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Gnesinka [82]
2 years ago
5

15. conjugated dienes routinely undergo 1,2 and 1,4 addition reactions with a variety of electrophilic reagents; this suggests t

hat ___ are likely intermediates during these reactions
Chemistry
1 answer:
son4ous [18]2 years ago
7 0

Conjugated dienes routinely undergo 1,2 and 1,4 addition reactions with a variety of electrophilic reagents; this suggests that electrophilic reagents are likely intermediates during these reactions.

Two double bonds and one single bond divide a conjugated diene into two halves. Nonconjugated (Isolated) Dienes have more than one single bond separating two double bonds. Two double bonds are joined to the same atom to form cumulated dienes.

Reagents that function by acquiring electrons or sharing electrons that once belonged to a foreign molecule are referred to as electrophilic reagents, or electrophiles, in some cases. Electrophiles are molecules with a positive charge and a lack of electrons that can react by exchanging electron pairs with nucleophiles, which have many electrons. Epoxides, hydroxy amines, nitroso and azoxy derivatives, nitrenium ions, and elemental sulfur are significant electrophiles.

To know more about Electrophiles refer to: brainly.com/question/21773561

#SPJ4

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If the density of a diamond is 3.15 g/cm3 then what is the mass of a 25cm diamond?
Yanka [14]

Answer:

78.75g

Explanation:

The density of diamond is the mass of diamond per unit of its volume. The density of a substance is calculated using the formula:

Density (g/cm^3) = mass (g) / volume (cm^3)

According to the question, the density of diamond= 3.15g/cm^3, volume of diamond= 25cm^3, mass = ?

Hence, to calculate the mass, we say;

Mass = Volume × density

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3 0
4 years ago
Using the following equation for the combustion of octane, calculate the heat associated with the combustion of 100.0 g of octan
Sonja [21]

Answer: B) -4819 kJ

Explanation:

The balanced reaction for combustion of octane is:

2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O  \Delta H^0_{rxn}=-11018kJ

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

given mass of octane = 100.0 g

Molar mass of octane = 114.33 g/mol

Putting in the values we get:

\text{Number of moles}=\frac{100.0g}{114.33g/mol}=0.8747moles

According to stoichiometry:

2 moles of octane give heat = -11018 kJ

Thus 0.8747 moles of octane give =\frac{-11018}{2}\times 0.8747=-4819kJ

Thus -4819 kJ of heat is released by 100.0 g of octane assuming complete combustion.

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1.554442238816872 moles

Explanation:

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