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The product of the reaction of excess benzene with dichloromethane is
diphenylmethane. The reaction that benzene undergoes with dichloromethane is a
Friedel-Crafts Alkylation, in which firstly Benzene is alkylated to Benzyl Chloride, and after that, the Benzyl Chloride reacts with another equivalent of Benzene to form the compound Diphenylmethane. The Reaction Scheme is shown in the image below:
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Potassium Iodide have a lot of strong bonds which requires a very high temperature (high energy) to break them and change into a different state of matter.
The names of the alkanes are 2,2- dimethylbutane, 2,4- dimethylhexane, 2,2,3,3- tetramethylbutane and 4- ethyl, 3,6- dimethyl heptane.
Alkanes are saturated hydrocarbon in the organic chemistry. These are organic compounds that consists of single bonded carbon and hydrogen atoms. The common formula for writing an alkane is given by CₙH₂ₙ₊₂.
Alkanes are further divided into three more types which are:
1. Chain alkanes
2. Cycloalkanes
3. Branched alkanes
Hydrogenation method is used for preparation of alkanes from alkene and alkyne.
The names of the given compounds are:
Part A:
2,2- dimethyl butane
Part B:
2,4- dimethyl hexane
Part C:
2,2,3,3- tetramethyl butane
Part D:
4- ethyl, 3,6- dimethyl heptane
Learn more about alkanes from the link given below.
brainly.com/question/16837399
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Answer: The potential of the following electrochemical cell is 1.08 V.
Explanation:
=-0.74V[/tex]
=0.34V[/tex]
The element with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.
Here Cr undergoes oxidation by loss of electrons, thus act as anode. copper undergoes reduction by gain of electrons and thus act as cathode.


Where both
are standard reduction potentials, when concentration is 1M.
![E^0=E^0_{[Cu^{2+}/Ni]}- E^0_{[Cr^{3+}/Cr]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BCu%5E%7B2%2B%7D%2FNi%5D%7D-%20E%5E0_%7B%5BCr%5E%7B3%2B%7D%2FCr%5D%7D)

Thus the potential of the following electrochemical cell is 1.08 V.
this is the answer of your question .
<em>Hope</em><em> </em><em>it</em><em> </em><em>helped </em><em>you</em>