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Dominik [7]
3 years ago
11

To gravimetrically analyze the silver content of a piece of jewelry made from an alloy of Ag and Cu, a student dissolves a small

, pre-weighed sample in HNO3(aq). Ag+(aq) and Cu2+(aq) ions form in the solution.Which of the following should be the next step in analytical process?Centrifuging the solution to isolate the heavier ionsAdding enough base solution to bring the pH up to 7.0Adding a solution containing an anion that forms an insoluble salt with only one of the metal ionsEvaporating the solution to recover the dissolved nitrates
Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
8 0

Answer:

Adding a solution containing an anion that forms an insoluble salt with only one of the metal ions.

Explanation:

The student have in solution Ag⁺ and Cu²⁺ ions but he just want to analyze the silver, that means he need to separate ions.

Centrifuging the solution to isolate the heavier ions <em>FALSE </em>Centrifugation allows the separation of a suspension but Ag⁺ and Cu²⁺ are both soluble in water.

Adding enough base solution to bring the pH up to 7.0 <em>FALSE </em>At pH = 7,0 these ions are soluble in water and its separation will not be possible.

Adding a solution containing an anion that forms an insoluble salt with only one of the metal ions <em>TRUE </em>For example, the addition of Cl⁻ will precipitate the Ag⁺ as AgCl(s) allowing its separation.

Evaporating the solution to recover the dissolved nitrates. <em>FALSE</em> . Thus, you will obtain the nitrates of these ions but will be mixed doing impossible its separation.

I hope it helps!

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An 80.0g sample of an unknown metal is at an initial temperature of 55.5oC. Afer 540 J of energy is absorbed by the metal, the t
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Answer:

Specific heat of metal = 0.26 j/g.°C

Explanation:

Given data:

Mass of sample = 80.0 g

Initial temperature = 55.5 °C

Final temperature = 81.75 °C

Amount of heat absorbed = 540 j

Specific heat of metal = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

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c = specific heat capacity of substance

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ΔT =  81.75 °C - 55.5 °C

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540 j / 2100 g.°C = c

c = 0.26 j/g.°C

7 0
2 years ago
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