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.
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Answer:

Explanation:
We have the reactions:
A: 
B: 
Our <u>target reaction</u> is:

We have
as a reactive in the target reaction and
is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.
A: 
Then if we add reactions A and B we can obtain the target reaction, so:
A: 
B: 
For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.


Answer:
C₄H₂N₂
Explanation:
First we<u> calculate the moles of the gas</u>, using PV=nRT:
P = 2670 torr ⇒ 2670/760 = 3.51 atm
V = 300 mL ⇒ 300/1000 = 0.3 L
T = 228 °C ⇒ 228 + 273.16 = 501.16 K
- 3.51 atm * 0.3 L = n * 0.082atm·L·mol⁻¹·K⁻¹ * 501.16 K
Now we<u> calculate the molar mass of the compound</u>:
- 2.00 g / 0.0256 mol = 78 g/mol
Finally we use the percentages given to<em> </em><u>calculate the empirical formula</u>:
- C ⇒ 78 g/mol * 61.5/100 ÷ 12g/mol = 4
- H ⇒ 78 g/mol * 2.56/100 ÷ 1g/mol = 2
- N ⇒ 78 g/mol * 35.9/100 ÷ 14g/mol = 2
So the empirical formula is C₄H₂N₂
<h3><u>Answer</u>;</h3>
B. 3/2
<h3><u> Explanation;</u></h3>
Balance the chemical equation
2Al + 3Cl2 → 2AlCl3
We want to convert moles of AlCl3 to moles of Cl2
The conversion factor is 2 mol AlCl3/3 mol Cl2.
We choose the one that makes the units cancel:
x mol AlCl3 x (3 mol Cl3)/(2mol AlCl3) = x mol Al
The fraction for the molar ratio is 3/2.