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3 years ago

When doing a recrystallization what observations would prompt you to perform a hot filtration?

Chemistry

2 answers:

viva [34]3 years ago

7 0

Answer:

I would do a hot filtering if I observed that I was facing a situation where crystallization occurs in a hot solution, but decrystalization occurs in a cold solution.

Explanation:

The hot filtration method is generally employed in solids purification processes (solids mixing), also called recrystallization purification. In general, hot filtration is associated with the vacuum filtration method.

The hot filtration method uses the principle of different solubility at different temperatures, ie a substance is hot soluble in one solvent but not soluble in the same solvent at room temperature. In the hot solvent, only one substance in the mixture dissolves completely, the other substance remains insoluble and is retained by the filter. When the filtered solution cools, the insoluble substance is separated from the solvent.

In this case, it is only favorable to use hot filtration if you observe that crystallization of a particular material occurs in a hot solution and its dissolution occurs in cold solution.

KATRIN_1 [288]3 years ago

3 0

Answer is: <span>because dissolved compounds can crystallizing from solution during filtration and forming crystals on the filter paper or funnel.
</span>Recrystallization<span> is a technique used to purify chemicals by dissolving both impurities and a compound in an appropriate solvent, either compound or impurities can be removed from the solution, leaving the other behind.</span>

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The reaction 2NO(g)+Cl2(g)→2NOCl(g) is carried out in a closed vessel. If the partial pressure of NO is decreasing at the rate o

Luba_88 [7]

Answer : The rate of change of the total pressure of the vessel is, 10.5 torr/min.

Explanation : Given,

$\frac{d[NO]}{dt}$ =21 torr/min

The balanced chemical reaction is,

$2NO(g)+Cl_2(g)\rightarrow 2NOCl(g)$

The rate of disappearance of $NO$ = $-\frac{1}{2}\frac{d[NO]}{dt}$

The rate of disappearance of $Cl_2$ = $-\frac{d[Cl_2]}{dt}$

The rate of formation of $NOCl$ = $\frac{1}{2}\frac{d[NOCl]}{dt}$

As we know that,

$\frac{d[NO]}{dt}$ =21 torr/min

So,

$-\frac{d[Cl_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}$

$\frac{d[Cl_2]}{dt}=\frac{1}{2}\times 21torr/min=10.5torr/min$

And,

$\frac{1}{2}\frac{d[NOCl]}{dt}=\frac{1}{2}\frac{d[NO]}$

$\frac{d[NOCl]}{dt}=\frac{d[NO]}=21torr/min$

Now we have to calculate the rate change.

Rate change = Reactant rate - Product rate

Rate change = (21 + 10.5) - 21 = 10.5 torr/min

Therefore, the rate of change of the total pressure of the vessel is, 10.5 torr/min.

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