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:
Element Atomic Number Atomic Mass
Nickel 27 58.6934
Cobalt 28 58.9332
Copper 29 63.546
Zinc 30 65.39
Explanation:
Answer:
70 mL of 5% HCl and 30 mL of 15% HCl
Explanation:
We will designate x to be the fraction of the final solution that is composed of 5% HCl, and y to be the fraction of the final solution that is composed of 15% HCl. Since the percentage of the final solution is 8%, we can write the following expression:
5x + 15y = 8
Since x and y are fractions of a total, they must equal one:
x + y = 1
This is a system of two equations with two unknowns. We will proceed to solve for x. First, an expression for y is found:
y = 1 - x
This expression is substituted into the first equation and we solve for x.
5x + 15(1 - x) = 8
5x+ 15 - 15x = 8
-10x = -7
x = 7/10 = 0.7
We then calculate the value of y:
y = 1 - x = 1 - 0.7 = 0.3
Thus 0.7 of the 100 mL will be the 5% HCl solution, so the volume of 5% HCl we need is:
(100 mL)(0.7) = 70 mL
Similarly, the volume of 15% HCl we need is:
(100 mL)(0.3) = 30 mL
<u>Given information:</u>
Mass of NaCl (m) = 87.75 g
Volume of solution (V) = 500 ml = 0.5 L
Molar mass of NaCl (M) = 58.44 g/mol
<u>To determine:</u>
The molarity of NaCl solution
<u>Explanation:</u>
Molarity is defined as the number of moles of solute(n) dissolved per liter of solution (V)
i.e. M = moles of solute/liters of solution = n/V
Moles of solute (n) = mass of solute (m)/molar mass (M)
moles of NaCl = 87.75 g/58.55 g.mol-1 = 1.499 moles
Therefore,
Molarity of NaCl = 1.499 moles/0.5 L = 2.998 moles/lit ≅ 3 M
<u>Ans: (D)</u>
