Chemistry, 21.01.2021 23:00 babyraerae06

Identify each described physical separation technique. A. A solid mixture is heated. One component transitions directly between solid and gas.
B. A liquid is carefully poured out, leaving the solid behind in the container.
C. A solvent is added to dissolve only one mixture component and then the liquids are separated.
D. A solid is caught on paper or a membrane while the liquid passes through.

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Answer from: lindseyklewis1p56uvi

A. A solid mixture is heated. One component transitions directly between solid and gas ---> Sublimation

B. A liquid is carefully poured out, leaving the solid behind in the container > Decantation

C. A solvent is added to dissolve only one mixture component and then the liquids are separated > solvent extraction

D. A solid is caught on paper or a membrane while the liquid passes through ---> Filtration using filter paper

Explanation:

A. Sublimation is a process of separating solids which sublimes (solids turning directly to gases) from a a mixture of other solids which do not sublime. When the mixture is heated, the solids which sublime are separated and those which do not are left behind. examples of solids which sublime are naphthalene, Iodine crystals, etc.

B. Decantation is the process of carefully pouring out a liquid from a mixture of it with a solid, leaving the solid behind. Example, a mixture of sand and water where water can be easily decanted, leaving the sand behind.

C. In the process of solvent extraction, a mixture of solids which are soluble in different solvent, one of the components is dissolved in a solvent in which it is soluble while the others are not. The solution is then filtered out and the solid is then separated from the solvent by evaporation.

D. In filtration using a filter paper, the solid-liquid mixture is passed through a filter paper which has pores the size through which liquids can pass through but the solid can not. therefore, the solid remains behind on the paper while the liquid passes through.

Answer from: Quest

C. 1s22s22p63s23p43d104s24p5

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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Identify each described physical separation technique. A. A solid mixture is heated. One component...

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