What additions must you make to the cell for it to generate a standard emf? Select all that apply. 1. You should add enough salt containing A2+ into the glass containing the electrode B(s) to produce a 1 M A2+ solution.
2. You should add enough salt containing B2+ into the glass containing the electrode B(s) to produce a 1 M B solution.
3. You should apply the initial emf to the electrodes.
4. You should add the salt bridge.
5. You should add enough salt containing A2+ into the glass containing the electrode A(s) to produce a 1 M A2 solution.
6. You should add enough salt containing B2+ into the glass containing the electrode A(s) to produce a 1 M B2+ solution.

60.0 - there are 3 significant numbers

No because it not possibly

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.