You are trying to determine the volume of the balloon needed to match the density of the air in the lab. you know that if you can get the balloon's density below this value, it will float. you measure both the temperature in the room as well as the mass of the balloon you will use and find they are 20.0°c and 0.570 grams, respectively. what volume of balloon will generate a density equal to that of the air around it? (hint: use the relationship you calculated above to determine the air density at the temperature given)

To solve this problem, we must be given first the density of air at 20 degrees Celsius. Looking up online, this is equal to:

density air (20C) = 0.0012041 g/mL

so that the volume is:

volume balloon = 0.57 g / (0.0012041 g/mL)

volume balloon = 473.38 mL

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.

C, small, hard spheres with insignificant volumes