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Setler [38]
3 years ago
11

Oxidation of a dithiol such as 2,5-hexanedithiol forms a six-membered ring containing a disulfide group as part of the ring. Dra

w the structure of this cyclic disulfide (Hint: Draw the starting compound in line structure format first).
Chemistry
1 answer:
rosijanka [135]3 years ago
6 0

Answer:

See picture below

Explanation:

In general terms, oxidation of thiol are rather different than oxidations with alcohols.

A dithiol is a molecule that has two thiols in an organic compound. In the case of the 2,5 hexanedithiol, we have the thiols in the carbon 2 and carbon 5.

When oxidation of thiols occurs, and depending of the number of thiols in the molecule, the final compound will have a sulfide group.

This is because the oxydation makes that the hydrogen atom of the sulfur is substracted. So, in the case of the dithiols, we have that both hydrogen atoms are substracted and then, they are available to join. Therefore, As the thiols are in the same molecule, it will occur a self condensation reaction.

The picture below will show the final product.

Hope this helps

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Answer:

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Explanation:

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~Hope it helps:).

5 0
3 years ago
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Nata [24]
Ok, after doing an immense amount of research I came up with the most logical answer.

A. Is indicated by a negative enthrall sign.

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When I searched up enthalpy, it said “When a substance changes at constant pressure, enthalpy tells how much heat and work was added or removed from the substance.” Which is similar to c, right? Yeah, meaning both a and c are similar in that aspect.

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5 0
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Firlakuza [10]

Answer:

\large \boxed{1.64\times 10^{-5}\text{ mol/L }}

Explanation:

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                           2x      0.007 50 + x

K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}

 

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