In a lab experiment for the dissolution of borax: na2[b4o5(oh)4] x 8 h2o(s) \rightleftharpoons 2 na+(aq) + b4o5(oh)) + 8 h2? o(l)5.00 ml of borax solution was prepared at 19.2 ^{\circ} ? c (292.35 k). it was diluted in a flask and titrated with 3.14 ml of 0.4760 m hcl. ksp? = 0.013355797 or 1.34 x 10-2? for this trial\deltag^{\circ}? = -rtln(k) for this trial was determined to be 10.48999 kj1.) what does the value of \delta g^{\circ} tell you about ksp? at this temperature? 2.) under standard conditions does the solvation reaction favor products or reactants? 3.) since \delta g^{\circ} is positive, why does some of the borax still dissolve at room temperature? (hint: what is the difference between \delta g and \delta g^{\circ}? )4.) does your value for \delta h^{\circ}? suggest that the solvation of borax is endothermic or exothermic? explain why is this result consistent with the observation that more borax dissolved at higher temperatures.

Hello thereone of the following properties was originally used to arrange elements on the periodic table, but is no longer used to organize the modern version. which property fits this description? answer: atomic mass  was originally used to arrange elements on the periodic table, but is no longer used to organize the modern version.

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.