All categories

2 years ago

Is there any industrial process that use evaporation to separate a solution?

Chemistry

2 answers:

sergiy2304 [10]2 years ago

8 0

Yes, it is known as <u>distillation</u>.

Misha Larkins [42]2 years ago

6 0

The industrial process that utilizes that strategy of evaporation to separate a solution is known as Distillation.

<h3>What is Evaporation?</h3>

Evaporation may be defined as the methodology of liquid transforming into vapors even below its boiling point.

Distillation is an influential technique to isolate mixtures that are constituted of two or more refined liquids.

Distillation is a purification procedure where the elements of a liquid combination are vaporized and then compacted and sequestered.

Therefore, it is well described above.

To learn more about Distillation, refer to the link:

brainly.com/question/24553469

#SPJ1

You might be interested in

How many molecules are in 3.6 grams of NaCl? Question options:

Levart [38]

Answer:

$\boxed {\boxed {\sf 3.7 * 10^{22} \ molecules \ NaCl}}$

Explanation:

We are asked to find how many molecules are in 3.6 grams of sodium chloride.

<h3>1. Convert Grams to Moles </h3>

First, we convert grams to moles using the molar mass. These values are equivalent to atomic masses on the Periodic Table, but the units are grams per moles instead of atomic mass units. Look up the molar masses of the individual elements: sodium and chlorine.

Na: 22.9897693 g/mol
Cl: 35.45 g/mol

There are no subscripts in the chemical formula (NaCl), so we simply add the 2 molar masses.

NaCl: 22.9897693 + 35.45 = 58.4397693 g/mol

Now we will convert using dimensional analysis. First, set up a ratio using the molar mass.

$\frac {58.4397693 \ g \ NaCl}{ 1 \ mol \ NaCl}$

We are converting 3.6 grams to moles, so we must multiply the ratio by this value.

$3.6 \ g \ NaCl *\frac {58.4397693 \ g \ NaCl}{ 1 \ mol \ NaCl}$

Flip the ratio so the units of grams of sodium chloride cancel.

$3.6 \ g \ NaCl *\frac { 1 \ mol \ NaCl}{58.4397693 \ g \ NaCl}$

$3.6 *\frac { 1 \ mol \ NaCl}{58.4397693}$

$\frac { 3.6}{58.4397693} \ mol \ NaCl$

$0.06160188589 \ mol \ NaCl$

<h3>2. Convert Moles to Molecules </h3>

Next, we convert moles to molecules using Avogadro's Number. This is 6.022 × 10²³ and it tells us the number of particles (atoms, molecules, formula units, etc). In this case, the particles are molecules of sodium chloride. Let's set up another ratio.

$\frac {6.022 \times 10^{23} \ molecules \ NaCl}{ 1 \ mol \ NaCl}$

Multiply by the number of moles we calculated.

$0.06160188589 \ mol \ NaCl * \frac{6.022 \times 10^{23} \ molecules \ NaCl}{1 \ mol \ NaCl}$

The units of moles of sodium chloride cancel.

$0.06160188589 * \frac{6.022 \times 10^{23} \ molecules \ NaCl}{1 }$

$3.70966557*10^{22} \ molecules \ NaCl$

<h3>3. Round </h3>

The original measurement of grams (3.6) has 2 significant figures, so our answer must have the same. For the number we found, that is the tenths place. The 0 in the hundredth place tells us to leave the 7 in the tenth place.

$3.7 * 10^{22} \ molecules \ NaCl$

There are $3.7 * 10^{22} \ molecules \ NaCl$ in 3.6 grams and the correct answer is choice D.

5 0

3 years ago

Find the potential energy of a 2 kg ball 15 m in the air.

amm1812

Answer:

294,3

Explanation:

$E=m*g*h$

While g = 9,81 m/s^2 then:

$E=2[kg]*9,81[\frac{m}{s^2} ]*15[m]=294,3 [J]$ [J]

7 0

3 years ago

What is the difference between acute and chronic radiation poisoning?

Andreas93 [3]

Acute radiation poisoning will be in the first 24 hours and sometimes even within the next 48 hours. Chronic radiation poisoning is often classified as something that is an affect ongoing for months.

4 0

3 years ago

do you think that creativity can contribute to the development of a scientific experiment or investigation ?

Bingel [31]

I would say yes because you need to be creative with an scientific experiment but still follow the rules of the experiment or else it won't be accurate. Be creative and think of something that is possible to investigate or test. I hope that makes sense and helps
:)

4 0

3 years ago

Many industrial reactions, like the reaction of nitrogen gas (N2) and hydrogen gas (H2) to produce ammonia for fertilizers, have

Jobisdone [24]

Catalysts always quicken the experiment

8 0

3 years ago