This is due tomorrow, so if i could get , i would greatly appreciate it!

1. calculate the moles of nacl contained in one level teaspoon.
moles of nacl = g nacl x 1 mol nacl/58.5g

2. repeat step 1 for the remaining compounds (h2o, caco3). use the periodic table to calculate the molar mass of water and calcium carbonate.

3. calculate the number of moles in each element present in the teaspoon-sized sample of h2o.
moles of h = mol h2o x 2 mol h/1 mol h2o

repeat for the other compounds in your table (nacl, caco3).

4. calculate the number of atoms of each element present in the teaspoon-sized sample of h2o.
atoms of h = mol h x 6.02x10^23 atoms h/ 1 mol h2o

repeat for the other compounds in your table (nacl, caco3).

5. which of the three teaspoon-sized samples contains the greatest number of moles?

6. which of the three compounds contains the most atoms?

Ametal with a large number of free flowing electrons most likely has good conductivity

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.