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mote1985 [20]
2 years ago
13

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

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What is physical weathering NOT caused by? (pls i really need help with this!!)
Anarel [89]

Answer:

the changing tempure in rocks causing it to back apart

3 0
3 years ago
Describe the relationship between the radius of a cation and that of the atom from which forms.
vichka [17]

The radius of the cation is much smaller than the corresponding neutral atom.(b) The radius of an anion is much larger than the corresponding neutral atom.Explanation:The size of the atom or ion is inversely proportional to the nuclear charge experienced by the electrons.(a)The size of the cation is smaller than the size of the corresponding neutral atom. This is because after removal of an electron from the highest principle energy level the nuclear charge experienced by the valence electrons increases resulting in the decrease in size.(b)The size of an anion is larger than the size of the corresponding neutral atom. In an anion, an extra electron is added to the highest principle energy level but the effective nuclear charge pulling the electrons towards the nucleus is still same. The net effective nuclear charge experienced by the electrons present in the outermost shell decrease. Moreover, due to the added electron, the repulsion between the electrons also increases resulting in the increase in size

Make since? i hope this helps

4 0
3 years ago
The thallium Subscript 81 Superscript 208 Baseline Tl nucleus is radioactive, with a half-life of 3.053 min. At a given instant,
Genrish500 [490]

Answer: (E) 300 bq

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.

Half life is represented by t_{\frac{1}{2}

Half life of Thallium-208 = 3.053 min

Thus after 9 minutes , three half lives will be passed, after ist half life, the activity would be reduced to half of original i.e. \frac{2400}{2}=1200, after second  half life, the activity would be reduced to half of 1200 i.e. \frac{1200}{2}=600,  and after third half life, the activity would be reduced to half of 600 i.e. \frac{600}{2}=300,

Thus the activity 9 minutes later is 300 bq.

7 0
3 years ago
A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________
gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

From the question  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s

b. the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

M₁ = molar mass sulfur dioxide = 64

M₂ =  molar mass nitrogen triodide = 395

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0
3 years ago
This is not a homework question
Dovator [93]
What state you live in? because I know someone who does it for free
8 0
3 years ago
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