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2 years ago

What is the name of this hydrocarbon?

Chemistry

2 answers:

UkoKoshka [18]2 years ago

8 0

Hydrocarbon tells you that it is a <span>six carbon compound containing a triple bond.</span>

lana66690 [7]2 years ago

3 0

Answer : The name of the given hydrocarbon is, 3,3-dimethylhexane

Explanation :

The basic rules for naming of organic compounds are :

First select the longest possible carbon chain.
The longest possible carbon chain should include the carbons of double or triple bonds.
The naming of alkane is done by adding the suffix -ane, alkene by adding the suffix -ene, alkyne by adding the suffix -yne and carboxylic acid by adding the suffix -oic acid.
The numbering is done in such a way that first carbon of double or triple bond gets the lowest number.
The carbon atoms of the double or triple bond get the preference over the other substituents present in the parent chain.
If two or more similar alkyl groups are present in a compound, the words di-, tri-, tetra- and so on are used to specify the number of times of the alkyl groups in the chain.

From the given structure we conclude that, the longest carbon contains 4 carbons and two methyl group are present at 3rd position.

So, the name of the given hydrocarbon is, 3,3-dimethylhexane

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The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate

Cloud [144]

Answer : The $\Delta G$ for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,

$Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : $Cu\rightarrow Cu^{2+}+2e^-$

Reduction : $2Ag^++2e^-\rightarrow 2Ag$

Now we have to calculate the Gibbs free energy.

Formula used :

$\Delta G^o=-nFE^o$

where,

$\Delta G^o$ = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

$E^o$ = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

$\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole$

Therefore, the $\Delta G$ for this reaction is, -88780 J/mole.

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3 years ago

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bagirrra123 [75]

Answer:

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kipiarov [429]

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