Chemistry, 01.07.2020 15:01 mewings

Tenormin, a member of the group of drugs known as beta-blockers, is used to treat high blood pressure and improve survival after a heart attack. It works by slowing down the heart to reduce its workload. Which atom in Tenormin is the most basic?

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Answer from: keidyhernandezm

Oxygen

Explanation:

In the compound tenormin, there are two highly electronegative atoms capable of accepting electrons; oxygen and nitrogen. Oxygen is more electronegative than nitrogen.

However, the oxygen atom in tenormin is bonded to carbon in a carbonyl bond. Recall that the carbonyl bond is polar and the direction of the dipole is towards the oxygen atom. Looking at the structure of tenormin, it is clear that the electron density of the bond tends towards the oxygen atom of the carbonyl group. Electron density is withdrawn from the adjacent nitrogen atom of the amine group via mesomeric and inductive mechanism towards the more electronegative oxygen atom.

On the other side of the structure, there are two oxygen atoms. These oxygen atoms are more electronegative than nitrogen thus they are more basic.

Hence the oxygen atom is the most basic atom in the compound tenormin.

Answer from: Quest

It is sure ( overimitation)

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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