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igomit [66]
4 years ago
12

Draw the aromatic compound toluene (methylbenzene). show all hydrogen atoms, including those on the ring.

Chemistry
1 answer:
Otrada [13]4 years ago
7 0

Aromatic compound has continuous cyclic structure with( 4n+2)π electrons (Huckels rule), where n = 0,1,2…

Here number of pi electron are 6, where 4 from two double bond and 2 from nitrogen non-bonding electrons, hence it has total 6 pi electrons, therefore

6= ( 4n+2)π

4 = 4n  

n =1

 

Hence it is an aromatic compound  

 


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When lewis structures are drawn, do you only show valence electrons??
Lynna [10]
Yes you do cause its more explanation to
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8 0
3 years ago
Sulfonation of benzene has the following mechanism: (1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3 [fast] (2) SO3 + C6H6 → H(C6H5+)SO3− [slow]
ziro4ka [17]

Question is incomplete, complete question is as follows :

Complete Question : .Sulfonation of benzene has the following mechanism:

(1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3

[fast]

(2) SO3 + C6H6 → H(C6H5+)SO3−

[slow]

(3) H(C6H5+)SO3− + HSO4− → C6H5SO3− + H2SO4

[fast]

(4) C6H5SO3− + H3O+ → C6H5SO3H + H2O

[fast]

write the overall rate law for the initial rate of the reaction as a fraction.

Rate=k(________/_________)

Answer:

The overall rate law for the initial reaction is = k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]

Explanation :

Frist of all, all the common terms are cancelled out and written the overall reaction.

As we know that the rate depednant step is the slowest step of the reaction, rate law is :

                        rate = k_{2} [SO_{3}][C_{6}H_{6}]

But the problem is that SO3 cannot be written in the overall rate law because it is an intermediate.

Rate law for synthesis of S03 is as follows :

                       rate = k_{1}[H_{2}SO_{4}]^{2}

Hence when we substitute equation 2 in equation one,

                   Rate comes out to be =  k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]

6 0
3 years ago
When FeC13 is ignited in an atmosphere of pure oxygen, this reaction takes place. 4FeCl3(sJ 30lgJ ~ 2F~0 (sJ 6Cl2(gJ If 3.00 mol
oksano4ka [1.4K]

Answer : The reagent present in excess and remains unreacted is, O_2

Solution : Given,

Moles of FeCl_3 = 3.00 mole

Moles of O_2 = 2.00 mole

Excess reagent : It is defined as the reactants not completely used up in the reaction.

Limiting reagent : It is defined as the reactants completely used up in the reaction.

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2FeCl_3(s)+O_2(g)\rightarrow 2FeO(s)+3Cl_2(g)

From the balanced reaction we conclude that

As, 2 moles of FeCl_3 react with 1 mole of O_2

So, 3.00 moles of FeCl_3 react with \frac{3.00}{2}=1.5 moles of O_2

From this we conclude that, O_2 is an excess reagent because the given moles are greater than the required moles and FeCl_3 is a limiting reagent and it limits the formation of product.

Hence, the reagent present in excess and remains unreacted is, O_2

4 0
3 years ago
If nitrogen (N) has 2 naturally occurring isotopes, nitrogen-14 (78.3%) and nitrogen-16 (21.7%), what is its average r.a.m.?
leva [86]

Answer:

14.434 r.a.m.

Explanation:

  • The atomic mass of an element is a weighted average of its isotopes in which the sum of the abundance of each isotope is equal to 1 or 100%.

∵ The atomic mass of N = ∑(atomic mass of each isotope)(its abundance)

∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16)

atomic mass of N-14 = 14.0 r.a.m, abundance of N-14 = percent of N-14/100 = 78.3/100 = 0.783.

atomic mass of N-16 = 16.0 r.a.m, abundance of N-16 = percent of N-16/100 = 21.7/100 = 0.217.

∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16) = (14.0 r.a.m)(0.783) + (16.0 r.a.m)(0.217) = 14.434 r.a.m.

5 0
3 years ago
Which factors influence the salinity of ocean surface water?
In-s [12.5K]
C and D

surface water is evaporated by the sun and concentrated the ocean salt and oceans are diluted when rivers flow into them
4 0
3 years ago
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