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3 years ago

Can aromatic compounds belong to other classes of compounds

Chemistry

2 answers:

bagirrra123 [75]3 years ago

8 0

Answer:

Yes

Explanation:

First of all, aromatic compounds are part of a group of compounds known as hydrocarbons. Hydrocarbons are compounds containing only gmhydrogen and carbon. Some aromatic compounds contain more than one benzene ring. These belong to another subgroup known as poly nuclear aromatic hydrocarbons.

34kurt3 years ago

3 0

Yes , it is possible for aromatic compounds to belong to other classes of compounds.

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How many significant figures are in 0.0034 cm? a 2 b 3 c 1 d 4

Annette [7]

Answer:

A. 2 sig figs

Explanation:

In the number 0.0034, the zeros are called leading zeros. The only function of leading zeros is to fix the decimal point. Leading zeros are not counted as significant figures.

8 0

4 years ago

Problem page gaseous ethane ch3ch3 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o

iVinArrow [24]

$m(\text{CO}_2) = 2.24 \; \text{g}$

Ethene react with oxygen at a $2 : 7$ molar ratio:

$2\; \text{C}_2 \text{H}_6 (g) + 7\; \text{O}_2 (g) \to 6\; \text{H}_2{O} (g) + 4\; \text{CO}_2 (g)$

Convert the quantity of each reactant supplied to number of moles of particles:

$n(\text{C}_2\text{H}_6) = 0.60 \; \text{g} / 28.05 \; \text{g} \cdot \text{mol}^{-1} = 0.0214 \; \text{mol}$
$n(\text{O}_2) = 3.27 \; \text{g} / 32.00 \; \text{g} \cdot \text{mol}^{-1} = 0.102 \; \text{mol}$

The question stated not whether both reactants were used up in this process. Thus start by testing the assumption that e.g., ethene was used up while some oxygen gas were left unreacted (ethene as the <em>limiting </em>reagent.) Under this assumption, the relative availability of the two species, $n(\text{C}_2 \text{H}_6) /2$ and $n(\text{O}_2) /7$ (as seen in the balanced chemical equation) shall satisfy the relationship

$n(\text{O}_2) / 7 - n(\text{C}_2 \text{H}_6) / 2 > 0$

In other words,

$n(\text{O}_2)/7 > n(\text{C}_2 \text{H}_6)/2$

$n(\text{C}_2 \text{H}_6) / n(\text{O}_2) < 2/7 \approx 0.286$

Evaluating the expression $n(\text{C}_2 \text{H}_6) / n(\text{O}_2)$ with data given in the question yields approximately $0.210 < 0.286$ , which does satisfy the relationship. Hence the assumption holds and ethene is the limiting reactant.

The quantity of a reactant produced in a chemical reaction is related to its stoichiometric (of relating to proportions) relationship with the limiting reactant (or any of the reactants in case of more than one limiting reactant.) For this scenario, given the molar ratio $n(\text{C}_2\text{H}_6) : n( \text{CO}_2) = 2:4$ ,

$n(\text{CO}_2) = n(\text{C}_2\text{H}_6) \cdot (2 / 4) = 0.0510 \; \text{mol}$

$m(\text{CO}_2) = 0.0510 \; \text{mol} \times 44.01 \; \text{g} \cdot \text{mol}^{-1} = 2.24 \; \text{g}$

4 0

3 years ago

What slows down heat energy from escaping the Earth?

jeka57 [31]

There are some gases in the atmosphere which trap the heat escaping from the Earth and stop it from travelling back into space. These gases are called greenhouse gases. The glass in a greenhouse has a similar effect on the Sun's rays and so it is called the Greenhouse Effect.

4 0

3 years ago

Read 2 more answers

True or False? Not all compounds will dissolve in water.

In-s [12.5K]

Answer:

False

Explanation:

Polar molecules and ionic compounds dissolve in water, not nonpolar molecules

4 0

3 years ago

For an enzyme that displays Michaelis-Menten kinetics, what is the reaction velocity, v(as a percentage of Vmax), observed at:a)

fiasKO [112]

Answer:

a) 50% of the maximum velocity

b) 33.33% of the maximum velocity

c) 9.09% of the maximum velocity

d) 66.66% of the maximum velocity

e) 90.9% of the maximum velocity

Explanation:

The Michaelis-Menten kinetis is represented by

v = Vmax*S/(Km+S)

where

v= reaction rate

S= Substrate's concentration

Vmax= maximum rate of reaction

Km= constant

a) for S=Km

v = Vmax*Km/(2Km) = Vmax/2

v/Vmax = 1/2= 50% of the maximum velocity

b) for S=Km/2

v = Vmax*(Km/2)/(3/2Km) = Vmax/3

v/Vmax = 1/3= 33.33% of the maximum velocity

c) for S= 0.1*Km=Km/10

v = Vmax*(Km/10)/(11/10Km) = Vmax/11

v/Vmax = 1/11= 9.09% of the maximum velocity

d) for S=2*Km

v = Vmax*(2*Km)/(3*Km) = (2/3)* Vmax

v/Vmax = 2/3 = 66.66% of the maximum velocity

d) for S=10*Km

v = Vmax*(10*Km)/(11*Km) = (10/11)* Vmax

v/Vmax = 10/11 = 90.9 % of the maximum velocity

7 0

4 years ago