Chemistry, 28.01.2020 05:31, magalya01

Consider a pure crystalline solid that is heated from absolute zero to a temperature above the boiling point of the liquid. which of the following processes produces the greatest increase in the entropy of a substance?

a) melting the solid
b) heating the liquid
c) heating the gas
d) heating the solid
e) vaporizing the liquid

Answers: 2

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Answers

answered: ava789

Option E) vaporizing the liquid.

Explanation:

Entropy is a measure of the disorder or randomness of a system.The solid, liquid and gas state are characterized by the degree of freedom of their particles. The particles in the solid state are commonly arranged in fixed positions formins structures called crystals. So, they are the most ordered structures.In the liquids, compared with solids, the particles gain some motion, so they are not as much ordered, but yet they are enough attached to neighbor particles as to keep a fixed volume.The particles in gases are in constant, rapid and random motion, which means that they have the greatest level of disorder, compared to liquids and solids. That is why gases have not a fixed voiume but take the shape and form of the vessel that contains them.

Hence, you conclude that, among the choices, when a pure crystalline solid that is heated from absolute zero to a temperature above the boiling point of the liquid, the process of vaporizing (converting into gas) a liquid produces the greatest increase in the entropy of a substance.

answered: Guest

a solution is always transparent, light passes through with no scattering from solute particles which are molecule in size. the solution is homogeneous and does not settle out. a colloid is intermediate between asolution and a suspension. while asuspension will separate out a colloid will not.

What’s is a suspension, and how is it different from a solution?

answered: Guest

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.