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Thepotemich [5.8K]
3 years ago
11

Separate a mixture of sand, common salt, copper pieces and iron fillings. Also measure the amount of common salt in the mixture.

Write the steps and the procedure of separation. Also what safety measures you will take during the experiment.
Chemistry
1 answer:
Neporo4naja [7]3 years ago
5 0

Answer:

Explanation:

This question seeks to test the knowledge of separation techniques.

From the narration in the question, the first separation to be done is the removal of Iron fillings by the use of magnet (magnetic separation). Since Iron is magnetic, the iron fillings will be attracted by the magnet hence removing the iron fillings from the mixture.

The second constituent to be removed will be the copper pieces by the use of a sieve (sieving). Copper pieces have relatively larger sizes than sand and common salt, hence a sieve (which separates particles based on size) can be used to remove the copper pieces from the mixture.

What will be left in the mixture after the processes above will be salt and water. This mixture will have to be dissolved in water; the salt will dissolve in water while the sand will not. After which, filtration will be done to remove the sand which will be collected on the filter paper as filtride and the salt solution will pass through the filter paper as filtrate.

The salt solution can then be evaporated to dryness to retrieve the solid salt from the solution.

The amount of salt in the mixture can then be measured using a weighing balance.

Some of safety measures to be taken during the course of this experiment includes performing the experiment in an airtight and controlled environment. Lab coat and hand gloves should be worn during the course of the experiment. The evaporation to dryness should not be done close to an inflammable material/substance

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The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

rate=k[H_2]^2[NH_3]

At a certain concentration of H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]

As we are given that:

Initial rate = 0.120 M/s

Expression for rate law for first observation:

0.120=k[H_2]^2[NH_3] ....(1)

Expression for rate law for second observation:

R=k(\frac{[H_2]}{2})^2[NH_3] ....(2)

Dividing 2 by 1, we get:

\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}

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Therefore, the initial rate of the reaction will be, 0.03 M/s

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Okay so we are given these requirements:

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likoan [24]

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Mass of isotope Ar- 40 = 39.96 amu

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Formula used for average atomic mass of an element :

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Hope it helped!
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3 years ago
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