases from 0.473atm to 0.376atm over 5.6hours. Calculate the initial rate of decomposition of X, that is, the rate at which Xwas disappearing at the start of the experiment.

Sorry i think the answer is 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.

Precipitate, and gas

Radon-222 or rn-222 has a half-life of 3.82 days. to determine the fraction of an original sample of rn-222 that remains unchanged after 7.646 days, we calculate the amount of rn-222 remaining after 7.646 days and divide it with the original amount. using 1 gram as a basis and using half-life calculation methods, we can determine that  0.249 or 0.25  grams of rn-222 will remain, and thus the fraction remaining is 0.25 or 1/4.