For 2-3 months it is impossible to learn (repeat, pull up) such a complex discipline as chemistry.

There are no changes in KIM USE 2020 in chemistry.

Don't delay your preparation.

1. Before starting the analysis of tasks, first study theory. The theory on the site is presented for each task in the form of recommendations that you need to know when completing the task. guides in the study of the main topics and determines what knowledge and skills will be required when completing the USE tasks in chemistry. For successful delivery USE in chemistry - theory is the most important thing.
2. Theory needs to be backed up practice constantly solving problems. Since most of the errors are due to the fact that I read the exercise incorrectly, I did not understand what is required in the task. The more often you solve thematic tests, the faster you will understand the structure of the exam. Training tasks developed on the basis of demos from FIPI give them the opportunity to decide and find out the answers. But do not rush to peek. First, decide for yourself and see how many points you have scored.

Points for each task in chemistry

• 1 point - for 1-6, 11-15, 19-21, 26-28 tasks.
• 2 points - 7-10, 16-18, 22-25, 30, 31.
• 3 points - 35.
• 4 points - 32, 34.
• 5 points - 33.

Total: 60 points.

The structure of the examination paper consists of two blocks:

1. Questions that require a short answer (in the form of a number or word) - tasks 1-29.
2. Tasks with detailed answers - tasks 30-35.

3.5 hours (210 minutes) are allotted to complete the examination paper in chemistry.

There will be three cheat sheets on the exam. And they need to be dealt with.

This is 70% of the information that will help you successfully pass the exam in chemistry. The remaining 30% is the ability to use the provided cheat sheets.

• If you want to get more than 90 points, you need to spend a lot of time on chemistry.
• To successfully pass the exam in chemistry, you need to solve a lot: training tasks, even if they seem easy and of the same type.
• Correctly distribute your strength and do not forget about the rest.

Dare, try and you will succeed!

In our last article, we talked about the general USE codifier in chemistry 2018 and how to properly start preparing for the USE in chemistry 2018. Now, we have to analyze the preparation for the exam in more detail. In this article, we will look at simple tasks (formerly called parts A and B) that are worth one and two points.

Simple tasks, in the USE codifier in chemistry in 2018 called Basic, make up the largest part of the exam (20 tasks) in terms of the maximum primary score - 22 primary scores(tasks 9 and 17 are now worth 2 points).

Therefore, we must pay special attention to preparing for simple tasks in chemistry in the USE 2018, given that many of them, with proper preparation, can be correctly done by spending from 10 to 30 seconds, instead of the 2-3 minutes suggested by the organizers, which will allow save time to complete those tasks that are more difficult for the student.

to basic USE assignments in chemistry 2018 are No. 1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 13, 14.15, 16, 17, 20, 21, 27, 28, 29.

We would like to draw your attention to the fact that at the Hodograph Training Center you will find qualified tutors in preparation for the OGE in chemistry for students, and. We practice individual and collective classes for 3-4 people, we provide discounts for training. Our students average 30 points more!

Task topics 1, 2, 3 and 4 in the exam in chemistry 2018

Aimed at testing knowledge related to the structure of atoms and molecules, the properties of atoms (electronegativity, metallic properties and the radius of an atom), the types of bonds formed during the interaction of atoms with each other to form molecules (covalent non-polar and polar bonds, ionic bonds, hydrogen bonds, etc. .) the ability to determine the oxidation state and valency of an atom. To successfully complete these tasks in the Unified State Examination in Chemistry 2018, you need:

• Navigate in the Periodic Table of Dmitry Ivanovich Mendeleev;
• Study classical atomic theory;
• Know the rules for constructing the electronic configuration of an atom (Hund's rule, Pauli's principle) and be able to read electronic configurations of various forms of notation;
• Understand the differences in the formation of various types of bonds (covalent NOT polar is formed only between identical atoms, covalent polar between atoms of different chemical elements);
• Be able to determine the oxidation state of each atom in any molecule (oxygen always has an oxidation state of minus two (-2), and hydrogen plus one (+1))

Task 5 in the exam in chemistry 2018

It will require the student to have knowledge of the nomenclature of inorganic chemical compounds (rules for the formation of names of chemical compounds), both classical (nomenclature) and trivial (historical).

The structure of tasks 6, 7, 8 and 9 of the exam in chemistry

Aimed at testing knowledge about inorganic compounds and their chemical properties. To successfully complete these tasks in the Unified State Examination in Chemistry 2018, you need:

• Know the classification of all inorganic compounds(non-salt-forming and salt-forming oxides (basic, amphoteric and acid), etc.);

Tasks 12, 13, 14, 15 16 and 17 in the exam

Test knowledge of organic compounds and their chemical properties. To successfully complete these tasks in the Unified State Examination in Chemistry 2018, you need:

• Know all classes of organic compounds (alkanes, alkenes, alkynes, arenes, etc.);
• Be able to give the name of the compound according to the trivial and international nomenclature;
• To study the relationship of various classes of organic compounds, their chemical properties and methods of laboratory preparation.

Tasks 20 and 21 in the USE 2018

Requires the student to know about a chemical reaction, the types of chemical reactions, and how chemical reactions are controlled.

Assignments 27, 28 and 29 in chemistry

These are calculation tasks. They contain the simplest chemical processes in their composition, which are aimed only at shaping the student's understanding of what happened in the task. The rest of the assignment is strictly mathematical. Therefore, to solve these tasks in the Unified State Examination in Chemistry 2018, you need to learn three basic formulas (mass fraction, molar fraction by mass and volume) and be able to use a calculator.

Average tasks, in the USE codifier in chemistry in 2018, called Increased (see table 4 in the codifier - Distribution of tasks by difficulty levels), make up the smallest part of the exam in terms of points (9 tasks) in terms of the maximum primary score - 18 primary scores or 30 %. Despite the fact that this is the smallest part of the exam, 5-7 minutes are planned for solving tasks, with high preparation it is quite possible to solve them in 2-3 minutes, thereby saving time on tasks that are difficult for the student to solve.

Increased tasks include tasks No.: 10, 11, 18, 19, 22, 23, 24, 25, 26.

Task 10 in chemistry 2018

These are redox reactions. To successfully complete this task in the Unified State Examination in Chemistry 2018, you need to know:

• What are an oxidizing agent and a reducing agent and how do they differ;
• How to correctly determine the oxidation states of atoms in molecules and trace which atoms changed the oxidation state as a result of the reaction.

Task 11 Unified State Examination in Chemistry 2018

Properties of inorganic substances. One of the most difficult tasks for a student to complete, due to the large volume of possible combinations of answers. Students often begin to write down ALL reactions, and there are hypothetically from forty (40) to sixty (60) in each task, which takes a very long time. To successfully complete this task in the Unified State Examination in Chemistry 2018, you need:

• Accurately determine which compound is in front of you (oxide, acid, base, salt);
• Know the basic principles of interclass interaction (acid will not react with acidic oxide, etc.);

Since this is one of the most problematic tasks, let's look at the solution of task No. 11 from USE demos in chemistry 2018:

Eleventh task: Establish a correspondence between the formula of a substance and the reagents, with each of which this substance can interact: for each position indicated by a letter, select the corresponding position indicated by a number.

SUBSTANCE FORMULA	REAGENTS
A) S	1) AgNO 3, Na 3 PO 4, Cl 2
B) SO 3	2) BaO, H 2 O, KOH
B) Zn (OH) 2	3) H 2, Cl 2, O 2
D) ZnBr 2 (solution)	4) HBr, LiOH, CH 3 COOH
	5) H 3 PO 4, BaCl 2, CuO

Write in the table the selected numbers under the corresponding letters.

The solution of task 11 in the exam in chemistry 2018

First of all, we need to determine what is offered to us as reagents: substance A is a pure sulfur substance, B is sulfur oxide VI is an acid oxide, C is zinc hydroxide is an amphoteric hydroxide, D is zinc bromide is an average salt. It turns out that in this task there are 60 hypothetical reactions. It is very important for solving this task to reduce the possible answers, the main tool for this is the student's knowledge of the main classes of inorganic substances and their interaction with each other, I propose to build the following table and cross out the possible answers as the task is logically evaluated:

A) S	1	2	3	4	5
B) SO 3	1	2	3	4	5
B) Zn (OH) 2	1	2	3	4	5
D) ZnBr 2 (solution)	1	2	3	4	5

And now, applying knowledge about the nature of substances and their interactions, we remove answer options that are definitely not correct, for example, answer B- acid oxide, which means it does NOT react with acids and acid oxides, which means the answer options are not suitable for us - 4.5, since sulfur oxide VI is the highest oxide, which means it will not react with oxidizing agents, pure oxygen and chlorine - we remove answers 3, four. There remains only answer 2, which suits us perfectly.

Answer B- here you need to apply the reverse technique, to whom amphoteric hydroxides react - both with bases and with acids, and we see an answer option consisting only of these compounds - answer 4.

Answer D- the average salt containing the anion of bromine, which means that adding a similar anion is meaningless - we remove answer 4 containing hydrobromic acid. We will also remove answer 5 - since the reaction with bromine chloride is meaningless, two soluble salts will be formed (zinc chloride and barium bromide), which means that the reaction is completely reversible. Answer option 2 is also not suitable, since we already have a salt solution, which means adding water will not lead to anything, and answer option 3 is also not suitable due to the presence of hydrogen, which is not able to restore zinc, which means that the answer option remains 1. Remains an option

answer A- which can cause the most difficulties, so we left it for last, which should also be done by the student, if difficulties arise, since two points are given for an advanced level task, and we allow one mistake (in this case, the student will receive one point for exercise). For the correct solution of this element of the task, it is necessary to have a good understanding of the chemical properties of sulfur and simple substances, respectively, so as not to describe the entire course of the solution, the answer will be 3 (where all answers are also simple substances).

Reactions:

BUT)S + H 2 à H 2 S

S + Cl 2 à SCl 2

S + O 2 à SO 2

B)SO 3 + BaO à BaSO 4

SO 3 + H 2 O à H 2 SO 4

SO 3 + KOH à KHSO 4 // SO 3 + 2 KOH à K 2 SO 4 + H 2 O

AT) Zn(OH) 2 + 2HBrà ZnBr 2 + 2H 2 O

Zn(OH) 2 + 2LiOHà Li 2 ZnO 2 + 2H 2 O // Zn(OH) 2 + 2LiOHà Li 2

Zn(OH) 2 + 2CH 3 COOHà (CH 3 COO) 2 Zn + 2H 2 O

G) ZnBr 2 + 2AgNO 3à 2AgBr↓ + Zn(NO 3) 2

3ZnBr 2 + 2Na 3 PO 4à Zn 3 (PO 4) 2 ↓ + 6NaBr

ZnBr 2 + Cl 2à ZnCl 2 + Br 2

Tasks 18 and 19 in the exam in chemistry

A more complex format, including all the knowledge needed to solve basic tasks №12-17 . Separately, we can highlight the need for knowledge Markovnikov's rules.

Task 22 in the exam in chemistry

Electrolysis of melts and solutions. To successfully complete this task in the Unified State Examination in Chemistry 2018, you need to know:

• The difference between solutions and melts;
• Physical foundations of electric current;
• Differences between melt electrolysis and solution electrolysis;
• The main regularities of the products obtained as a result of solution electrolysis;
• Features of the electrolysis of a solution of acetic acid and its salts (acetates).

Task 23 in chemistry

Salt hydrolysis. To successfully complete this task in the Unified State Examination in Chemistry 2018, you need to know:

• Chemical processes occurring during the dissolution of salts;
• Due to what it forms the solution medium (acidic, neutral, alkaline);
• Know the color of the main indicators (methyl orange, litmus and phenolphthalein);
• Learn strong and weak acids and bases.

Task 24 in the exam in chemistry

Reversible and irreversible chemical reactions. To successfully complete this task in the Unified State Examination in Chemistry 2018, you need to know:

• Be able to determine the amount of a substance in a reaction;
• Know the main factors influencing the reaction (pressure, temperature, concentration of substances)

Task 25 in chemistry 2018

Qualitative reactions to inorganic substances and ions.

To successfully complete this task in the Unified State Examination in Chemistry 2018, you need to learn these reactions.

Task 26 in chemistry

Chemical laboratory. The concept of metallurgy. Production. chemical pollution environment. Polymers. To successfully complete this task in the Unified State Examination in Chemistry 2018, you need to have an idea about all the elements of the task, regarding a variety of substances (it is best to study in conjunction with chemical properties, etc.)

Once again, I would like to note that the necessary for the successful passing of the exam in chemistry in 2018 theoretical bases have remained virtually unchanged, which means that all the knowledge that your child received at school will help him in passing the exam in chemistry in 2018.

In ours, your child will receive all necessary for the preparation of theoretical materials, and in the classroom will consolidate the knowledge gained for successful implementation all exam assignments. The best teachers who have passed a very large competition and difficult entrance tests will work with him. Classes are held in small groups, which allows the teacher to devote time to each child and form his individual strategy for completing the examination work.

We have no problems with the lack of tests of a new format, our teachers write them themselves, based on all the recommendations of the codifier, specifier and demo version of the Unified State Examination in Chemistry 2018.

Call today and tomorrow your child will thank you!

In the next article, we will talk about the features of solving complex USE tasks in chemistry and how to get maximum number points at passing the exam 2018.

The work consists of two parts:
- part 1 - tasks with a short answer (26 - basic level, 9 increased),
- part 2 - tasks with a detailed answer (5 high-level tasks).
Maximum number primary points remained the same: 64.
However, some changes will be made.:

1. In tasks of the basic level of complexity(former part A) will include:
a) 3 tasks (6,11,18) with multiple choice (3 out of 6, 2 out of 5)
b) 3 tasks with an open answer (calculation problems), the correct answer here will be the result of calculations, written with a specified degree of accuracy;
Like other basic level assignments, these assignments will be worth 1 primary point.

2. Advanced level tasks (former part B) will be presented in one type: compliance assignments. They will be evaluated at 2 points (if there is one mistake - 1 point);

3. From the tasks of the basic level to the advanced one, the question was moved on the topic: "Reversible and irreversible chemical reactions. Chemical equilibrium. Shift in equilibrium under the influence of various factors".
However, the issue of nitrogen-containing compounds will be tested at a basic level.

4. Time spending single exam in chemistry will be increased from 3 hours to 3.5 hours(from 180 to 210 minutes).

To complete tasks 1-3, use the following row of chemical elements. The answer in tasks 1-3 is a sequence of numbers, under which the chemical elements in this row.

1) Na 2) K 3) Si 4) Mg 5) C

Task number 1

Determine which atoms of the elements indicated in the series have four electrons on the external energy level.

Answer: 3; 5

The number of electrons in the outer energy level (electronic layer) of the elements of the main subgroups is equal to the group number.

Thus, from the presented answers, silicon and carbon are suitable, because. they are in the main subgroup of the fourth group of the table D.I. Mendeleev (IVA group), i.e. Answers 3 and 5 are correct.

Task number 2

From the chemical elements indicated in the series, select three elements that are in the Periodic Table of Chemical Elements of D.I. Mendeleev are in the same period. Arrange the selected elements in ascending order of their metallic properties.

Write in the answer field the numbers of the selected elements in the desired sequence.

Answer: 3; four; one

Three of the presented elements are in the same period - sodium Na, silicon Si and magnesium Mg.

When moving within one period of the Periodic Table, D.I. Mendeleev (horizontal lines) from right to left, the return of electrons located on the outer layer is facilitated, i.e. the metallic properties of the elements are enhanced. Thus, the metallic properties of sodium, silicon and magnesium are enhanced in the series Si

Task number 3

From among the elements listed in the row, select two elements that exhibit the lowest oxidation state, equal to -4.

Write down the numbers of the selected elements in the answer field.

Answer: 3; 5

According to the octet rule, the atoms of chemical elements tend to have 8 electrons in their outer electronic level, like the noble gases. This can be achieved either by donating electrons of the last level, then the previous one, containing 8 electrons, becomes external, or, conversely, by adding additional electrons up to eight. Sodium and potassium are alkali metals and are in the main subgroup of the first group (IA). This means that on the outer electron layer of their atoms there is one electron each. In this regard, the loss of a single electron is energetically more favorable than the addition of seven more. With magnesium, the situation is similar, only it is in the main subgroup of the second group, that is, it has two electrons on the outer electronic level. It should be noted that sodium, potassium and magnesium are metals, and for metals, in principle, a negative oxidation state is impossible. The minimum oxidation state of any metal is zero and is observed in simple substances.

The chemical elements carbon C and silicon Si are non-metals and are in the main subgroup of the fourth group (IVA). This means that there are 4 electrons on their outer electron layer. For this reason, for these elements, both the return of these electrons and the addition of four more up to a total of 8 are possible. Silicon and carbon atoms cannot attach more than 4 electrons, therefore the minimum oxidation state for them is -4.

Task number 4

From the proposed list, select two compounds in which there is an ionic chemical bond.

• 1. Ca(ClO 2) 2
• 2. HClO 3
• 3.NH4Cl
• 4. HClO 4
• 5.Cl2O7

Answer: 1; 3

In the vast majority of cases, the presence of an ionic type of bond in a compound can be determined by the fact that its structural units simultaneously include atoms of a typical metal and non-metal atoms.

On this basis, we establish that there is an ionic bond in compound number 1 - Ca(ClO 2) 2, because in its formula, one can see atoms of a typical calcium metal and atoms of non-metals - oxygen and chlorine.

However, there are no more compounds containing both metal and non-metal atoms in this list.

In addition to the above feature, the presence of an ionic bond in a compound can be said if its structural unit contains an ammonium cation (NH 4 +) or its organic analogs - cations of alkylammonium RNH 3 +, dialkylammonium R 2 NH 2 + , trialkylammonium R 3 NH + and tetraalkylammonium R 4 N + , where R is some hydrocarbon radical. For example, the ionic type of bond takes place in the compound (CH 3) 4 NCl between the cation (CH 3) 4 + and the chloride ion Cl - .

Among the compounds indicated in the assignment there is ammonium chloride, in which the ionic bond is realized between the ammonium cation NH 4 + and the chloride ion Cl − .

Task number 5

Establish a correspondence between the formula of a substance and the class / group to which this substance belongs: for each position indicated by a letter, select the corresponding position from the second column, indicated by a number.

Write down the numbers of the selected connections in the answer field.

Answer: A-4; B-1; AT 3

Explanation:

Acid salts are called salts resulting from the incomplete replacement of mobile hydrogen atoms by a metal cation, ammonium cation or alkyl ammonium.

In inorganic acids, which take place as part of the school curriculum, all hydrogen atoms are mobile, that is, they can be replaced by a metal.

Examples of acidic inorganic salts among the presented list is ammonium bicarbonate NH 4 HCO 3 - the product of replacing one of the two hydrogen atoms in carbonic acid with an ammonium cation.

In fact, an acid salt is a cross between a normal (medium) salt and an acid. In the case of NH 4 HCO 3 - the average between the normal salt (NH 4) 2 CO 3 and carbonic acid H 2 CO 3.

In organic substances, only hydrogen atoms that are part of carboxyl groups (-COOH) or hydroxyl groups of phenols (Ar-OH) can be replaced by metal atoms. That is, for example, sodium acetate CH 3 COONa, despite the fact that not all hydrogen atoms in its molecule are replaced by metal cations, is an average, not an acid salt (!). Hydrogen atoms in organic substances, attached directly to the carbon atom, are practically never able to be replaced by metal atoms, with the exception of hydrogen atoms in the triple C≡C bond.

Non-salt-forming oxides - oxides of non-metals that do not form salts with basic oxides or bases, that is, they either do not react with them at all (most often), or give a different product (not a salt) in reaction with them. It is often said that non-salt-forming oxides are oxides of non-metals that do not react with bases and basic oxides. However, for the detection of non-salt-forming oxides, this approach does not always work. So, for example, CO, being a non-salt-forming oxide, reacts with basic iron (II) oxide, but with the formation of a free metal rather than a salt:

CO + FeO = CO 2 + Fe

Non-salt-forming oxides from the school chemistry course include non-metal oxides in the oxidation state +1 and +2. In total, they are found in the USE 4 - these are CO, NO, N 2 O and SiO (I personally never met the last SiO in assignments).

Task number 6

From the proposed list of substances, select two substances, with each of which iron reacts without heating.

1. zinc chloride
2. copper(II) sulfate
3. concentrated nitric acid
4. dilute hydrochloric acid
5. aluminium oxide

Answer: 2; four

Zinc chloride is a salt, and iron is a metal. The metal reacts with the salt only if it is more reactive than the one in the salt. The relative activity of metals is determined by a series of metal activity (in other words, a series of metal stresses). Iron is located to the right of zinc in the activity series of metals, which means that it is less active and is not able to displace zinc from salt. That is, the reaction of iron with substance No. 1 does not go.

Copper (II) sulfate CuSO 4 will react with iron, since iron is located to the left of copper in the activity series, that is, it is a more active metal.

Concentrated nitric acid, as well as concentrated sulfuric acid, are not able to react with iron, aluminum and chromium without heating due to such a phenomenon as passivation: on the surface of these metals, under the action of these acids, an insoluble salt is formed without heating, which acts as a protective shell. However, when heated, this protective shell dissolves and the reaction becomes possible. Those. since it is indicated that there is no heating, the reaction of iron with conc. HNO 3 does not leak.

Hydrochloric acid, regardless of concentration, refers to non-oxidizing acids. Metals that are in the activity series to the left of hydrogen react with non-oxidizing acids with the release of hydrogen. Iron is one of these metals. Conclusion: the reaction of iron with hydrochloric acid proceeds.

In the case of a metal and a metal oxide, the reaction, as in the case of a salt, is possible if the free metal is more active than that which is part of the oxide. Fe, according to the activity series of metals, is less active than Al. This means that Fe does not react with Al 2 O 3.

Task number 7

From the proposed list, select two oxides that react with a solution of hydrochloric acid, but do not react with sodium hydroxide solution.

• 1. CO
• 2 SO 3
• 3. CuO
• 4. MgO
• 5. ZnO

Write down the numbers of the selected substances in the answer field.

Answer: 3; four

CO is a non-salt-forming oxide; it does not react with an aqueous solution of alkali.

(It should be remembered that, nevertheless, under harsh conditions - high pressure and temperature - it still reacts with solid alkali, forming formates - salts of formic acid.)

SO 3 - sulfur oxide (VI) - acid oxide, which corresponds to sulfuric acid. Acid oxides do not react with acids and other acid oxides. That is, SO 3 does not react with hydrochloric acid and reacts with a base - sodium hydroxide. Not suitable.

CuO - copper (II) oxide - is classified as an oxide with predominantly basic properties. Reacts with HCl and does not react with sodium hydroxide solution. Fits

MgO - magnesium oxide - is classified as a typical basic oxide. Reacts with HCl and does not react with sodium hydroxide solution. Fits

ZnO - an oxide with pronounced amphoteric properties - easily reacts with both strong bases and acids (as well as acidic and basic oxides). Not suitable.

Task number 8

• 1.KOH
• 2.HCl
• 3. Cu(NO 3) 2
• 4.K2SO3
• 5. Na 2 SiO 3

Answer: 4; 2

In the reaction between two salts of inorganic acids, gas is formed only when hot solutions of nitrites and ammonium salts are mixed due to the formation of thermally unstable ammonium nitrite. For example,

NH 4 Cl + KNO 2 \u003d t o \u003d\u003e N 2 + 2H 2 O + KCl

However, both nitrites and ammonium salts are not on the list.

This means that one of the three salts (Cu (NO 3) 2, K 2 SO 3 and Na 2 SiO 3) reacts with either an acid (HCl) or an alkali (NaOH).

Among the salts of inorganic acids, only ammonium salts emit gas when interacting with alkalis:

NH 4 + + OH \u003d NH 3 + H 2 O

Ammonium salts, as we have already said, are not on the list. The only option left is the interaction of the salt with the acid.

Salts among these substances include Cu(NO 3) 2, K 2 SO 3 and Na 2 SiO 3. The reaction of copper nitrate with hydrochloric acid does not proceed, because no gas, no precipitate, no low-dissociating substance (water or weak acid) is formed. Sodium silicate reacts with hydrochloric acid, however, due to the release of a white gelatinous precipitate of silicic acid, and not gas:

Na 2 SiO 3 + 2HCl \u003d 2NaCl + H 2 SiO 3 ↓

The last option remains - the interaction of potassium sulfite and hydrochloric acid. Indeed, as a result of the ion exchange reaction between sulfite and almost any acid, unstable sulfurous acid is formed, which instantly decomposes into colorless gaseous sulfur oxide (IV) and water.

Task number 9

• 1. KCl (solution)
• 2.K2O
• 3.H2
• 4. HCl (excess)
• 5. CO 2 (solution)

Write in the table the numbers of the selected substances under the corresponding letters.

Answer: 2; 5

CO 2 is an acidic oxide and must be treated with either a basic oxide or a base to convert it to a salt. Those. to obtain potassium carbonate from CO 2, it must be treated with either potassium oxide or potassium hydroxide. Thus, substance X is potassium oxide:

K 2 O + CO 2 \u003d K 2 CO 3

Potassium bicarbonate KHCO 3, like potassium carbonate, is a salt of carbonic acid, with the only difference being that the bicarbonate is a product of incomplete substitution of hydrogen atoms in carbonic acid. To obtain an acid salt from a normal (medium) salt, one must either act on it with the same acid that formed this salt, or else act on it with an acid oxide corresponding to this acid in the presence of water. Thus reactant Y is carbon dioxide. When it is passed through an aqueous solution of potassium carbonate, the latter turns into potassium bicarbonate:

K 2 CO 3 + H 2 O + CO 2 \u003d 2KHCO 3

Task number 10

Establish a correspondence between the reaction equation and the property of the nitrogen element that it exhibits in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.

Write in the table the numbers of the selected substances under the corresponding letters.

Answer: A-4; B-2; IN 2; G-1

A) NH 4 HCO 3 - salt, which includes the ammonium cation NH 4 +. In the ammonium cation, nitrogen always has an oxidation state of -3. As a result of the reaction, it turns into ammonia NH 3. Hydrogen almost always (except for its compounds with metals) has an oxidation state of +1. Therefore, for the ammonia molecule to be electrically neutral, nitrogen must have an oxidation state of -3. Thus, there is no change in the degree of nitrogen oxidation; it does not exhibit redox properties.

B) As already shown above, nitrogen in ammonia NH 3 has an oxidation state of -3. As a result of the reaction with CuO, ammonia is converted into a simple substance N 2. In any simple substance, the oxidation state of the element with which it is formed is equal to zero. Thus, the nitrogen atom loses its negative charge, and since electrons are responsible for the negative charge, this means that they are lost by the nitrogen atom as a result of the reaction. An element that loses some of its electrons in a reaction is called a reducing agent.

C) As a result of the reaction, NH 3 with an oxidation state of nitrogen equal to -3 turns into nitric oxide NO. Oxygen almost always has an oxidation state of -2. Therefore, in order for the nitric oxide molecule to be electrically neutral, the nitrogen atom must have an oxidation state of +2. This means that the nitrogen atom changed its oxidation state from -3 to +2 as a result of the reaction. This indicates the loss of 5 electrons by the nitrogen atom. That is, nitrogen, as in the case of B, is a reducing agent.

D) N 2 is a simple substance. In all simple substances, the element that forms them has an oxidation state of 0. As a result of the reaction, nitrogen is converted into lithium nitride Li3N. The only oxidation state of an alkali metal other than zero (any element has an oxidation state of 0) is +1. Thus, for the Li3N structural unit to be electrically neutral, nitrogen must have an oxidation state of -3. It turns out that as a result of the reaction, nitrogen acquired a negative charge, which means the addition of electrons. Nitrogen is the oxidizing agent in this reaction.

Task number 11

Establish a correspondence between the formula of a substance and the reagents, with each of which this substance can interact: for each position indicated by a letter, select the corresponding position indicated by a number.

SUBSTANCE FORMULA	REAGENTS
D) ZnBr 2 (solution)	1) AgNO 3, Na 3 PO 4, Cl 2 2) BaO, H 2 O, KOH 3) H 2, Cl 2, O 2 4) HBr, LiOH, CH 3 COOH 5) H 3 PO 4, BaCl 2, CuO

Write in the table the numbers of the selected substances under the corresponding letters.

Answer: A-3; B-2; AT 4; G-1

Explanation:

A) When hydrogen gas is passed through a sulfur melt, hydrogen sulfide H 2 S is formed:

H 2 + S \u003d t o \u003d\u003e H 2 S

When chlorine is passed over crushed sulfur at room temperature, sulfur dichloride is formed:

S + Cl 2 \u003d SCl 2

To pass the exam, you do not need to know exactly how sulfur reacts with chlorine and, accordingly, be able to write this equation. The main thing is to remember at a fundamental level that sulfur reacts with chlorine. Chlorine is a strong oxidizing agent, sulfur often exhibits a dual function - both oxidizing and reducing. That is, if a strong oxidizing agent acts on sulfur, which is molecular chlorine Cl 2, it will oxidize.

Sulfur burns with a blue flame in oxygen to form a gas with a pungent odor - sulfur dioxide SO 2:

B) SO 3 - sulfur oxide (VI) has pronounced acidic properties. For such oxides, the most characteristic reactions are interactions with water, as well as with basic and amphoteric oxides and hydroxides. In the list at number 2, we just see water, and the basic oxide BaO, and hydroxide KOH.

When an acidic oxide reacts with a basic oxide, a salt of the corresponding acid and a metal that is part of the basic oxide is formed. An acidic oxide corresponds to an acid in which the acid-forming element has the same oxidation state as in the oxide. The oxide SO 3 corresponds to sulfuric acid H 2 SO 4 (both there and there the oxidation state of sulfur is +6). Thus, when SO 3 interacts with metal oxides, sulfuric acid salts will be obtained - sulfates containing the sulfate ion SO 4 2-:

SO 3 + BaO = BaSO 4

When interacting with water, the acid oxide turns into the corresponding acid:

SO 3 + H 2 O \u003d H 2 SO 4

And when acid oxides interact with metal hydroxides, a salt of the corresponding acid and water are formed:

SO 3 + 2KOH \u003d K 2 SO 4 + H 2 O

C) Zinc hydroxide Zn (OH) 2 has typical amphoteric properties, that is, it reacts both with acidic oxides and acids, and with basic oxides and alkalis. In list 4, we see both acids - hydrobromic HBr and acetic, and alkali - LiOH. Recall that water-soluble metal hydroxides are called alkalis:

Zn(OH) 2 + 2HBr = ZnBr 2 + 2H 2 O

Zn (OH) 2 + 2CH 3 COOH \u003d Zn (CH 3 COO) 2 + 2H 2 O

Zn(OH) 2 + 2LiOH \u003d Li 2

D) Zinc bromide ZnBr 2 is a salt, soluble in water. For soluble salts, ion exchange reactions are the most common. A salt can react with another salt provided that both starting salts are soluble and a precipitate forms. Also ZnBr 2 contains bromide ion Br-. Metal halides are characterized by the fact that they are able to react with Hal 2 halogens, which are higher in the periodic table. In this way? the described types of reactions proceed with all substances of list 1:

ZnBr 2 + 2AgNO 3 \u003d 2AgBr + Zn (NO 3) 2

3ZnBr 2 + 2Na 3 PO 4 = Zn 3 (PO 4) 2 + 6NaBr

ZnBr 2 + Cl 2 = ZnCl 2 + Br 2

Task number 12

Establish a correspondence between the name of the substance and the class / group to which this substance belongs: for each position indicated by a letter, select the corresponding position indicated by a number.

Write in the table the numbers of the selected substances under the corresponding letters.

Answer: A-4; B-2; IN 1

Explanation:

A) Methylbenzene, also known as toluene, has the structural formula:

As you can see, the molecules of this substance consist only of carbon and hydrogen, therefore methylbenzene (toluene) refers to hydrocarbons

B) The structural formula of aniline (aminobenzene) is as follows:

As can be seen from the structural formula, the aniline molecule consists of an aromatic hydrocarbon radical (C 6 H 5 -) and an amino group (-NH 2), thus, aniline belongs to aromatic amines, i.e. correct answer 2.

C) 3-methylbutanal. The ending "al" indicates that the substance belongs to aldehydes. The structural formula of this substance:

Task number 13

From the proposed list, select two substances that are structural isomers of butene-1.

1. butane
2. cyclobutane
3. butin-2
4. butadiene-1,3
5. methylpropene

Write down the numbers of the selected substances in the answer field.

Answer: 2; 5

Explanation:

Isomers are substances that have the same molecular formula and different structural, i.e. Substances that differ in the order in which atoms are combined, but with the same composition of molecules.

Task number 14

From the proposed list, select two substances, the interaction of which with a solution of potassium permanganate will cause a change in the color of the solution.

1. cyclohexane
2. benzene
3. toluene
4. propane
5. propylene

Write down the numbers of the selected substances in the answer field.

Answer: 3; 5

Explanation:

Alkanes, as well as cycloalkanes with a ring size of 5 or more carbon atoms, are very inert and do not react with aqueous solutions of even strong oxidizing agents, such as, for example, potassium permanganate KMnO 4 and potassium dichromate K 2 Cr 2 O 7 . Thus, options 1 and 4 disappear - when cyclohexane or propane is added to an aqueous solution of potassium permanganate, a color change will not occur.

Among the hydrocarbons of the homologous series of benzene, only benzene is passive to the action of aqueous solutions of oxidizing agents, all other homologues are oxidized, depending on the medium, either to carboxylic acids or to their corresponding salts. Thus, option 2 (benzene) is eliminated.

The correct answers are 3 (toluene) and 5 (propylene). Both substances decolorize the purple solution of potassium permanganate due to the reactions taking place:

CH 3 -CH=CH 2 + 2KMnO 4 + 2H 2 O → CH 3 -CH(OH)–CH 2 OH + 2MnO 2 + 2KOH

Task number 15

From the proposed list, select two substances with which formaldehyde reacts.

• 1. Cu
• 2. N 2
• 3.H2
• 4. Ag 2 O (NH 3 solution)
• 5. CH 3 DOS 3

Write down the numbers of the selected substances in the answer field.

Answer: 3; four

Explanation:

Formaldehyde belongs to the class of aldehydes - oxygen-containing organic compounds that have an aldehyde group at the end of the molecule:

Typical reactions of aldehydes are oxidation and reduction reactions proceeding along the functional group.

Among the list of responses for formaldehyde, reduction reactions are typical, where hydrogen is used as a reducing agent (cat. - Pt, Pd, Ni), and oxidation - in this case, the silver mirror reaction.

When reduced with hydrogen on a nickel catalyst, formaldehyde is converted to methanol:

The silver mirror reaction is the reduction of silver from an ammonia solution of silver oxide. When dissolved in an aqueous solution of ammonia, silver oxide turns into a complex compound - diammine silver (I) OH hydroxide. After the addition of formaldehyde, a redox reaction occurs in which silver is reduced:

Task number 16

From the proposed list, select two substances with which methylamine reacts.

1. propane
2. chloromethane
3. hydrogen
4. sodium hydroxide
5. hydrochloric acid

Write down the numbers of the selected substances in the answer field.

Answer: 2; 5

Explanation:

Methylamine is the simplest organic compound of the amine class. A characteristic feature of amines is the presence of a lone electron pair on the nitrogen atom, as a result of which amines exhibit the properties of bases and act as nucleophiles in reactions. Thus, in this regard, from the proposed answers, methylamine as a base and nucleophile reacts with chloromethane and hydrochloric acid:

CH 3 NH 2 + CH 3 Cl → (CH 3) 2 NH 2 + Cl -

CH 3 NH 2 + HCl → CH 3 NH 3 + Cl -

Task number 17

The following scheme of transformations of substances is given:

Determine which of the given substances are substances X and Y.

• 1.H2
• 2. CuO
• 3. Cu(OH) 2
• 4. NaOH (H 2 O)
• 5. NaOH (alcohol)

Write in the table the numbers of the selected substances under the corresponding letters.

Answer: 4; 2

Explanation:

One of the reactions for obtaining alcohols is the hydrolysis of haloalkanes. Thus, ethanol can be obtained from chloroethane by acting on the latter with an aqueous solution of alkali - in this case, NaOH.

CH 3 CH 2 Cl + NaOH (aq.) → CH 3 CH 2 OH + NaCl

The next reaction is the oxidation reaction of ethyl alcohol. The oxidation of alcohols is carried out on a copper catalyst or using CuO:

Task number 18

Establish a correspondence between the name of the substance and the product that is mainly formed during the interaction of this substance with bromine: for each position indicated by a letter, select the corresponding position indicated by a number.

Answer: 5; 2; 3; 6

Explanation:

For alkanes, the most characteristic reactions are free radical substitution reactions, during which a hydrogen atom is replaced by a halogen atom. Thus, by brominating ethane, one can obtain bromoethane, and by brominating isobutane, 2-bromoisobutane can be obtained:

Since the small cycles of cyclopropane and cyclobutane molecules are unstable, during bromination the cycles of these molecules are opened, thus the addition reaction proceeds:

In contrast to the cycles of cyclopropane and cyclobutane, the cyclohexane cycle is large, resulting in the replacement of a hydrogen atom by a bromine atom:

Task #19

Establish a correspondence between the reacting substances and the carbon-containing product that is formed during the interaction of these substances: for each position indicated by a letter, select the corresponding position indicated by a number.

Write in the table the selected numbers under the corresponding letters.

Answer: 5; four; 6; 2

Task number 20

From the proposed list of types of reactions, select two types of reactions, which include the interaction of alkali metals with water.

1. catalytic
2. homogeneous
3. irreversible
4. redox
5. neutralization reaction

Write down the numbers of the selected types of reactions in the answer field.

Answer: 3; four

Alkali metals (Li, Na, K, Rb, Cs, Fr) are located in the main subgroup of group I of the table D.I. Mendeleev and are reducing agents, easily donating an electron located at the outer level.

If we denote the alkali metal with the letter M, then the reaction of the alkali metal with water will look like this:

2M + 2H 2 O → 2MOH + H 2

Alkali metals are very active towards water. The reaction proceeds violently with the release of a large amount of heat, is irreversible and does not require the use of a catalyst (non-catalytic) - a substance that accelerates the reaction and is not part of the reaction products. It should be noted that all highly exothermic reactions do not require the use of a catalyst and proceed irreversibly.

Since metal and water are substances that are in different states of aggregation, this reaction proceeds at the interface, therefore, it is heterogeneous.

The type of this reaction is substitution. Reactions between inorganic substances are classified as substitution reactions if a simple substance interacts with a complex one and as a result other simple and complex substances are formed. (The neutralization reaction proceeds between an acid and a base, as a result of which these substances exchange their constituents and form a salt and a low-dissociating substance).

As mentioned above, alkali metals are reducing agents, donating an electron from the outer layer, therefore, the reaction is redox.

Task number 21

From the proposed list of external influences, select two influences that lead to a decrease in the rate of the reaction of ethylene with hydrogen.

1. temperature drop
2. increase in ethylene concentration
3. use of a catalyst
4. decrease in hydrogen concentration
5. pressure increase in the system

Write in the answer field the numbers of the selected external influences.

Answer: 1; four

The following factors influence the rate of a chemical reaction: changes in temperature and concentration of reagents, as well as the use of a catalyst.

According to Van't Hoff's empirical rule, for every 10 degrees increase in temperature, the rate constant of a homogeneous reaction increases by 2-4 times. Therefore, a decrease in temperature also leads to a decrease in the reaction rate. The first answer is correct.

As noted above, the reaction rate is also affected by a change in the concentration of reagents: if the concentration of ethylene is increased, the reaction rate will also increase, which does not meet the requirements of the problem. And a decrease in the concentration of hydrogen - the initial component, on the contrary, reduces the reaction rate. Therefore, the second option is not suitable, but the fourth one is.

A catalyst is a substance that speeds up the rate of a chemical reaction but is not part of the products. The use of a catalyst accelerates the ethylene hydrogenation reaction, which also does not correspond to the condition of the problem, and therefore is not the right answer.

When ethylene reacts with hydrogen (on Ni, Pd, Pt catalysts), ethane is formed:

CH 2 \u003d CH 2 (g) + H 2 (g) → CH 3 -CH 3 (g)

All components involved in the reaction and the product are gaseous substances, therefore, the pressure in the system will also affect the reaction rate. From two volumes of ethylene and hydrogen, one volume of ethane is formed, therefore, the reaction proceeds to a decrease in pressure in the system. By increasing the pressure, we will speed up the reaction. The fifth answer does not fit.

Task #22

Establish a correspondence between the formula of salt and the products of electrolysis aqueous solution this salt, which stood out on inert electrodes: to each position,

SALT FORMULA

ELECTROLYSIS PRODUCTS

Write in the table the selected numbers under the corresponding letters.

Answer: 1; four; 3; 2

Electrolysis is a redox process that occurs on electrodes when a direct electric current passes through an electrolyte solution or melt. At the cathode, the reduction occurs predominantly of those cations that have the highest oxidizing activity. At the anode, those anions are oxidized first of all, which have the greatest reduction ability.

Electrolysis of aqueous solution

1) The process of electrolysis of aqueous solutions on the cathode does not depend on the material of the cathode, but depends on the position of the metal cation in the electrochemical series of voltages.

For cations in a row

Li + - Al 3+ recovery process:

2H 2 O + 2e → H 2 + 2OH - (H 2 is released at the cathode)

Zn 2+ - Pb 2+ recovery process:

Me n + + ne → Me 0 and 2H 2 O + 2e → H 2 + 2OH - (H 2 and Me will be released at the cathode)

Cu 2+ - Au 3+ reduction process Me n + + ne → Me 0 (Me is released at the cathode)

2) The process of electrolysis of aqueous solutions at the anode depends on the material of the anode and on the nature of the anion. If the anode is insoluble, i.e. inert (platinum, gold, coal, graphite), the process will depend only on the nature of the anions.

For anions F -, SO 4 2-, NO 3 -, PO 4 3-, OH - the oxidation process:

4OH - - 4e → O 2 + 2H 2 O or 2H 2 O - 4e → O 2 + 4H + (oxygen is released at the anode) halide ions (except F-) oxidation process 2Hal - - 2e → Hal 2 (free halogens are released ) organic acids oxidation process:

2RCOO - - 2e → R-R + 2CO 2

The overall electrolysis equation is:

A) Na 3 PO 4 solution

2H 2 O → 2H 2 (at the cathode) + O 2 (at the anode)

B) KCl solution

2KCl + 2H 2 O → H 2 (at the cathode) + 2KOH + Cl 2 (at the anode)

C) CuBr2 solution

CuBr 2 → Cu (at the cathode) + Br 2 (at the anode)

D) Cu(NO3)2 solution

2Cu(NO 3) 2 + 2H 2 O → 2Cu (at the cathode) + 4HNO 3 + O 2 (at the anode)

Task #23

Establish a correspondence between the name of the salt and the ratio of this salt to hydrolysis: for each position indicated by a letter, select the corresponding position indicated by a number.

Write in the table the selected numbers under the corresponding letters.

Answer: 1; 3; 2; four

Hydrolysis of salts - the interaction of salts with water, leading to the addition of the hydrogen cation H + of the water molecule to the anion of the acid residue and (or) the hydroxyl group OH - of the water molecule to the metal cation. Salts formed by cations corresponding to weak bases and anions corresponding to weak acids undergo hydrolysis.

A) Ammonium chloride (NH 4 Cl) - a salt formed by strong hydrochloric acid and ammonia (weak base), undergoes hydrolysis by the cation.

NH 4 Cl → NH 4 + + Cl -

NH 4 + + H 2 O → NH 3 H 2 O + H + (formation of ammonia dissolved in water)

The solution medium is acidic (pH< 7).

B) Potassium sulfate (K 2 SO 4) - a salt formed by strong sulfuric acid and potassium hydroxide (alkali, i.e. strong base), does not undergo hydrolysis.

K 2 SO 4 → 2K + + SO 4 2-

C) Sodium carbonate (Na 2 CO 3) - a salt formed by a weak carbonic acid and sodium hydroxide (an alkali, i.e. a strong base), undergoes anion hydrolysis.

CO 3 2- + H 2 O → HCO 3 - + OH - (formation of a weakly dissociating hydrocarbonate ion)

The solution is alkaline (pH > 7).

D) Aluminum sulfide (Al 2 S 3) - a salt formed by a weak hydrosulfide acid and aluminum hydroxide (weak base), undergoes complete hydrolysis with the formation of aluminum hydroxide and hydrogen sulfide:

Al 2 S 3 + 6H 2 O → 2Al(OH) 3 + 3H 2 S

The solution medium is close to neutral (pH ~ 7).

Task #24

Establish a correspondence between the equation of a chemical reaction and the direction of displacement of chemical equilibrium with increasing pressure in the system: for each position indicated by a letter, select the corresponding position indicated by a number.

REACTION EQUATION A) N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) B) 2H 2 (g) + O 2 (g) ↔ 2H 2 O (g) C) H 2 (g) + Cl 2 (g) ↔ 2HCl (g) D) SO 2 (g) + Cl 2 (g) ↔ SO 2 Cl 2 (g)

DIRECTION OF SHIFT OF CHEMICAL EQUILIBRIUM 1) shifts towards a direct reaction 2) shifts towards the back reaction 3) there is no shift in equilibrium

Write in the table the selected numbers under the corresponding letters.

Answer: A-1; B-1; AT 3; G-1

A reaction is in chemical equilibrium when the rate of the forward reaction is equal to the rate of the reverse. The shift of equilibrium in the desired direction is achieved by changing the reaction conditions.

Factors that determine the position of equilibrium:

- pressure: an increase in pressure shifts the equilibrium towards a reaction leading to a decrease in volume (conversely, a decrease in pressure shifts the equilibrium towards a reaction leading to an increase in volume)

- temperature: an increase in temperature shifts the equilibrium towards an endothermic reaction (conversely, a decrease in temperature shifts the equilibrium towards an exothermic reaction)

- concentrations of starting substances and reaction products: an increase in the concentration of the starting substances and the removal of products from the reaction sphere shift the equilibrium towards the direct reaction (on the contrary, a decrease in the concentration of the starting substances and an increase in the reaction products shift the equilibrium towards the reverse reaction)

- Catalysts do not affect the equilibrium shift, but only accelerate its achievement

A) In the first case, the reaction proceeds with a decrease in volume, since V (N 2) + 3V (H 2) > 2V (NH 3). By increasing the pressure in the system, the equilibrium will shift to the side with a smaller volume of substances, therefore, in the forward direction (in the direction of the direct reaction).

B) In the second case, the reaction also proceeds with a decrease in volume, since 2V (H 2) + V (O 2) > 2V (H 2 O). By increasing the pressure in the system, the equilibrium will also shift in the direction of the direct reaction (in the direction of the product).

C) In the third case, the pressure does not change during the reaction, because V (H 2) + V (Cl 2) \u003d 2V (HCl), so there is no equilibrium shift.

D) In ​​the fourth case, the reaction also proceeds with a decrease in volume, since V (SO 2) + V (Cl 2) > V (SO 2 Cl 2). By increasing the pressure in the system, the equilibrium will shift towards the formation of the product (direct reaction).

Task #25

Establish a correspondence between the formulas of substances and a reagent with which you can distinguish between their aqueous solutions: for each position indicated by a letter, select the corresponding position indicated by a number.

SUBSTANCE FORMULA A) HNO 3 and H 2 O C) NaCl and BaCl 2 D) AlCl 3 and MgCl 2

Write in the table the selected numbers under the corresponding letters.

Answer: A-1; B-3; AT 3; G-2

A) Nitric acid and water can be distinguished using salt - calcium carbonate CaCO 3. Calcium carbonate does not dissolve in water, and when interacting with nitric acid forms a soluble salt - calcium nitrate Ca (NO 3) 2, while the reaction is accompanied by the release of colorless carbon dioxide:

CaCO 3 + 2HNO 3 → Ca(NO 3) 2 + CO 2 + H 2 O

B) Potassium chloride KCl and alkali NaOH can be distinguished by a solution of copper (II) sulfate.

When copper (II) sulfate interacts with KCl, the exchange reaction does not proceed, the solution contains K +, Cl -, Cu 2+ and SO 4 2- ions, which do not form poorly dissociating substances with each other.

When copper (II) sulfate interacts with NaOH, an exchange reaction occurs, as a result of which copper (II) hydroxide precipitates (blue base).

C) Sodium chloride NaCl and barium BaCl 2 are soluble salts, which can also be distinguished by a solution of copper (II) sulfate.

When copper (II) sulfate interacts with NaCl, the exchange reaction does not proceed, the solution contains Na +, Cl -, Cu 2+ and SO 4 2- ions, which do not form poorly dissociating substances with each other.

When copper (II) sulfate interacts with BaCl 2, an exchange reaction occurs, as a result of which barium sulfate BaSO 4 precipitates.

D) Aluminum chloride AlCl 3 and magnesium MgCl 2 dissolve in water and behave differently when interacting with potassium hydroxide. Magnesium chloride with alkali forms a precipitate:

MgCl 2 + 2KOH → Mg(OH) 2 ↓ + 2KCl

When alkali interacts with aluminum chloride, a precipitate first forms, which then dissolves to form a complex salt - potassium tetrahydroxoaluminate:

AlCl 3 + 4KOH → K + 3KCl

Task #26

Establish a correspondence between the substance and its scope: for each position indicated by a letter, select the corresponding position indicated by a number.

Write in the table the selected numbers under the corresponding letters.

Answer: A-4; B-2; AT 3; G-5

A) Ammonia is the most important product of the chemical industry, its production is more than 130 million tons per year. Ammonia is mainly used in the production of nitrogen fertilizers (ammonium nitrate and sulfate, urea), medicines, explosives, nitric acid, and soda. Among the proposed answers, the area of ​​application of ammonia is the production of fertilizers (Fourth answer option).

B) Methane is the simplest hydrocarbon, the most thermally stable representative of a number of saturated compounds. It is widely used as a domestic and industrial fuel, as well as a raw material for industry (Second answer). Methane is 90-98% a component of natural gas.

C) Rubbers are materials that are obtained by polymerization of compounds with conjugated double bonds. Isoprene just belongs to this type of compounds and is used to obtain one of the types of rubbers:

D) Low molecular weight alkenes are used to make plastics, in particular ethylene is used to make a plastic called polyethylene:

n CH 2 \u003d CH 2 → (-CH 2 -CH 2 -) n

Task number 27

Calculate the mass of potassium nitrate (in grams) that should be dissolved in 150 g of a solution with a mass fraction of this salt of 10% to obtain a solution with a mass fraction of 12%. (Write down the number to tenths.)

Answer: 3.4 g

Explanation:

Let x g be the mass of potassium nitrate, which is dissolved in 150 g of the solution. Calculate the mass of potassium nitrate dissolved in 150 g of solution:

m(KNO 3) \u003d 150 g 0.1 \u003d 15 g

In order for the mass fraction of salt to be 12%, x g of potassium nitrate was added. In this case, the mass of the solution was (150 + x) g. We write the equation in the form:

(Write down the number to tenths.)

Answer: 14.4 g

Explanation:

As a result of the complete combustion of hydrogen sulfide, sulfur dioxide and water are formed:

2H 2 S + 3O 2 → 2SO 2 + 2H 2 O

A consequence of Avogadro's law is that the volumes of gases under the same conditions are related to each other in the same way as the number of moles of these gases. Thus, according to the reaction equation:

ν(O 2) = 3/2ν(H 2 S),

therefore, the volumes of hydrogen sulfide and oxygen are related to each other in exactly the same way:

V (O 2) \u003d 3 / 2V (H 2 S),

V (O 2) \u003d 3/2 6.72 l \u003d 10.08 l, hence V (O 2) \u003d 10.08 l / 22.4 l / mol \u003d 0.45 mol

Calculate the mass of oxygen required for the complete combustion of hydrogen sulfide:

m(O 2) \u003d 0.45 mol 32 g / mol \u003d 14.4 g

Task number 30

Using the electron balance method, write the equation for the reaction:

Na 2 SO 3 + ... + KOH → K 2 MnO 4 + ... + H 2 O

Determine the oxidizing agent and reducing agent.

Mn +7 + 1e → Mn +6 │2 reduction reaction

S +4 − 2e → S +6 │1 oxidation reaction

Mn +7 (KMnO 4) - oxidizing agent, S +4 (Na 2 SO 3) - reducing agent

Na 2 SO 3 + 2KMnO 4 + 2KOH → 2K 2 MnO 4 + Na 2 SO 4 + H 2 O

Task number 31

Iron was dissolved in hot concentrated sulfuric acid. The resulting salt was treated with an excess of sodium hydroxide solution. The brown precipitate formed was filtered off and dried. The resulting substance was heated with iron.

Write the equations for the four described reactions.

1) Iron, like aluminum and chromium, does not react with concentrated sulfuric acid, becoming covered with a protective oxide film. The reaction occurs only when heated with the release of sulfur dioxide:

2Fe + 6H 2 SO 4 → Fe 2 (SO 4) 2 + 3SO 2 + 6H 2 O (on heating)

2) Iron (III) sulfate - a salt soluble in water, enters into an exchange reaction with alkali, as a result of which iron (III) hydroxide precipitates (brown compound):

Fe 2 (SO 4) 3 + 3NaOH → 2Fe(OH) 3 ↓ + 3Na 2 SO 4

3) Insoluble metal hydroxides decompose upon calcination to the corresponding oxides and water:

2Fe(OH) 3 → Fe 2 O 3 + 3H 2 O

4) When iron (III) oxide is heated with metallic iron, iron (II) oxide is formed (iron in the FeO compound has an intermediate oxidation state):

Fe 2 O 3 + Fe → 3FeO (on heating)

Task #32

Write the reaction equations that can be used to carry out the following transformations:

When writing reaction equations, use the structural formulas of organic substances.

1) Intramolecular dehydration occurs at temperatures above 140 o C. This occurs as a result of the elimination of a hydrogen atom from the carbon atom of the alcohol, located one through to the alcohol hydroxyl (in the β-position).

CH 3 -CH 2 -CH 2 -OH → CH 2 \u003d CH-CH 3 + H 2 O (conditions - H 2 SO 4, 180 o C)

Intermolecular dehydration proceeds at a temperature below 140 o C under the action of sulfuric acid and ultimately comes down to the elimination of one water molecule from two alcohol molecules.

2) Propylene refers to unsymmetrical alkenes. When hydrogen halides and water are added, a hydrogen atom is attached to a carbon atom at a multiple bond associated with a large number of hydrogen atoms:

CH 2 \u003d CH-CH 3 + HCl → CH 3 -CHCl-CH 3

3) Acting with an aqueous solution of NaOH on 2-chloropropane, the halogen atom is replaced by a hydroxyl group:

CH 3 -CHCl-CH 3 + NaOH (aq.) → CH 3 -CHOH-CH 3 + NaCl

4) Propylene can be obtained not only from propanol-1, but also from propanol-2 by the reaction of intramolecular dehydration at temperatures above 140 o C:

CH 3 -CH(OH)-CH 3 → CH 2 \u003d CH-CH 3 + H 2 O (conditions H 2 SO 4, 180 o C)

5) In an alkaline environment, acting with a dilute aqueous solution of potassium permanganate, hydroxylation of alkenes occurs with the formation of diols:

3CH 2 \u003d CH-CH 3 + 2KMnO 4 + 4H 2 O → 3HOCH 2 -CH (OH) -CH 3 + 2MnO 2 + 2KOH

Task number 33

Determine the mass fractions (in%) of iron (II) sulfate and aluminum sulfide in the mixture, if during the treatment of 25 g of this mixture with water a gas was released that completely reacted with 960 g of a 5% solution of copper (II) sulfate.

In response, write down the reaction equations that are indicated in the condition of the problem, and give all the necessary calculations (indicate the units of measurement of the required physical quantities).

Answer: ω(Al 2 S 3) = 40%; ω(CuSO 4) = 60%

When a mixture of iron (II) sulfate and aluminum sulfide is treated with water, the sulfate is simply dissolved, and the sulfide is hydrolyzed to form aluminum (III) hydroxide and hydrogen sulfide:

Al 2 S 3 + 6H 2 O → 2Al(OH) 3 ↓ + 3H 2 S (I)

When hydrogen sulfide is passed through a solution of copper (II) sulfate, copper (II) sulfide precipitates:

CuSO 4 + H 2 S → CuS↓ + H 2 SO 4 (II)

Calculate the mass and amount of substance of dissolved copper(II) sulfate:

m (CuSO 4) \u003d m (p-ra) ω (CuSO 4) \u003d 960 g 0.05 \u003d 48 g; ν (CuSO 4) \u003d m (CuSO 4) / M (CuSO 4) \u003d 48 g / 160 g \u003d 0.3 mol

According to the reaction equation (II) ν (CuSO 4) = ν (H 2 S) = 0.3 mol, and according to the reaction equation (III) ν (Al 2 S 3) = 1/3ν (H 2 S) = 0, 1 mol

Calculate the masses of aluminum sulfide and copper (II) sulfate:

m(Al 2 S 3) \u003d 0.1 mol 150 g / mol \u003d 15 g; m(CuSO4) = 25 g - 15 g = 10 g

ω (Al 2 S 3) \u003d 15 g / 25 g 100% \u003d 60%; ω (CuSO 4) \u003d 10 g / 25 g 100% \u003d 40%

Task number 34

When burning a sample of some organic compound weighing 14.8 g, 35.2 g of carbon dioxide and 18.0 g of water were obtained.

It is known that the relative hydrogen vapor density of this substance is 37. In the course of studying the chemical properties of this substance, it was found that when this substance interacts with copper (II) oxide, a ketone is formed.

Based on these conditions of the assignment:

1) make the calculations necessary to establish the molecular formula of organic matter (indicate the units of measurement of the required physical quantities);

2) write down the molecular formula of the original organic matter;

3) make a structural formula of this substance, which unambiguously reflects the order of bonding of atoms in its molecule;

4) write the equation for the reaction of this substance with copper(II) oxide using the structural formula of the substance.

Secondary general education

Getting ready for the USE-2018 in chemistry: analysis of the demo

We bring to your attention an analysis of the demo version of the USE 2018 in chemistry. This article contains explanations and detailed algorithms for solving tasks. To help prepare for the exam, we recommend our selection of reference books and manuals, as well as several articles on a topical topic published earlier.

Exercise 1

Determine the atoms of which of the elements indicated in the row in the ground state have four electrons at the external energy level.

1) Na
2) K
3) Si
4) Mg
5)C

Answer: The Periodic Table of Chemical Elements is a graphical representation of the Periodic Law. It consists of periods and groups. A group is a vertical column of chemical elements, consists of the main and secondary subgroups. If the element is in the main subgroup of a certain group, then the group number indicates the number of electrons in the last layer. Therefore, in order to answer this question, it is necessary to open the periodic table and see which elements from those presented in the task are located in the same group. We come to the conclusion that such elements are: Si and C, therefore the answer will be: 3; 5.

Task 2

Of the chemical elements listed in the series

1) Na
2) K
3) Si
4) Mg
5)C

select three elements that are in the same period in the Periodic Table of Chemical Elements of D.I. Mendeleev.

Arrange the elements in ascending order of their metallic properties.

Write in the answer field the numbers of the selected chemical elements in the desired sequence.

Answer: The Periodic Table of Chemical Elements is a graphical representation of the Periodic Law. It consists of periods and groups. A period is a horizontal row of chemical elements arranged in order of increasing electronegativity, which means decreasing metallic properties and strengthening non-metallic ones. Each period (with the exception of the first) begins with an active metal, which is called an alkali, and ends with an inert element, i.e. an element that does not form chemical compounds with other elements (with rare exceptions).

Looking at the table of chemical elements, we note that from the data in the element task, Na, Mg and Si are located in the 3rd period. Next, you need to arrange these elements in ascending order of metallic properties. From the above, we determine if the metallic properties decrease from left to right, then they increase on the contrary, from right to left. Therefore, the correct answers will be 3; four; one.

Task 3

From among the elements indicated in the row

1) Na
2) K
3) Si
4) Mg
5)C

choose the two elements that exhibit the lowest oxidation state -4.

Answer: The highest oxidation state of a chemical element in a compound is numerically equal to the number of the group in which the chemical element is located with a plus sign. If an element is located in group 1, then its highest oxidation state is +1, in the second group +2, and so on. The lowest oxidation state of a chemical element in compounds is 8 (the highest oxidation state that a chemical element can exhibit in a compound) minus the group number, with a minus sign. For example, the element is in the 5th group, the main subgroup; therefore, its highest oxidation state in compounds will be +5; the lowest oxidation state, respectively, 8 - 5 \u003d 3 with a minus sign, i.e. -3. For elements of 4 periods, the highest valency is +4, and the lowest is -4. Therefore, we are looking for two elements located in the 4th group of the main subgroup from the list of data elements in the task. This will be the C and Si numbers of the correct answer 3; 5.

Task 4

From the proposed list, select two compounds in which there is an ionic bond.

1) Ca(ClO 2) 2
2) HClO 3
3) NH4Cl
4) HClO 4
5) Cl 2 O 7

Answer: Under chemical bond understand such an interaction of atoms that binds them into molecules, ions, radicals, crystals. There are four types of chemical bonds: ionic, covalent, metallic and hydrogen.

Ionic bond - a bond resulting from the electrostatic attraction of oppositely charged ions (cations and anions), in other words, between a typical metal and a typical non-metal; those. elements with very different electronegativity. (> 1.7 on the Pauling scale). An ionic bond is present in compounds containing metals of groups 1 and 2 of the main subgroups (with the exception of Mg and Be) and typical non-metals; oxygen and elements of the 7th group of the main subgroup. The exception is ammonium salts, they do not contain a metal atom, instead an ion, but in ammonium salts between the ammonium ion and the acid residue, the bond is also ionic. Therefore, the correct answers will be 1; 3.

Task 5

Establish a correspondence between the formula of a substance and the classes / group to which this substance belongs: for each position indicated by a letter, select the corresponding position indicated by a number.

Write in the table the selected numbers under the corresponding letters.

Answer:

Answer: To answer this question, we must remember what oxides and salts are. Salts are complex substances consisting of metal ions and acid residue ions. The exception is ammonium salts. These salts have an ammonium ion instead of metal ions. Salts are medium, acidic, double, basic and complex. Medium salts are products of the complete replacement of the hydrogen of an acid with a metal or an ammonium ion; for example:

H 2 SO 4 + 2Na \u003d H 2 + Na 2 SO 4 .

This salt is average. Acid salts are the product of incomplete replacement of the hydrogen of the salt with a metal; for example:

2H 2 SO 4 + 2Na \u003d H 2 + 2 NaHSO 4 .

This salt is acidic. Now let's look at our task. It contains two salts: NH 4 HCO 3 and KF. The first salt is acidic because it is the product of incomplete hydrogen replacement in the acid. Therefore, in the plate with the answer under the letter "A" we put the number 4; the other salt (KF) does not contain hydrogen between the metal and the acid residue, therefore, in the plate with the answer under the letter “B”, we put the number 1. Oxides are a binary compound that includes oxygen. It is in second place and exhibits an oxidation state of -2. Oxides are basic (i.e. metal oxides, for example Na 2 O, CaO - they correspond to bases; NaOH and Ca (OH) 2), acidic (i.e. oxides of non-metals P 2 O 5, SO 3 - they correspond to acids ; H 3 PO 4 and H 2 SO 4), amphoteric (oxides, which, depending on the circumstances, may exhibit basic and acidic properties - Al 2 O 3, ZnO) and non-salt-forming. These are non-metal oxides that exhibit neither basic, nor acidic, nor amphoteric properties; these are CO, N 2 O, NO. Therefore, NO oxide is a non-salt-forming oxide, so in the answer plate under the letter “B” we put the number 3. And the completed table will look like this:

Answer:

Task 6

From the proposed list, select two substances, with each of which iron reacts without heating.

1) calcium chloride (solution)
2) copper (II) sulfate (solution)
3) concentrated nitric acid
4) dilute hydrochloric acid
5) aluminum oxide

Answer: Iron is an active metal. Reacts with chlorine, carbon and other non-metals when heated:

2Fe + 3Cl 2 = 2FeCl 3

Displaces from salt solutions metals that are in the electrochemical series of voltages to the right of iron:

For example:

Fe + CuSO 4 \u003d FeSO 4 + Cu

It dissolves in dilute sulfuric and hydrochloric acids with the release of hydrogen,

Fe + 2НCl \u003d FeCl 2 + H 2

with nitric acid solution

Fe + 4HNO 3 \u003d Fe (NO 3) 3 + NO + 2H 2 O.

Concentrated sulfuric and hydrochloric acid do not react with iron under normal conditions, they passivate it:

Based on this, the correct answers will be: 2; four.

Task 7

A strong acid X was added to one of the test tubes with a precipitate of aluminum hydroxide, and a solution of substance Y was added to the other. As a result, the precipitate was observed to dissolve in each of the test tubes. From the proposed list, select substances X and Y that can enter into the described reactions.

1) hydrobromic acid.
2) sodium hydrosulfide.
3) hydrosulfide acid.
4) potassium hydroxide.
5) ammonia hydrate.

Write in the table the numbers of the selected substances under the corresponding letters.

Answer: Aluminum hydroxide is an amphoteric base, therefore it can interact with solutions of acids and alkalis:

1) Interaction with an acid solution: Al(OH) 3 + 3HBr = AlCl 3 + 3H 2 O.

In this case, the precipitate of aluminum hydroxide dissolves.

2) Interaction with alkalis: 2Al(OH) 3 + Ca(OH) 2 = Ca 2.

In this case, the aluminum hydroxide precipitate also dissolves.

Answer:

Task 8

Establish a correspondence between the formula of a substance and the reagents, with each of which this substance can interact: for each position indicated by a letter, select the corresponding position indicated by a number

SUBSTANCE FORMULA	REAGENTS
D) ZnBr 2 (solution)	1) AgNO 3, Na 3 PO 4, Cl 2 2) BaO, H 2 O, KOH 3) H 2, Cl 2, O 2 4) НBr, LiOH, CH 3 COOH (solution) 5) H 3 PO 4 (solution), BaCl 2, CuO

Answer: Under the letter A is sulfur (S). As a simple substance, sulfur can enter into redox reactions. Most reactions occur with simple substances, metals and non-metals. It is oxidized with solutions of concentrated sulfuric and hydrochloric acids. Interacts with alkalis. Of all the reagents located under the numbers 1-5, simple substances under the number 3 are most suitable for the properties described above.

S + Cl 2 \u003d SCl 2

The next substance is SO 3, letter B. Sulfur oxide VI is a complex substance, acidic oxide. This oxide contains sulfur in the +6 oxidation state. This is the highest oxidation state of sulfur. Therefore, SO 3 will react, as an oxidizing agent, with simple substances, for example, with phosphorus, with complex substances, for example, with KI, H 2 S. At the same time, its oxidation state can drop to +4, 0 or -2, it also enters in reaction without changing the oxidation state with water, metal oxides and hydroxides. Based on this, SO 3 will react with all reagents under the number 2, that is:

SO 3 + BaO = BaSO 4

SO 3 + H 2 O \u003d H 2 SO 4

SO 3 + 2KOH \u003d K 2 SO 4 + H 2 O

Zn (OH) 2 - amphoteric hydroxide is located under the letter B. It has unique properties - it reacts with both acids and alkalis. Therefore, from all the reagents presented, you can safely choose the reagents under the number 4.

Zn(OH) 2 + HBr = ZnBr 2 + H 2 O

Zn (OH) 2 + LiOH \u003d Li 2

Zn(OH) 2 + CH 3 COOH = (CH 3 COO) 2 Zn + H 2 O

And finally, under the letter G is the substance ZnBr 2 - salt, zinc bromide. Salts react with acids, alkalis, other salts, and salts of anoxic acids, like this salt, can interact with non-metals. In this case, the most active halogens (Cl or F) can displace the less active ones (Br and I) from solutions of their salts. These criteria are met by reagents under the number 1.

ZnBr 2 + 2AgNO 3 \u003d 2AgBr + Zn (NO 3) 2

3ZnBr 2 + 2Na 3 PO 4 = Zn 3 (PO 4) 2 + 6NaBr

ZnBr 2 + Cl 2 = ZnCl 2 + Br 2

The response options are as follows:

The new guide contains all theoretical material in the course of chemistry required to pass the exam. It includes all elements of the content, checked by control and measuring materials, and helps to generalize and systematize knowledge and skills for the course of the secondary (complete) school. The theoretical material is presented in a concise and accessible form. Each topic is accompanied by examples of test tasks. Practical tasks correspond to the USE format. Answers to the tests are given at the end of the manual. The manual is addressed to schoolchildren, applicants and teachers.

Task 9

Establish a correspondence between the starting substances that enter into the reaction and the products of this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.

STARTING SUBSTANCES	REACTION PRODUCTS
A) Mg and H 2 SO 4 (conc) B) MgO and H 2 SO 4 B) S and H 2 SO 4 (conc) D) H 2 S and O 2 (ex.)	1) MgSO 4 and H 2 O 2) MgO, SO 2, and H 2 O 3) H 2 S and H 2 O 4) SO 2 and H 2 O 5) MgSO 4 , H 2 S and H 2 O 6) SO 3 and H 2 O

Answer: A) Concentrated sulfuric acid is a strong oxidizing agent. It can also interact with metals standing in the electrochemical series of voltages of metals after hydrogen. In this case, hydrogen, as a rule, is not released in a free state, it is oxidized into water, and sulfuric acid is reduced to various compounds, for example: SO 2 , S and H 2 S, depending on the activity of the metal. When interacting with magnesium, the reaction will have the following form:

4Mg + 5H 2 SO 4 (conc) = 4MgSO 4 + H 2 S + H 2 O (answer number 5)

B) When sulfuric acid reacts with magnesium oxide, salt and water are formed:

MgO + H 2 SO 4 \u003d MgSO 4 + H 2 O (Answer number 1)

C) Concentrated sulfuric acid oxidizes not only metals, but also non-metals, in this case sulfur, according to the following reaction equation:

S + 2H 2 SO 4 (conc) = 3SO 2 + 2H 2 O (answer digit 4)

D) During the combustion of complex substances with the participation of oxygen, oxides of all elements that make up the complex substance are formed; for example:

2H 2 S + 3O 2 \u003d 2SO 2 + 2H 2 O (answer number 4)

So the general answer would be:

Determine which of the given substances are substances X and Y.

1) KCl (solution)
2) KOH (solution)
3) H2
4) HCl (excess)
5) CO2

Answer: Carbonates react chemically with acids to form weak carbonic acid, which at the time of formation decomposes into carbon dioxide and water:

K 2 CO 3 + 2HCl (excess) \u003d 2KCl + CO 2 + H 2 O

When excess carbon dioxide is passed through a solution of potassium hydroxide, potassium bicarbonate is formed.

CO 2 + KOH \u003d KHCO 3

We write the answer in the table:

Answer: A) Methylbenzene belongs to the homologous series of aromatic hydrocarbons; its formula is C 6 H 5 CH 3 (number 4)

B) Aniline belongs to the homologous series of aromatic amines. Its formula is C 6 H 5 NH 2 . The NH 2 group is a functional group of amines. (number 2)

C) 3-methylbutanal belongs to the homologous series of aldehydes. Since aldehydes end in -al. Its formula:

Task 12

From the proposed list, select two substances that are structural isomers of butene-1.

1) butane
2) cyclobutane
3) butin-2
4) butadiene-1,3
5) methylpropene

Answer: Isomers are substances that have the same molecular formula but different structures and properties. Structural isomers are a type of substances that are identical to each other in quantitative and qualitative compositions, but the order of atomic binding (chemical structure) is different. To answer this question, let's write the molecular formulas of all substances. The formula for butene-1 will look like this: C 4 H 8

1) butane - C 4 H 10
2) cyclobutane - C 4 H 8
3) butin-2 - C 4 H 6
4) butadiene-1, 3 - C 4 H 6
5) methylpropene - C 4 H 8

Cyclobutane No. 2 and methylpropene No. 5 have the same formulas. They will be the structural isomers of butene-1.

Write the correct answers in the table:

Task 13

From the proposed list, select two substances, when interacting with a solution of potassium permanganate in the presence of sulfuric acid, a change in the color of the solution will be observed.

1) hexane
2) benzene
3) toluene
4) propane
5) propylene

Answer: Let's try to answer this question by elimination. Saturated hydrocarbons are not subject to oxidation by this oxidizing agent, therefore we cross out hexane No. 1 and propane No. 4.

Cross out number 2 (benzene). In benzene homologues, alkyl groups are readily oxidized by oxidizing agents such as potassium permanganate. Therefore, toluene (methylbenzene) will undergo oxidation at the methyl radical. Propylene (an unsaturated hydrocarbon with a double bond) is also oxidized.

Correct answer:

Aldehydes are oxidized by various oxidizing agents, including an ammonia solution of silver oxide (the famous silver mirror reaction)

The book contains materials for the successful passing of the exam in chemistry: brief theoretical information on all topics, tasks of different types and levels of complexity, methodological comments, answers and evaluation criteria. Students do not have to search for additional information on the Internet and buy other manuals. In this book, they will find everything they need for independent and effective training to the exam. In the publication, in a concise form, the basics of the subject are set out in accordance with the current educational standards and the most difficult exam questions of an increased level of complexity are analyzed in the most detailed way. In addition, training tasks are given, with the help of which you can check the level of assimilation of the material. The appendix of the book contains the necessary reference materials on the subject.

Task 15

From the proposed list, select two substances with which methylamine reacts.

1) propane
2) chloromethane
3) hydrogen
4) sodium hydroxide
5) hydrochloric acid.

Answer: Amines, being derivatives of ammonia, have a structure similar to it and exhibit properties similar to it. They are also characterized by the formation of a donor-acceptor bond. Like ammonia, they react with acids. For example, with hydrochloric acid to form methylammonium chloride.

CH 3 -NH 2 + HCl \u003d Cl.

From organic substances, methylamine enters into alkylation reactions with haloalkanes:

CH 3 -NH 2 + CH 3 Cl \u003d [(CH 3) 2 NH 2] Cl

Amines do not react with other substances from this list, so the correct answer is:

Task 16

Establish a correspondence between the name of the substance and the product that is mainly formed during the interaction of this substance with bromine: for each position indicated by a letter, select the corresponding position indicated by a number.

3) Br–CH 2 –CH 2 –CH 2 –Br

Answer: A) ethane is a saturated hydrocarbon. It is not characterized by addition reactions, therefore, the hydrogen atom is replaced by bromine. And it turns out bromoethane:

CH 3 -CH3 + Br 2 \u003d CH 3 -CH 2 -Br + HBr (answer 5)

B) Isobutane, like ethane, is a representative of saturated hydrocarbons, therefore, it is characterized by reactions of substitution of hydrogen for bromine. Unlike ethane, isobutane contains not only primary carbon atoms (attached to three hydrogen atoms), but also one primary carbon atom. And since the replacement of a hydrogen atom by a halogen is easiest at the less hydrogenated tertiary carbon atom, then at the secondary and lastly at the primary, bromine will attach to it. As a result, we get 2-bromine, 2-methylpropane:

	C	H3	C	H3
	│		│
CH 3 -	C	-CH 3 + Br 2 \u003d CH 3 -	C	–CH3 + HBr	(answer 2)
	│		│
	H		B	r

C) Cycloalkanes, which include cyclopropane, differ greatly in terms of cycle stability: the least stable are three-membered and the most stable are five- and six-membered rings. During bromination of 3- and 4-membered cycles, they break with the formation of alkanes. In this case, 2 bromine atoms are added at once.

D) The reaction of interaction with bromine in five and six-membered rings does not lead to ring rupture, but is reduced to the reaction of substitution of hydrogen for bromine.

So the general answer would be:

Task 17

Establish a correspondence between the reacting substances and the carbon-containing product that is formed during the interaction of these substances: for each position indicated by a letter, select the corresponding position indicated by a number.

Answer: A) The reaction between acetic acid and sodium sulfide refers to exchange reactions when complex substances exchange their constituent parts.

CH 3 COOH + Na 2 S \u003d CH 3 COONa + H 2 S.

Salts of acetic acid are called acetates. This salt, respectively, is called sodium acetate. The answer is number 5

B) The reaction between formic acid and sodium hydroxide also refers to exchange reactions.

HCOOH + NaOH \u003d HCOONa + H 2 O.

Salts of formic acid are called formates. In this case, sodium formate is formed. The answer is number 4.

C) Formic acid, unlike other carboxylic acids, is an amazing substance. In addition to the functional carboxyl group -COOH, it also contains the aldehyde group COH. Therefore, they enter into reactions characteristic of aldehydes. For example, in the reaction of a silver mirror; reduction of copper (II) hydroxide, Cu (OH) 2 when heated to copper (I) hydroxide, CuOH, decomposing at high temperature to copper (I) oxide, Cu 2 O. A beautiful orange precipitate forms.

2Cu(OH) 2 + 2HCOOH = 2СO 2 + 3H 2 O + Cu 2 O

Formic acid itself is oxidized to carbon dioxide. (correct answer is 6)

D) When ethanol reacts with sodium, hydrogen gas and sodium ethoxide are formed.

2C 2 H 5 OH + 2Na \u003d 2C 2 H 5 ONa + H 2 (answer 2)

So the answers to this question will be:

The attention of schoolchildren and applicants is offered a new manual for USE preparation, which contains 10 options for standard examination papers in chemistry. Each option is compiled in full accordance with the requirements of the unified state exam, includes tasks of different types and levels of complexity. At the end of the book, answers are given for self-examination of all tasks. Suggested training options will help the teacher to organize preparation for the final certification, and students to independently test their knowledge and readiness for the final exam. The manual is addressed to senior students, applicants and teachers.

Task 18

The following scheme of transformation of substances is given:

Alcohols at high temperatures in the presence of oxidizing agents can be oxidized to the corresponding aldehydes. In this case, copper II oxide (CuO) serves as an oxidizing agent according to the following reaction:

CH 3 CH 2 OH + CuO (t) = CH 3 COH + Cu + H 2 O (answer: 2)

The general answer of this number:

Task 19

From the proposed list of types of reactions, select two types of reactions, which include the interaction of alkali metals with water.

1) catalytic
2) homogeneous
3) irreversible
4) redox
5) neutralization reaction

Answer: Let's write the reaction equation, for example, sodium with water:

2Na + 2H 2 O \u003d 2NaOH + H 2.

Sodium is a very active metal, so it will interact vigorously with water, in some cases even with an explosion, so the reaction proceeds without catalysts. Sodium is a metal, a solid, water and sodium hydroxide solution are liquids, hydrogen is a gas, so the reaction is heterogeneous. The reaction is irreversible because hydrogen leaves the reaction medium as a gas. During the reaction, the oxidation states of sodium and hydrogen change,

therefore, the reaction is classified as redox, since sodium acts as a reducing agent, and hydrogen as an oxidizing agent. It does not apply to neutralization reactions, since as a result of the neutralization reaction, substances are formed that have a neutral reaction of the medium, and here alkali is formed. From this we can conclude that the correct answers will be

Task 20

From the proposed list of external influences, select two influences that lead to a decrease in the rate of the chemical reaction of ethylene with hydrogen:

1) lowering the temperature
2) increase in ethylene concentration
3) the use of a catalyst
4) decrease in hydrogen concentration
5) pressure increase in the system.

Answer: The rate of a chemical reaction is a value that shows how the concentrations of the starting substances or reaction products change per unit of time. There is a concept of the rate of homogeneous and heterogeneous reactions. In this case, a homogeneous reaction is given, therefore, for homogeneous reactions, the rate depends on the following interactions (factors):

1. concentration of reactants;
2. temperature;
3. catalyst;
4. inhibitor.

This reaction takes place at an elevated temperature, so lowering the temperature will reduce its rate. Answer number 1. Next: if you increase the concentration of one of the reactants, the reaction will go faster. It doesn't suit us. A catalyst - a substance that increases the rate of a reaction - is also not suitable. Reducing the concentration of hydrogen will slow down the reaction, which is what we want. So, another correct answer is number 4. To answer point 4 of the question, let's write the equation for this reaction:

CH 2 \u003d CH 2 + H 2 \u003d CH 3 -CH 3.

It can be seen from the reaction equation that it proceeds with a decrease in volume (2 volumes of substances entered into the reaction - ethylene + hydrogen), and only one volume of the reaction product was formed. Therefore, with increasing pressure, the reaction rate should increase - also not suitable. Summarize. The correct answers were:

The manual contains tasks that are as close as possible to the real ones used in the exam, but distributed by topic in the order they are studied in grades 10-11 of high school. Working with the book, you can consistently work out each topic, eliminate gaps in knowledge, and also systematize the material being studied. This structure of the book will help to prepare more effectively for the exam. This publication is addressed to high school students preparing for the exam in chemistry. Training tasks will allow you to systematically, with the passage of each topic, prepare for the exam.

Task 21

Establish a correspondence between the reaction equation and the property of the nitrogen element that it exhibits in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.

Answer: Let's see how the oxidation states change in the reactions:

in this reaction, nitrogen does not change the oxidation state. It is stable in his reaction 3–. So the answer is 4.

in this reaction, nitrogen changes its oxidation state from 3– to 0, that is, it is oxidized. So he is a restorer. Answer 2.

Here nitrogen changes its oxidation state from 3– to 2+. The reaction is redox, nitrogen is oxidized, which means it is a reducing agent. Correct answer 2.

General answer:

Task 22

Establish a correspondence between the salt formula and the products of electrolysis of an aqueous solution of this salt, which stood out on inert electrodes: for each position indicated by a letter, select the corresponding position indicated by a number.

SALT FORMULA	ELECTROLYSIS PRODUCTS

Answer: Electrolysis is a redox reaction that occurs on electrodes when a direct electric current passes through an electrolyte solution or melt. At the cathode always the recovery process is underway; at the anode always there is an oxidation process. If the metal is in the electrochemical series of voltages of metals up to manganese, then water is reduced at the cathode; from manganese to hydrogen, water and metal can be released, if to the right of hydrogen, then only the metal is reduced. Processes occurring at the anode:

If the anode inert, then in the case of oxygen-free anions (except for fluorides), anions are oxidized:

In the case of oxygen-containing anions and fluorides, the process of water oxidation occurs, while the anion is not oxidized and remains in solution:

During the electrolysis of alkali solutions, hydroxide ions are oxidized:

Now let's look at this task:

A) Na 3 PO 4 dissociates in solution into sodium ions and an acid residue of an oxygen-containing acid.

The sodium cation rushes to the negative electrode - the cathode. Since the sodium ion in the electrochemical series of voltages of metals is before aluminum, it will not be restored from, water will be restored according to the following equation:

2H 2 O \u003d H 2 + 2OH -.

Hydrogen is released at the cathode.

The anion rushes to the anode - a positively charged electrode - and is located in the near-anode space, and water is oxidized on the anode according to the equation:

2H 2 O - 4e \u003d O 2 + 4H +

Oxygen is released at the anode. Thus, the overall reaction equation will have the following form:

2Na 3 PO 4 + 8H 2 O \u003d 2H 2 + O 2 + 6NaOH + 2 H 3 PO 4 (answer 1)

B) during the electrolysis of a solution of KCl at the cathode, water will be reduced according to the equation:

2H 2 O \u003d H 2 + 2OH -.

Hydrogen will be evolved as a reaction product. At the anode, Cl will be oxidized to a free state according to the following equation:

2CI - - 2e \u003d Cl 2.

The overall process on the electrodes is as follows:

2KCl + 2H 2 O \u003d 2KOH + H 2 + Cl 2 (answer 4)

C) During the electrolysis of the CuBr 2 salt, copper is reduced at the cathode:

Cu 2+ + 2e = Cu 0 .

Bromine is oxidized at the anode:

The overall reaction equation will have the following form:

Correct answer 3.

D) The hydrolysis of the Cu(NO 3) 2 salt proceeds as follows: copper is released at the cathode according to the following equation:

Cu 2+ + 2e = Cu 0 .

Oxygen is released at the anode:

2H 2 O - 4e \u003d O 2 + 4H +

Correct answer 2.

General answer to this question:

All materials of the school course in chemistry are clearly structured and divided into 36 logical blocks (weeks). The study of each block is designed for 2-3 independent lessons per week during the academic year. The manual contains all the necessary theoretical information, tasks for self-control in the form of diagrams and tables, as well as in the form of the exam, forms and answers. The unique structure of the manual will help structure the preparation for the exam and study all topics step by step throughout the academic year. The publication contains all the topics of the school course in chemistry required to pass the exam. All material is clearly structured and divided into 36 logical blocks (weeks), including the necessary theoretical information, tasks for self-control in the form of diagrams and tables, as well as in the form of the exam. The study of each block is designed for 2-3 independent lessons per week during the academic year. In addition, the manual provides training options, the purpose of which is to assess the level of knowledge.

Task 23

Establish a correspondence between the name of the salt and the ratio of this salt to hydrolysis: for each position indicated by a letter, select the corresponding position indicated by a number.

Answer: Hydrolysis is the reaction of the interaction of salt ions with water molecules, leading to the formation of a weak electrolyte. Any salt can be thought of as the reaction product of an acid and a base. According to this principle, all salts can be divided into 4 groups:

1. Salts formed by a strong base and a weak acid.
2. Salts formed from a weak base and a strong acid.
3. Salts formed from a weak base and a weak acid.
4. Salts formed by a strong base and a strong acid.

Let's now analyze this task from this point of view.

A) NH 4 Cl - a salt formed by a weak base NH 4 OH and a strong acid HCl - undergoes hydrolysis. The result is a weak base and a strong acid. This salt hydrolyzes at the cation, since this ion is part of a weak base. The answer is number 1.

B) K 2 SO 4 is a salt formed by a strong base and a strong acid. Such salts do not undergo hydrolysis, since no weak electrolyte is formed. Answer 3.

C) Sodium carbonate Na 2 CO 3 - a salt formed by a strong base NaOH and a weak carbonic acid H 2 CO 3 - undergoes hydrolysis. Since the salt is formed by a dibasic acid, the hydrolysis can theoretically proceed in two stages. as a result of the first stage, an alkali and an acid salt are formed - sodium bicarbonate:

Na 2 CO 3 + H 2 O ↔NaHCO 3 + NaOH;

as a result of the second stage, weak carbonic acid is formed:

NaHCO 3 + H 2 O ↔ H 2 CO 3 (H 2 O + CO 2) + NaOH -

this salt is hydrolyzed at the anion (answer 2).

D) Aluminum sulfide salt Al 2 S 3 is formed by a weak base Al (OH) 3 and a weak acid H 2 S. Such salts undergo hydrolysis. The result is a weak base and a weak acid. Hydrolysis proceeds by cation and anion. Correct answer 4.

Thus, the general answer to the task is:

Task 24

Establish a correspondence between the equation of a reversible reaction and the direction of the shift in chemical equilibrium with increasing pressure: for each position indicated by a letter, select the corresponding position indicated by a number.

REACTION EQUATION	DIRECTION OF SHIFT OF CHEMICAL EQUILIBRIUM
A) N 2 (g) + 3H 2 (g) \u003d 2NH 3 (g) B) 2H 2 (g) + O 2 (g) \u003d 2H 2 O (g) C) H 2 (g) + CI 2 (g) = 2HCl (g) D) SO 2 (g) + CI 2 (g) \u003d SO 2 Cl 2 (g)	1) shifts towards a direct reaction 2) shifts towards the back reaction 3) practically does not move.

Answer: Reversible reactions are called reactions that can simultaneously go in two opposite directions: in the direction of a direct and reverse reaction, therefore, in the equations of reversible reactions, instead of equality, the sign of reversibility is put. Every reversible reaction ends in a chemical equilibrium. This is a dynamic process. In order to bring the reaction out of the state of chemical equilibrium, it is necessary to apply certain external influences to it: change the concentration, temperature or pressure. This is done according to the Le Chatelier principle: if a system in a state of chemical equilibrium is acted upon from the outside, the concentration, temperature or pressure is changed, then the system tends to take a position that counteracts this action.

Let's analyze this with examples of our task.

A) The homogeneous reaction N 2 (g) + 3H 2 (g) \u003d 2NH 3 (g) is also exothermic, that is, it goes with the release of heat. Then 4 volumes of reactants entered into the reaction (1 volume of nitrogen and 3 volumes of hydrogen), and as a result, one volume of ammonia was formed. Thus, we determined that the reaction proceeds with a decrease in volume. According to the Le Chatelier principle, if the reaction proceeds with a decrease in volume, then an increase in pressure shifts the chemical equilibrium towards the formation of a reaction product. Correct answer 1.

B) The reaction 2H 2 (g) + O 2 (g) \u003d 2H 2 O (g) is similar to the previous reaction, it also goes with a decrease in volume (3 volumes of gas entered, and 2 were formed as a result of the reaction), so an increase in pressure will shift the equilibrium to direction of formation of the reaction product. Answer 1.

C) This reaction H 2 (g) + Cl 2 (g) \u003d 2HCl (g) proceeds without changing the volume of reactants (2 volumes of gases entered and 2 volumes of hydrogen chloride were formed). Reactions proceeding without a change in volume are not affected by pressure. Answer 3.

D) The reaction of interaction of sulfur oxide (IV) and chlorine SO 2 (g) + Cl 2 (g) \u003d SO 2 Cl 2 (g) is a reaction that proceeds with a decrease in the volume of substances (2 volumes of gases entered into the reaction, and one volume was formed SO 2 Cl 2). Answer 1.

The answer to this task will be the following set of letters and numbers:

In the book, solutions to all types of problems of basic, advanced and high levels difficulties in all topics tested at the exam in chemistry. Regular work with this manual will allow students to learn how to quickly and without errors solve problems in chemistry of different levels of complexity. The manual analyzes in detail the solutions to all types of tasks of basic, advanced and high levels of complexity in accordance with the list of content elements tested at the exam in chemistry. Regular work with this manual will allow students to learn how to quickly and without errors solve problems in chemistry of different levels of complexity. The publication will provide invaluable assistance to students in preparing for the exam in chemistry, and can also be used by teachers in organizing the educational process.

Task 25

Establish a correspondence between the formulas of substances and a reagent with which you can distinguish between aqueous solutions of these substances: for each position indicated by a letter, select the corresponding position indicated by a number.

SUBSTANCE FORMULA
A) HNO 3 and NaNO 3 B) KCI and NaOH C) NaCI and BaCI 2 D) AICI 3 and MgCI 2

Answer: a) Two substances are given, an acid and a salt. Nitric acid is a strong oxidizing agent and interacts with metals standing in the electrochemical series of metal voltages both before and after hydrogen, and both concentrated and diluted interact. For example, nitric acid HNO 3 reacts with copper to form a copper salt, water and nitric oxide. In this case, in addition to gas evolution, the solution acquires a blue color characteristic of copper salts, for example:

8HNO 3 (p) + 3Cu \u003d 3Cu (NO 3) 2 + 2NO + 4H 2 O,

and NaNO 3 salt does not react with copper. Answer 1.

B) Salt and hydroxide of active metals are given, in which almost all compounds are soluble in water, therefore, we select a substance from the column of reagents, which, when interacting with one of these substances, precipitates. This substance is copper sulfate. The reaction will not go with potassium chloride, but with sodium hydroxide a beautiful blue precipitate will fall out, according to the reaction equation:

CuSO 4 + 2NaOH \u003d Cu (OH) 2 + Na 2 SO 4.

C) Two salts are given, sodium and barium chlorides. If all sodium salts are soluble, then with barium salts, on the contrary, many barium salts are insoluble. According to the solubility table, we determine that barium sulfate is insoluble, so copper sulfate will be the reagent. Answer 5.

D) Again, 2 salts are given - AlCl 3 and MgCl 2 - and again chlorides. When these solutions are drained with HCl, KNO 3 CuSO 4 do not form any visible changes, they do not react with copper at all. Remains KOH. With it, both salts precipitate, with the formation of hydroxides. But aluminum hydroxide is an amphoteric base. When an excess of alkali is added, the precipitate dissolves to form a complex salt. Answer 2.

The general answer to this question looks like this:

Task 26

Establish a correspondence between the substance and its main field of application: for each position indicated by a letter, select the corresponding position indicated by a number.

Answer: A) Methane, when burned, releases a large amount of heat, so it can be used as a fuel (answer 2).

B) Isoprene, being a diene hydrocarbon, forms rubber during polymerization, which is then converted into rubber (answer 3).

C) Ethylene is an unsaturated hydrocarbon that enters into polymerization reactions, therefore it can be used as plastics (answer 4).

Task 27

Calculate the mass of potassium nitrate (in grams) that should be dissolved in 150.0 g of a solution with a mass fraction of this salt of 10% to obtain a solution with a mass fraction of 12%. (Write down the number to tenths).

Let's solve this problem:

1. Determine the mass of potassium nitrate contained in 150 g of a 10% solution. Let's use the magic triangle:

Hence the mass of matter is equal to: ω · m(solution) \u003d 0.1 150 \u003d 15 g.

2. Let the mass of added potassium nitrate be x g. Then the mass of all salt in the final solution will be equal to (15 + x) g, mass of solution (150 + x), and the mass fraction of potassium nitrate in the final solution can be written as: ω (KNO 3) \u003d 100% - (15 + x)/(150 + x)

100% – (15 + x)/(150 + x) = 12%

(15 + x)/(150 + x) = 0,12

15 + x = 18 + 0,12x

0,88x = 3

x = 3/0,88 = 3,4

Answer: To obtain a 12% salt solution, 3.4 g of KNO 3 must be added.

The handbook contains detailed theoretical material on all topics tested by the Unified State Examination in Chemistry. After each section, multi-level tasks are given in the form of the exam. For the final control of knowledge at the end of the handbook, training options are given that correspond to the exam. Students do not have to search for additional information on the Internet and buy other manuals. In this guide, they will find everything they need to independently and effectively prepare for the exam. The reference book is addressed to high school students to prepare for the exam in chemistry.

Task 28

As a result of the reaction, the thermochemical equation of which

2H 2 (g) + O 2 (g) \u003d H 2 O (g) + 484 kJ,

1452 kJ of heat were released. Calculate the mass of the resulting water (in grams).

This task can be solved in one step.

According to the reaction equation, as a result of it, 36 grams of water were formed and 484 kJ of energy were released. And 1454 kJ of energy will be released during the formation of X year of water.

Answer: With the release of 1452 kJ of energy, 108 g of water is formed.

Task 29

Calculate the mass of oxygen (in grams) required for the complete combustion of 6.72 liters (N.O.) of hydrogen sulfide.

To solve this problem, we write the reaction equation for the combustion of hydrogen sulfide and calculate the masses of oxygen and hydrogen sulfide that have entered into the reaction, according to the reaction equation

1. Determine the amount of hydrogen sulfide contained in 6.72 liters.

2. Determine the amount of oxygen that will react with 0.3 mol of hydrogen sulfide.

According to the reaction equation, 3 mol O 2 reacts with 2 mol H 2 S.

According to the reaction equation, with 0.3 mol H 2 S will react with X mol O 2.

Hence X = 0.45 mol.

3. Determine the mass of 0.45 mol of oxygen

m(O2) = n · M\u003d 0.45 mol 32 g / mol \u003d 14.4 g.

Answer: the mass of oxygen is 14.4 grams.

Task 30

From the proposed list of substances (potassium permanganate, potassium bicarbonate, sodium sulfite, barium sulfate, potassium hydroxide), select substances between which a redox reaction is possible. In your answer, write down the equation for only one of the possible reactions. Make an electronic balance, indicate the oxidizing agent and reducing agent.

Answer: KMnO 4 is a well-known oxidizing agent that oxidizes substances containing elements in lower and intermediate oxidation states. Its actions can take place in neutral, acidic and alkaline environments. In this case, manganese can be reduced to various degrees of oxidation: in an acidic environment - to Mn 2+, in a neutral environment - to Mn 4+, in an alkaline environment - to Mn 6+. Sodium sulfite contains sulfur in the 4+ oxidation state, which can be oxidized to 6+. Finally, potassium hydroxide will determine the reaction of the medium. We write the equation for this reaction:

KMnO 4 + Na 2 SO 3 + KOH \u003d K 2 MnO 4 + Na 2 SO 4 + H 2 O

After placing the coefficients, the formula takes the following form:

2KMnO 4 + Na 2 SO 3 + 2KOH \u003d 2K 2 MnO 4 + Na 2 SO 4 + H 2 O

Therefore, KMnO 4 is an oxidizing agent, and Na 2 SO 3 is a reducing agent.

All the information necessary for passing the exam in chemistry is presented in visual and accessible tables, after each topic there are training tasks for knowledge control. With the help of this book, students will be able to improve their knowledge in the shortest possible time, remember all the most important topics in a matter of days before the exam, practice completing assignments in the USE format and become more confident in their abilities. After repeating all the topics presented in the manual, the long-awaited 100 points will be much closer! The manual contains theoretical information on all topics tested at the exam in chemistry. After each section, training tasks of different types with answers are given. A visual and accessible presentation of the material will allow you to quickly find the information you need, eliminate gaps in knowledge and repeat a large amount of information in the shortest possible time.

Task 31

From the proposed list of substances (potassium permanganate, potassium bicarbonate, sodium sulfite, barium sulfate, potassium hydroxide), select substances between which an ion exchange reaction is possible. In your answer, write down the molecular, full and abbreviated ionic equation of only one of the possible reactions.

Answer: Consider the exchange reaction between potassium bicarbonate and potassium hydroxide

KHCO 3 + KOH \u003d K 2 CO 3 + H 2 O

If, as a result of a reaction in electrolyte solutions, an insoluble or gaseous, or low-dissociating substance is formed, then such a reaction proceeds irreversibly. In accordance with this, this reaction is possible, since one of the reaction products (H 2 O) is a low-dissociating substance. Let us write down the complete ionic equation.

Since water is a low-dissociating substance, it is written as a molecule. Next, we compose an abbreviated ionic equation. Those ions that have passed from the left side of the equation to the right without changing the sign of the charge are crossed out. We rewrite the rest into a reduced ionic equation.

This equation will be the answer to this task.

Task 32

During the electrolysis of an aqueous solution of copper (II) nitrate, a metal was obtained. The metal was treated with concentrated sulfuric acid when heated. The resulting gas reacted with hydrogen sulfide to form a simple substance. This substance was heated with a concentrated solution of potassium hydroxide. Write the equations for the four described reactions.

Answer: Electrolysis is a redox process that takes place on electrodes by passing a direct electric current through an electrolyte solution or melt. The task refers to the electrolysis of a solution of copper nitrate. In the electrolysis of salt solutions, water can also take part in electrode processes. When salt dissolves in water, it breaks down into ions:

Reduction processes take place at the cathode. Depending on the activity of the metal, metal, metal and water can be reduced. Since copper in the electrochemical series of voltages of metals is to the right of hydrogen, copper will be reduced at the cathode:

Cu 2+ + 2e = Cu 0 .

The process of water oxidation will take place at the anode.

Copper does not react with solutions of sulfuric and hydrochloric acids. But concentrated sulfuric acid is a strong oxidizing agent, so it can react with copper according to the following reaction equation:

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

Hydrogen sulfide (H 2 S) contains sulfur in the oxidation state 2–, therefore it acts as a strong reducing agent and reduces sulfur in sulfur oxide IV to a free state

2H 2 S + SO 2 \u003d 3S + 2H 2 O.

The resulting substance, sulfur, reacts with a concentrated solution of potassium hydroxide when heated to form two salts: sulfur sulfide and sulfur sulfite and water.

S + KOH \u003d K 2 S + K 2 SO 3 + H 2 O

Task 33

Write the reaction equations that can be used to carry out the following transformations:

When writing reaction equations, use the structural formulas of organic substances.

Answer: In this chain, it is proposed to fulfill 5 reaction equations, according to the number of arrows between substances. In reaction equation No. 1, sulfuric acid plays the role of a water-removing liquid, therefore, as a result of it, an unsaturated hydrocarbon should be obtained.

The next reaction is interesting because it proceeds according to Markovnikov's rule. According to this rule, when hydrogen halides are combined with asymmetrically built alkenes, the halogen is attached to the less hydrogenated carbon atom at the double bond, and hydrogen, vice versa.

The new handbook contains all the theoretical material on the course of chemistry required to pass the exam. It includes all elements of the content, checked by control and measuring materials, and helps to generalize and systematize knowledge and skills for the course of the secondary (complete) school. The theoretical material is presented in a concise, accessible form. Each section is accompanied by examples of training tasks that allow you to test your knowledge and the degree of preparedness for the certification exam. Practical tasks correspond to the USE format. At the end of the manual, answers to tasks are given that will help you objectively assess the level of your knowledge and the degree of preparedness for the certification exam. The manual is addressed to senior students, applicants and teachers.

Task 34

When a sample of calcium carbonate was heated, part of the substance decomposed. At the same time, 4.48 l (n.o.) of carbon dioxide were released. The weight of the solid residue was 41.2 g. This residue was added to 465.5 g of hydrochloric acid solution taken in excess. Determine the mass fraction of salt in the resulting solution.

In your answer, write down the reaction equations that are indicated in the condition of the problem, and give all the necessary calculations (indicate the units of measurement of the quantities you are looking for).

Answer: Let us write a brief condition of this problem.

After all the preparations are given, we proceed to the solution.

1) Determine the amount of CO 2 contained in 4.48 liters. his.

n(CO 2) \u003d V / Vm \u003d 4.48 l / 22.4 l / mol \u003d 0.2 mol

2) Determine the amount of calcium oxide formed.

According to the reaction equation, 1 mol of CO 2 and 1 mol of CaO are formed

Consequently: n(CO2) = n(CaO) and equals 0.2 mol

3) Determine the mass of 0.2 mol CaO

m(CaO) = n(CaO) M(CaO) = 0.2 mol 56 g/mol = 11.2 g

Thus, the solid residue weighing 41.2 g consists of 11.2 g of CaO and (41.2 g - 11.2 g) 30 g of CaCO 3

4) Determine the amount of CaCO 3 contained in 30 g

n(CaCO3) = m(CaCO3) / M(CaCO 3) \u003d 30 g / 100 g / mol \u003d 0.3 mol

CaO + HCl \u003d CaCl 2 + H 2 O

CaCO 3 + HCl \u003d CaCl 2 + H 2 O + CO 2

5) Determine the amount of calcium chloride formed as a result of these reactions.

0.3 mol of CaCO 3 and 0.2 mol of CaO, only 0.5 mol, entered into the reaction.

Accordingly, 0.5 mol CaCl 2 is formed

6) Calculate the mass of 0.5 mol of calcium chloride

M(CaCl2) = n(CaCl 2) M(CaCl 2) \u003d 0.5 mol 111 g / mol \u003d 55.5 g.

7) Determine the mass of carbon dioxide. 0.3 mol of calcium carbonate participated in the decomposition reaction, therefore:

n(CaCO3) = n(CO 2) \u003d 0.3 mol,

m(CO2) = n(CO2) · M(CO 2) \u003d 0.3 mol 44g / mol \u003d 13.2 g.

8) Find the mass of the solution. It consists of the mass of hydrochloric acid + the mass of the solid residue (CaCO 3 + CaO) min the mass of the released CO 2 . Let's write this as a formula:

m(r-ra) = m(CaCO 3 + CaO) + m(HCl) - m(CO 2) \u003d 465.5 g + 41.2 g - 13.2 g \u003d 493.5 g.

9) And finally, we will answer the question of the problem. Find the mass fraction in % of salt in the solution using the following magic triangle:

ω%(CaCI 2) = m(CaCl 2) / m(solution) \u003d 55.5 g / 493.5 g \u003d 0.112 or 11.2%

Answer: ω% (СaCI 2) = 11.2%

Task 35

Organic substance A contains 11.97% nitrogen, 9.40% hydrogen and 27.35% oxygen by mass and is formed by the reaction of organic substance B with propanol-2. It is known that substance B is of natural origin and is able to interact with both acids and alkalis.

Based on these conditions, complete the tasks:

1) Carry out the necessary calculations (indicate the units of measurement of the required physical quantities) and establish the molecular formula of the original organic substance;

2) Make a structural formula of this substance, which will unambiguously show the order of bonding of atoms in its molecule;

3) Write the reaction equation for obtaining substance A from substance B and propanol-2 (use the structural formulas of organic substances).

Answer: Let's try to solve this problem. Let's write a short condition:

ω(C) = 100% - 11.97% - 9.40% - 27.35% = 51.28% (ω(C) = 51.28%)

2) Knowing the mass fractions of all the elements that make up the molecule, we can determine its molecular formula.

Let us take the mass of substance A for 100 g. Then the masses of all the elements that make up its composition will be equal to: m(C) = 51.28 g; m(N) = 11.97 g; m(H) = 9.40 g; m(O) = 27.35 g. Determine the amount of each element:

n(C)= m(C) · M(C) = 51.28 g / 12 g/mol = 4.27 mol

n(N) = m(N) · M(N) = 11.97 g / 14 g/mol = 0.855 mol

n(H) = m(H) M(H) = 9.40 g / 1 g/mol = 9.40 mol

n(O) = m(O) M(O) = 27.35 g / 16 g/mol = 1.71 mol

x : y : z : m = 5: 1: 11: 2.

Thus, the molecular formula of substance A is: C 5 H 11 O 2 N.

3) Let's try to make a structural formula of substance A. We already know that carbon in organic chemistry is always tetravalent, hydrogen is monovalent, oxygen is bivalent and nitrogen is trivalent. The condition of the problem also says that substance B is able to interact with both acids and alkalis, that is, it is amphoteric. From natural amphoteric substances, we know that amino acids are highly amphoteric. Therefore, it can be assumed that substance B refers to amino acids. And of course, we take into account that it is obtained by interacting with propanol-2. By counting the number of carbon atoms in propanol-2, we can boldly conclude that substance B is aminoacetic acid. After some number of attempts, the following formula was obtained:

4) In conclusion, we write the equation for the reaction of the interaction of aminoacetic acid with propanol-2.

For the first time, the attention of schoolchildren and applicants is offered tutorial to prepare for the exam in chemistry, which contains training tasks collected by topic. The book contains tasks of different types and levels of complexity on all the topics of the chemistry course being tested. Each section of the manual includes at least 50 tasks. The tasks correspond to the modern educational standard and the regulation on holding a unified state exam in chemistry for graduates of secondary educational institutions. The implementation of the proposed training tasks on topics will allow you to prepare well for passing the exam in chemistry. The manual is addressed to senior students, applicants and teachers.