Methods for the separation of heterogeneous mixtures. Chemical and physical methods for separating mixtures

I. New material

In preparing the lesson, materials of the author were used: N.K.Cheremisina,

chemistry teacher of secondary school No. 43

(Kaliningrad),

We live among chemicals. We inhale air, and this is a mixture of gases ( nitrogen, oxygen and others), exhale carbon dioxide. We wash ourselves water- This is another substance, the most common on Earth. We drink milk- mixture water with tiny drops of milk fat, and not only: there is still milk protein casein, mineral salt, vitamins and even sugar, but not the one with which they drink tea, but a special, milky one - lactose. We eat apples, which consist of a whole range of chemicals - here and sugar, and Apple acid, and vitamins... When the chewed pieces of an apple enter the stomach, human digestive juices begin to act on them, which help to absorb all the tasty and healthy substances not only of the apple, but also of any other food. We not only live among chemicals, but we ourselves are made of them. Every person - his skin, muscles, blood, teeth, bones, hair are built of chemicals, like a house of bricks. Nitrogen, oxygen, sugar, vitamins are substances of natural, natural origin. Glass, rubber, steel is also a substance, more precisely, materials(mixtures of substances). Both glass and rubber are of artificial origin; they did not exist in nature. Completely pure substances are not found in nature or are very rare.

What is the difference between pure substances and mixtures of substances?

An individual pure substance has a certain set of characteristic properties (constant physical properties). Only pure distilled water has tmelt = 0 °С, tboil = 100 °С, and has no taste. Sea water freezes at a lower temperature, and boils at a higher temperature, its taste is bitter-salty. The water of the Black Sea freezes at a lower temperature and boils at a higher temperature than the water of the Baltic Sea. Why? The fact is that sea water contains other substances, for example, dissolved salts, i.e. it is a mixture of various substances, the composition of which varies over a wide range, but the properties of the mixture are not constant. The concept of "mixture" was defined in the 17th century. English scientist Robert Boyle : "A mixture is an integral system consisting of heterogeneous components."

Comparative characteristics of a mixture and a pure substance

Signs of comparison	pure substance	Mixture
Compound	Constant	fickle
Substances	Same	Various
Physical Properties	Permanent	Fickle
Energy change during formation	going on	Not happening
Separation	Through chemical reactions	Physical methods

Mixtures differ from each other in appearance.

The classification of mixtures is shown in the table:

Here are examples of suspensions (river sand + water), emulsions (vegetable oil + water) and solutions (air in a flask, salt + water, small change: aluminum + copper or nickel + copper).

In suspensions, solid particles are visible, in emulsions - liquid droplets, such mixtures are called heterogeneous (heterogeneous), and in solutions the components are not distinguishable, they are homogeneous (homogeneous) mixtures.

Methods for separating mixtures

In nature, substances exist in the form of mixtures. For laboratory research, industrial production, for the needs of pharmacology and medicine, pure substances are needed.

Various methods of separation of mixtures are used to purify substances.

These methods are based on differences in the physical properties of the components of the mixture.

Consider waysseparationheterogeneous and homogeneous mixtures .

Blend example	Separation method
Suspension - a mixture of river sand with water	settling Separation upholding based on different densities of substances. Heavier sand settles to the bottom. You can also separate the emulsion: to separate oil or vegetable oil from water. In the laboratory, this can be done using a separating funnel. Petroleum or vegetable oil forms an upper, lighter layer.As a result of settling, dew falls out of the fog, soot is deposited from smoke, cream is settled in milk. Separation of a mixture of water and vegetable oil by settling
A mixture of sand and table salt in water	Filtration What is the basis for the separation of heterogeneous mixtures using filtering? On various solubility of substances in water and on various sizes of particles. Through the pores of the filter pass only commensurate particles of substances, while larger particles are retained on the filter. This is how you can separate a heterogeneous mixture of table salt and river sand.Various porous substances can be used as filters: cotton wool, coal, fired clay, pressed glass, and others. The filtering method is the basis for the operation of household appliances, such as vacuum cleaners. It is used by surgeons - gauze bandages; drillers and workers of elevators - respiratory masks. With the help of a tea strainer for filtering tea leaves, Ostap Bender, the hero of the work of Ilf and Petrov, managed to take one of the chairs from Ellochka Ogre (“The Twelve Chairs”).
A mixture of iron powder and sulfur	Action by magnet or water Iron powder was attracted by a magnet, but sulfur powder was not.. The non-wettable sulfur powder floated to the surface of the water, while the heavy wettable iron powder settled to the bottom.. Separation of a mixture of sulfur and iron using a magnet and water
A solution of salt in water is a homogeneous mixture	Evaporation or crystallization The water evaporates and salt crystals remain in the porcelain cup. When water is evaporated from lakes Elton and Baskunchak, table salt is obtained. This separation method is based on the difference in the boiling points of the solvent and the solute. If a substance, such as sugar, decomposes when heated, then the water is not completely evaporated - the solution is evaporated, and then sugar crystals are precipitated from a saturated solution. Sometimes it is required to remove impurities from solvents with a lower temperature boiling, such as water from salt. In this case, the vapors of the substance must be collected and then condensed upon cooling. This method of separating a homogeneous mixture is called distillation or distillation. In special devicesdistillers produce distilled water , whichused for the needs of pharmacology, laboratories, car cooling systems . At home, you can design such a distiller: If, however, a mixture of alcohol and water is separated, then the first to be distilled off (collected in a receiving test tube) is alcohol with t bp = 78 ° C, and water will remain in the test tube. Distillation is used to obtain gasoline, kerosene, gas oil from oil. Separation of homogeneous mixtures

A special method of separating components, based on their different absorption by a certain substance, is chromatography.

At home, you can do the following experiment. Hang a strip of filter paper over the bottle of red ink, dipping only the end of the strip into it. The solution is absorbed by the paper and rises along it. But the border of the rise of the paint lags behind the border of the rise of the water. This is how the separation of two substances occurs: water and the coloring matter in the ink.

With the help of chromatography, the Russian botanist M. S. Tsvet was the first to isolate chlorophyll from the green parts of plants. In industry and laboratories, instead of filter paper for chromatography, starch, coal, limestone, and aluminum oxide are used. Are substances always required with the same degree of purification?

For different purposes, substances with different degrees of purification are needed. Cooking water is sufficiently settled to remove impurities and chlorine used to disinfect it. Drinking water must first be boiled. And in chemical laboratories for the preparation of solutions and experiments, in medicine, distilled water is needed, as purified as possible from the substances dissolved in it. Highly pure substances, the content of impurities in which does not exceed one millionth of a percent, are used in electronics, semiconductor, nuclear technology and other precision industries.

Read L. Martynov's poem "Distilled Water":

Water
Favored
pour!
She is
shone
So pure
Whatever to drink
Don't wash.
And it was no accident.
She missed
Willows, tala
And the bitterness of flowering vines,
She missed seaweed
And fish oily from dragonflies.
She missed being wavy
She missed flowing everywhere.
She didn't have enough life.
Clean -
Distilled water!

Application of distilled water

II. Tasks for fixing

1) Work with machines #1-4(necessarydownload the simulator, it will open in the Internet Explorer browser)

theoretical block.

The concept of "mixture" was defined in the 17th century. English scientist Robert Boyle: "A mixture is an integral system consisting of heterogeneous components."

Comparative characteristics of a mixture and a pure substance

Signs of comparison	pure substance	Mixture
	Constant	fickle
Substances	Same	Various
Physical Properties	Permanent	Fickle
Energy change during formation	going on	Not happening
Separation	Through chemical reactions	Physical methods

Mixtures differ from each other in appearance.

The classification of mixtures is shown in the table:

Here are examples of suspensions (river sand + water), emulsions (vegetable oil + water) and solutions (air in a flask, salt + water, small change: aluminum + copper or nickel + copper).

Methods for separating mixtures

In nature, substances exist in the form of mixtures. For laboratory research, industrial production, for the needs of pharmacology and medicine, pure substances are needed.

Various methods of separation of mixtures are used to purify substances.

Evaporation - the separation of solids dissolved in a liquid by converting it into vapor.

Distillation- distillation, separation of substances contained in liquid mixtures according to boiling points, followed by cooling of the vapor.

In nature, water in its pure form (without salts) does not occur. Oceanic, sea, river, well and spring water are varieties of salt solutions in water. However, often people need clean water that does not contain salts (used in car engines; in chemical production to obtain various solutions and substances; in the manufacture of photographs). Such water is called distilled, and the method of obtaining it is called distillation.

Filtration is the filtering of liquids (gases) through a filter in order to purify them from solid impurities.

These methods are based on differences in the physical properties of the components of the mixture.

Consider ways to separate heterogeneousand homogeneous mixtures.

Blend example	Separation method
Suspension - a mixture of river sand with water	settling Separation upholding based on different densities of substances. Heavier sand settles to the bottom. You can also separate the emulsion: to separate oil or vegetable oil from water. In the laboratory, this can be done using a separating funnel. Oil or vegetable oil forms the top, lighter layer. As a result of settling, dew falls out of the fog, soot is deposited from smoke, cream is settled in milk. Separation of a mixture of water and vegetable oil by settling
A mixture of sand and table salt in water	Filtration What is the basis for the separation of heterogeneous mixtures using filtering? On various solubility of substances in water and on various sizes of particles. Only particles of substances commensurate with them pass through the pores of the filter, while larger particles are retained on the filter. So you can separate a heterogeneous mixture of table salt and river sand. Various porous substances can be used as filters: cotton wool, coal, fired clay, pressed glass, and others. The filtering method is the basis for the operation of household appliances, such as vacuum cleaners. It is used by surgeons - gauze bandages; drillers and workers of elevators - respiratory masks. With the help of a tea strainer for filtering tea leaves, Ostap Bender, the hero of the work of Ilf and Petrov, managed to take one of the chairs from Ellochka Ogre (“The Twelve Chairs”). Separation of a mixture of starch and water by filtration
A mixture of iron powder and sulfur	Action by magnet or water Iron powder was attracted by a magnet, but sulfur powder was not. The non-wettable sulfur powder floated to the surface of the water, while the heavy wettable iron powder settled to the bottom. Separation of a mixture of sulfur and iron using a magnet and water
A solution of salt in water is a homogeneous mixture	Evaporation or crystallization The water evaporates and salt crystals remain in the porcelain cup. When water is evaporated from lakes Elton and Baskunchak, table salt is obtained. This separation method is based on the difference in the boiling points of the solvent and the solute. If a substance, such as sugar, decomposes when heated, then the water is not completely evaporated - the solution is evaporated, and then sugar crystals are precipitated from a saturated solution. Sometimes it is required to remove impurities from solvents with a lower boiling point, for example, water from salt. In this case, the vapors of the substance must be collected and then condensed upon cooling. This method of separating a homogeneous mixture is called distillation or distillation. In special devices - distillers, distilled water is obtained, which is used for the needs of pharmacology, laboratories, and car cooling systems. At home, you can design such a distiller: If, however, a mixture of alcohol and water is separated, then the first to be distilled off (collected in a receiving test tube) is alcohol with tboil = 78 °C, and water will remain in the test tube. Distillation is used to obtain gasoline, kerosene, gas oil from oil. Separation of homogeneous mixtures

A special method of separating components, based on their different absorption by a certain substance, is chromatography.

Using chromatography, the Russian botanist was the first to isolate chlorophyll from the green parts of plants. In industry and laboratories, instead of filter paper for chromatography, starch, coal, limestone, and aluminum oxide are used. Are substances always required with the same degree of purification?

Methods for expressing the composition of mixtures.

· Mass fraction of the component in the mixture- the ratio of the mass of the component to the mass of the entire mixture. Usually the mass fraction is expressed in %, but not necessarily.

ω ["omega"] = mcomponent / mmixture

· Mole fraction of a component in a mixture- the ratio of the number of moles (amount of substance) of the component to the total number of moles of all substances in the mixture. For example, if the mixture includes substances A, B and C, then:

χ [“chi”] component A \u003d n component A / (n (A) + n (B) + n (C))

· Molar ratio of components. Sometimes in tasks for a mixture, the molar ratio of its components is indicated. For example:

ncomponent A: ncomponent B = 2: 3

· Volume fraction of the component in the mixture (only for gases)- the ratio of the volume of substance A to the total volume of the entire gas mixture.

φ ["phi"] = Vcomponent / Vmixture

Practice block.

Consider three examples of problems in which mixtures of metals react with hydrochloric acid:

Example 1When a mixture of copper and iron weighing 20 g was exposed to an excess of hydrochloric acid, 5.6 liters of gas (n.a.) were released. Determine the mass fractions of metals in the mixture.

In the first example, copper does not react with hydrochloric acid, that is, hydrogen is released when the acid reacts with iron. Thus, knowing the volume of hydrogen, we can immediately find the amount and mass of iron. And, accordingly, the mass fractions of substances in the mixture.

Example 1 solution.

n \u003d V / Vm \u003d 5.6 / 22.4 \u003d 0.25 mol.

2. According to the reaction equation:

3. The amount of iron is also 0.25 mol. You can find its mass:
mFe = 0.25 56 = 14 g.

Answer: 70% iron, 30% copper.

Example 2Under the action of an excess of hydrochloric acid on a mixture of aluminum and iron weighing 11 g, 8.96 liters of gas (n.a.) were released. Determine the mass fractions of metals in the mixture.

In the second example, the reaction is both metal. Here, hydrogen is already released from the acid in both reactions. Therefore, direct calculation cannot be used here. In such cases, it is convenient to solve using a very simple system of equations, taking for x - the number of moles of one of the metals, and for y - the amount of substance of the second.

Example 2 solution.

1. Find the amount of hydrogen:
n \u003d V / Vm \u003d 8.96 / 22.4 \u003d 0.4 mol.

2. Let the amount of aluminum be x mol, and iron y mol. Then we can express in terms of x and y the amount of hydrogen released:


	2HCl = FeCl2 +

4. We know the total amount of hydrogen: 0.4 mol. Means,
1.5x + y = 0.4 (this is the first equation in the system).

5. For a mixture of metals, you need to express masses through quantities of substances.
m = Mn
So the mass of aluminum
mAl = 27x,
mass of iron
mFe = 56y,
and the mass of the whole mixture
27x + 56y = 11 (this is the second equation in the system).

6. So, we have a system of two equations:

7. Solving such systems is much more convenient by subtracting by multiplying the first equation by 18:
27x + 18y = 7.2
and subtracting the first equation from the second:

8. (56 - 18)y \u003d 11 - 7.2
y \u003d 3.8 / 38 \u003d 0.1 mol (Fe)
x = 0.2 mol (Al)

mFe = n M = 0.1 56 = 5.6 g
mAl = 0.2 27 = 5.4 g
ωFe = mFe / mmixture = 5.6 / 11 = 0.50.91%),

respectively,
ωAl \u003d 100% - 50.91% \u003d 49.09%

Answer: 50.91% iron, 49.09% aluminum.

Example 316 g of a mixture of zinc, aluminum and copper was treated with an excess of hydrochloric acid solution. In this case, 5.6 liters of gas (n.a.) were released and 5 g of the substance did not dissolve. Determine the mass fractions of metals in the mixture.

In the third example, two metals react, but the third metal (copper) does not react. Therefore, the remainder of 5 g is the mass of copper. The quantities of the remaining two metals - zinc and aluminum (note that their total mass is 16 - 5 = 11 g) can be found using a system of equations, as in example No. 2.

Answer to Example 3: 56.25% zinc, 12.5% aluminum, 31.25% copper.

Example 4A mixture of iron, aluminum and copper was treated with an excess of cold concentrated sulfuric acid. At the same time, part of the mixture dissolved, and 5.6 liters of gas (n.a.) were released. The remaining mixture was treated with an excess of sodium hydroxide solution. 3.36 liters of gas evolved and 3 g of undissolved residue remained. Determine the mass and composition of the initial mixture of metals.

In this example, remember that cold concentrated sulfuric acid does not react with iron and aluminum (passivation), but reacts with copper. In this case, sulfur oxide (IV) is released.
With alkali reacts only aluminum- amphoteric metal (in addition to aluminum, zinc and tin also dissolve in alkalis, and beryllium can still be dissolved in hot concentrated alkali).

Example 4 solution.

1. Only copper reacts with concentrated sulfuric acid, the number of moles of gas:
nSO2 = V / Vm = 5.6 / 22.4 = 0.25 mol


	2H2SO4 (conc.) = CuSO4 +

2. (do not forget that such reactions must be equalized using an electronic balance)

3. Since the molar ratio of copper and sulfur dioxide is 1:1, then copper is also 0.25 mol. You can find the mass of copper:
mCu \u003d n M \u003d 0.25 64 \u003d 16 g.

4. Aluminum reacts with an alkali solution, and an aluminum hydroxocomplex and hydrogen are formed:
2Al + 2NaOH + 6H2O = 2Na + 3H2

Al0 − 3e = Al3+

5. Number of moles of hydrogen:
nH2 = 3.36 / 22.4 = 0.15 mol,
the molar ratio of aluminum and hydrogen is 2:3 and, therefore,
nAl = 0.15 / 1.5 = 0.1 mol.
Aluminum weight:
mAl \u003d n M \u003d 0.1 27 \u003d 2.7 g

6. The remainder is iron, weighing 3 g. You can find the mass of the mixture:
mmix \u003d 16 + 2.7 + 3 \u003d 21.7 g.

7. Mass fractions of metals:

ωCu = mCu / mmixture = 16 / 21.7 = 0.7.73%)
ωAl = 2.7 / 21.7 = 0.1.44%)
ωFe = 13.83%

Answer: 73.73% copper, 12.44% aluminum, 13.83% iron.

Example 521.1 g of a mixture of zinc and aluminum was dissolved in 565 ml of a nitric acid solution containing 20 wt. % HNO3 and having a density of 1.115 g/ml. The volume of the released gas, which is a simple substance and the only product of the reduction of nitric acid, amounted to 2.912 l (n.a.). Determine the composition of the resulting solution in mass percent. (RCTU)

The text of this problem clearly indicates the product of nitrogen reduction - "simple substance". Since nitric acid does not produce hydrogen with metals, it is nitrogen. Both metals dissolved in acid.
The problem asks not the composition of the initial mixture of metals, but the composition of the solution obtained after the reactions. This makes the task more difficult.

Example 5 solution.

1. Determine the amount of gas substance:
nN2 = V / Vm = 2.912 / 22.4 = 0.13 mol.

2. Determine the mass of the nitric acid solution, the mass and amount of the dissolved HNO3 substance:

msolution \u003d ρ V \u003d 1.115 565 \u003d 630.3 g
mHNO3 = ω msolution = 0.2 630.3 = 126.06 g
nHNO3 = m / M = 126.06 / 63 = 2 mol

Please note that since the metals have completely dissolved, it means - just enough acid(these metals do not react with water). Accordingly, it will be necessary to check Is there too much acid?, and how much of it remains after the reaction in the resulting solution.

3. Compose reaction equations ( do not forget about the electronic balance) and, for convenience of calculations, we take for 5x - the amount of zinc, and for 10y - the amount of aluminum. Then, in accordance with the coefficients in the equations, nitrogen in the first reaction will be x mol, and in the second - 3y mol:


	12HNO3 = 5Zn(NO3)2 +

Zn0 − 2e = Zn2+


	36HNO3 = 10Al(NO3)3 +

Al0 − 3e = Al3+

5. Then, given that the mass of the mixture of metals is 21.1 g, their molar masses are 65 g/mol for zinc and 27 g/mol for aluminum, we obtain the following system of equations:

6. It is convenient to solve this system by multiplying the first equation by 90 and subtracting the first equation from the second.

7. x \u003d 0.04, which means nZn \u003d 0.04 5 \u003d 0.2 mol
y \u003d 0.03, which means that nAl \u003d 0.03 10 \u003d 0.3 mol

8. Check the mass of the mixture:
0.2 65 + 0.3 27 \u003d 21.1 g.

9. Now let's move on to the composition of the solution. It will be convenient to rewrite the reactions again and write down over the reactions the amounts of all reacted and formed substances (except water):

10. The next question is: did nitric acid remain in the solution and how much is left?
According to the reaction equations, the amount of acid that reacted:
nHNO3 = 0.48 + 1.08 = 1.56 mol,
i.e. the acid was in excess and you can calculate its remainder in the solution:
nHNO3res. \u003d 2 - 1.56 \u003d 0.44 mol.

11. So, in final solution contains:

zinc nitrate in the amount of 0.2 mol:
mZn(NO3)2 = n M = 0.2 189 = 37.8 g
aluminum nitrate in the amount of 0.3 mol:
mAl(NO3)3 = n M = 0.3 213 = 63.9 g
an excess of nitric acid in an amount of 0.44 mol:
mHNO3res. = n M = 0.44 63 = 27.72 g

12. What is the mass of the final solution?
Recall that the mass of the final solution consists of those components that we mixed (solutions and substances) minus those reaction products that left the solution (precipitates and gases):

13.
Then for our task:

14. new solution \u003d mass of acid solution + mass of metal alloy - mass of nitrogen
mN2 = n M = 28 (0.03 + 0.09) = 3.36 g
new solution \u003d 630.3 + 21.1 - 3.36 \u003d 648.04 g

ωZn(NO3)2 \u003d mv-va / mr-ra \u003d 37.8 / 648.04 \u003d 0.0583
ωAl(NO3)3 \u003d mv-va / mr-ra \u003d 63.9 / 648.04 \u003d 0.0986
ωHNO3res. \u003d mv-va / mr-ra \u003d 27.72 / 648.04 \u003d 0.0428

Answer: 5.83% zinc nitrate, 9.86% aluminum nitrate, 4.28% nitric acid.

Example 6When processing 17.4 g of a mixture of copper, iron and aluminum with an excess of concentrated nitric acid, 4.48 liters of gas (n.a.) were released, and when this mixture was exposed to the same mass of excess hydrochloric acid, 8.96 l of gas (n.a.). u.). Determine the composition of the initial mixture. (RCTU)

When solving this problem, we must remember, firstly, that concentrated nitric acid with an inactive metal (copper) gives NO2, while iron and aluminum do not react with it. Hydrochloric acid, on the other hand, does not react with copper.

Answer for example 6: 36.8% copper, 32.2% iron, 31% aluminum.

Tasks for independent solution.

1. Simple problems with two mixture components.

1-1. A mixture of copper and aluminum weighing 20 g was treated with a 96% solution of nitric acid, and 8.96 liters of gas (n.a.) were released. Determine the mass fraction of aluminum in the mixture.

1-2. A mixture of copper and zinc weighing 10 g was treated with a concentrated alkali solution. In this case, 2.24 liters of gas (n. y.) were released. Calculate the mass fraction of zinc in the initial mixture.

1-3. A mixture of magnesium and magnesium oxide weighing 6.4 g was treated with a sufficient amount of dilute sulfuric acid. At the same time, 2.24 liters of gas (n.a.) were released. Find the mass fraction of magnesium in the mixture.

1-4. A mixture of zinc and zinc oxide weighing 3.08 g was dissolved in dilute sulfuric acid. Zinc sulfate weighing 6.44 g was obtained. Calculate the mass fraction of zinc in the initial mixture.

1-5. Under the action of a mixture of iron and zinc powders weighing 9.3 g on an excess of copper (II) chloride solution, 9.6 g of copper was formed. Determine the composition of the initial mixture.

1-6. What mass of a 20% hydrochloric acid solution will be required to completely dissolve 20 g of a mixture of zinc with zinc oxide, if hydrogen is released in the amount of 4.48 liters (n.a.)?

1-7. When dissolved in dilute nitric acid, 3.04 g of a mixture of iron and copper releases nitric oxide (II) with a volume of 0.896 l (n.a.). Determine the composition of the initial mixture.

1-8. When dissolving 1.11 g of a mixture of iron and aluminum filings in a 16% hydrochloric acid solution (ρ = 1.09 g / ml), 0.672 liters of hydrogen (n.a.) were released. Find the mass fractions of metals in the mixture and determine the volume of hydrochloric acid consumed.

2. Tasks are more complex.

2-1. A mixture of calcium and aluminum weighing 18.8 g was calcined without access to air with an excess of graphite powder. The reaction product was treated with dilute hydrochloric acid, and 11.2 liters of gas (n.a.) were released. Determine the mass fractions of metals in the mixture.

2-2. To dissolve 1.26 g of an alloy of magnesium with aluminum, 35 ml of a 19.6% sulfuric acid solution (ρ = 1.1 g/ml) was used. The excess acid reacted with 28.6 ml of a 1.4 mol/L potassium hydrogen carbonate solution. Determine the mass fractions of metals in the alloy and the volume of gas (n.a.) released during the dissolution of the alloy.

Topic: "Methods for separating mixtures" (Grade 8)

theoretical block.

The concept of "mixture" was defined in the 17th century. English scientist Robert Boyle: "A mixture is an integral system consisting of heterogeneous components."

Comparative characteristics of a mixture and a pure substance

Signs of comparison	pure substance	Mixture
	Constant	fickle
Substances	Same	Various
Physical Properties	Permanent	Fickle
Energy change during formation	going on	Not happening
Separation	Through chemical reactions	Physical methods

Mixtures differ from each other in appearance.

The classification of mixtures is shown in the table:

Here are examples of suspensions (river sand + water), emulsions (vegetable oil + water) and solutions (air in a flask, salt + water, small change: aluminum + copper or nickel + copper).

Methods for separating mixtures

In nature, substances exist in the form of mixtures. For laboratory research, industrial production, for the needs of pharmacology and medicine, pure substances are needed.

Various methods of separation of mixtures are used to purify substances.

Evaporation is the separation of solids dissolved in a liquid by converting it into vapor.

Distillation- distillation, separation of substances contained in liquid mixtures according to boiling points, followed by cooling of the vapor.

Filtration is the filtering of liquids (gases) through a filter in order to purify them from solid impurities.

These methods are based on differences in the physical properties of the components of the mixture.

Consider ways to separate heterogeneous and homogeneous mixtures.

Blend example	Separation method
Suspension - a mixture of river sand with water	settling Separation upholding based on different densities of substances. Heavier sand settles to the bottom. You can also separate the emulsion: to separate oil or vegetable oil from water. In the laboratory, this can be done using a separating funnel. Oil or vegetable oil forms the top, lighter layer. As a result of settling, dew falls out of the fog, soot is deposited from smoke, cream is settled in milk. Separation of a mixture of water and vegetable oil by settling
A mixture of sand and table salt in water	Filtration What is the basis for the separation of heterogeneous mixtures using filtering? On various solubility of substances in water and on various sizes of particles. Only particles of substances commensurate with them pass through the pores of the filter, while larger particles are retained on the filter. So you can separate a heterogeneous mixture of table salt and river sand. Various porous substances can be used as filters: cotton wool, coal, fired clay, pressed glass, and others. The filtering method is the basis for the operation of household appliances, such as vacuum cleaners. It is used by surgeons - gauze bandages; drillers and workers of elevators - respiratory masks. With the help of a tea strainer for filtering tea leaves, Ostap Bender, the hero of the work of Ilf and Petrov, managed to take one of the chairs from Ellochka Ogre (“The Twelve Chairs”). Separation of a mixture of starch and water by filtration
A mixture of iron powder and sulfur	Action by magnet or water Iron powder was attracted by a magnet, but sulfur powder was not. The non-wettable sulfur powder floated to the surface of the water, while the heavy wettable iron powder settled to the bottom. Separation of a mixture of sulfur and iron using a magnet and water
A solution of salt in water is a homogeneous mixture	Evaporation or crystallization The water evaporates and salt crystals remain in the porcelain cup. When water is evaporated from lakes Elton and Baskunchak, table salt is obtained. This separation method is based on the difference in the boiling points of the solvent and the solute. If a substance, such as sugar, decomposes when heated, then the water is not completely evaporated - the solution is evaporated, and then sugar crystals are precipitated from a saturated solution. Sometimes it is required to remove impurities from solvents with a lower temperature boiling, such as water from salt. In this case, the vapors of the substance must be collected and then condensed upon cooling. This method of separating a homogeneous mixture is called distillation or distillation. In special devices - distillers, distilled water is obtained, which is used for the needs of pharmacology, laboratories, and car cooling systems. At home, you can design such a distiller: If, however, a mixture of alcohol and water is separated, then the first to be distilled off (collected in a receiving test tube) is alcohol with t bp = 78 ° C, and water will remain in the test tube. Distillation is used to obtain gasoline, kerosene, gas oil from oil. Separation of homogeneous mixtures

A special method of separating components, based on their different absorption by a certain substance, is chromatography.

Methods for expressing the composition of mixtures.

Mass fraction of the component in the mixture- the ratio of the mass of the component to the mass of the entire mixture. Usually the mass fraction is expressed in %, but not necessarily.

ω ["omega"] = m component / m mixture

Mole fraction of a component in a mixture- the ratio of the number of moles (amount of substance) of the component to the total number of moles of all substances in the mixture. For example, if the mixture includes substances A, B and C, then:

χ [“chi”] component A \u003d n component A / (n (A) + n (B) + n (C))

Molar ratio of components. Sometimes in tasks for a mixture, the molar ratio of its components is indicated. For example:

n component A: n component B = 2: 3

Volume fraction of the component in the mixture (only for gases)- the ratio of the volume of substance A to the total volume of the entire gas mixture.

φ ["phi"] = V component / V mixture

Practice block.

Consider three examples of problems in which mixtures of metals react with hydrochloric acid:

Example 1When a mixture of copper and iron weighing 20 g was exposed to an excess of hydrochloric acid, 5.6 liters of gas (n.o.) were released. Determine the mass fractions of metals in the mixture.

Example 1 solution.

Finding the amount of hydrogen:
n \u003d V / V m \u003d 5.6 / 22.4 \u003d 0.25 mol.

According to the reaction equation:

The amount of iron is also 0.25 mol. You can find its mass:
m Fe \u003d 0.25 56 \u003d 14 g.

Answer: 70% iron, 30% copper.

Example 2Under the action of an excess of hydrochloric acid on a mixture of aluminum and iron weighing 11 g, 8.96 liters of gas (n.o.) were released. Determine the mass fractions of metals in the mixture.

Example 2 solution.

Finding the amount of hydrogen:
n \u003d V / V m \u003d 8.96 / 22.4 \u003d 0.4 mol.

Let the amount of aluminum be x mol, and iron y mol. Then we can express in terms of x and y the amount of hydrogen released:

2HCl \u003d FeCl 2 +

We know the total amount of hydrogen: 0.4 mol. Means,
1.5x + y = 0.4 (this is the first equation in the system).

For a mixture of metals, you need to express masses through quantities of substances.
m = Mn
So the mass of aluminum
m Al = 27x,
mass of iron
m Fe = 56y,
and the mass of the whole mixture
27x + 56y = 11 (this is the second equation in the system).

So we have a system of two equations:

It is much more convenient to solve such systems by the subtraction method, multiplying the first equation by 18:
27x + 18y = 7.2
and subtracting the first equation from the second:
(56 - 18)y \u003d 11 - 7.2
y \u003d 3.8 / 38 \u003d 0.1 mol (Fe)
x = 0.2 mol (Al)

m Fe = n M = 0.1 56 = 5.6 g
m Al = 0.2 27 = 5.4 g
ω Fe = m Fe / m mixture = 5.6 / 11 = 0.50909 (50.91%),

respectively,
ω Al \u003d 100% - 50.91% \u003d 49.09%

Answer: 50.91% iron, 49.09% aluminum.

Example 316 g of a mixture of zinc, aluminum and copper was treated with an excess of hydrochloric acid solution. In this case, 5.6 l of gas (n.o.) was released and 5 g of the substance did not dissolve. Determine the mass fractions of metals in the mixture.

Answer to Example 3: 56.25% zinc, 12.5% aluminum, 31.25% copper.

Example 4A mixture of iron, aluminum and copper was treated with an excess of cold concentrated sulfuric acid. At the same time, part of the mixture dissolved, and 5.6 liters of gas (n.o.) were released. The remaining mixture was treated with an excess of sodium hydroxide solution. 3.36 liters of gas evolved and 3 g of undissolved residue remained. Determine the mass and composition of the initial mixture of metals.

Example 4 solution.

Only copper reacts with concentrated sulfuric acid, the number of moles of gas:
n SO2 \u003d V / Vm \u003d 5.6 / 22.4 \u003d 0.25 mol


	2H 2 SO 4 (conc.) = CuSO 4 +

(do not forget that such reactions must be equalized using an electronic balance)
Since the molar ratio of copper and sulfur dioxide is 1:1, then copper is also 0.25 mol. You can find the mass of copper:
m Cu \u003d n M \u003d 0.25 64 \u003d 16 g.
Aluminum reacts with an alkali solution, and an aluminum hydroxocomplex and hydrogen are formed:
2Al + 2NaOH + 6H 2 O = 2Na + 3H 2

Al 0 − 3e = Al 3+

2H + + 2e = H 2
Number of moles of hydrogen:
n H3 \u003d 3.36 / 22.4 \u003d 0.15 mol,
the molar ratio of aluminum and hydrogen is 2:3 and, therefore,
nAl \u003d 0.15 / 1.5 \u003d 0.1 mol.
Aluminum weight:
m Al \u003d n M \u003d 0.1 27 \u003d 2.7 g
The remainder is iron, weighing 3 g. You can find the mass of the mixture:
m mixture \u003d 16 + 2.7 + 3 \u003d 21.7 g.
Mass fractions of metals:

ω Cu \u003d m Cu / m mixture \u003d 16 / 21.7 \u003d 0.7373 (73.73%)
ω Al = 2.7 / 21.7 = 0.1244 (12.44%)
ω Fe = 13.83%

Answer: 73.73% copper, 12.44% aluminum, 13.83% iron.

Example 521.1 g of a mixture of zinc and aluminum was dissolved in 565 ml of a nitric acid solution containing 20 wt. % HNO 3 and having a density of 1.115 g/ml. The volume of the released gas, which is a simple substance and the only product of the reduction of nitric acid, was 2.912 l (n.o.). Determine the composition of the resulting solution in mass percent. (RCTU)

Example 5 solution.

Determine the amount of gas substance:
n N2 \u003d V / Vm \u003d 2.912 / 22.4 \u003d 0.13 mol.

We determine the mass of the nitric acid solution, the mass and amount of the dissolved HNO3 substance:

m solution \u003d ρ V \u003d 1.115 565 \u003d 630.3 g
m HNO3 \u003d ω m solution \u003d 0.2 630.3 \u003d 126.06 g
n HNO3 \u003d m / M \u003d 126.06 / 63 \u003d 2 mol

We compose the reaction equations ( do not forget about the electronic balance) and, for convenience of calculations, we take for 5x - the amount of zinc, and for 10y - the amount of aluminum. Then, in accordance with the coefficients in the equations, nitrogen in the first reaction will be x mol, and in the second - 3y mol:


	12HNO 3 \u003d 5Zn (NO 3) 2 +

Zn 0 − 2e = Zn 2+
2N+5+10e=N2


	36HNO 3 \u003d 10Al (NO 3) 3 +

It is convenient to solve this system by multiplying the first equation by 90 and subtracting the first equation from the second.

x \u003d 0.04, which means n Zn \u003d 0.04 5 \u003d 0.2 mol
y \u003d 0.03, which means that n Al \u003d 0.03 10 \u003d 0.3 mol

Let's check the mass of the mixture:
0.2 65 + 0.3 27 \u003d 21.1 g.

Now let's move on to the composition of the solution. It will be convenient to rewrite the reactions again and write down over the reactions the amounts of all reacted and formed substances (except water):

The next question is: did nitric acid remain in the solution and how much is left?
According to the reaction equations, the amount of acid that reacted:
n HNO3 \u003d 0.48 + 1.08 \u003d 1.56 mol,
those. the acid was in excess and you can calculate its remainder in solution:
n HNO3 rest. \u003d 2 - 1.56 \u003d 0.44 mol.

So in final solution contains:

zinc nitrate in the amount of 0.2 mol:
m Zn(NO3)2 = n M = 0.2 189 = 37.8 g
aluminum nitrate in the amount of 0.3 mol:
m Al(NO3)3 = n M = 0.3 213 = 63.9 g
an excess of nitric acid in an amount of 0.44 mol:
m HNO3 rest. = n M = 0.44 63 = 27.72 g

What is the mass of the final solution?
Recall that the mass of the final solution consists of those components that we mixed (solutions and substances) minus those reaction products that left the solution (precipitates and gases):

Then for our task:

m new solution \u003d mass of acid solution + mass of metal alloy - mass of nitrogen
m N2 = n M = 28 (0.03 + 0.09) = 3.36 g
m new solution \u003d 630.3 + 21.1 - 3.36 \u003d 648.04 g

ωZn (NO 3) 2 \u003d m in-va / m solution \u003d 37.8 / 648.04 \u003d 0.0583
ωAl (NO 3) 3 \u003d m in-va / m solution \u003d 63.9 / 648.04 \u003d 0.0986
ω HNO3 rest. \u003d m in-va / m solution \u003d 27.72 / 648.04 \u003d 0.0428

Answer: 5.83% zinc nitrate, 9.86% aluminum nitrate, 4.28% nitric acid.

Example 6When processing 17.4 g of a mixture of copper, iron and aluminum with an excess of concentrated nitric acid, 4.48 liters of gas (n.o.) were released, and when this mixture was exposed to the same mass of excess hydrochloric acid, 8.96 l of gas (n.o.). u.). Determine the composition of the initial mixture. (RCTU)

When solving this problem, we must remember, firstly, that concentrated nitric acid with an inactive metal (copper) gives NO 2, and iron and aluminum do not react with it. Hydrochloric acid, on the other hand, does not react with copper.

Answer for example 6: 36.8% copper, 32.2% iron, 31% aluminum.

Explanatory note

Pure substances and mixtures. Ways separation mixtures. To form an understanding of pure substances and mixtures. Ways purification substances: ... substances to various classes organic compounds. Characterize: basic classes organic compounds...

Order from 2013 No. Work program on the subject "Chemistry" Grade 8 (basic level 2 hours)

Working programm

Assessing students' knowledge of the possibility and ways separation mixtures substances; the formation of relevant experimental skills ... classification and chemical properties of basic substances classes inorganic compounds, the formation of ideas about ...

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... mixtures, ways separation mixtures. Tasks: Give the concept of pure substances and mixtures; Consider classification mixtures; Introduce students to ways separation mixtures... student and raises before class a card with the formula of an inorganic substance ...

In our article, we will consider what pure substances and mixtures are, methods for separating mixtures. Each of us uses them in everyday life. Do pure substances occur in nature at all? And how to distinguish them from mixtures?

Pure substances and mixtures: ways to separate mixtures

Pure substances are substances that contain particles of only a certain type. Scientists believe that they practically do not exist in nature, since all of them, albeit in negligible proportions, contain impurities. Absolutely all substances are also soluble in water. Even if, for example, a silver ring is immersed in this liquid, the ions of this metal will go into solution.

A sign of pure substances is the constancy of composition and physical properties. In the process of their formation, a change in the amount of energy occurs. Moreover, it can both increase and decrease. A pure substance can be separated into its individual components only by a chemical reaction. For example, only distilled water has a typical boiling and freezing point for this substance, the absence of taste and smell. And its oxygen and hydrogen can be decomposed only by electrolysis.

And how do they differ from pure substances in their totality? Chemistry will help us answer this question. Methods for separating mixtures are physical, since they do not lead to a change in the chemical composition of substances. Unlike pure substances, mixtures have variable composition and properties, and they can be separated by physical methods.

What is a mixture

A mixture is a collection of individual substances. An example is sea water. Unlike distilled, it has a bitter or salty taste, boils at a higher temperature, and freezes at a lower temperature. Methods for separating mixtures of substances are physical. So, pure salt can be obtained from sea water by evaporation and subsequent crystallization.

Types of mixtures

If you add sugar to water, after a while its particles will dissolve and become invisible. As a result, they cannot be distinguished with the naked eye. Such mixtures are called homogeneous or homogeneous. Air, gasoline, broth, perfume, sweet and salt water, and an alloy of copper and aluminum are also examples of these. As you can see, they can be in different states of aggregation, but liquids are most common. They are also called solutions.

In heterogeneous, or heterogeneous mixtures, particles of individual substances can be distinguished. Iron and wood filings, sand and table salt are typical examples. Heterogeneous mixtures are also called suspensions. Among them, suspensions and emulsions are distinguished. The former consists of a liquid and a solid. So, an emulsion is a mixture of water and sand. An emulsion is a combination of two liquids with different densities.

There are heterogeneous mixtures with special names. So, an example of foam is foam, and aerosols include fog, smoke, deodorants, air fresheners, antistatic agents.

Methods for separating mixtures

Of course, many mixtures have more valuable properties than individual individual substances that make up their composition. But even in everyday life there are situations when they need to be separated. And in industry, entire industries are based on this process. For example, from oil as a result of its processing, gasoline, gas oil, kerosene, fuel oil, solar oil and machine oil, rocket fuel, acetylene and benzene are obtained. Agree, it is more profitable to use these products than mindlessly burning oil.

Now let's see if there is such a thing as chemical methods for separating mixtures. Suppose we need to obtain pure substances from an aqueous solution of salt. To do this, the mixture must be heated. As a result, the water will turn into steam, and the salt will crystallize. But at the same time, there will be no transformation of one substance into another. This means that the basis of this process are physical phenomena.

Methods for separating mixtures depend on the state of aggregation, the ability to dissolve, the difference in boiling point, the density and composition of its components. Let's consider each of them in more detail with specific examples.

Filtration

This separation method is suitable for mixtures containing a liquid and an insoluble solid. For example, water and river sand. This mixture must be passed through a filter. As a result, clean water will freely pass through it, and the sand will remain.

settling

Some methods of separating mixtures are based on the action of gravity. In this way, suspensions and emulsions can be decomposed. If vegetable oil gets into the water, the mixture must first be shaken. Then leave it for a while. As a result, the water will be at the bottom of the vessel, and the oil will cover it in the form of a film.

In laboratory conditions, they are used for settling. As a result of its work, a denser liquid is drained into a vessel, and a light one remains.

Settling is characterized by a low speed of the process. It takes a certain amount of time for the precipitate to form. In industrial conditions, this method is carried out in special structures called sedimentation tanks.

Magnet action

If the mixture contains metal, then it can be separated using a magnet. For example, to separate iron and But do all metals have such properties? Not at all. For this method, only mixtures containing ferromagnets are suitable. In addition to iron, these include nickel, cobalt, gadolinium, terbium, dysprosium, holmium, and erbium.

Distillation

This name, translated from Latin, means "draining drops." Distillation is a method of separating mixtures based on the difference in boiling points of substances. Thus, even at home, alcohol and water can be separated. The first substance begins to evaporate already at a temperature of 78 degrees Celsius. Touching the cold surface, the alcohol vapor condenses, turning into a liquid state.

In industry, oil refining products, aromatic substances, and pure metals are obtained in this way.

Evaporation and crystallization

These separation methods are suitable for liquid solutions. The substances that make up their composition differ in their boiling point. Thus, it is possible to obtain crystals of salt or sugar from the water in which they are dissolved. To do this, the solutions are heated and evaporated to a saturated state. In this case, the crystals are deposited. If it is necessary to obtain pure water, then the solution is brought to a boil, followed by condensation of the vapors on a colder surface.

Methods for separating gas mixtures

Gaseous mixtures are separated by laboratory and industrial methods, since this process requires special equipment. The raw material of natural origin is air, coke, generator, associated and natural gas, which is a combination of hydrocarbons.

The physical methods for separating mixtures in the gaseous state are as follows:

Condensation is the process of gradual cooling of a mixture, during which the condensation of its constituents occurs. In this case, first of all, high-boiling substances, which are collected in separators, pass into the liquid state. In this way, hydrogen is obtained from and also ammonia is separated from the unreacted part of the mixture.
Sorption is the absorption of some substances by others. This process has opposite components, between which equilibrium is established during the reaction. The forward and reverse processes require different conditions. In the first case, it is a combination of high pressure and low temperature. This process is called sorption. Otherwise, the opposite conditions are used: low pressure at high temperature.
Membrane separation is a method in which the property of semipermeable partitions is used to selectively pass molecules of various substances.
Reflux - the process of condensation of high-boiling parts of mixtures as a result of their cooling. In this case, the temperature of the transition to the liquid state of the individual components should differ significantly.

Chromatography

The name of this method can be translated as "I write with color." Imagine that ink is added to the water. If you lower the end of the filter paper into such a mixture, it will begin to be absorbed. In this case, water will be absorbed faster than ink, which is associated with a different degree of sorption of these substances. Chromatography is not only a method for separating mixtures, but also a method for studying such properties of substances as diffusion and solubility.

So, we got acquainted with such concepts as "pure substances" and "mixtures". The first are elements or compounds consisting only of particles of a certain type. Their examples are salt, sugar, distilled water. Mixtures are a collection of individual substances. A number of methods are used to separate them. The way they are separated depends on the physical properties of its constituents. The main ones are settling, evaporation, crystallization, filtration, distillation, magnetization and chromatography.

Pure substances and mixtures. Methods for separating mixtures.

In order to establish the properties of a substance, it is necessary to have it in its pure form, but substances in nature do not occur in a pure form. Each substance always contains a certain amount of impurities. A substance that contains almost no impurities is called pure. They work with such substances in a scientific laboratory, a school chemistry room. Note that absolutely pure substances do not exist.

Almost all natural substances, foodstuffs (except for salt, sugar, and some others), building materials, household chemicals, many medicines and cosmetics are mixtures.

Natural substances are mixtures, sometimes consisting of a very large number of different substances. For example, natural water always contains salts and gases dissolved in it. Sometimes a very small impurity content can lead to a very strong change in some properties of a substance. For example, the content in zinc of only hundredths of iron or copper accelerates its interaction with hydrochloric acid hundreds of times. When one of the substances is in the mixture in a predominant amount, the entire mixture usually bears its name.

A component is each substance contained in a mixture.

A pure substance is always homogeneous, but mixtures can be homogeneous and heterogeneous.

uniform mixtures.

Add a small portion of sugar to a glass of water and stir until all the sugar is dissolved. The liquid will taste sweet. Thus, the sugar did not disappear, but remained in the mixture. But we will not see its crystals, even when examining a drop of liquid through a powerful microscope.

Rice. 3. Homogeneous mixture (water solution of sugar)

The prepared mixture of sugar and water is homogeneous (Fig. 3); the smallest particles of these substances are evenly mixed in it.

Mixtures in which components cannot be detected with the naked eye are called homogeneous.

Most metal alloys are also homogeneous mixtures. For example, an alloy of gold and copper (used in jewelry) lacks red copper particles and yellow gold particles.

Water mixed with sand, chalk or clay freezes at 0 0 C and boils at 100 0 C.

Some types of heterogeneous mixtures have special names: foam (for example, foam, soap suds), suspension (a mixture of water with a small amount of flour), emulsion (milk, well-shaken vegetable oil with water), aerosol (smoke, fog).

Rice. 5. Heterogeneous mixtures:
a - a mixture of water and sulfur;
b - a mixture of vegetable oil and water;
c - a mixture of air and water

There are different ways to separate mixtures. The choice of method for separating a mixture is influenced by the properties of the substances that form this mixture.

Let's take a closer look at each method:

settling- a common method of purifying or liquids from water-insoluble mechanical impurities, or liquid substances that are insoluble in each other, having different densities.

Imagine that you have a mixture of vegetable oil and water. Determine the type of mixture. ( heterogeneous). Compare the physical properties of oil and water. (These are liquid substances that are insoluble in each other, having different densities). Suggest a way to separate this mixture ( upholding). It is carried out using a separating funnel.

Settling is used in the preparation of water for technological and domestic needs, the treatment of sewage, the dehydration and desalination of crude oil, and in many processes of chemical technology. It is an important step in the natural self-purification of natural and artificial reservoirs.

Filtration- separation of liquid from solid insoluble impurities in it; liquid molecules pass through the pores of the filter, and large particles of impurities are retained.

Filtration can be done not only with a paper filter. Other loose or porous materials can also be used for filtering. Bulk materials used in this method include, for example, quartz sand. And to the porous - burnt clay and glass wool.

Imagine that you have a mixture of river sand and water. Determine the type of mixture. ( heterogeneous). Compare the physical properties of river sand and water. (These are substances that are insoluble in each other, having different densities). Suggest a method for separating this mixture ( filtering).

Magnet action- this is a method of separating inhomogeneous mixtures, when one of the substances in the mixture is able to be attracted by a magnet

Imagine that you have a mixture of iron and sulfur in front of you. Determine the type of mixture. ( heterogeneous). Compare the physical properties of iron and sulfur. This mixture can be divided upholding, since sulfur and iron are solid substances that are insoluble in water. If you pour this mixture into water, sulfur will float to the surface, and iron will sink. Also, this mixture can be divided with magnet, since iron is attracted by a magnet, but sulfur is not.

Evaporation - this is a method of separating homogeneous mixtures, in this case, a solid soluble substance is released from a solution, when heated, water evaporates, and solid crystals remain.

Imagine that you have a mixture of table salt and water. Determine the type of mixture. ( homogeneous). This mixture can be divided evaporation, since when boiled, the water evaporates, and table salt remains in the cup for.

Distillation (Latin meaning "dropping") – This is a method of separating homogeneous mixtures, in which case liquid mixtures are separated into fractions differing in composition. It is carried out by partial evaporation of the liquid, followed by vapor condensation. The distilled fraction (distillate) is enriched with relatively more volatile (low-boiling) substances, and the non-distilled liquid (distillation residue) is enriched with relatively less volatile (high-boiling) substances.

Distillation allows you to purify natural water from impurities. The resulting pure (distilled) water is used in research laboratories, in the production of substances for modern technology, in medicine for the preparation of medicines.

In the laboratory, distillation is carried out on a special installation (Fig. 6). When a mixture of liquids is heated, the substance with the lowest boiling point boils first. Its vapor leaves the vessel, cools, condenses1, and the resulting liquid flows into the receiver. When this substance is no longer in the mixture, the temperature will begin to rise, and over time, another liquid component will boil. Non-volatile liquids remain in the vessel.

Rice. 6. Laboratory installation for distillation: a - conventional; b - simplified
1 - a mixture of liquids with different boiling points;
2 - thermometer;
3 - water cooler;
4 - receiver

Consider how some methods separation of mixtures.

The filtering process underlies the operation of a respirator, a device that protects the lungs of a person working in a heavily dusty environment. The respirator has filters that prevent dust from entering the lungs (Fig. 7). The simplest respirator is a bandage made of several layers of gauze. A filter that extracts dust from the air is also in the vacuum cleaner.

Rice. 7. Worker in a respirator

Draw a conclusion by what methods it is possible to separate a mixture of soluble and insoluble substances in water.

Portal for the student. Self-training

Order from 2013 No. Work program on the subject "Chemistry" Grade 8 (basic level 2 hours)

Pure substances and mixtures: ways to separate mixtures

What is a mixture

Types of mixtures

Methods for separating mixtures

Filtration

settling

Magnet action

Distillation

Evaporation and crystallization

Methods for separating gas mixtures

Chromatography

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