Question

The total number of cationic radicals that
give black precipitate with H2S in acid
solution is :
Pb2+, Zn2+, Cu2+, Mn²+, Hg2+, Cr3+, Cd2+​

Answer 1

Answer:

I think the number is 5

Explanation:

1. Pb²+
2. Zn²+
3. Cu²+
4. Mn²+
5. Cr3+

Answer 2

Answer:

There are 4 cationic radicals that give a black precipitate with H₂S in acid solution.

Explanation:

Given,

The cationic radicals:

Pb⁺², Zn⁺², Cu⁺², Mn⁺², Hg⁺², Cr⁺³, Cd⁺².

To find,

The total number of cationic radicals that give a black precipitate with H₂S. in acid solution is.

Calculation,

In the given cationic radicals only the ions Pb⁺², Cu⁺², Hg⁺², and Cr⁺³ will give a black precipitate with H₂S.

The reaction of H₂S with Pb⁺²:

$Pb^{+2}+H_2S \rightarrow PbS + 2H^+$

The PbS formed on the product's side is a black precipitate.

The reaction of H₂S with Cu⁺²:

$Cu^{+2}+H_2S \rightarrow CuS + 2H^+$

The CuS formed on the product's side is a black precipitate.

The reaction of H₂S with Hg⁺²:

$Hg^{+2} + H_2S \rightarrow HgS + 2H^+$

The HgS formed on the product's side is a black precipitate.

The reaction of H₂S with Cr⁺³:

$2Cr^{+3}+3H_2S \rightarrow Cr_2S_3 + 6H^+$

The Cr₂S₃ formed on the product's side is a black precipitate.

Therefore, there are 4 cationic radicals that give a black precipitate with H₂S in acid solution.

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