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?

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.

you did not give any options thus this is the best answer i can give to you with this little information you have given

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the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.