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 .
Answer:
30%
Explanation:
<em>This is the chemical formula for zinc bromate: Zn(BrO₃)₂. Calculate the mass percent of oxygen in zinc bromate. Round your answer to the nearest percentage.</em>
Step 1: Determine the mass of 1 mole of Zn(BrO₃)₂
M(Zn(BrO₃)₂) = 1 × M(Zn) + 2 × M(Br) + 6 × M(O)
M(Zn(BrO₃)₂) = 1 × 65.38 g/mol + 2 × 79.90 g/mol + 6 × 16.00 g/mol
M(Zn(BrO₃)₂) = 321.18 g/mol
Step 2: Determine the mass of oxygen in 1 mole of Zn(BrO₃)₂
There are 6 moles of atoms of oxygen in 1 mole of Zn(BrO₃)₂.
6 × m(O) = 6 × 16.00 g = 96.00 g
Step 3: Calculate the mass percent of oxygen in Zn(BrO₃)₂
%O = mO/mZn(BrO₃)₂ × 100%
%O = 96.00 g/321.18 g × 100% ≈ 30%
Answer:
Because , the advantage of steam distillation over simple distillation is that the lower boiling point reduces decomposition of temperature-sensitive compounds. Steam distillation is useful for the purification of organic compounds, although vacuum distillation is more common
Explanation:
