All categories

2 years ago

How do aromatic hydrocarbons differ from alkenes?

Chemistry

2 answers:

ra1l [238]2 years ago

7 0

C that's the answer

Alex Ar [27]2 years ago

7 0

Answer: Option (C) is the correct answer.

Explanation:

A molecule whose all the atoms lie in the same plane is known as a planar molecule. For example, benzene is a planar molecule.

A compound which have ring structure is known as aromatic compound like benzene is an aromatic compound.

Thus, we can conclude that out of the given options aromatic hydrocarbons differ from alkenes as aromatic hydrocarbons are flat, planar molecules, and alkenes are not.

You might be interested in

A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %

kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

$H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)$

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and $y_{C-6}$ and $y_{N_2}$ are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

$H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol$

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

$Q=-40036+(-672.21)=-40708.21J$

Finally we convert this result to kJ:

$Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ$

Learn more:

6 0

2 years ago

The atomic mass of hydrogen is one. True False

mafiozo [28]

Answer:

It's true :) Hope that helps

8 0

3 years ago

Read 2 more answers

Why do we need to know these organic molecules? Are they important to living things?

laila [671]

Answer:

<h2>Organic molecules are important to living things because life is based on the properties of carbon. Carbon is an important element because it can </h2><h2>form four covalent bonds. ... The carbon skeletons contain the functional groups that are involved in biochemical reactions.</h2>

Explanation:

<h2>PLEASE MARK ME BRAINLIEST AND FOLLOW ME LOTS OF LOVE FROM MY HEART AND SOUL DARLING TEJASWINI SINHA HERE ❤️</h2>

7 0

3 years ago

Read 2 more answers

The equilibrium constant for the gas-phase isomerization of borneol (c10h17oh) to isoborneol at 503 k is 0.106. a mixture consis

Dimas [21]

The solution is as follows:

K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106

The molar mass of isoborneol/borneol is 154.25 g/mol

Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol

Use the ICE approach

borneol → isoborneol
I 0.0972 0.0486
C -x +x
E 0.0972 - x 0.0486 + x

Total moles = 0.1458

Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P

0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832

Thus,
Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904
Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096

6 0

3 years ago

Can someone please help me with this....?

N76 [4]

Answer:

Explanation:

The answer is D

4 0

2 years ago