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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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The percentage error is given by multiplying relative error by 100%.
Therefore, to get the percentage error we need relative error which is given by dividing the absolute error with the actual value. 
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3 years ago
Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)
allsm [11]

Answer:

\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}

Explanation:

1. Density from mass and volume

\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}

2. Volume from density and mass

V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}

3. Mass from density and volume

\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water                =<u> 12.8 mL</u>

Volume of object               = 11.8 mL

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Your drawing showing water displacement using a graduated cylinder should resemble the figure below.

 

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