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Nata [24]
3 years ago
14

A student has a flask containing two immiscible liquids. One of the liquids is a solution of a solid in water. Describe how you

would separate the mixture into three separate components.
Chemistry
2 answers:
Oksanka [162]3 years ago
6 0

The mixture can be separated into three separate components by using a centrifuge and a separating funnel.

A centrifuge is used to separate a solid that is held in suspension by a liquid. In this case a solid dissolved in water. In contrast, a separating funnel is used to separate two immiscible liquids.

-First take the flask containing the two immiscible liquids and pour contents in to a closed separating funnel.

- The most dense liquid will settle at the bottom, then open the separating funnel valve to collect contents of dense liquid into a flask.

- The remaining liquid on the separating funnel can be collected in a different flask.

-Now take the liquid that is a solution of a solid in water and pour in a test tube. Place the contents in a centrifuge.

- Centrifuge at high speed until the solid sinks to the bottom of the test tube, then decant the liquid into a flask and the solid remains in the test tube

I am Lyosha [343]3 years ago
3 0

Answer:

It can be separated using a separating funnel followed by crystallization.

Explanation:

First the mixture is put into a separating funnel. This will separate the two immiscible liquids. When the two immiscible liquids are separated, we recall that one is a solution of a solid solute and we wish to recover the solid from the solution whilst recovering the solvent also.

The solid can only be recovered by crystalization if we do not want to loose the solvent. Seeds of already crystalized solute may be added to system to induce crystalization.

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3 years ago
Chloroform, CHCl3, reacts with chlorine, Cl2, to form carbon tetrachloride, CCl4, and hydrogen chloride, HCl. In an experiment 2
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Answer:

Chloroform= limiting reactant

0.209mol of CCl4 is formed

And 32.186g of CCl4 is formed

Explanation:

The equation of reaction

CHCl3 + Cl2= CCl4 + HCl

From the equation 1 mol of

CHCl3 reacts with 1mol Cl2 to yield 1mol of CCl4

From the question

25g of CHCl3 really with Cl2

Molar mass of CHCl3= 119.5

Molar mass of Cl2 = 71

Hence moles of CHCl3= 25/119.5 = 0.209mol

Moles of Cl2 = 25/71 = 0.352mol

Hence CHCl3 is the limiting reactant

Since 1 mole of CHCl3 gave 1mol of CCl4

It implies that 0.209moles of CHCl3 will also give 0.209mol of CCl4

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