Calculate e degree for the following half-reaction, fes(s) + 2e^- fe(s) + s^2^-(aq) given that the solubility product for fes at 25 degree c is 8.0 times 10^-19 and the standard reduction potential of the half-reaction shown is -0.440 v.fe^2+(aq) + 2e^- fe(s) construct the line notation of a cell that consists of a saturated calomel reference electrode, the fes half-reaction shown at the top of the problem, and an iron indicator electrode, which can be used to measure the s^2- concentration of a solution if the cell described by the line notation above has a potential of -1.035 v, what is the ps of the solution?

The given parameter for the solubility of NaNO₃ in H₂O are;

The maximum solubility of NaNO₃ in H₂O at 10°C = 78 g

1. If we have 50 g of NaNO₃ in 100 g of water at 10°C, the solution is;

Undersaturated

2. If we have exactly 78 g of NaNO₃ in 100 g of water at 10°C the solution is;

A saturated solution

3. If will add 80 g of NaNO₃ in 100 g of water at 10°C,

The excess NaNO₃ which cannot be dissolved will be observed as solids in the mixture

4) The quantity of NaNO₃ that can be dissolved in H₂O at 40°C = 94 g

If the solution containing 94 g of NaNO₃ at 40°C is cooled to 10C about 16 g of NaNO₃ will precipitate out of the solution and exist as solids in the mixture

Explanation:

1. An undersaturated is a solution that holds smaller amount of solute that it can hold at a given temperature

2. A saturated solution holds as much solute as it can dissolve at a given temperature

3. When more solutes are added to a saturated solution, the excess solution will remain in the solid form in the mixture

4) Cooling a saturated solution below the saturation temperature will result in the precipitation or crystallization of the excess solutes in the solution.