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IgorLugansk [536]
3 years ago
8

Explane factor affecting stability of alkene?

Chemistry
2 answers:
egoroff_w [7]3 years ago
8 0

Answer:

The stability of alkene can be determined by measuring the amount of energy associated with the hydrogenation of the molecule. Since the double bond is breaking in this reaction, the energy released in hydrogenation is proportional to the energy in the double bond of the molecule.

Explanation:

n chemistry, an alkene is a hydrocarbon that contains a carbon–carbon double bond. The term is often used as synonym of olefin, that is, any hydrocarbon containing one or more double bonds. However, the IUPAC recommends using the name "alkene" only for acyclic hydrocarbons with just one double bond; alkadiene, alkatriene, etc., or polyene for acyclic hydrocarbons with two or more double bonds; cycloalkene, cycloalkadiene, etc. for cyclic ones; and "olefin" for the general class — cyclic or acyclic, with one or more double bonds.

lions [1.4K]3 years ago
7 0

Answer:

Alkenes have substituents, hydrogen atoms attached to the carbons in the double bonds. The more substituents the alkenes have, the more stable they are. Thus, a tetra substituted alkene is more stable than a tri-substituted alkene, which is more stable than a di-substituted alkene or an unsubstituted one.

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A large balloon is initially filled to a volume of 25.0 L at 353 K and a pressure of 2575 mm Hg. What volume of gas will the bal
bija089 [108]

Answer:

45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.

Explanation:

Using Ideal gas equation for same mole of gas as

\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}

Given ,  

V₁ = 25.0 L

V₂ = ?

P₁ = 2575 mm Hg

Also, P (atm) = P (mm Hg) / 760

P₁ = 2575 / 760 atm = 3.39 atm

P₂ = 1.35 atm

T₁ = 353 K

T₂ = 253 K

Using above equation as:

\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}

\frac{{3.39}\times {25.0}}{353}=\frac{{1.35}\times {V_2}}{253}

\frac{1.35V_2}{253}=\frac{3.39\times \:25}{353}

Solving for V₂ , we get:

<u>V₂ = 45.0 L</u>

45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.

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