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Aleksandr [31]
3 years ago
13

Vanillin, C8H8O3, is the active ingredient in vanilla flavoring. It contains a six-membered aromatic ring with an aldehyde group

on carbon 1, an alkoxy group (ether) on carbon 3, and a hydroxyl group (alcohol) on carbon 4. Draw the structure of vanillin.

Chemistry
1 answer:
dlinn [17]3 years ago
3 0

Answer:

The answer is in the picture attached

Explanation:

A six-membered aromatic ring is one of the most common structures in natural compounds.

The aldehyde group in position one consist in one carbon double-bonded to oxygen and bonded to hydrogen.

The alkoxy group in position three consist in a oxygen bonded to an alkyl group (The simplest alkyl group is methyl -CH₃-).

There is, in position four, an hydroxyl group -OH. bonded to this six-membered aromatic ring.

I hope it helps!

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<u>Answer:</u> The concentration of Sn^{2+} in the cell is 9.0\times 10^{-3}M

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<u>Oxidation half reaction:</u>  Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-   E^o_{Zn^{2+}/Zn}=-0.76V

<u>Reduction half reaction:</u>  Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)   E^o_{Sn^{2+}/Sn}=-0.136V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

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To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

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To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}

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E_{cell} = electrode potential of the cell = 0.660 V

E^o_{cell} = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

[Zn^{2+}]=2.5\times 10^{-3}M

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Putting values in above equation, we get:

0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})

[Sn^{2+}]=9.0\times 10^{-3}M

Hence, the concentration of Sn^{2+} ions is 9.0\times 10^{-3}M

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