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frosja888 [35]
2 years ago
11

In the reaction of a carbonyl compound with silver cyanide, the lone pair of electrons on the nitrogen are the one that attack t

he carbonyl carbon. But if a carbonyl compound reacts with hydrogen cyanide, the cyanide ions directly attacks the carbonyl carbon. Explain why this happens??​
Chemistry
2 answers:
Marina CMI [18]2 years ago
7 0

Answer:

For Silver cyanide: Silver is more electropositive. Therefore no need for silver cyanide to split into silver ion and cyanide ion, only the lone pair of electrons from nitrogen atom will be attracted to the carbonyl carbon of the aldehyde or ketone.

For hydrogen cyanide: Hydrogen is less electropositive. Therefore hydrogen cyanide splits into hydrogen ion and cyanide ion, since cyanide ion is negative, It will be directly attracted to the carbonyl compound.

Equations: [ considering a ketone and aldehyde ]

RCHO + HCN → RCHCN + H2O

R'COCH3 + AgCN → R'C(CN)CH3 + Ag + H2O

.

.

never [62]2 years ago
3 0

Answer:

The larvae of mosquitoes live in water and provide food for fish and other wildlife, including larger larvae of other species such as dragonflies. The mosquito larvae themselves consume a lot of organic matter in wetlands, helping recycle nutrients back into the ecosystem.Ramabai Bhimrao Ambedkar was the first wife of B. R. Ambedkar, who said her support was instrumental in helping him pursue his higher education and his true potential. She has been the subject of a number of biographical movies and books. A number of landmarks across India have been named after her.Hydrochloric acid: HCl. Nitric acid: HNO. Phosphoric acid: H3PO. Sulfuric acid: H2SO.

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Answer:

Ka3 for the triprotic acid is 7.69*10^-11

Explanation:

Step 1: Data given

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Step 2: Calculate Ka3

pKa = -log (Ka2) = 6.824

The pH at the second equivalence point (8.469) will be the average of pKa2  and pKa3. So,

8.469 = (6.824 + pKa3) / 2

pKa3 = 10.114

Ka3 = 10^-10.114 = 7.69*10^-11

Ka3 for the triprotic acid is 7.69*10^-11

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