Chemistry, 11.02.2020 02:59 samueltaye

Complete the mechanism for the Hofmann elimination reaction by adding the missing curved arrow notation. Be sure to add any missing lone pairs and nonzero formal charges. Then in part two sort the common elementary steps in the order in which they are used in this mechanism. 1st attempt Part 1 (2 points) l See Periodic Table See Hint Add lone pairs and curved arrows to complete the mechanistic step
Part 2 (1 point) Identify the two unique common elementary steps in the above mechanism, by sorting them in the order in which they take place in the mechanism. Items (10 items) (Drag and drop into the appropriate area below) Nucleophile elimination Bimolecular nucleophilic substitution (SN2) Heterolysis Proton transfer Electrophilic addition Coordination Electrophil elimination Categories Step 1 Step 2 Drag and drop here Drag and drop here
Electrophile elimination(E2) Nucleophilic addition Simolecular Carbocation elimination rearrangemen

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Organisms are more likely to be more preserved in mud than sand because mud is moist and wet, while sand is very dry.

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in scientific notation, the value of the speed of light in a vacuum

will be written as 2.995 * 10^8. when this value is represented in the form of

a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.

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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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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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