An experiment is designed to test what color of light will be activated photoelectric cell best. the photocell is set in circuit that "clicks" in response to current. the faster the current, the more clicks per minute. in the experiment, the number of clicks in one minute is recorded for each color of light shining on the photocell. to change the color of light, a different color of cellophane is place over the same flashlight and the flashlight is then located a specific distance from the photocell.
which of the following would be the best hypothesis for this experiment?
a.) red light will activate the photocell to the highest degree.
b.) if photoelectric cell respond better to short wavelengths of light, the red light should activate it the least and blue the most.
c.) photoelectric cells are affected by different wavelengths of light.

Because a model would make the investigation easier to understand

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.