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Slav-nsk [51]
3 years ago
10

To your unknown you add HCl to precipitate the Group A ions (Ag and Hg). This precipitate is removed and NH3 is added to it to r

edissolve the Ag and check for the presence of Hg. From the picture below which was taken right after the NH3 was added, please select a choice that best represents what you are seeing.
Chemistry
1 answer:
amm18123 years ago
5 0

Options are not given, however, the following reactions occurs when Group A ions are reacted with HCl followed by NH3

Answer:

Gray precipitate is seen, which confirms the presence of mercury ions

Explanation:

Selective precipitation is a qualitative analysis, which involves addition of a carefully selected reagents to an aqueous mixture of ions. This results in the precipitation of one or more ions, while leaving the rest in solution. Later, a reaction specific to that ion is carried out separately to determine its identity.

HCl react with both Ag+ and Hg+ ions to form the following precipitates:

Ag+(aq) + Cl-(aq) → AgCl(s)

Hg_{2}^{2+}(aq) + 2Cl- → Hg_{2}Cl_{2}(s)

The precipitate, i.e silver chloride and mercury(I) chloride is removed and solution of NH3 is added.

Silver chloride will dissolve since its forms a soluble complex ion:

 AgCl(s) + 2NH_{3}(aq) → Ag(NH_{3})_{2}^{+}(aq) + Cl^{-}(aq)

However, Mercury(I) chloride will react with ammonia to form a gray solid which is actually a mixture of black mercury and white AgNH_{2}Cl :

Hg_{2}Cl_{2}(s) + 2NH_{3}(aq) → Hg(l) + HgNH_{2}Cl(s) + NH_{4}^{+}(aq) + Cl^{-}(aq)

The presence of gray solid is the confirmation of the presence of Hg_{2}^{2+} ion

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Answer: a) Frequency=666\times10^{12}Hz-789\times10^{12}Hz

b) Wavelength=380nm-450nm

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