In this section, I systematize the analysis of tasks from OGE in chemistry. Similarly, the section, you will find detailed analysis with the instructions for solving typical tasks in chemistry in OGE grade 9. Before analyzing each block of typical tasks, I give theoretical help, without which the solution of this task is impossible. Theories are exactly as much as it is enough to know to successfully perform the task on the one hand. On the other hand, I tried to paint theoretical material with an interesting and understandable language. I am sure that having passed training according to my materials, you will not only successfully surrender OGE in chemistry, but also love this subject.

General Exam Information

OGE in chemistry consists of three Parts.

In the first part 15 tasks with one answer - This is the first level and tasks in it are simple, if there is, of course, basic knowledge of chemistry. These tasks do not require calculations, with the exception of 15 tasks.

The second part consists of four questions - In the first two - 16 and 17 it is necessary to choose two correct responses, and in 18 and 19 to relate the values \u200b\u200bor statements from the right column with the left.

The third part is solving tasks. In 20, it is necessary to level the reaction and determine the coefficients, and in 21 decide the estimated task.

Fourth part - Practical, simple, but it is necessary to be attentive and careful, as always when working with chemistry.

Everything is given to work 140 minutes.

Below are sorted by typical options for tasks, accompanied by the theory necessary for solving. All tasks thematic - opposite each task, a topic is indicated for a general understanding.

Working lessons (abstract lessons)

Attention! Site Administration Website is not responsible for the content methodological developments, as well as for the compliance of the development of GEF.

Question number 21 of the examination materials of OGE in chemistry is a task by equation chemical reaction. In the specification of control measuring materials for the main main state exam In chemistry, the following verifiable skills and action methods are specified when performing this task: « Calculation of the mass fraction of the dissolved substance in solution. Calculation of the amount of substance, mass or volume of substance in the amount of substance, mass or volume of one of the reagents or reaction products. " Analysis demonstration work and tasks open Bank allowed to distinguish three varieties of tasks used in exam work. When preparing for OGE, we decide with students, examples of the tasks of each type and suggest similar tasks for an independent decision. When solving problems according to the equations of chemical reactions, I use the algorithm submitted in the textbook in chemistry of grade 8 O.S. Gabrielevina.

1 View

The mass of the solution of the product or one of the initial substances of the reaction is given. Calculate the mass (volume) of the source substance or the reaction product.

1 Action: We calculate the mass of the product or one of the initial substances of the reaction.

2 Action: We calculate the mass or volume of the starting material according to the algorithm.

Example problem: TO solvent Aluminum chloride Weighing 53.2 g and a mass fraction of 5% increased overweight of silver nitrate. Calculate the mass of the precipitate formed.

Place a decision

1. TO solvent Aluminum sulfate Weighing 34.2 g and a mass fraction of 10% wasched by an excess solution of barium nitrate. Calculate the mass of the precipitate formed.
2. Through the calcium hydroxide solution, carbon dioxide was missed. 324 g solocalcium bicarbonate with a mass fraction of 1%. Calculate the volume of reacted gas.

2 View

The mass of the solution of the substance or product of the reaction is given. Calculate the mass fraction of a substance or product reaction.

1 Action: According to the algorithm, we calculate the mass of the initial substance (product) of the reaction. It is not paying attention to the mass of its solution.

2 Action: We know the mass of the initial substance (product) - found in the first action. We know the mass of the solution - given on the condition. We find a mass fraction.

Example problem: 73 g solohydrochloric acid was mixed with a portion of calcium carbonate. At the same time, 0.896 liters of gas was separated. Calculate the mass fraction of the original solo of hydrochloric acid.

Place a decision

2. Ω \u003d M (V-BA) / M (P-RG) · 100%

ω \u003d 2.92 / 73 · 100 \u003d 4%

Tasks for self solutions.

1. To 200 g solo Calcium chloride was added to sodium carbonate solution until the precipitate is stopped. The mass of the sediment was 12.0 g. Calculate the mass fraction of calcium chloride in the initial solution. (Relative atomic mass of chlorine Take equal to 35.5)
2. After passing 4.4 g of carbon dioxide through 320 g solo Potassium hydroxide was obtained by a solution of middle salts. Calculate the mass fraction of the lump in the solution

3 Type

The mass fraction of the solution of the source substance is given. Determine the mass of the starting material.

1 action. According to the algorithm to find a mass of the starting material.

2 action. We know the mass of the starting material (according to the first action). We know a mass fraction (from the condition). We find a mass solution.

Example of the task: To a solution of potassium carbonate with a mass fraction of 6%, an excess solution of barium chloride was added. As a result, a precipitate for a mass of 9.85 g. Determine the mass of the initial solution of potassium carbonate.

Place a decision

2. Ω \u003d M (V-BA) / M (P-RG) · 100%

m (p-ra) \u003d 6.9 / 6 ▪100% \u003d 115

Tasks for self solutions

1. After passing 11.2 liters (n. Y.) Ammonia after a 10% solution of sulfuric acid, a solution of middle salt was obtained. Determine the mass of the initial solution of sulfuric acid.
2. When 4.48 liters of carbon dioxide (N.U.) through a solution of barium hydroxide solution with a mass fraction of 12%, barium carbonate was formed. Calculate the mass of the initial solution of barium hydroxide.

Algorithm for solving problems according to chemical reaction equations

1. Summary Task condition.
2. Record chemical reaction equation.
3. Recording known and unknown values \u200b\u200babove formulas of substances.
4. Recording under the formulas of substances of the amount, molar masses and masses (or molar volumes and volumes) of substances.
5. Drawing up and solution proportion.
6. Record answer task.

Methods of solving tasks in chemistry

When solving tasks, you must be guided by several simple rules:

1. Carefully read the condition of the problem;
2. Write down that is given;
3. Translate if necessary, units physical quantities in units of the SI system (some non-system units are allowed, for example liters);
4. Write, if necessary, the reaction equation and place the coefficients;
5. Solve the problem using the concept of the amount of substance, and not the method of drawing up proportions;
6. Record the answer.

In order to successfully prepare for chemistry, it is possible to carefully consider solutions to the tasks in the text, as well as to solve their independently enough. It is in the process of solving problems that the main theoretical provisions of the Chemistry Course will be enshrined. Relieving tasks are necessary throughout the time of study of chemistry and preparation for the exam.

You can use tasks on this page, or you can download a good collection of tasks and exercises with a solution of typical and complicated tasks (M. I. Lebedeva, I. A. Ankudimova): Download.

Mole, molar mass

Molar mass - this is the ratio of mass of matter to the amount of substance, i.e.

M (x) \u003d m (x) / ν (x), (1)

where M (x) is the molar mass of the substance X, M (x) - the mass of the substance X, ν (x) is the amount of the substance X. The unit of the molar mass - kg / mol, however, it is usually used a unit of g / mol. Unit of mass - g, kg. The unit of the amount of substance is mole.

Any the task in chemistry is solved Through the amount of substance. It is necessary to remember the main formula:

ν (x) \u003d m (x) / m (x) \u003d v (x) / v m \u003d n / n a, (2)

where V (x) is the volume of the substance X (L), V M is the molar volume of the gas (l / mol), N is the number of particles, N A is constant avogadro.

1. Determine the mass Sodium iodide NAI with a substance of 0.6 mol.

Dano: ν (NAI) \u003d 0.6 mol.

To find: M (NAI) \u003d?

Decision. The molar mass of sodium iodide is:

M (NAI) \u003d M (Na) + m (i) \u003d 23 + 127 \u003d 150 g / mol

We determine the mass of NAI:

m (NAI) \u003d ν (NAI) M (NAI) \u003d 0.6 150 \u003d 90

2. Determine the amount of substance Atomic boron contained in sodium tetraborate Na 2 B 4 O 7 weighing 40.4 g

Dano: M (Na 2 B 4 O 7) \u003d 40.4

To find: ν (b) \u003d?

Decision. The molar mass of the sodium tetraborate is 202 g / mol. Determine the amount of substance Na 2 B 4 O 7:

ν (Na 2 B 4 O 7) \u003d M (Na 2 B 4 O 7) / M (Na 2 B 4 O 7) \u003d 40.4 / 202 \u003d 0.2 mol.

Recall that 1 mol of the sodium tetragano molecule contains 2 mols of sodium atoms, 4 mole of boron atoms and 7 mol of oxygen atoms (see sodium tetraborate formula). Then the amount of the atomic boron substance is: ν (b) \u003d 4 ν (Na 2 B 4 O 7) \u003d 4 0.2 \u003d 0.8 mol.

Calculations for chemical formulas. Mass fraction.

The mass fraction of the substance is the ratio of the mass of this substance in the system by the mass of the entire system, i.e. ω (x) \u003d m (x) / m, where ω (x) is the mass fraction of the substance x, m (x) - the mass of the substance X, M is the mass of the entire system. Mass fraction is a dimensionless value. It is expressed in fractions from one or percentage. For example, the mass fraction of atomic oxygen is 0.42, or 42%, i.e. Ω (O) \u003d 0.42. The mass fraction of atomic chlorine in sodium chloride is 0.607, or 60.7%, i.e. Ω (CL) \u003d 0.607.

3. Determine the mass share Crystallization Water in Barium Chloride Dihydrate BACL 2 2H 2 O.

Decision: The Molar Mass BACL 2 2H 2 O is:

M (BACL 2 2H 2 O) \u003d 137+ 2 35.5 + 2 18 \u003d 244 g / mol

From the BACL 2 2H 2H formula, it follows that 1 mole of the barium chloride dihydrate contains 2 mol H 2 O. From here, it is possible to determine the mass of water contained in BACL 2 2H 2 O:

m (H 2 O) \u003d 2 18 \u003d 36

We find the mass fraction of crystallization water in the dihydrate of barium chloride BACL 2 2H 2 O.

ω (H 2 O) \u003d M (H 2 O) / M (BACL 2 2H 2 O) \u003d 36/244 \u003d 0.1475 \u003d 14.75%.

4. From a rock sample with a mass of 25 g containing a mineral ARGEntitis AG 2 S, a silver mass of 5.4 g. Determine the mass share Argentita in the sample.

Dano: m (Ag) \u003d 5.4 g; m \u003d 25

To find: Ω (AG 2 S) \u003d?

Decision: Determine the amount of silver substance located in Argentita: ν (AG) \u003d M (Ag) / M (Ag) \u003d 5.4 / 108 \u003d 0.05 mol.

It follows from the AG 2 S formula that the amount of a substance of argentitis is two times less than the amount of silver substance. Determine the amount of the substance of the Argentita:

ν (Ag 2 S) \u003d 0.5 ν (Ag) \u003d 0.5 0.05 \u003d 0.025 mol

Calculate the mass of argentitis:

m (Ag 2 S) \u003d ν (Ag 2 S) M (Ag 2 S) \u003d 0.025 248 \u003d 6.2 g

Now we determine the mass fraction of argentitis in the sample of rock, weighing 25 g.

Ω (Ag 2 S) \u003d M (Ag 2 S) / m \u003d 6.2 / 25 \u003d 0.248 \u003d 24.8%.

Output of the formulas of compounds

5. Determine the simplest compound formula Potassium with manganese and oxygen, if the mass fractions of the elements in this substance are 24.7, 34.8 and 40.5%, respectively.

Dano: ω (k) \u003d 24.7%; Ω (Mn) \u003d 34.8%; Ω (O) \u003d 40.5%.

To find: Connection formula.

Decision: For calculations, we choose a mass of compound equal to 100 g, i.e. M \u003d 100 g. Mass of potassium, manganese and oxygen will be:

m (k) \u003d m Ω (K); m (k) \u003d 100 0.247 \u003d 24.7 g;

m (Mn) \u003d m Ω (Mn); M (Mn) \u003d 100 0.348 \u003d 34.8 g;

m (O) \u003d M Ω (O); M (O) \u003d 100 0.405 \u003d 40.5 g

We determine the amounts of substances of atomic potassium, manganese and oxygen:

ν (k) \u003d m (k) / m (k) \u003d 24.7 / 39 \u003d 0.63 mol

ν (Mn) \u003d M (Mn) / M (Mn) \u003d 34.8 / 55 \u003d 0.63 mol

ν (O) \u003d M (O) / M (O) \u003d 40.5 / 16 \u003d 2.5 mol

We find the ratio of the amounts of substances:

ν (k): ν (Mn): ν (O) \u003d 0.63: 0.63: 2.5.

Dividing the right-hand side of equality for a smaller number (0.63) we obtain:

ν (k): ν (Mn): ν (O) \u003d 1: 1: 4.

Consequently, the simplest formula of the KMNO 4 compound.

6. With a combustion of 1.3 g of substance, 4.4 g of carbon oxide (IV) and 0.9 g of water were formed. Find a molecular formula Substances, if its hydrogen density is 39.

Dano: m (V-Ba) \u003d 1.3 g; m (CO 2) \u003d 4.4 g; m (H 2 O) \u003d 0.9 g; D H2 \u003d 39.

To find: The formula of substance.

Decision: Suppose that the desired substance contains carbon, hydrogen and oxygen, because With its combustion, CO 2 and H 2 O is formed. Then it is necessary to find the amounts of substances CO 2 and H 2 O to determine the amounts of substances of atomic carbon, hydrogen and oxygen.

ν (CO 2) \u003d M (CO 2) / M (CO 2) \u003d 4.4 / 44 \u003d 0.1 mol;

ν (H 2 O) \u003d M (H 2 O) / M (H 2 O) \u003d 0.9 / 18 \u003d 0.05 mol.

We determine the amounts of substances of atomic carbon and hydrogen:

ν (c) \u003d ν (CO 2); ν (c) \u003d 0.1 mol;

ν (n) \u003d 2 ν (H 2 O); ν (n) \u003d 2 0.05 \u003d 0.1 mol.

Consequently, carbon and hydrogen mass will be equal:

m (c) \u003d ν (c) m (c) \u003d 0.1 12 \u003d 1.2 g;

m (n) \u003d ν (n) m (n) \u003d 0.1 1 \u003d 0.1 g

Determine qualitative composition Substances:

m (V-BA) \u003d M (C) + m (n) \u003d 1.2 + 0.1 \u003d 1.3 g

Consequently, the substance consists only of carbon and hydrogen (see the condition of the task). We will now define its molecular weight based on this tasks Density of substance on hydrogen.

M (V-BA) \u003d 2 D H2 \u003d 2 39 \u003d 78 g / mol.

ν (c): ν (n) \u003d 0.1: 0,1

Sharing the right-hand side of the equality by the number 0.1, we get:

ν (C): ν (H) \u003d 1: 1

We will take the number of carbon atoms (or hydrogen) for "x", then, multiplying "x" to atomic weights of carbon and hydrogen and equating this amount of the molecular weight of the substance, solving the equation:

12x + x \u003d 78. Hence x \u003d 6. Therefore, the formula of the substance from 6 H 6 is benzene.

Molar volume of gases. Laws of perfect gases. Volumetric share.

The molar volume of gas is equal to the ratio of the volume of gas to the amount of substance of this gas, i.e.

V m \u003d v (x) / ν (x),

where V m is the molar volume of the gas - a constant value for any gas under these conditions; V (x) - the volume of gas x; ν (x) - the amount of the substance of gas x. The molar volume of gases under normal conditions (normal pressure p \u003d 101 325 Pa ≈ 101.3 kPa and temperature TN \u003d 273.15 K ≈ 273 K) is v m \u003d 22.4 l / mole.

In the calculations associated with gases, often have to move from these conditions to normal or vice versa. At the same time, it is convenient to use the formula following from the combined gas law of Boyl-Mariott and Gay Loursak:

──── = ─── (3)

Where p is pressure; V - volume; T- temperature in the Kelvin scale; The "H" index indicates normal conditions.

The composition of gas mixtures is often expressed using a bulk fraction - the ratio of the volume of this component to the total volume of the system, i.e.

where φ (x) is the volume fraction of the component X; V (x) - the volume of the component X; V is the volume of the system. The volume fraction is a dimensionless value, it is expressed in fractions from one or in percent.

7. What volume It takes at a temperature of 20 o C and a pressure of 250 kPa ammonia weighing 51 g?

Dano: M (NH 3) \u003d 51 g; p \u003d 250 kPa; T \u003d 20 o C.

To find: V (NH 3) \u003d?

Decision: Determine the amount of ammonia substance:

ν (NH 3) \u003d M (NH 3) / M (NH 3) \u003d 51/17 \u003d 3 mole.

Ammonia volume under normal conditions is:

V (NH 3) \u003d V M ν (NH 3) \u003d 22.4 3 \u003d 67.2 liters.

Using formula (3), give the volume of ammonia to these conditions [Temperature T \u003d (273 +20) K \u003d 293 K]:

p H TV H (NH 3) 101.3 293 67.2

V (NH 3) \u003d ──────── \u003d ───────── \u003d 29.2 liters.

8. Determine volumewhich will take a gas mixture under normal conditions containing hydrogen, weighing 1.4 g and nitrogen, weighing 5.6 g.

Dano: m (n 2) \u003d 5.6 g; M (H 2) \u003d 1.4; Well.

To find: V (mixtures) \u003d?

Decision: We find the amount of substance of hydrogen and nitrogen:

ν (n 2) \u003d m (n 2) / m (n 2) \u003d 5.6 / 28 \u003d 0.2 mol

ν (H 2) \u003d M (H 2) / M (H 2) \u003d 1.4 / 2 \u003d 0.7 mol

Since under normal conditions, these gases do not interact with each other, the volume of the gas mixture will be equal to the amount of gases, i.e.

V (mixtures) \u003d V (N 2) + V (H 2) \u003d V M ν (n 2) + V M ν (H 2) \u003d 22.4 0.2 + 22.4 0.7 \u003d 20,16 l.

Calculations for chemical equations

Calculations for chemical equations (stoichiometric calculations) are based on the law of preserving the mass of substances. However, in real chemical processes, due to the incomplete flow of reaction and various losses of substances, the mass of products formed often is less than that which should be formed in accordance with the law of preserving the mass of substances. The yield of the reaction product (or the mass fraction of the output) is pronounced in percentage of the mass ratio of the actually obtained product to its mass, which should be formed in accordance with theoretical calculation, i.e.

η \u003d / m (x) (4)

Where η is the product output,%; M p (x) - the mass of the product X, obtained in the real process; M (x) - the calculated mass of the substance X.

In those tasks where the product yield is not specified, it is assumed that it is a quantitative (theoretical), i.e. η \u003d 100%.

9. What a mass of phosphorus should be burned for getting Phosphorus oxide (V) weighing 7.1 g?

Dano: m (P 2 O 5) \u003d 7.1 g

To find: m (p) \u003d?

Decision: Record the combustion equation of phosphorus burning and set stoichiometric coefficients.

4p + 5O 2 \u003d 2p 2 O 5

Determine the amount of substance P 2 O 5, which has been in the reaction.

ν (P 2 O 5) \u003d M (P 2 O 5) / M (P 2 O 5) \u003d 7.1 / 142 \u003d 0.05 mol.

It follows from the reaction equation that ν (P 2 O 5) \u003d 2 ν (P), therefore, the amount of phosphorus substance required in the reaction is:

ν (P 2 O 5) \u003d 2 ν (p) \u003d 2 0.05 \u003d 0.1 mol.

From here we find the mass of phosphorus:

m (p) \u003d ν (p) m (p) \u003d 0.1 31 \u003d 3.1 g

10. In the excess of hydrochloric acid, magnesium mass was dissolved with a mass of 6 g and zinc weighing 6.5 g. What volume hydrogen measured under normal conditions stand out wherein?

Dano: m (Mg) \u003d 6 g; m (zn) \u003d 6.5 g; Well.

To find: V (H 2) \u003d?

Decision: Record the equations of the reaction of the interaction of magnesium and zinc with hydrochloric acid and arrange the stoichiometric coefficients.

Zn + 2 HCl \u003d ZnCl 2 + H 2

Mg + 2 HCl \u003d MgCl 2 + H 2

Determine the amounts of magnesium and zinc substances that have joined the reaction with hydrochloric acid.

ν (Mg) \u003d M (Mg) / m (Mg) \u003d 6/24 \u003d 0.25 mol

ν (zn) \u003d m (zn) / m (zn) \u003d 6.5 / 65 \u003d 0.1 mol.

From the equations of the reaction it follows that the amount of substance of metal and hydrogen is equal, i.e. ν (Mg) \u003d ν (H 2); ν (zn) \u003d ν (H 2), determine the amount of hydrogen, resulting from two reactions:

ν (H 2) \u003d ν (Mg) + ν (zn) \u003d 0.25 + 0.1 \u003d 0.35 mol.

We calculate the volume of hydrogen highlighted as a result of the reaction:

V (H 2) \u003d V M ν (H 2) \u003d 22.4 0.35 \u003d 7.84 liters.

11. When the sulfide of 2.8 l (normal conditions) is passed through an excess of copper sulfate solution (II), a sediment was formed by a mass of 11.4 g. Determine the output Reaction product.

Dano: V (H 2 S) \u003d 2.8 l; m (precipitate) \u003d 11.4 g; Well.

To find: η =?

Decision: Record the equation of reacting the reaction of hydrogen sulfide and copper sulfate (II).

H 2 S + CUSO 4 \u003d CUS ↓ + H 2 SO 4

Determine the amount of the substance of the hydrogen sulfide participating in the reaction.

ν (H 2 S) \u003d V (H 2 S) / V m \u003d 2.8 / 22.4 \u003d 0.125 mol.

It follows from the reaction equation that ν (H 2 S) \u003d ν (Cus) \u003d 0.125 mol. So you can find the theoretical mass of CUS.

m (CUS) \u003d ν (CUS) M (CUS) \u003d 0.125 96 \u003d 12

Now we determine the product output, using formula (4):

η \u003d / m (x) \u003d 11.4 100/12 \u003d 95%.

12. What weight Ammonium chloride is formed in the interaction of chloride produce weighing 7.3 g with ammonia weighing 5.1 g? What gas will remain in excess? Determine the mass of excess.

Dano: M (HCl) \u003d 7.3 g; M (NH 3) \u003d 5.1 g

To find: M (NH 4 CL) \u003d? M (excess) \u003d?

Decision: Record the reaction equation.

HCl + NH 3 \u003d NH 4 Cl

This task for "excess" and "disadvantage". We calculate the amounts of the substance of the chloride farmer and ammonia and determine which gas is in excess.

ν (HCl) \u003d M (HCl) / M (HCl) \u003d 7.3 / 36.5 \u003d 0.2 mol;

ν (NH 3) \u003d M (NH 3) / M (NH 3) \u003d 5.1 / 17 \u003d 0.3 mol.

Ammonia is in excess, so the calculation is carried out by deficiency, i.e. By chloride. It follows from the reaction equation that ν (HCl) \u003d ν (NH 4 Cl) \u003d 0.2 mol. We determine the mass of ammonium chloride.

m (NH 4 Cl) \u003d ν (NH 4 Cl) M (NH 4 Cl) \u003d 0.2 53.5 \u003d 10.7 g

We determined that ammonia is in excess (in the amount of substance an excess is 0.1 mol). We calculate the mass of excess ammonia.

m (NH 3) \u003d ν (NH 3) M (NH 3) \u003d 0.1 17 \u003d 1.7 g

13. The technical carbide of calcium weighing 20 g was treated with an excess of water, having received acetylene, with a passage of which, through an excess of bromine water, formed 1,1,2,2 -thetrabrometan weighing 86.5 g. Determine mass share CAC 2 in technical carbide.

Dano: m \u003d 20 g; M (C 2 H 2 Br 4) \u003d 86.5

To find: Ω (SAC 2) \u003d?

Decision: Record the equations of interaction of calcium carbide with water and acetylene with bromine water and arrange the stoichiometric coefficients.

CAC 2 +2 H 2 O \u003d Ca (OH) 2 + C 2 H 2

C 2 H 2 +2 BR 2 \u003d C 2 H 2 BR 4

We find the amount of substance of the tetrabromethane.

ν (C 2 H 2 BR 4) \u003d M (C 2 H 2 Br 4) / M (C 2 H 2 Br 4) \u003d 86.5 / 346 \u003d 0.25 mol.

From the equations of the reaction it follows that ν (C 2 H 2 br 4) \u003d ν (C 2 H 2) \u003d ν (SAC 2) \u003d 0.25 mol. From here we can find a mass of pure calcium carbide (without impurities).

m (SAC 2) \u003d ν (SAC 2) M (SAC 2) \u003d 0.25 64 \u003d 16

We determine the mass fraction of SC 2 in technical carbide.

Ω (SAC 2) \u003d M (SAC 2) / M \u003d 16/20 \u003d 0.8 \u003d 80%.

Solutions. Mass fraction of the component of the solution

14. In benzene, 170 ml was dissolved in a mass of 1.8 g. The density of benzene is 0.88 g / ml. Determine mass share sulfur in solution.

Dano: V (C 6 H 6) \u003d 170 ml; M (S) \u003d 1.8 g; ρ (C 6 C 6) \u003d 0.88 g / ml.

To find: Ω (s) \u003d?

Decision: To find the mass fraction of sulfur in the solution it is necessary to calculate the mass of the solution. We determine the mass of benzene.

m (C 6 C 6) \u003d ρ (C 6 C 6) V (C 6 H 6) \u003d 0.88 170 \u003d 149.6

We find a total mass of the solution.

m (p-ra) \u003d m (C 6 C 6) + m (S) \u003d 149,6 + 1.8 \u003d 151.4 g.

Calculate the mass fraction of sulfur.

ω (s) \u003d m (s) / m \u003d 1.8 / 151,4 \u003d 0.0119 \u003d 1.19%.

15. In water weighing 40 g dissolved iron cuneition FESO 4 7H 2 O weighing 3.5 g. Determine mass fraction of iron sulfate (II) In the resulting solution.

Dano: m (H 2 O) \u003d 40 g; M (FESO 4 7H 2 O) \u003d 3.5 g

To find: Ω (FESO 4) \u003d?

Decision: We will find a mass FESO 4 contained in FESO 4 7H 2 O. For this, we calculate the amount of substance FESO 4 7H 2 O.

ν (FESO 4 7H 2 O) \u003d M (FESO 4 7H 2 O) / M (FESO 4 7H 2 O) \u003d 3.5 / 278 \u003d 0,0125mol

It follows from the formula of the iron mood that ν (feso 4) \u003d ν (FESO 4 7H 2 O) \u003d 0.0125 mol. Let's calculate the mass of FESO 4:

m (FESO 4) \u003d ν (FESO 4) M (FESO 4) \u003d 0.0125 152 \u003d 1.91

Considering that the mass of the solution is consisted of the mass of iron vitriol (3.5 g) and the mass of water (40 g), we calculate the mass fraction of iron sulfate in solution.

Ω (FESO 4) \u003d M (FESO 4) / M \u003d 1.91 / 43,5 \u003d 0.044 \u003d 4.4%.

Tasks for self solutions

1. At 50 g of iodide methyl in hexane, the metallic sodium was carried out, while 1.12 liters of gas measured under normal conditions was separated. Determine the mass fraction of iodide methyl in solution. Answer: 28,4%.
2. Some alcohol was oxidized, while a monosular carboxylic acid was formed. When burning 13.2 g of this acid, carbon dioxide was obtained, for the complete neutralization of which 192 ml of a solution of KOV with a mass fraction of 28% was required. Kon's solution density is 1.25 g / ml. Determine the alcohol formula. Answer: Butanol.
3. The gas obtained by the interaction of 9.52 g of copper with 50 ml of 81% solution of nitric acid, 1.45 g / ml density was passed through 150 ml of 20% NaOH solution with a density of 1.22 g / ml. Determine the mass fractions of solutes. Answer: 12.5% \u200b\u200bNaOH; 6.48% Nano 3; 5.26% Nano 2.
4. Determine the volume of the surrounding gases in the explosion of 10 g of nitroglycerin. Answer: 7.15 liters.
5. A sample of organic matter weighing 4.3 g burned in oxygen. Reaction products are carbon monoxide (IV) with a volume of 6.72 liters (normal conditions) and water weighing 6.3 g. The density of the steam of the starting material according to hydrogen is 43. Determine the formula of substance. Answer: C 6 H 14.

Solution of school challenges in chemistry can represent some difficulties for schoolchildren, so we post a number of examples of solutions. The main types of school chemistry tasks with a detailed analysis.

To solve the tasks in chemistry, it is necessary to know a number of formulas specified in the table below. Competently using this simple set, you can solve almost any task from the course of chemistry.

Calculations of the amount of substance	Share calculations	Reaction Product Output Calculations
ν \u003d m / m, ν \u003d v / v m, ν \u003d N / N A, ν \u003d PV / RT	ω \u003d m h / m about, φ \u003d V h / v about, χ \u003d ν h / ν	η \u003d m. / m Theore. . η \u003d v. / v Theorem. . η \u003d ν Ave. / ν Theorem.
ν - the amount of substance (mole); ν h - the amount of the substance is private (mol); ν OB - the amount of substance is general (mole); m - mass (g); m h - Private mass (d); m about - the mass of the total (g); V - volume (l); V M - volume 1 mol (l); V h - the volume of private (L); V about - the total volume (L); N is the number of particles (atoms, molecules, ions); N A - Number of Avogadro (number of particles in 1 mol of substance) N a \u003d 6.02 × 10 23; Q - the amount of electricity (CL); F - Permanent Faraday (F "96500 CB); P - pressure (PA) (1ATM "10 5 PA); R - universal gas constant R »8.31 J / (mol × K); T - absolute temperature (K); ω is a mass fraction; φ is a volume fraction; χ is a molar share; η - the yield of the product of the reaction; m pr., v., ν. - Mass, volume, the amount of substance practical; m TheOR., V Theorem., ν Theorem. - Weight, volume, amount of substance theoretical.

Calculating the mass of a certain amount of substance

The task:

Determine the mass of 5 mol of water (H 2 O).

Decision:

1. Calculate the molar mass of the substance using the periodic table of D. I. Mendeleev. The masses of all atoms are rounded to units, chlorine - up to 35.5.
   M (H 2 O) \u003d 2 × 1 + 16 \u003d 18 g / mol
2. Find a lot of water by the formula:
   m \u003d ν × m (H 2 O) \u003d 5 mol × 18 g / mol \u003d 90 g
3. Record the answer:
   Answer: Mass of 5 mol of water is equal to 90 g

Calculation of the mass fraction of the dissolved substance

The task:

Calculate the mass fraction of salt (NaCl) in the solution obtained by dissolving in 475 g of water 25 g of salt.

Decision:

1. Write a formula for finding a mass fraction:
   Ω (%) \u003d (M V-V / M Rr) × 100%
2. Find a mass solution.
   m p-ra \u003d m (H 2 O) + M (NaCl) \u003d 475 + 25 \u003d 500 g
3. Calculate a mass fraction, substituting the values \u200b\u200bin the formula.
   Ω (NaCl) \u003d (M V-V / M Rr) × 100% = (25/500) × 100% \u003d 5%
4. Record the answer.
   Answer: The mass fraction of NaCl is 5%

Calculation of the mass of the substance in the solution by its mass fraction

The task:

How many grams of sugar and water must be taken to receive 200 g 5% solution?

Decision:

1. Write the formula to determine the mass fraction of the dissolved substance.
   ω \u003d m V-V / M R-RA → M V-V \u003d M p-ra × Ω
2. Calculate the mass of the salt.
   m B-VA (salt) \u003d 200 × 0.05 \u003d 10 g
3. Determine the mass of water.
   m (H 2 O) \u003d M (p-ra) - m (salts) \u003d 200 - 10 \u003d 190 g
4. Record the answer.
   Answer: It is necessary to take 10 g of sugar and 190 g of water

Determining the release of the reaction product in% of theoretically possible

The task:

Calculate the yield of ammonium nitrate (NH 4 NO 3) in% of theoretically possible, if with a passage of 85 g of ammonia (NH 3) into a solution of nitric acid (HNO 3), 380 g of fertilizer was obtained.

Decision:

1. Write the chemical reaction equation and place the coefficients
   NH 3 + HNO 3 \u003d NH 4 NO 3
2. Data from the condition of the task to write over the reaction equation.
   

   m \u003d 85 g				m pr. \u003d 380 g
   NH 3.	+	HNO 3.	=	NH 4 NO 3

3. Under the formulas of substances, calculate the amount of substance according to coefficients as a product of the amount of substance on the molar mass of the substance:
   
4. The practically obtained mass of ammonium nitrate is known (380 g). In order to determine the theoretical mass of ammonium nitrate to draw up proportion
   85/17 \u003d x / 380
   
5. Solve equation, determine x.
   x \u003d 400 g Theoretical mass of ammonium nitrate
6. Determine the yield of the reaction product (%), taken by the practical mass to theoretical and multiply by 100%
   η \u003d m. / m Theore. \u003d (380/400) × 100% \u003d 95%
7. Record the answer.
   Answer: Ammonium nitrate yield amounted to 95%.

Calculation of the mass of the product at a well-known mass of the reagent containing a certain fraction of impurities

The task:

Calculate the mass of calcium oxide (CAO), which was obtained under the burning of 300 g of limestone (CACO 3) containing 10% impurities.

Decision:

1. Write the chemical reaction equation, put the coefficients.
   Saso 3 \u003d Cao + CO 2
2. Calculate the mass of pure Camia 3 contained in limestone.
   ω (clean) \u003d 100% - 10% \u003d 90% or 0.9;
   M (Caco 3) \u003d 300 × 0.9 \u003d 270 g
3. The resulting mass of Saco 3 is to write over the CACO 3 formula in the reaction equation. The desired mass of SAO designate through x.
   

   270 g		x G.
   Saco 3.	=	SAO	+	CO 2

4. Under the formulas of substances in the equation to write the amount of substance (according to coefficients); Production of the amounts of substances on their molar mass (molecular weight of Saco 3 \u003d 100 , Sao \u003d. 56 ).
   
5. Create a proportion.
   270/100 \u003d x / 56
6. Solve equation.
   x \u003d 151.2 g
7. Record the answer.
   Answer: The mass of calcium oxide will be 151, 2 g

Calculation of the mass of the reaction product if the reaction product yield is known

The task:

How many ammonium nitrates (NH 4 NO 3) can be obtained by interaction of 44.8 liters of ammonia (n. Y.) With nitric acid, if it is known that the practical yield is 80% of theoretically possible?

Decision:

1. Record the chemical reaction equation, lay out the coefficients.
   NH 3 + HNO 3 \u003d NH 4 NO 3
2. These terms of the task write over the reaction equation. The mass of ammonia nitrate designated through x.
   
3. For the reaction equation, write:
   a) the amount of substances according to coefficients;
   b) the product of the molar volume of ammonia to the amount of substance; The product of the molar mass of NH 4 NO 3 by the amount of substance.
4. Make a proportion.
   44.4 / 22.4 \u003d x / 80
5. Solve the equation, finding x (theoretical mass of ammonia nitrate):
   x \u003d 160
6. Find the practical mass of NH 4 NO 3, by changing the theoretical mass on the practical output (in fractions from one)
   m (NH 4 NO 3) \u003d 160 × 0.8 \u003d 128 g
7. Write down the answer.
   Answer: Mass of ammonium nitrate will be 128 g.

Determining the mass of the product if one of the reagents are taken in excess

The task:

14 g of calcium oxide (CAO) was treated with a solution containing 37.8 g of nitric acid (HNO 3). Calculate the mass of the reaction product.

Decision:

1. Write down the reaction equation, arrange the coefficients
   Cao + 2HNO 3 \u003d Ca (NO 3) 2 + H 2 O
2. Determine the mole of reagents by the formula: ν \u003d m / m
   ν (Cao) \u003d 14/56 \u003d 0.25 mol;
   ν (HNO 3) \u003d 37.8 / 63 \u003d 0.6 mol.
   
3. Over the reaction equation, write the calculated amounts of the substance. Under the equation - the amount of substance according to stoichiometric coefficients.
   
4. Determine the substance taken in a disadvantage by comparing the relations of the amounts taken amounts to stoichiometric coefficients.
   0,25/1 < 0,6/2
   Consequently, nitric acid is taken in the disadvantage. According to it, we will determine the mass of the product.
5. Under the calcium nitrate formula (Ca (NO 3) 2) in the Equation equation:
   a) the amount of substance according to the stoichiometric coefficient;
   b) The product of the molar mass by the amount of substance. Above the formula (Ca (NO 3) 2) -

   0.25 mol		0.6 mol		X G.
   Cao.	+	2hno 3.	=	SA (NO 3) 2	+	H 2 O.
   1 mol		2 mol		1 mol
   				m \u003d 1 × 164 g

6. Make a proportion
   0.25 / 1 \u003d x / 164
7. Determine H.
   x \u003d 41 g
8. Write down the answer.
   Answer: Salt weight (Ca (NO 3) 2) will be 41

Calculations on thermochemical reaction equations

The task:

How much heat is separated when dissolved 200 g of copper (II) oxide (Cuo) in hydrochloric acid (an aqueous solution of HCl), if the thermochemical reaction equation:

Cuo + 2HCl \u003d CUCl 2 + H 2 O + 63.6 kJ

Decision:

1. Data from the condition of the problem Write over the reaction equation
   
2. Under the formula of copper oxide write its amount (according to the coefficient); The product of the molar mass by the amount of substance. Above the amount of heat in the reaction equation to put x.
   

   200 g
   Cuo.	+	2hcl.	=	CUCL 2.	+	H 2 O.	+	63.6 kJ
   1 mol
   m \u003d 1 × 80 g

3. Create a proportion.
   200/80 \u003d x / 63.6
4. Calculate x.
   x \u003d 159 kj
5. Record the answer.
   Answer: When dissolved 200 g of Cuo in hydrochloric acid, 159 CJD heat outdoor.

Compilation of the thermochemical equation

The task:

When burning 6 g of magnesium, 152 KJ heat is highlighted. Create a thermochemical equation for the formation of magnesium oxide.

Decision:

1. Write the chemical reaction equation, showing the release of heat. Plaintive coefficients.
   2mg + O 2 \u003d 2MGO + Q
2. 
   

   6 g						152
   2mg.	+	O 2.	=	2mgo.	+	Q.

3. Under the formulas of substances to write:
   a) the amount of substance (according to coefficients);
   b) The product of the molar mass by the amount of substance. Under the thermal effect of the reaction to put x.
   
4. Create a proportion.
   6 / (2 × 24) \u003d 152 / x
5. Calculate x (the amount of heat, according to the equation)
   x \u003d 1216 kj
6. Record the thermochemical equation in response.
   Answer: 2mg + O 2 \u003d 2mgo + 1216 kJ

Calculation of gases for chemical equations

The task:

When oxidizing ammonia (NH 3) oxygen in the presence of a catalyst, nitrogen oxide (II) and water is formed. What volume of oxygen will come into reaction from 20 liters of ammonia?

Decision:

1. Write the reaction equation and place the coefficients.
   4NH 3 + 5O 2 \u003d 4NO + 6H 2 O
2. Data from the condition of the problem to write on the reaction equation.
   

   20 L.		x.
   4NH 3.	+	5O 2.	=	4no.	+	6H 2 O.

3. Under the equation of reaction, write down the amount of substances according to the coefficients.
   
4. Create a proportion.
   20/4 \u003d x / 5
5. Find x.
   x \u003d 25 l
6. Record the answer.
   Answer: 25 l oxygen.

Determining the volume of the gaseous product according to the known mass of the reagent containing impurities

The task:

What volume (N.U) of carbon dioxide (CO 2) is released during dissolution of 50 g of marble (CACO 3) containing 10% impurities in hydrochloric acid?

Decision:

1. Write the chemical reaction equation, place the coefficients.
   Caco 3 + 2HCl \u003d CaCl 2 + H 2 O + CO 2
2. Calculate the amount of pure Camia 3 contained in 50 g marble.
   Ω (Saco 3) \u003d 100% - 10% \u003d 90%
   To transfer to a share from a unit to divide by 100%.
   w (sasso 3) \u003d 90% / 100% \u003d 0.9
   m (Caco 3) \u003d M (marble) × w (sasso 3) \u003d 50 × 0.9 \u003d 45 g
3. The resulting value to write over calcium carbonate in the reaction equation. Over CO 2 put x l.
   

   45 g								x.
   Caco 3.	+	2hcl.	=	CaCl 2.	+	H 2 O.	+	CO 2.

4. Under the formulas of substances to write:
   a) the amount of substance according to coefficients;
   b) The product of the molar mass on the number of substances, if the mass of the substance, and the product of the molar volume on the amount of substance, if the volume of the substance says.

   Calculation of the composition of the mixture according to the chemical reaction equation

   The task:

   The total combustion of the mixture of methane and carbon oxide (II) was required as the same volume of oxygen. Determine the composition of the gas mixture in volume shares.

   Decision:

   1. Write the reaction equations, arrange the coefficients.
      CO + 1 / 2O 2 \u003d CO 2
      CH 4 + 2O 2 \u003d CO 2 + 2N 2
   2. Identify the amount of carbon monoxide (CO) - x, and the amount of methane for
      

   45 g								x.
   Caco 3.	+	2hcl.	=

   h.
   SO	+	1 / 2O 2	=	CO 2
   w.
   CH 4.	+	2O 2.	=	CO 2	+	2N 2 O.

5. Determine the amount of oxygen, which will be spent on the combustion of X mol CO and Mol CH 4.
   

   h.		0.5 x
   SO	+	1 / 2O 2	=	CO 2
   w.		2ow
   CH 4.	+	2O 2.	=	CO 2	+	2N 2 O.

6. Make a conclusion about the ratio of the amount of substance of oxygen and gas mixture.
   Equality of gases indicates the equality of the amount of matter.
7. Make an equation.
   x + y \u003d 0,5x + 2
8. Simplify equation.
   0.5 x \u003d y
9. Take the amount of CO for 1 mol and determine the required amount of CH 4.
   If x \u003d 1, then y \u003d 0.5
10. Find the total amount of substance.
    x + y \u003d 1 + 0.5 \u003d 1.5
11. Determine the volume fraction of carbon monoxide oxide (CO) and methane in the mixture.
    φ (CO) \u003d 1 / 1,5 \u003d 2/3
    φ (CH 4) \u003d 0.5 / 1,5 \u003d 1/3
12. Record the answer.
    Answer: Volume share of CO is 2/3, and CH 4 - 1/3.

Reference material:

Mendeleev table

Solubility table

We discussed the overall algorithm for solving problems No. 35 (C5). It's time to disassemble specific examples and offer you a selection of tasks for an independent solution.

Example 2.. 4.48 liters of hydrogen are consumed on complete hydrogenation of 5.4 g of some alkina (n.) Determine the molecular formula of this alkina.

Decision. We will act in accordance with the general plan. Let the unknown alkina molecule contain N carbon atoms. The total formula of the homologous series C n h 2N-2. Alkin hydrogenation proceeds in accordance with the equation:

C n h 2N-2 + 2H 2 \u003d C n h 2n + 2.

The amount of hydrogen entered into the reaction can be found according to the formula N \u003d V / VM. In this case, n \u003d 4.48 / 22.4 \u003d 0.2 mol.

The equation shows that 1 mole of alkina attaches 2 mol of hydrogen (we recall that the task is about full Hydrogenation), therefore, n (C n h 2N-2) \u003d 0.1 mol.

By weight and quantity of alkina, we find its molar mass: M (C n H 2N-2) \u003d M (weight) / N (number) \u003d 5.4 / 0.1 \u003d 54 (g / mol).

The relative molecular weight of alkina is made of N atomic carbon masses and 2N-2 atomic mass of hydrogen. We get the equation:

12N + 2N - 2 \u003d 54.

We solve the linear equation, we obtain: n \u003d 4. Alkina formula: C 4 H 6.

Answer: C 4 H 6.

I would like to pay attention to one substantive point: the molecular formula C 4 H 6 corresponds to several isomers, including, two alkina (boudo-1 and buttine-2). Relying on these tasks, we will not be able to unambiguously install structural formula The substance under study. However, in this case, this is not required!

Example 3.. When combustion 112 l (n. Y.) Unknown cycloalkane in excess oxygen forms 336 liters of 2. Set the structural formula of cycloalkane.

Decision. The general formula of the homologous series of cycloalkanes: with N H 2N. With full combustion of cycloalkanes, as in combustion of any hydrocarbons, carbon dioxide and water are formed:

C n H 2N + 1.5N O 2 \u003d N CO 2 + N H 2 O.

Note: The coefficients in the reaction equation in this case depends on N!

During the reaction, 336 / 22.4 \u003d 15 mol of carbon dioxide was formed. 112 / 22.4 \u003d 5 mole of hydrocarbon joined the reaction.

Further reasoning is obvious: if 15 mol CO 2 is formed at 5 mol, then 15 carbon dioxide molecules are formed, i.e., one cycloalka molecule gives 3 CO 2 molecules. Since each carbon oxide molecule (IV) contains one at one carbon atom, it can be concluded: in one cycloalka molecule contains 3 carbon atoms.

Conclusion: N \u003d 3, Cycloalka Formula - C 3 H 6.

As you can see, the solution of this task does not "fit" into the general algorithm. We have not searched here a molar mass of the compound, no equation was. According to formal criteria, this example is not similar to the standard C5 task. But above, I have already emphasized that it is important not to drive the algorithm, but to understand the meaning of the actions produced. If you understand the meaning, you will be able to make changes to the general scheme on the exam, choose the most rational decision path.

In this example, there is another "oddity": it is necessary to find not only the molecular, but also the structural formula of the compound. In the previous task, we could not do this, and in this example, please! The fact is that the formula C 3 H 6 corresponds to only one isomer - cyclopropane.

Answer: Cyclopropane.

Example 4.. 116 g of some limit aldehyde was heated for a long time with ammonia solution of silver oxide. During the reaction, 432 g of metallic silver was formed. Install the molecular formula of aldehyde.

Decision. The general formula of the homologous series of limit aldehydes: C n H 2N + 1 COH. Aldehydes are easily oxidized to carboxylic acids, in particular, under the action of ammonia solutions of silver oxide:

C n h 2n + 1 COH + AG 2 O \u003d C n H 2N + 1 COOH + 2AG.

Note. In fact, the reaction is described by a more complex equation. When the AG 2 O is added to an ammonia aqueous solution, an integrated OH compound is a diaminuseside hydroxide. This is the connection and acts as an oxidizing agent. During the reaction, ammonium carboxylic acid salt is formed:

C n H 2N + 1 COH + 2OH \u003d C n H 2N + 1 Coonh 4 + 2Ag + 3NH 3 + H 2 O.

Another important moment! Oxidation of formaldehyde (HCOH) is not described by the equation. With the interaction of Nson with an ammonium solution of silver oxide, 4 mole AG is released per 1 moth aldehyde:

HCOH + 2AG 2 O \u003d CO 2 + H 2 O + 4AG.

Be careful, solving problems associated with the oxidation of carbonyl compounds!

Let's go back to our example. By the mass of the silver distinguished, the amount of this metal can be found: n (Ag) \u003d m / m \u003d 432/108 \u003d 4 (mol). In accordance with the equation, 2 mole of silver is formed on 1 moth of aldehyde, therefore N (aldehyde) \u003d 0.5N (Ag) \u003d 0.5 * 4 \u003d 2 mol.

Molar mass of aldehyde \u003d 116/2 \u003d 58 g / mol. Further actions Try to do it yourself: you need to make an equation to solve it and draw conclusions.

Answer: C 2 H 5 COH.

Example 5.. When the interaction of 3.1 g of some primary amine with a sufficient amount of HBr, 11.2 g of salts are formed. Install the formula of the amine.

Decision. Primary amines (with N H 2N + 1 NH 2) When interacting with acids, alkyllammonium salts are formed:

With N H 2N + 1 NH 2 + HBr \u003d [C n H 2N + 1 NH 3] + BR -.

Unfortunately, by the mass of the amine and the resulting salt, we will not be able to find their quantities (since the molar masses are unknown). Let's go on another way. Recall the law of preserving the mass: m (amine) + m (HBr) \u003d M (salts), therefore, M (HBr) \u003d M (salts) - M (amine) \u003d 11.2 - 3.1 \u003d 8.1.

Pay attention to this reception, very often used when solving C. 5. If even a mass of the reagent is not given in an explicit form in the task condition, you can try to find it through the masses of other connections.

So, we returned to the standard algorithm. By weight of bromomodorod, we find quantities, N (HBr) \u003d N (amine), M (amine) \u003d 31 g / mol.

Answer: CH 3 NH 2.

Example 6.. A certain amount of alkene x with an extension with an excess of chlorine forms 11.3 g of dichloride, and with a reaction with an excess of bromine - 20.2 g of dibromide. Determine the molecular formula X.

Decision. Alkenes join chlorine and bromine with the formation of digalogen derivatives:

With n h 2n + Cl 2 \u003d C n h 2n Cl 2,

With n h 2n + br 2 \u003d with n h 2n br 2.

It is pointless in this task to try to find the amount of dichloride or dibromide (their molar masses) or the amount of chlorine or bromine (their masses are unknown).

We use one non-standard reception. The molar mass with n h 2n Cl 2 is 12N + 2N + 71 \u003d 14N + 71. M (with N H 2N BR 2) \u003d 14N + 160.

The masses of digaloids are also known. You can find the amount of substances obtained: n (with n h 2n Cl 2) \u003d m / m \u003d 11.3 / (14n + 71). n (with n h 2n br 2) \u003d 20.2 / (14n + 160).

By condition, the amount of dichloride is equal to the number of dibromide. This fact gives us the opportunity to make the equation: 11.3 / (14N + 71) \u003d 20.2 / (14N + 160).

This equation has a single solution: n \u003d 3.

Answer: C 3 H 6

In the final part, I offer you a selection of tasks of the type C5 of different complexity. Try to solve them yourself - it will be excellent training before surveying ege in chemistry!