Answer:
Explanation:
In the following reaction we have shown an example of aromatic substitution reaction .
C₆H₆ + RCl = C₆H₅R + HCl
This reaction takes place in the presence of catalyst like AlCl₃ which is a lewis acid .
First of all formation of carbocation is made as follows .
RCl + AlCl₃ = R⁺ + AlCl₄⁻
This R⁺ is carbocation which is also called electrophile . It attacks the ring to get attached with it .
C₆H₆ + R⁺ = C₆H₅R⁺H.
The complex formed is unstable , though it is stabilized by resonance effect . In the last step H⁺ is kicked out of the ring . The driving force that does it is the steric hindrance due to presence of two adjacent group of H and R⁺ at the same place . Second driving force is attack by the base AlCl₄⁻ that had been formed earlier . It acts as base and it extracts proton ( H⁺ ) from the ring .
C₆H₅R⁺H + AlCl₄⁻ = C₆H₆ + AlCl₃ + HCl .
The formation of a stable product C₆H₆ also drives the reaction to form this product .
Substances have different tendencies to donate or accept electrons. When a really good donor meets a great acceptor, the chemical reaction releases a lot of energy. Oxygen (O2) is the best electron acceptor and is used in many aerobic reactions (reactions with oxygen). Hydrogen gas (H2) is a good electron donor.
When O2 and H2 are combined, along with a catalyst, water (H2O) is formed. This example of a redox reaction can be written like this:

Answer is: pH of aniline is 9.13.<span>
Chemical reaction: C</span>₆H₅NH₂(aq)+
H₂O(l) ⇌ C₆H₅NH₃⁺(aq) + OH⁻(aq).
pKb(C₆H₅NH₂) = 9.40.
Kb(C₆H₅NH₂) = 10∧(-9.4) = 4·10⁻¹⁰.
c₀(C₆H₅NH₂) = 0.45 M.
c(C₆H₅NH₃⁺) = c(OH⁻) = x.
c(C₆H₅NH₂) = 0.45 M - x.
Kb = c(C₆H₅NH₃⁺) · c(OH⁻) / c(C₆H₅NH₂).
4·10⁻¹⁰ = x² / (0.45 M - x).
Solve quadratic equation: x = c(OH⁻) = 0.0000134 M.
pOH = -log(0.0000134 M.) = 4.87.
pH = 14 - 4.87 = 9.13.
Answer:
6 moles of CuCl₂ will produced 4 moles of AlCl₃ .
Explanation:
Given data:
Moles of CuCl₂ = 6 mole
Moles of AlCl₃ produced = ?
Solution:
3CuCl₂ + 2Al → 2AlCl₃ + 3Cu
Now we will compare the moles of CuCl₂ with AlCl₃ .
CuCl₂ : AlCl₃
3 : 2
6 : 2/3 ×6 = 4 mol
So, 6 moles of CuCl₂ will produced 4 moles of AlCl₃ .
Answer:2817.8
Explanation:multiply the value by 365
