Chemistry, 02.03.2020 16:52, Jc8777

Describe the difference in energy involved between two carbon atoms in a single bond, double bond and triple bond. How much energy is involved and why is there a difference?

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answered: galaxyworld36

Methane - 349 kJ mol^-1

, Ethene - 681 kJ mol^-1

, and Ethyne - 815 kJ mol^-1

Explanation:

The difference in the energy involved in carbon single bond, double bond, and triple bond are as mentioned below-

Methane - 349 kJ mol^-1

Ethene - 681 kJ mol^-1

Ethyne - 815 kJ mol^-1

These differences in energy depend on the number of bonds formed, bond lengths, sizes of atoms that are involved in the bond formation, affinities of the electron, electronegativity differences.

Hence, the shorter the bond length, the higher is the bond energy .

The bonds in double and triple bonds that are formed in ethene and ethyne respectively are shorter than the bonds formed in single bonds formed in methane. Ethyne has the highest bond energy and methane has the lowest bond energy.

answered: Guest

the correct answer would be b. amelia, because the particals of gas move really fast (charged particals) and gas also has a lot of room (particals move independently) so the only answer that has both of those is amelia's.

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.