Chemistry, 08.10.2019 18:00 aminak185

18 explain what happens to the average kinetic energy, thermal energy, and temperature of a substance when the particles in the substance slow down.

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Answer from: siddhi50

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the particles in a substance slow down when the average kinetic energy of the particles decreases. as the average kinetic energy decreases, the internal energy decreases, and so the thermal energy decreases. as the thermal energy of the substance decreases, the temperature decreases.

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Answer from: Quest

Read the sentences below in 1962, i tried to enroll as the first black student at the university of mississippi. the governor of mississippi blocked my entrance which best describes the verbs used in the sentences? o they are past tense verbs, which are often used in narrative essays o they are past tense verbs, which are not often used in narrative essays. o they are future tense verbs, which are often used in narrative essays o they are future tense verbs, which are not often used in narrative essays.

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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18 explain what happens to the average kinetic energy, thermal energy, and temperature of a substan...