Chemistry, 21.01.2020 21:31 19thomasar

Imagine a solution of two liquids in whichthe molecules interact less favorably than they do in theindividual liquids. will this solution deviate positively from, deviate negativelyfrom, or ideally follow raoult's law? a. it will deviate positively. b. it will deviate negatively. c. it will be an ideal solution.

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Answer from: hamnah83

Option (A) is the correct answer.

Explanation:

It is known that according to Raoult's law, the vapor pressure of an ideal solution is dependent on the vapor pressure of each chemical component and the mole fraction of the component present in the solution.

And, total vapor pressure "p" of the solution once the present components in the solution have reached to the equilibrium is as follows.

$p = p^{o}_{A} x_{A} + p^{o}_{B} x_{B} + .....$

Hence, when in between the molecules cohesive forces are greater than the adhesive forces then, dissimilarities of polarity or internal pressure will lead both the components to escape out from the solution more easily.

As a result, the vapor pressure will be greater than expected from the Raoult's law, and it will show a positive deviation.

Thus, we can conclude that the given solution will deviate positively.

Answer from: Quest

And a) oxygen i believe

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ 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: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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