You are about to perform some intricate electrical studies on single skeletal muscle fibers from a gastronemius muscle. but first, you must prepare 6l of a 170 mm nacl solution in which to bath the isolated muscle during your studies. how many grams of nacl must you weigh out on the lab balance to prepare this bath solution? (hint: the molecular weight of nacl is 58.44 grams/mole).

I'm quite confused with the given unit for concentration reported as 'mm'. Usually that would just be 'm' for molality or 'M' for molarity. Based on the given data, I think that would be mM which is 1/1000 of M. So, 170 mM is equal to 0.17 M. With that being said, the solution is as follows:

0.17 M = Moles solute/ 6 L
Moles solute = 1.02 moles
Mass of solute = 1.02 moles * 58.44 g/mole = 59.6088 grams NaCl

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.

Answer it is also known as the completion state is untrue of the activation complex;