Chemistry, 25.04.2021 16:10 tdyson3p6xvtu

Please help me with thi Write a molecular equation, name of atom and number used in the chemical reaction for Iron(III) nitrate and
sodium thiosulfate reaction.

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The teacher is mostly talking about the distillation of a mixture

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definition

entropy is a measure of the randomness or the extent of disorder of a chemical process.

enthalpy is a measure of the heat change of a reaction occurring at a constant pressure.

measurement units:

entropy is measured in joule per kelvin (jk⁻¹) whereas enthalpy is measured in joule per mole (jmol⁻¹)

requirements:

entropy has no requirements or limits, and its change is measured by the division between the heat change of the chemical process and the temperature.

in contrast, enthalpy can only be related to the heat change of the reaction under standard conditions.

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Idon't hope someone you

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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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