The electrophilic bromination or chlorination of benzene requires Lewis acid along with the halogen.
<h3>
What is bromination of benzene?</h3>
The bromination or chlorination of benzene is an example of an electrophilic aromatic substitution reaction.
During the reaction, the bromine forms a sigma bond to the benzene ring, yielding an intermediate. Subsequently a a proton is removed from the intermediate to form a substituted benzene ring.
This reaction is achieved with the help of Lewis acid as catalysts.
Thus, the electrophilic bromination or chlorination of benzene requires Lewis acid along with the halogen.
Learn more about bromination of benzene here: brainly.com/question/26428023
Answer:
nonmetal
Explanation:
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(missing part of your question):
when we have K = 1 x 10^-2 and [A] = 2 M & [B] = 3M & m= 2 & i = 1
So when the rate = K[A]^m [B]^i
and when we have m + i = 3 so the order of this reaction is 3 So the unit of K is L^2.mol^-2S^-1
So by substitution:
∴ the rate = (1x 10 ^-2 L^-2.mol^-2S^-1)*(2 mol.L^-1)^2*(3mol.L^-1)
= 0.12 mol.L^-1.S^-1
Answer:
Volume of acid, Va=250mL; Volume of quinine,Vb=20mL; Molarity of acid, Ma=0.05M.
Molar mass of acid= H2+S+O4= 2+32+(16X4)= 2+32+64=98g
Concentration of acid, Ca= Molar mass of acid/ Ma =98/0.05=1960g/mol
Explanation: To calculate concentration of quinine, Cb is as follow
Va*Ca=Vb*Cb
∴ Cb=Va*Ca/Vb =250*1960/20 =24500g/mol
<span>Answer:
The HCl and KOH will react until one or the other is gone. As you have a larger volume of an equal concentration of HCl, the KOH will go first.
moles HCl = 0.04000 L * 0.100 M = 0.00400 moles
moles KOH = 0.02500 L * 0.100 M = 0.00250 moles
moles HCl left = 0.00400 - 0.00250 = 0.00150 moles
Your total volume is now 65.00 mL, so the [HCl] = 0.00150 moles / 0.06500 L = 0.0231 M = [H+]
pH = -log [H+] = -log (0.0231) = 1.64</span>