The electrophilic bromination or chlorination of benzene requires Lewis acid along with the halogen.
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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:
Total mass of the reactant = 2+2.5 =4.5 mg
Total mass of product = 4.15 mg
therefore, mass of unreacted oxygen = 4.50-4.15 = 0.35 g
Answer:
D. It is the sharing of electrons between atoms with an electronegativity difference below 1.7
Covalent bonds share electrons, whereas ionic bonds exchange electrons. Covalent bonds have an electronegativity of 0.0-1.7 (0.0-0.3 is a nonpolar covalent bond and 0.3-1.7 is a polar covalent bond). Ionic bonds are bonds that go beyond the electronegativity of 1.7 to 4.0 (1.7-4.0).
Answer:
CICUO3
Explanation:
copper 1 chlorate is made up of copper ion and a polyatomic ion of chlorine and oxygen
NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa
NaOH + CH3COOH → CH3COONa + H2O
Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH
Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH
These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L
Molarity of CH3COOH = 0.0106/0.071 = 0.1493M
CH3COONa = 0.0076 / 0.071 = 0.1070M
pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.
pH using Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.1070/0.1493)
pH = 4.74 + log 0.717
pH = 4.74 + (-0.14)
pH = 4.60.
