Follow these steps to complete the table. 1. Reuse the same test tubes from part C, labeled Fe2+ and Fe3+. Be sure they're clean.
2. Add 4 milliliters of Iron(II) sulfate to the test tube labeled Fe2+,
3. Add 4 milliliters of iron(III) nitrate to the test tube labeled Fe3+,
4. Add 4 milliliters of sodium hydroxide to each test tube.
5. Observe the contents of the test tubes for about 10 minutes. Record your observations, noting any evidence of a
chemical reaction. If there is no evidence of a reaction, write "no reaction."

Because electrons have negligible mass, to account for the mass of the isotope, there must be 16 neutrons, 16 neutrally charged, massive, fundamental particles present in the sulfur nucleus. and thus finally, if we have the 32s2− ion, there are 18 electrons, 16 protons, and 16 neutrons.

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.

Ithink carbon dioxide, i am not completely sure.tell me if i am wrong.