Chemistry, 04.01.2020 02:31 StevencoolGG

During which method of separation is one of the components of the mixture lost?

a) distillation
b) magnetic separation
c) screening
d) evaporation

Answers: 3

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Answers

I am quite confused with your question. You are asking about the method of separation wherein one of the components is 'lost'. However, these methods are only physical changes that separates components of the mixtures. But nothing gets lost according to the Law of Conservation of Mass. The components are separated but both can still be recovered.

In distillation, the more volatile component vaporizes and allowed to condense again to gather the volatile-rich solution. The less volatile stays on the bottoms of the tower. In magnetic separation, you just collect the metals from non-metals. In screening, you separate the smaller pieces from the larger pieces with the use of screens. In evaporation, water is separated from the solute by vaporizing it. You will be left with a more concentrated solution. However, the water is not lost because it could be condensed again back to liquid water.

However, to provide an answer, I think the closest would be evaporation. This applies to natural evaporation like a cup of coffee left in open air. You can't recover water anymore because it evaporized to the atmosphere. But still, technically, it comes back as precipitated rain.

Answer from: babas97

Explanation:

because

Answer from: Quest

when the volume of gas is changed from 717.73 ml to 852 ml the temperature will change from 315 c to 452 c

calculation

this is calculated using the charles law formula

that is, v1/t1 = v2/t2

v1=?

t1 = 315 c into kelvin = 315 +273 = 588 k

v2 = 852 ml

t2 = 425c + 273 =698 k

make v1 the subject of the formula ( by multiplying both side by t1)

v1=( v2 xt1) / t2

v1= (852 ml x 588 k)/ 698 k= 717.73 ml

Answer from: Quest

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

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During which method of separation is one of the components of the mixture lost?

a) dist...

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