Draw the structures of the products of the neutralization reaction between amide ion (nh2-) and acetic acid.
1 answer:
The reaction between strong acid and strong base which results in the formation of salt and water is said to be a neutralization reaction.
The given acid and base are acetic acid and amide ion. Since, acetic acid is a weak acid and amide ion is strong base.
The amide ion is a strong base so it will extract the proton from acetic acid and results in the formation of ammonia and acetate ion.
The reaction will takes place as shown in the image.
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Answer:
0.0890 M
Explanation:
Since the concentration of KCl is irrelevant in this case, the concentration of Na2S2O3 can be determined using a simple dilution equation:
C1V1 = C2V2, where C1 = 0.149 M, V1 = 150 mL, V2 = 250 mL
C2 = 0.149 x 150/250
= 0.089 M
To determine the concentration of S2O32- (aq), consider the equation:
The concentration of Na2S2O3 and S2O32- (aq) is 1:1
Hence, the concentration in molarity of S2O32- (aq) is 0.089 M.
To 3 significant figures = 0.0890 M