Theoretical block.

The definition of the concept of "mixture" was given in the XVII century. english scientist Robert Boyl: "The mixture is a holistic system consisting of heterogeneous components."

Comparative characteristics of the mixture and pure substance

Signs of comparison	Clean substance	Mixture
	Constant	Unstable
Substances	Same	Various
Physical properties	Permanent	Non-permanent
Energy change in education	Occurs	Not happening
Separation	Via chemical reactions	Physical methods

The mixtures differ from each other in appearance.

The mixture classification is shown in the table:

We give examples of suspensions (river sand + water), emulsions (vegetable oil + water) and solutions (air in the flask, salt + Water, translating coin: aluminum + copper or nickel + copper).

Methods for separating mixtures

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

Different mixtures separation methods are applied to clean the substances

Evaporation is the release of solids dissolved in liquid by the method of transformation into steam.

Distillation- Distillation, separation of substances contained in liquid mixtures for boiling temperatures followed by steam cooling.

In nature, water in its pure form (without salts) is not found. Oceanic, marine, river, well and spring water is a variety of solutions of salts in water. However, often people need clean water that does not contain salts (used in car engines; in chemical production to produce various solutions and substances; in the manufacture of photos). Such water is called distilled, and the method of obtaining it is distillation.

Filming - Filling 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 mixture components.

Consider ways of separation heterogeneousand homogeneous mixes.

Sample mix	Method of separation
Suspension - a mixture of river sand with water	Adjustion Separation defending Based on various densities of substances. Heavy sand settles on the bottom. You can also divide the emulsion: separate oil or vegetable oil from water. In the laboratory this can be done using a dividing funnel. Oil or vegetable oil forms upper, lighter layer. As a result of settling, dew from the fog falls, the soot deposit from smoke, cream in milk is defended. Separation of a mixture of water and vegetable oil with defending
Sand and cook salt in water	Filtration What is the separation of heterogeneous mixtures using filtering? On different solubility of substances in water and on various particle sizes. Through the filter pores, only the particles of substances are commensurate with them, while larger particles are delayed on the filter. So you can divide the heterogeneous mixture of the cooking salt and river sand. Various porous substances can be used as filters: cotton, coal, burned clay, extruded glass and others. The filtering method is the basis of the work of household appliances, such as vacuum cleaners. It uses surgeons - gauze bandages; Buroviki and workers elevators - respiratory masks. With the help of a teahouse to filter Caulook Osta Bender - the hero of the work of Ilf and Petrov - managed to pick up one of the chairs at the Ellochka of the cannibals ("Twelve chairs"). Separation of starch and water filtration mixture
Iron and sulfur powder mixture	Magnet or water action Iron powder attracted a magnet, and sulfur powder is not. The non-compatible sulfur powder surfaced to the surface of the water, and a heavy wetting iron powder saved to the bottom. Separation of a mixture of sulfur and iron with a magnet and water
Salt solution in water - a homogeneous mixture	Wrout or crystallization Water evaporates, and in the porcelain cup remains salt crystals. When evaporation of water from Lakes, Elton and Baskunchak receive a cook salt. This separation method is based on the difference in the boiling points of the solvent and the dissolved substance. If a substance, such as sugar, decomposes when heated, will evaporate the water with incompleteness - the solution is evaporated, and then the sugar crystals are precipitated from the saturated solution. Sometimes it is necessary to clean with impurities solvents with a smaller boiling point, such as salt water. In this case, the pair of substances must be assembled and then condensed when cooled. This method of separation of a homogeneous mixture is called distillation, or distillation. In special devices - distillers are obtained by distilled water, which is used for the needs of pharmacology, laboratories, car cooling systems. At home, you can construct such a distiller: If we separate the mixture of alcohol and water, then the first will be distilled off (gather in the test tube - receiver) alcohol with Tkip \u003d 78 ° C, and water will remain in the tube. Distillation is used to produce gasoline, kerosene, gas oil from oil. Separation of homogeneous mixtures

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

With the help of chromatography, the Russian botany first allocated chlorophyl from green parts of plants. In the industry and laboratories, starch, coal, limestone, aluminum oxide are used instead of filter paper. Are there always substances with the same degree of cleaning?

For different purposes, substances are needed with varying degrees of cleaning. Water for cooking is enough to remove impurities and chlorine used to disinfect it. Water for drinking must be pre-boiled. And in chemical laboratories for the preparation of solutions and conducting experiments, distilled water is needed in medicine, the most purified substances dissolved in it. Particularly clean substances, the content of impurities in which does not exceed one million percent, are used in electronics, in semiconductor, nuclear technology and other accurate industries.

Methods for expressing the composition of the 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 a mass fraction is expressed in%, but not necessarily.

ω [«omega»] \u003d MKOMPONENT / MMSE

· The molar proportion of the component in the mixture - The ratio of the number of mole (amount of substance) of the component to the total number of mole of all substances in the mixture. For example, if the mixture includes substances A, B and C, then:

χ ["hee"] component A \u003d Ncompensant A / (n (a) + n (b) + n (c))

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

nKOMPONENT A: Ncompensant B \u003d 2: 3

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

φ ["FI"] \u003d VKOMPONENT / VSEM

Practical block.

Consider three examples of tasks in which metals mixtures react with salo Acid:

Example 1.Under action on a mixture of copper and iron, a mass of 20 g of an excess of hydrochloric acid was distinguished by 5.6 liters of gas (n. Y.). Determine the massive lobes of metals in the mixture.

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

Solution of Example 1.

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

2. By the reaction equation:

3. The amount of iron is also 0.25 mol. You can find it:
mFE \u003d 0.25 56 \u003d 14

Answer: 70% iron, 30% copper.

Example 2.Under action on a mixture of aluminum and iron weighing 11 g of an excess of hydrochloric acid, 8.96 liters of gas was distinguished (n. Y.). Determine the massive lobes of metals in the mixture.

In the second example, the reaction takes both Metal. Here already hydrogen from the acid is allocated in both reactions. Therefore, the direct calculation cannot be used here. In such cases, it is convenient to solve with a very simple system of equations, adopting a number of mole of one of the metals, and for y - the amount of the substance is the second.

Solution of Example 2.

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 - x mol, and the iron at mol. Then it can be expressed through x and in the amount of hydrogen distinguished:

	2HCl \u003d FECL2 +

4. We know the total amount of hydrogen: 0.4 mol. It means
1.5x + y \u003d 0.4 (this is the first equation in the system).

5. For a mixture of metals you need to express masses Through the amounts of substances.
m \u003d m n
So, the mass of aluminum
mAL \u003d 27X,
weight Mass
mFE \u003d 56U,
and the mass of the whole mixture
27x + 56u \u003d 11 (this is the second equation in the system).

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

7. Solve such systems is much more convenient by subtracting, the first 18 equation:
27x + 18u \u003d 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 \u003d 0.2 mol (AL)

mFE \u003d N m \u003d 0.1 56 \u003d 5.6 g
mAL \u003d 0.2 27 \u003d 5.4 g
ωFE \u003d MFE / MSE \u003d 5.6 / 11 \u003d 0.50.91%),

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

Answer: 50.91% iron, 49.09% aluminum.

Example 3.16 g of zinc mixture, aluminum and copper was treated with an excess solution of hydrochloric acid. At the same time, 5.6 liters of gas was separated (n. Y.) And 5 g of substance was dissolved. Determine the massive lobes of metals in the mixture.

In the third example, two metal react, and the third metal (copper) does not react. Therefore, the residue is 5 g is a mass of copper. The number of other two metals is zinc and aluminum (note that their total weight is 16 - 5 \u003d 11 g) can be found using the system of equations, as in Example No. 2.

Answer for example 3: 56.25% zinc, 12.5% \u200b\u200baluminum, 31.25% copper.

Example 4.The mixture of iron, aluminum and copper has affected the excess of cold concentrated sulfuric acid. In this case, a part of the mixture was dissolved, and 5.6 liters of gas was separated (n. Y.). The remaining mixture was treated with an excessive solution of caustic soda. There were 3.36 liters of gas and there were 3 g of a non-dissolved residue. Determine the mass and composition of the initial mixture of metals.

In this example, you must remember that cold concentrated sulfuric acid Does not react with iron and aluminum (passivation), but reacts with copper. This distinguishes sulfur oxide (IV).
With alkaliste Reacts only aluminum - amphoter metal (except aluminum, alkalis is dissolved in alkalis more zinc and tin, in hot concentrated alkalis - you can still dissolve beryllium).

Solution of Example 4.

1. With concentrated sulfuric acid, only copper reacts, the number of gas moles:
nSO2 \u003d V / VM \u003d 5.6 / 22.4 \u003d 0.25 mol

	2H2SO4 (conc.) \u003d CUSO4 +

2. Do not forget that such reactions must necessarily equalize with the help of electronic balance)

3. Since the molar ratio of copper and sulfur gas 1: 1, then copper is also 0.25 mol. You can find a mass of copper:
mCU \u003d N m \u003d 0.25 64 \u003d 16

4. Aluminum and hydrox complex of aluminum and hydrogen are applied to the reaction with alkali solution, while aluminum hydrox complex is formed:
2AL + 2NAOH + 6H2O \u003d 2NA + 3H2

AL0 - 3E \u003d AL3 +

5. Number of mooss of hydrogen:
nH2 \u003d 3.36 / 22.4 \u003d 0.15 mol,
the molar ratio of aluminum and hydrogen 2: 3 and, therefore,
nAL \u003d 0.15 / 1.5 \u003d 0.1 mol.
Aluminum mass:
mAL \u003d N m \u003d 0.1 27 \u003d 2.7 g

6. The residue is iron, weigh 3 g. You can find a mass of the mixture:
msmes \u003d 16 + 2.7 + 3 \u003d 21.7 g.

7. Mass shares of metals:

ωCU \u003d MCU / MSEM \u003d 16 / 21.7 \u003d 0.7.73%)
ωal \u003d 2.7 / 21.7 \u003d 0.1.44%)
ωFE \u003d 13.83%

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

Example 5.21.1 g of zinc and aluminum mixtures were dissolved in 565 ml of a solution of nitric acid containing 20 wt. % NNO3 and having a density of 1.115 g / ml. The volume of highlighted gas, which is simple substance and the only product of the reduction of nitric acid was 2.912 liters (n.). Determine the composition of the resulting solution in mass percent. (PCTU)

In the text of this problem, the product of nitrogen recovery is clearly indicated - "simple substance". Since nitric acid with metals does not give hydrogen, it is nitrogen. Both metal dissolved in acid.
The problem is asked not the composition of the initial mixture of metals, and the composition of the solution obtained after the reactions. This makes the task more complicated.

Solution of Example 5.

1. Determine the amount of gas substance:
nn2 \u003d v / vm \u003d 2.912 / 22.4 \u003d 0.13 mol.

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

missor \u003d ρ v \u003d 1,115 565 \u003d 630.3 g
mhnO3 \u003d Ω Mussere \u003d 0.2 630.3 \u003d 126.06 g
nhno3 \u003d m / m \u003d 126,06 / 63 \u003d 2 mole

Note that since the metals are completely dissolved, it means acids are just enough (With water, these metals do not react). Accordingly, it will be necessary to check was not an acid in excessand how much it remains after the reaction in the resulting solution.

3. Compile the equation of reactions ( do not forget about the electronic balance) And, for the convenience of calculations, we accept for 5x - the amount of zinc, and for 10th - the amount of aluminum. Then, in accordance with the coefficients in equations, nitrogen in the first reaction it will turn out x mol, and in the second - 3th mol:

	12hnO3 \u003d 5ZN (NO3) 2 +

Zn0 - 2e \u003d zn2 +

	36hnO3 \u003d 10Al (NO3) 3 +

AL0 - 3E \u003d AL3 +

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

6. Solve this system is convenient, the first equation for 90 and the first equation of their second.

7. x \u003d 0.04, it means NZN \u003d 0.04 5 \u003d 0.2 mol
y \u003d 0.03, it means 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

9. Now we turn to the composition of the solution. It will be convenient to rewrite the reactions once again and record over the reactions of the number of all reacted and formed substances (except for water):

10. The following question is whether nitric acid remained in solution and how much it remains?
According to the reaction equations, the amount of acid entered into the reaction:
nhno3 \u003d 0.48 + 1.08 \u003d 1.56 mol,
i.e. the acid was in excess and it is possible to calculate its residue in the solution:
nhno3ost. \u003d 2 - 1.56 \u003d 0.44 mol.

11. So, in total solution Contains:

zinc nitrate in an amount of 0.2 mol:
mZN (NO3) 2 \u003d n m \u003d 0.2 189 \u003d 37.8 g
aluminum nitrate in an amount of 0.3 mol:
mAL (NO3) 3 \u003d n m \u003d 0.3 213 \u003d 63.9 g
excess nitric acid in an amount of 0.44 mol:
mhno3ost. \u003d n m \u003d 0.44 63 \u003d 27.72 g

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

13.
Then for our task:

14. Mn. Solution \u003d Acid solution mass + Mass alloy mass - Nitrogen mass
mn2 \u003d n m \u003d 28 (0.03 + 0.09) \u003d 3.36 g
mn. The solution \u003d 630.3 + 21.1 - 3.36 \u003d 648.04 g

ωzn (NO3) 2 \u003d MB / Mr - 37.8 / 648.04 \u003d 0.0583
ωal (NO3) 3 \u003d MB / Mr - RA \u003d 63.9 / 648.04 \u003d 0,0986
ωhno3ost. \u003d MB / Mr - Ra \u003d 27.72 / 648.04 \u003d 0.0428

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

Example 6.When processing 17.4 g of copper, iron and aluminum mixture with an excess of concentrated nitric acid, 4.48 liters of gas was distinguished (n. Y.), And under the action to this mixture, the same mass of excess of chloride acid is 8.96 liters of gas (H. u.). Determine the composition of the initial mixture. (PCTU)

When solving this problem, it is necessary to recall, firstly, that concentrated nitric acid with inactive metal (copper) gives NO2, and iron and aluminum do not react with it. Salonic acid, on the contrary, does not react with copper.

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

Tasks for self solutions.

1. Uncomplicated tasks with two mixture components.

1-1. A mixture of copper and aluminum weighing 20 g was treated with a 96% solution of nitric acid, with 8.96 liters of gas (n. Y.). 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. It was released 2.24 liters of gas (n. Y.). Calculate the mass fraction of zinc in the source 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. Y.). Find a 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. Received zinc sulfate weighing 6.44 g. 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 to an excess solution of copper (II chloride), 9.6 g of copper was formed. Determine the composition of the initial mixture.

1-6. What a mass of a 20% solution of hydrochloric acid will be required for the total dissolution of 20 g of zinc mixture with zinc oxide, if the hydrogen was 4.48 liters (n.)?

1-7. When dissolved in dilute nitric acid 3.04 g of the mixture of iron and copper, nitrogen oxide (Ii) with a volume of 0.896 l (n is also) was released. Determine the composition of the initial mixture.

1-8. When dissolved 1.11 g of a mixture of iron and aluminum sawdust in a 16% hydrochloric acid solution (ρ \u003d 1.09 g / ml), 0.672 liters of hydrogen was distinguished (n. Y.). Find mass fractions of metals in the mixture and determine the volume of consumed hydrochloric acid.

2. Tasks are more complex.

2-1. A mixture of calcium and aluminum weighing 18.8 g was carried out without air access with an excess of graphite powder. The reaction product was treated with dilute hydrochloric acid, while 11.2 liters of gas (n. Y.) Was separated. Determine the mass fractions of metals in the mixture.

2-2. For dissolution of 1.26 g of magnesium alloy with aluminum used 35 ml of 19.6% sulfuric acid solution (ρ \u003d 1.1 g / ml). Excess an acid reacted with a 28.6 ml of potassium bicarbonate solution with a concentration of 1.4 mol / l. Determine the mass fractions of metals in the alloy and the volume of gas (n. Y.), Highlighted when dissolving the alloy.

Studying chemistry, I learned that clean substances in nature, technique, everyday life. Much more often meet mixtures - combinations of two or more components, chemically unfinished with each other. The mixtures differ in the magnitude of the particles of substances included in their composition, as well as the aggregate state of the components. Clean substances are needed for chemical studies. And how to get them or highlight from the mixture? I tried to answer this question in my work.

IN everyday life We are surrounded by mixtures of substances. The air we breathe food, which we consume, water - which we drink, and even we ourselves - all this in terms of the chemistry of the mixture containing from 2-3 to many thousands of substances.

The mixtures are systems consisting of several components, chemically unfinished with each other. The mixtures are distinguished by the magnitude of the particles of substances included in their composition. Sometimes these particles are so great that they can be seen. non-equipped eye. To similar mixtures, for example, a washing powder can be attributed, culinary mixes for baking, building mixtures. Sometimes the particles of the components in the mixtures are smaller, indistinguishable by the eye. For example, the flour includes graft and protein graft, which cannot be distinguished by a naked eye. Milk is also a water mixture, which contains small droplets of fat, protein, lactose and other substances. You can see the droplets of fat in milk if you look at a drop of milk under a microscope. The aggregate state of substances in mixtures may be different. Toothpaste, for example, is a mixture of solid and liquid components. There are mixtures, with the formation of which substances so "penetrate each other", which are divided into the smallest particles, not distinguishable even under the microscope. No matter how we peered into the air, it will not be possible to distinguish between the components of its gases.

Thus, the mixtures are classified:

The mixtures in which the particles of substances constituting the mixture are visible to the naked eye or under a microscope are called inhomogeneous or heterogeneous.

The mixtures in which even with the help of a microscope cannot be seen particles of substances constituting the mixture are called homogeneous or homogeneous.

Uniform mixtures in the aggregative state are divided into gaseous, liquid and solid. A mixture of any gases of homogeneous. For example, clean air is a homogeneous mixture of nitrogen, oxygen, carbon dioxide and noble gases. But dusty air is a heterogeneous mixture of the same gases, only containing dust particles. Liquid natural mixtures include oil. It consists of hundreds of various components. Of course, the most common liquid mixture, or rather solution, is the water of the seas and oceans. In 1 liter sea water It contains an average of 35 grams of various salts. With liquid mixtures in everyday life, we are constantly meeting. Shampoos and drinks, medicines and preparations of household chemicals - all this mixtures of substances. Even the water from the tap cannot be considered a pure substance: it contains dissolved salts, the smallest insoluble impurities, as well as microorganisms that are disinfected with chlorination. Solid mixtures are widespread. Mountain breeds are a mixture of several substances. Soil, sand, clay is solid mixtures. A glass, ceramics, alloys can be attributed to solid mixtures.

Chemists make up mixtures with simple stirring various substances - Composite parts whose properties may be different. It is important that the mixtures of their component parts are preserved in the mixtures. So, for example, gray paint is obtained by mixing black and white. Although we see gray, it does not mean that all particles such a gray paint have a gray color. Under the microscope, the particles of black and white colors are defined, from which black and white paint consisted.

The separation of mixtures into components (individual substances) is a more complex task than the preparation of mixtures, but no less important. The most important ways of separating mixtures can be reflected by the scheme:

Applying various methods for separating mixtures (settling, filtering, distillation, combustion, and others), oil from milk, gold - from river sand, alcohol - from the braga, purify water from insoluble and soluble impurities.

Clean substances are often required for chemical laboratories and industry. Pure called substances that possess constant physical properties, for example, distilled water. (Practically absolutely clean substances are not received.)

There are various ways of separation of mixtures. You will get acquainted in more detail with these ways.

Selection of inhomogeneous mixture.

1. Settlement.

a) Isolation of substances of an inhomogeneous mixture formed by substances insoluble in water with different density. For example, iron sawdusts from wood can be separated by shaking this mixture with water, and then defending. Iron sawdust fall at the bottom of the vessel, and the woods float, and together with water can be merged.

b) Some substances are deposited in water at different speeds. If you shake with water clay with an admixture of sand, then the sand settles much faster. This method is used in ceramic production to separate the sand from clay (production of red bricks, clay dishes, etc.) c) separation of a mixture of low-soluble in each other liquids with different density. Mixtures of gasoline with water, water with water, vegetable oil with water are quickly resolve, so they can be divided using a dividing funnel or column. Sometimes liquids with different density are separated by centrifugation, such as cream from milk.

2. Filtering.

The release of substances from an inhomogeneous mixture formed by soluble substances.

To highlight a cook salt, a mixture with sand is shaked in water. Cooking salt dissolves, and sand cares.

To speed up the separation of insoluble particles from the solution, the mixture is filtered. Sand remains on the filter paper, and the transparent solution of the cooking salt passes through the filter.

3. Action magnet.

Isolation from an inhomogeneous mixture of substances capable of magnetization. If there is, for example, a mixture of gland and sulfur powders, they can be divided by magnet.

Selection of substances from a homogeneous mixture.

4. Evaluation. Crystallization.

In order for the dissolved substance, for example, a table salt, to isolate from the solution, the latter is evaporated. Water evaporates, and a cook salt remains in the porcelain cup. Sometimes evaporation is used, i.e. partial evaporation of water. As a result, a concentrated solution is formed, when the dissolved substance is cooled in the form of crystals. This method of purification of substances is called crystallization.

5. Distillation.

This method of separation of mixtures is based on the difference in boiling temperatures of soluble in each other.

Distillation (distillation) - receiving the separation of homogeneous mixtures by evaporating volatile liquids with subsequent condensation of their vapors. For example, obtaining distilled water.

For this, water with substances dissolved in it are boiled in one vessel. The resulting water vapors are condensed in another vessel in the form of distilled water.

6. Chromatography.

This method is based on the fact that individual substances at different speeds are absorbed (binds) to the surface of another substance.

With the essence of this method, you can get acquainted with the following experience.

If the strip of filter paper hang over the vessel with red ink and immerse them in them only the end of the strip, then it can be noted that the solution will be absorbed by paper and climb on it. However, the paint lift boundary will fall behind the water lift border. Thus, two substances are separated: water and coloring agents that give the solution is red.

Experimental part.

Safety regulations in the home laboratory.

It is impossible to imagine chemistry without chemical experiments. Therefore, to study this science, understand its laws and, of course, it is possible to love it only through the experiment. There was an opinion that a chemical experiment is complex equipment and inaccessible reagents, poisonous compounds and terrible explosions and chemistry needs special conditions. Nevertheless, more than 300 chemical experiments with the most different substances can be performed at home. Due to the fact that in the home laboratory there is no exhaust cabinet and other special devices, it is necessary to strictly follow the safety regulations:

2. It is impossible to accumulate and store houses large quantities of reagents.

3. Chemical reagents and substances must have labels with titles, concentration and production time.

4. Chemical substances You can not taste.

5. To determine the smell, a vessel with a substance is close to face. We need a hand with a hand to make some smooth waves from the hole of the vessel to the nose.

6. If acid or alkali sheds, the substance is pre-neutralized or falling asleep and removed with a cloth or collected in the scoop.

7. Before conducting an experiment, no matter how easy it seems, you need to carefully read the description of the experience and understand the properties of the substances used. For this there are textbooks, reference books and other literature.

Experience number 1. Separation of heterogeneous mixtures.

A) Prepare a heterogeneous mixture of sand and iron powder.

The purpose of the experiment: learn to separating heterogeneous mixtures in different ways.

Equipment: river sand, iron powder, magnet, two chemical glasses.

In a chemical glass, add on one tablespoon of iron powder and river sand, gently mix the mixture to a uniformly colored product. Mark its color and experience its magnetic properties, bringing a magnet to the outside of the glass. Determine which substances give the mixture color and magnetic properties. We divide the prepared heterogeneous mixture with a magnet. To do this, we will bring a glass of glass to the outer wall, and tapping a magnet on the outer wall slightly, we collect the iron powder on the inner wall of the glass. Holding iron by the magnet on the inner wall of the glass, speck the sand to another glass. Survey data we enter the table.

B) Prepare a mixture of table salt, earth and chips, resulting after sharpening a pencil.

Equipment: Salt, Earth, chips after sharpening a pencil, a glass, water, filter, spoon, frying pan.

An experimental technique:

Prepare the mixture by stirring on one teaspoon of the cooking salt, earth and pencil chips. Dissolve the mixture in a glass of water, pop-up chips remove the noise and put for drying on a sheet of paper. Make the filter from the bandage or gauze, folding 3-4 layers, and not tighten it to another glass. Profiltrate the mixture. Filter from the remaining earth dry, then consider it from the filter. Filtered liquid (filtrate) Pour from a glass to an enameled bowl or frying pan and score. The distinguished crystalline salt collect. Compare the amounts of substances before and after the experiments done.

Experience number 2. Separation of homogeneous mixtures by paper chromatography.

A) split a homogeneous mixture of a red and green dye.

Equipment: Filter Paper Strip, Chemical Glass, Cork on a Glass, Flomasters Red and Green, Alcohol (70% aqueous solution).

An experimental technique:

Take a strip of filter paper, the length of which is 2-3 cm more than the height of the chemical glasses. In the middle of this strip, mark a simple pencil point, retreating from the edge of 1. 5 cm. In the marked point, apply the stains of the dyes with a diameter of no more than 5 mm. First, make a point with a size of 1-2 mm with a red felt-tip pen, and then on top of the red spindle to apply green so that the green speck across the red border of about 1 mm. Give a stain to dry the mixture (1-2 minutes) and then carefully, so as not to damage the paper, cover it with a simple pencil along the contour.

In a chemical glass, pour the alcohol with a layer of 0.5-1 cm. Place a vertically paper strip with a stain mixture of dyes into a glass and bend the protruding part of the strip to the outer surface of the glass. The dyes stain must be above the liquid at a distance of 0.5 cm. Cover the glass with a folded plug. Watch wetting paper strips and the movement of the painted stain up with its separation into two spots. For the full separation of the mixture of dyes, it will take about 20 minutes. After the paper is completely soaked with alcohol, take it out and let dry 5-10 minutes. Mark the colors of separation of spots. The results of observations will be in a table.

B) divide the following mixtures of chromatography on paper: alcohol solution "Green"; Aqueous solution of black carcass for drawing works.

The purpose of the experiment: to master the method of paper chromatography, learn to determine the difference between pure substances and mixtures.

Equipment: chemical glass, strip of filter or clock paper, alcohol solution "Green", aqueous solution of carcass for intensive works.

An experimental technique:

The strip of filter paper must be suspended above the vessel with a solution of "green" and a black carcass so that the paper only touches the solution.

The border of the rise of "zeral" and the coloring agent will lag behind the border of the lift of alcohol and water, respectively. Thus, two substances are separated in the composition of homogeneous mixtures: a) alcohol and diamond green, b) water and coloring matter.

Experience number 3. Diffusion.

The purpose of the experiment: to study the diffusion process in practice.

Equipment: food gelatin, mangartan, copper sipop, water, pan, stainless steel spoon for mixing, electric or gas tile, tweezers, two transparent bubbles.

An experimental technique:

Gelatin teaspoon lower in a glass with cold water And leave for an hour or another so that the powder could swell. Pour the mixture into a small saucepan. Heating the mixture on weak heat; Watch that it does not go in any way! Stir the contents of the saucepan until gelatin is completely dissolved. Hot solution Put two bubbles. When it cools down, in the middle of one of the bubbles, in a quick and cautious movement, enter the tweezers in which the crystalline of mangarteau will be trampled. Sleep slightly tweezers and quickly remove it. In another bubble, enter the crystal of copper mood. Gelatin slows down the diffusion process, and for several hours in a row you can watch a very interesting picture: a painted ball will grow around crystallists.

Experience number 4. Separation of homogeneous mixtures by crystallization.

Grow crystal or crystals from a saturated solution of cook salt, copper sulfate or alumokalia alum.

The purpose of the experiment: to learn how to prepare a saturated solution of salt or other substances, grow crystals of various sizes, consolidate skills and skills when working with substances and chemical equipment.

Equipment: a glass and a liter of a solution for making a solution, a wooden spoon or a wand for mixing, salt for an experiment - salt, copper sipop or alum, hot water, seed - Salt crystal, suspended on a thread, funnel and filter paper.

An experimental technique:

Prepare a saturated salt solution. To do this, first pour hot water into a half of it to half its volume, then add the appropriate salt by portions, constantly stirring. Add salts until it stops dissolve. Filter the resulting solution into a glass through a funnel with filter paper or cotton and leave the solution to cool for 2-3 hours. Make a seed in a cooled solution - a salt crystalline suspended on the thread, carefully cover the solution with a lid and leave for a long time (2-3 days or more).

Results of work and conclusions:

Explore your crystal and answer questions:

How many days have you grown crystal?

What is his form?

What color is crystal?

Is it transparent or not?

What are the sizes of the crystal: height, width, thickness?

What is the mass of the crystal?

Draw or take a picture of your crystal.

Experience number 5. Separation of homogeneous mixtures by distillation.

Get at home 50 ml of distilled water.

The purpose of the experiment: learn how to separate homogeneous mixtures by distillation.

Equipment: enameled kettle, two glass jars.

An experimental technique:

Pour into the enameled kettle on 1/3 of the water volume and put on the gas stove so that the kettle spout play the edge of the plate. When the water boils, drink a glass jar of the refrigerator on the kettle sprout, to fit the second can for collecting condensate. In order for the refrigerator bank not overheat, it is possible to put a napkin moistened with cold water.

Results of work and conclusions:

Answer the questions set:

What is a tap water?

What methods are homogeneous mixtures?

What is distilled water? Where and in what purpose is it used?

Draw your experience.

Experience number 6. Removing starch from potatoes.

Get at home a small amount of starch.

Equipment: 2-3 potatoes, twinks, march, small saucepan, water.

An experimental technique:

Purified Potatoes Sattail on a shallow grain and gradually obtained in water. Then pick it out through the gauze and squeeze. The residue of the mass in the gauze again mix with water. Give liquids to stand. Stachmal will fall on the bottom of the dishes. Drain the liquid, and starch starch again. Repeat the operation several times until the starch becomes completely clean and white. Profiltrate and dry the resulting starch.

What do you think, from what potato will turn out more starch: from the younger (who recently dug) or the old (who was in a vegetable store all winter)?

Experience number 7. Removing sugar from sugar coarse.

Get a small amount of sugar at home.

The purpose of the experiment: learn to extract substances from vegetable raw materials.

Equipment: large sugar, activated carbon, river sand, saucepan, two banks, wool, spoon, funnel, march.

An experimental technique:

Cut into small pieces of cooler, put them in a saucepan, pour a glass of water into it and boil for 15-20 minutes. Slices of welded beds thoroughly squeeze with a spoon or pestle. This mass of the dark color is filtered through a funnel in which you are wool. Then the resulting solution is filtered through a funnel prepared in a special way. In it, put a piece of gauze, for gauze - a thin layer of wool, then crushed activated carbon (4-5 tablets) and a thin layer (1 cm) of pure river sand (river sand is rinsed in advance and dry). The resulting solution (filtrate) Place in a saucepan. It is necessary to evaporate its part before the appearance of transparent crystals. This is sugar. Try it to taste!

What do you think, why do you need liquid filtration through a layer of activated carbon?

Experience number 8. Removing cottage cheese of milk.

Get a few grams of cottage cheese at home.

The purpose of the experiment: learn how to make cottage cheese at home.

Equipment: milk, vinegar, saucepan, gauze, gas stove.

An experimental technique:

There are protein in milk. If the milk boils, "runs away" through the edge, then at the same time the smell characteristic of the burning protein is immediately distributed. The appearance of the characteristic smell of burning milk indicates that the phenomenon of denaturation (the collapse of the protein and the transition to an insoluble form). Denaturation of protein occurs not only because of heating.

Let's do the following experience. Select the half of the milk so that it becomes a little warm, and add vinegar. The milk will come to the same time, forming large flakes. (If you leave the milk in a warm place, then the protein is also collapsed, but for another reason - this "work" with milk-sour bacteria). The contents of the saucepans are filtered through the gauze, holds it for the edges. If then connect the edges of the gauze, lift over a glass and squeeze, then it will remain a thick mass - cottage cheese.

Experience number 9. Getting cream oil.

Get a small amount of butter at home.

The purpose of the experiment: to learn at home to extract the creamy oil from milk.

Equipment: milk, glass jar, a small transparent bubble with a plug or tightly closing lid.

An experimental technique:

Pour into the glass jar of fresh milk, put it in the refrigerator. A few hours later, and it is better to look carefully the next day: what happened to milk? Explain the observed.

Carefully collect cream (upper milk layer) and transfer them to the bubble. If it is necessary to make oil from cream, you will have to forget for a long time and patiently shake them for at least half an hour in a bubble, closed with a lid, until an oil lump is formed.

Experience number 10. Extraction.

In practice, the extraction process.

The purpose of the experiment: to carry out practically the extraction process.

A) Equipment: sunflower seeds, gasoline, tube, saucer, mortar with pestle.

An experimental technique:

Grind several pieces of sunflower seeds into the mortar. Grinding seeds put in a test tube, and pour into a small amount of gasoline, shake well several times. We give the test tube to stand two hours (away from the fire), not forgetting it from time to shake it. Drain gasoline on the saucer and put on the balcony. When gasoline evaporates, there will be some oil on the bottom, which dissolved in gasoline.

B) Equipment: iodine tincture, water, gasoline, test tube.

An experimental technique:

Gasoline can also be removed iodine from pharmacy iodine tincture. To do this, pour into the water test tube for a third, add approximately 1 ml of iodine tincture and to the resulting brought solution, insert as much gasoline. Shake the test tube and leave it alone. When the mixture is stirred, the upper gasoline layer will become dark brown, and the lower, water, is almost colorless: after all, iodine dissolves badly in water, and in gasoline is good.

What is the extraction? The process of separating the mixture of liquid or solids using extraction - selective dissolution in certain fluids (extractants) of a mixture component. Most often extracted substances from aqueous solutions with organic solvents that are usually not mixed with water. Main requirements for extractants: selectivity (selectivity), non-toxicity, possibly small volatility, chemical inertness and low cost. Extraction enjoy in the chemical industry, refining, drug production and especially widely in non-ferrous metallurgy

Conclusion.

Conclusions for work.

When performing this work, I learned how to prepare heterogeneous and homogeneous mixtures, conducted a study of the properties of substances and found out that with a simple compilation of a mixture of two components, these substances do not transmit their properties to each other, and retain them with them. On the properties of the source components (such as: volatility, state of aggregation, the ability to magnetize, solubility in water, particle size and others) are based and methods for their separation. When performing a training study, I mastered the following methods of separation of heterogeneous mixtures: action by magnet, settling, filtering and homogeneous mixtures: evaporation, crystallization, distillation, chromatography, extraction. I managed to highlight clean substances from food: sugar sugar, starch from potatoes, cottage cheese and butter of milk. I realized that chemistry is a very interesting and informative science, and that the knowledge gained in the lessons of chemistry and after school hours will be very useful to me in life.

The results of the separation of the mixture of iron and sand.

experience №1 №1 №1 №2 №2

iron Sand Stuff Mixture Part 1 Part 2

color Gray yellow gray-yellow gray yellow attraction to magnet There is no. There is no output of the properties of iron and the properties of iron and the mixture of the selected substance - the selected substance -

sand Different Sand Different Properties and Iron, and Iron Sand Sand

The results of the separation of dyes on paper.

experience №1 №2 Substance Mixture of dyes before separating a mixture of dyes after separation Color Black Dye No. 1 - Red Dye No. 2 - Green conclusion This mixture is homogeneous. The mixture is divided into two source substances; This is red and green dyes.

Educational experiment
At the beginning of the course of chemistry

Separation of mixtures and purification of substances

Continued. Beginning See No. 19/2007

In nature, pure substances are rare, most often they are part of mixtures. And in everyday life we \u200b\u200bare mainly not the individual (individual) substances, but with mixtures or materials of complex composition. The subject of studying science of chemistry is substance and its transformations. Consequently, students must learn that one of essential tasks Chemistry is the preparation of individual (pure) substances. This problem has two solutions:

synthesis of substances in laboratories, plants, factories and combines from other substances and materials;

separation mixtures (natural or artificial) on separate components - individual substances.

We remind you that tasks for deepening and systematization of students' knowledge are printed in italics.

Experiments on the separation of mixtures
and cleaning substances by physical methods

Depending on the aggregate state and the properties of their components of the mixture there are uniform and heterogeneous. In any case, the substance in the mixture retain their properties.

The separation of the mixture by physical or chemical methods is possible when substances (components), their components, have sharply different properties. The choice of the method of separation of mixtures depends not only on the type of mixture (homogeneous or inhomogeneous) and the individual properties of the components, but also from the one substance or substance it is necessary to highlight in its pure form. It must be borne in mind that the substance obtained as a result of the separation will not be absolutely clean substancesand will contain a certain proportion of impurities.

Examine labels on packaging of various substances (chemical reagents) in the chemical office. Pay attention to the color and verbal designations of various purity substances and the content of impurities in them in accordance with the standard or technical condition of each reagent.

EXPERIENCE 1. Substances in the mixture retain their individual properties

Equipment and materials. Magnet, mortar with pestle, glasses, paper; Water, sulfur, iron (powder).

Conduct. Distribute sulfur in the mortar and pour (2-3 g) on \u200b\u200ba white paper sheet. On another sheet of paper, pour iron powder (2-3 g). Consider external signs of these substances. Hereinafter, in this experience, pay attention to the similarity and distinction of the individual properties of iron and sulfur (aggregate state, color, smell, water solubility, water wettability, density, magnet action, etc.). Add a pinch of sulfur and iron into cups with water. Cover portions of substances on paper sheets by other sheets and touch them from above the magnet.

Scroll into the mortar of iron powder (2 g) with gray (2 g) and consider the mixture. Throw the pinch of the resulting mixture into a glass with water. Pour another portion of the mixture on a sheet of paper, cover the other sheet and bring the magnet. Describe your observations in detail. Answer the questions.

1. Why finely chopped sulfur powder does not sink in water? Is this property of sulfur density or is another reason?

2. What properties of sulfur and iron are you installed in this experience?

3. Did the individual properties of the components in the mixture preserved?

4. What properties of sulfur and iron were used in this experiment to separate the mixture of iron with gray?

EXPERIENCE2-3. Inhomogeneous mixtures can be divided by upholding

Equipment and materials. Tripod, glasses, cylinders, divisory funnels; Muddy (clay and sand) water, a mixture of vegetable oil and water.

Conduct.Shake the turbid water in the glass and pour suspensionin a cylinder. Stir a thoroughly mixture of oil with water and break emulsion In a dividing funnel, fixed in the tripod.

Mark your observations after 1, 2, 5 minutes. Decant Liquid from the cylinder into a clean glass. Consider the residue in the cylinder and water in the glass.

Turning the tap of a dividing funnel, drain the lower layer of fluid into the glass from it.

1. What properties of the components allowed to divide the data of the mixture?

2. Is it possible to assert that the substance isolated from the mixture (which?) Are clean?

3. Give examples of the separation of mixtures by setting up in practice. What properties of substances is the difference between this method?

EXPERIENCE 4. Separation of inhomogeneous mixtures
You can speed up centrifugation

Equipment and materials. Centrifuge; Midget (clay) water.

Conduct. Pour a suspension into centrifuge tubes, install them in the centrifuge sockets and turn on the device according to the instructions (or use the manual centrifuge) by 3-5 minutes. Drain the water into a clean glass.

EXPERIENCE5-6. Suspension can be divided
on the components of filtration

Equipment and materials. Tripod with a ring, funnel for filtering, glasses, glass sticks, filtering paper, cotton wool, march; Muddy water, 3% solution of copper sulfate (II).

Conduct. Collect the filtering installation and filter the turbid water first through the gauze layer, then the wool and, finally, using filter paper with sufficiently small pores. Similar experience, with a solution of copper sulfate (II).

Mark your observations, compare the filtrate cleanliness when using various filtering materials and the use of various methods for separating mixtures. Make appropriate conclusions.

1. Is it possible to split the mixture of water and vegetable oil or other emulsions?

2. Give examples of practical separation of mixtures using filtering. What is the basis of this method of separation of mixtures?

3. What mixtures can be divided by filtration, and what mixtures cannot be divided by this method?

EXPERIENCE 7. Some mixtures can be divided by a magnet.

Equipment and materials. Magnet, paper leaves 10x10 cm; Blend of iron powder with sand, set (mix) coins of various advantages, mixture magnetite with empty breed.

Conduct. The mixture is placed on a sheet of paper, covered with another leaflet, make a magnet and, without removing it, turn over the upper sheet with a substance that attracts to the magnet.

Describe your observations. Check which other substances and materials are attracted by a magnet.

1. What substances or materials mediated from the mixtures using a magnet?

2. What is the basis of the magnetic separation method of mixtures? Give examples of using this method in practice.

EXPERIENCE 8. Flotation is applied
For enrichment of minerals

Equipment and materials. High chemical glass, spatula; A mixture of fine crushed sulfur with sand, water.

Conduct. Using the spatula, a mixture of sulfur with sand into a glass with water, mixing the contents of a glass every time, is well mixed.

Describe your observations. Specify the directory density of sand, sulfur and water and write down their values \u200b\u200bin the notebook.

1. Do you notice any contradiction between the properties of sulfur and the density of this substance?

2. Give examples of the practical application of flotation as a method of separation of substances when enriching minerals. What is this method based on?

EXPERIENCE9-10. Is it possible to evaporate solutions
Get salt and sugar sand?

Equipment and materials. Tripod with ring, grid, porcelain cups for evaporation, alcohol (burner); 30% solution of cook salt, 40% sugar solution.

Conduct. Collect installation for evaporation. Pour a 3-4 ml solution of the cook salt into the cup and evaporate the liquid almost to dryness. Cable tongs remove the cup from the fire and make sure the water is completely evaporated. Otherwise, carefully bring your experience to the end, preventing excessive overheating of salt. (O with t o r o w o! It is possible to splash the hot concentrated solution.) After the cup with salt cools, collect a dry residue on a blank sheet of paper. Similarly (O with t o r o w n o!) Spend evaporation of 3-4 ml of sugar solution. Try and in this case collect a dry residue.

Describe your observations and compare the results of evaporation of solutions of the table salt and sugar. pay attention to appearance Received substances. P o m n and t e, that trying substances to taste in the laboratory is categorically prohibited!

1. Is all solids dissolved in water, can be obtained in a pure form by evaporation of the solution under normal conditions?

2. Give examples of obtaining substances in a pure form by evaporation method in practice. What is this method based on?

EXPERIENCE 11. Is it possible to turn marine water into fresh water?

Equipment and materials. Installation for distillation of water, broken fayans, slide, pipettes, crucible tongs; 3% solution of cook salt (imitation of sea water).

Conduct. Wave a drop of "sea water" on the glass glass and prove that this sample of the liquid is a solution. (There will remain a "spot" of salt.) Assemble the installation for distillation of water or its simplified version, placing the flask for distillation to distillation (for uniform fluid boiling) and tear away
2-3 ml distillate. Check for cleanliness of the obtained portion of distilled water by evaporation on the slide.

Describe observations, compare the results of evaporation of the "marine" drops and distilled water, evaluate the effectiveness of this method of purification of substances.

1. What mixtures (homogeneous or inhomogeneous) can be separated by distillation?

2. Which components of mixtures can be, and which cannot be allocated by distillation?

3. Give examples of the practical application of distillation (distillation). What is this method of purification of substances based on?

EXPERIENCE 12. Beautiful crystals can be "growing" at home

Equipment and materials. Glasses, heating device, capening thread, glass wand; Copper sulphate, salt and other salts, water.

Conduct. Prepare 250-300 ml of a solution saturated at 30 ° C (from available). If the solution contains visible impurities, pick it up into a large glass.

To the middle of the glass sticks, tie a thin cape thread. Put the wand on the top of the glass, and the free end of the thread lower in the solution almost to the bottom of the vessel. After 1-2 days, inspect the thread and remove all crystalline from it, except for one - the largest and most correct form. The solution can be heated again before dissolving the crystals dropped and after cooling it is re-lowering the thread with a crystal. The operation is carried out until the large crystal is obtained. Grown crystals are better stored in transparent closed vessels, providing them with labels.

Draw the obtained crystals, compare the forms of large and small crystals of the same substance and shape of crystals of various substances. Make appropriate conclusions.

Give examples of the practical application of crystallization and recrystallization as a method of purification of substances. What is this method based on?

EXPERIENCE 13. The solubility of iodine in hexane is higher than in water

Equipment and materials. Detacious funnel, glass; Iodine water, hexane (you can take non-colored gasoline or kerosene direct distillation).

Conduct. Pour into a dividing funnel of 5-10 ml of iodine water and carefully on the wall of the vessel, add 2-3 ml of solvent. Please note that the solvent is lighter than water. Close the cork funnel and carefully, holding the plug, mix the mixture. Please note that iodine moved from the water layer in the solvent layer.

Describe your observations, compare the coloring of the initial and the resulting solutions. Explain these changes. According to the dictionary, find the interpretation of the concept of "extraction".

Give examples of practical use of extraction as a method for cleaning and separating substances. What is this method based on?

EXPERIENCE 14. Black coal bleaches ink

Equipment and materials. Conical flask, filtering accessories; Water, ink, activated carbon tablets.

Conduct. Pour in flask 40-50 ml of water and add 1-3 drops ink to get a weakly colored solution. Add 3-5 activated carbon tablets and circular movements of the flask intensively mix the mixture. Give the mixture to stand out. If the discoloration did not happen, add a few more coal tablets and repeat the mixing. Making sure that adsorption It happened completely, filter the mixture.

What is the basis of adsorption phenomenon and where does it find practical application?

EXPERIENCE 15. We "write" paints

Equipment and materials. Filter paper, pipettes, water, markers of various colors.

Conduct. Several touch of color marker in the same point, get a small one on the filter paper, but intensively painted stain. Drip a drop of alcohol or water into the center of the spot and, as you break, add the following drops of solvent. If the dye is uniform, then the color ring will be homogeneous. If the dye of the felt-meter consists of a mixture of several paints, then you will get chromatogramof several colors corresponding to the composition of the dye. The method of separation of complex painted mixtures into components in this case is called paper chromatography. The painted stain can also be obtained on paper as well with two or more marksters and repeat the experience.

Describe your observations in the mixture separation experiment by chromatography. The method is based on various degrees of adsorption of substances with special adsorbents.

Give examples of the separation of substances by chromatography using various adsorbents. What is this method based on?

Questions and tasks for systematization
and generalization of the concepts of the topic

1. Make a division plan for the following mixes:

a) sand, salt;

b) sand, clay, wood sawdust;

c) sand, iodine, cook salt;

d) small iron nails, household garbage;

e) iron sawdust, salt, sulfur.

2. If the cook reduced the soup, then it is recommended to lower a small linen bag with rice (20-30 g) for 10-15 minutes into the pan. What is the basis of this "grandmother's secret"? Can you suggest another way how to fix the case?

3. Flour before cooking the test is sifted through a sieve. Is it possible to refer to one of the methods of purification of substances? If so, what is this method based on?

4. In the famous tales of stepmother or other worst forced the heroine to separate some mixes into separate components. Recall what kind of mixtures were and based on what method they were divided?

G.I.Strepeller,
Professor of the Department of Chemistry
and learning techniques
Saratov state state
University

Printed with the continuation

The mixtures can be separated by different methods, among which the most common are settling, filtering, evaporation.

Settlement.The shortage is separated by the mixture, the components of which are easily separated, for example, a mixture of starch and water (Fig. 25, a).

Shortly after the preparation of the mixture, we see that starch settles on the bottom (Fig. 25, b), since it is insoluble and heavier than water. The water layer is located above the starch. In fig. 25, it is shown how this mixture is separated, carefully merging the water.

However, the total separation of the components of the mixture will not occur. Part of the water remains with starch either part of the collapse-small with water separated from the mixture.

We carry out the separation of the mixture of vegetable oil and water (Fig. 26). To separate the use of laboratory equipment, which is called a dividing funnel. As in the first case, these substances do not dissolve each other, but vegetable oil is lighter than water.

The mixture is placed in a dividious funnel. Soon the layer of vegetable oil is located on top above the water. Clearly visible line of separation of two liquids. By turning the krane-ka, open the hole in the funnel, through which water is poured into the glass. After water, the crane is closed. Through the top hole of the funnels, vegetable oil is switched into separate dishes.

Adjustion - One of the ways to divide the mixture. The components of the mixture as a result of the upset are smelling, so they are easy to divide.

Filtration. For the separation of a mixture of fluid and the soluble solvent in it is better to use the filtering method.

For filtering, it will take up to-full equipment - a common funnel, filter, glass wand. Filters are non-dense porous materials through which the liquid is missing, but particles of the solid component of the mixture do not penetrate. Such properties have paper, fabric, sand layer, cotton wool.

Filtration - This is a method of separating the mixture by passing it through filters capable of cutting particles of one of its components.

In fig. 27 shows how to divide the mixture of iron sawdust and water by filtration. A mixture of water and sawdust Caution with a glass stick, the bottom of the funnel, as shown in the figure, is poured onto the filter. Water quickly penetrates through the pores available in the filter and flows into the suduch receiving. We see, as a transparent clear water falls into the receiving-receiver. The size of iron sawdust is greater than the pores of the filter, so it is settled on it.

As in the previous two experiments, the mixture was separated, since one component of the mixture was not raised in another.

Evaporation.In nature and everyday life, quite a lot of mixtures, in which particles of substances are so re-mixed and small sizes, which is not divided into either filtration. For example, a mixture of water and cooking salts passes through the filter completely, none of its components remains on the filter. How to divide this mixture? In this case, another method is used - evaporation.

Evaporation - This is removal when the liquid component of the mixture is heated.

In fig. 28, but It is shown to prepare a mixture of a boiled salt and water, as well as its separation by you-steam. Material from site.

When evaporation, water evaporates and turns into water steam (Fig. 28, b). At the bottom of the ass, in which evaporation passed, a solid substance remains - a cook salt (Fig. 28, c).

In addition to those considered, there are also other ways to divide mixtures. For example, the property of the substances are attracted to the magnet. This method of separation of mixtures can be used if one of the substances respond to the magnet action, and the other is not.

Magnetization is characteristic of iron and is absent from sulfur. If you bring the magnet to the mixture of these substances (this can be done through a thin sheet of paper), then the mixture is sampled, the iron sawdust will be pulled to the magnet, then it can be easily cleansed from them.

Using large magnets at metals processing factories, iron scrap is separated from other components.

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