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प्रश्न
Give reason why the carbon compounds generally have low melting and boiling points.
उत्तर
Carbon compounds generally have low melting points and boiling points because the force of attraction between the molecules of carbon compounds is not very strong. These weak intermolecular forces make them very easy to pull apart from each other. Since they are easy to separate, carbon compounds have low melting and boiling points.
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संबंधित प्रश्न
List two reasons for carbon forming a large number of compounds. Name the type of bonding found in most of its compounds. Why does carbon form compounds mainly by this kind of bonding?
Select saturated hydrocarbons from the following: C3H6; C5H10; C4H10; C6H14; C2H4
What is meant by isomers?
Draw the structures for Butanone
How many isomers of the following hydrocharbons are possible?
C4H10
What is the molecular formula and structural formula of a cyclic hydrocarbon whose one molecule contains 8 hydrogen atoms?
The functional group which always occurs in the middle of a carbon chain is:
(a) alcohol group
(b) aldehyde group
(c) carboxyl group
(d) ketone group
The molecular formulae of some organic compounds are given below. Which of these compounds contains an aldehyde group?
(a) C3H8O
(b) C3H6O2
(c) C3H6O
(d) C3H7Cl
Which of the following are correct structural isomers of butane?
- \[\begin{array}{cc}
\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\phantom{..}\\
|\phantom{....}|\phantom{....}|\phantom{....}|\phantom{..}\\
\ce{H - C - C - C - C - H}\\
|\phantom{....}|\phantom{....}|\phantom{....}|\phantom{..}\\
\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\phantom{..}
\end{array}\] - \[\begin{array}{cc}
\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\\
|\phantom{....}|\phantom{....}|\\
\ce{H - C - C - C - H}\\
|\phantom{.....}\backslash\phantom{..}|\\
\phantom{.....}\ce{H}\phantom{.......}\ce{C - H}\\
\phantom{.........}|\\
\phantom{.........}\ce{H}
\end{array}\] - \[\begin{array}{cc}
\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\\
|\phantom{....}|\phantom{....}|\\
\ce{H - C - C - C - H}\\
|\phantom{....}\phantom{....}|\\
\ce{H}\phantom{........}\ce{H}\\
|\\
\ce{H - C - H}\\
|\\\ce{H}
\end{array}\] - \[\begin{array}{cc}
\ce{H}\phantom{...}\ce{H}\\
|\phantom{....}|\\
\ce{H - C - C - H}\\
|\phantom{....}|\\
\ce{H - C - C - H}\\
|\phantom{....}|\\
\ce{H}\phantom{...}\ce{H}
\end{array}\]
Draw two structural isomers of butane.
