Aromatic compounds are compounds that contain carbon-carbon multiple bonds.
The question did not mention that a heteroatom is present in the compound so we can assume that there is none of such. In that case, the compound contains only hydrogen and carbon.
So,
(CH)n = 78
where n is the number of each atom present.
(12 +1)n = 78
n = 78/13
n = 6
The molecular formula of the compound is C6H6
When C6H6 is treated with .conc.HNO3/conc.H2SO4 the compound shown in image 1 is formed. The reaction occurs at the C-C multiple bond.
When C6H6 is reacted with chlorine in the presence of sunlight, hexachlorobenzene (shown in image 2 attached) is formed.
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Explanation:
Organic compounds are defined as the compounds which contain carbon as their main element. For example, 
is an organic compound.
Generally, organic compounds are non-polar in nature and due to the presence of covalent bonding organic compounds have low melting point.
As compound A melts at 
and is soluble in water. This means it is an ionic compound as it has high melting point and it is also polar in nature.
Whereas compound B melts at 
and is insoluble in water. This means that this compound has covalent bonding and it is also non-polar in nature
. Hence, it is more likely to be organic in nature.
Thus, we can conclude that compound B is more likely to be an organic
compound.
Answer:
Explanation:
Covalent bonding occurs when pairs of electrons are shared by atoms. Atoms will covalently bond with other atoms to gain more stability, which is gained by forming a full electron shell. By sharing their outermost (valence) electrons, atoms can fill up their outer electron shell and gain stability.
Answer:
Explanation:
4
N
a
+
O
2
→
2
N
a
2
O
.
By the stoichiometry of this reaction if 5 mol natrium react, then 2.5 mol
N
a
2
O
should result.
Explanation:
The molecular mass of natrium oxide is
61.98
g
⋅
m
o
l
−
1
. If
5
m
o
l
natrium react, then
5
2
m
o
l
×
61.98
g
⋅
m
o
l
−
1
=
154.95
g
natrium oxide should result.
So what have I done here? First, I had a balanced chemical equation (this is the important step; is it balanced?). Then I used the stoichiometry to get the molar quantity of product, and converted this molar quantity to mass. If this is not clear, I am willing to have another go.
