In our last article, we talked about the general codifier of the exam in chemistry in 2018 and how to properly start preparing for the exam in chemistry in 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), which are graded at one and two points.

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

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

The basic tasks of the exam in chemistry in 2018 include No. 1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 13, 14,15, 16, 17, 20, 21, 27, 28, 29.

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Topics of assignments 1, 2, 3 and 4 in the exam in chemistry 2018

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

• Navigate the Periodic Table of Dmitry Ivanovich Mendeleev;
• Study the 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 the same atoms, covalent polar between atoms of different chemical elements);
• To 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 know the nomenclature of inorganic chemical compounds (the rules for forming the 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 exam in chemistry in 2018, you need:

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

Assignments 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 exam in chemistry in 2018, you need:

• Know all classes of organic compounds (alkanes, alkenes, alkynes, arenes, etc.);
• Be able to name 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.

Assignments 20 and 21 in the exam in 2018

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

Chemistry tasks 27, 28 and 29

These are computational tasks. They contain the simplest chemical processes, which are aimed only at forming an understanding in the student of what happened in the task. The rest of the assignment is strictly mathematical. Therefore, to solve these tasks in the USE in chemistry in 2018, you need to learn three basic formulas (mass fraction, molar fraction by mass and by volume) and be able to use a calculator.

Average tasks, in the codifier of the USE in chemistry of 2018 called Increased (see in the codifier table 4 - Distribution of tasks by difficulty levels), constitute the smallest part of the exam in terms of points (9 tasks) in terms of the maximum primary score - 18 primary points 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 number: 10, 11, 18, 19, 22, 23, 24, 25, 26.

Chemistry Challenge 10 2018

These are redox reactions. To successfully complete this task in the exam in chemistry in 2018, you need to know:

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

Assignment 11 Unified State Exam 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. Pupils often begin to describe ALL reactions, and their in each task is hypothetically from forty (40) to sixty (60), which takes a very long time. To successfully complete this task in the exam in chemistry in 2018, you need:

• Unmistakably 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 analyze the solution to task number 11 from the demo version of the USE in chemistry in 2018:

Eleventh task: Establish a correspondence between the formula of the 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.

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

Solving task 11 in the exam in chemistry in 2018

First of all, we need to determine what is proposed to us as reagents: substance A - pure sulfur substance, B - sulfur oxide VI - acid oxide, C - zinc hydroxide - amphoteric hydroxide, G - zinc bromide - average salt. It turns out that there are 60 hypothetical reactions in this task. It is very important for solving this task, it is to reduce the possible answer options, 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 answer options as the task is logically evaluated:

And now, applying knowledge about the nature of substances and their interactions, we remove the answer options that are definitely not correct, for example, answer B- acidic oxide, which means it does NOT react with acids and acidic oxides, which means that 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 oxidants, pure oxygen and chlorine - we remove answers 3, 4. There is only answer 2 which suits us perfectly.

Answer B- here you need to apply the opposite method, to whom the amphoteric hydroxides react - both with bases and with acids, and we see a variant of the answer, consisting only of these compounds - answer 4.

Answer D- an average salt containing a bromine anion, which means that adding a similar anion is meaningless - we remove answer 4, containing hydrobromic acid. Let's also remove answer 5 - since the reaction with bromine chloride is meaningless, two soluble salts will form (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 that 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 reduce zinc, which means that the answer remains 1. The option remains

Answer A- which can cause the greatest difficulties, therefore we left it to last, which should also be done by the student if difficulties arise, since for an advanced task they give two points, and we admit 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 the answers are also simple substances).

Reactions:

A)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

V) 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

Assignments 18 and 19 in the exam in chemistry

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

Task 22 in the exam in chemistry

Electrolysis of melts and solutions. To successfully complete this task in the exam in chemistry in 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 patterns of products obtained as a result of electrolysis of a solution;
• Features of electrolysis of a solution of acetic acid and its salts (acetates).

Chemistry Task 23

Salt hydrolysis. To successfully complete this task in the exam in chemistry in 2018, you need to know:

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

Assignment 24 in the exam in chemistry

Reversible and irreversible chemical reactions. To successfully complete this task in the exam in chemistry in 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)

Chemistry Challenge 25 2018

Qualitative reactions to inorganic substances and ions.

To successfully complete this task in the exam in chemistry in 2018, you need to learn these reactions.

Chemistry Task 26

Chemical laboratory. The concept of metallurgy. Production. Chemical pollution of the environment. Polymers. To successfully complete this task in the exam in chemistry in 2018, you need to have an idea of ​​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 theoretical bases necessary for successfully passing the exam in chemistry in 2018 have not practically changed, which means that all the knowledge that your child received at school will help him pass the exam in chemistry in 2018.

In ours, your child will receive all theoretical materials necessary for preparation, and in the classroom will consolidate the knowledge gained for successful implementation of all examination tasks. 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 performing the examination work.

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

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

We continue to discuss the solution to the problem of the type C1 (No. 30), which will surely be encountered by everyone who will take the exam in chemistry. In the first part of the article, we presented a general algorithm for solving Problem 30, in the second part we analyzed several rather complex examples.

We begin the third part by discussing typical oxidizing and reducing agents and their transformations in various environments.

Fifth step: discussing typical IDS that can be encountered in problem # 30

I would like to remind you of a few points related to the concept of the oxidation state. We have already noted that a constant oxidation state is characteristic only for a relatively small number of elements (fluorine, oxygen, alkali and alkaline earth metals, etc.). Most of the elements can exhibit different oxidation states. For example, for chlorine, all states from -1 to +7 are possible, although odd values ​​are the most stable. Nitrogen exhibits oxidation states from -3 to +5, etc.

There are two important rules to remember clearly.

1. The highest oxidation state of an element - a non-metal, in most cases coincides with the number of the group in which this element is located, and the lowest oxidation state = group number - 8.

For example, chlorine is in group VII, therefore, its highest oxidation state = +7, and the lowest - 7 - 8 = -1. Selenium is in group VI. The highest oxidation state = +6, the lowest - (-2). Silicon is located in group IV; the corresponding values ​​are +4 and -4.

Remember that there are exceptions to this rule: the highest oxidation state of oxygen = +2 (and even it appears only in oxygen fluoride), and the highest oxidation state of fluorine = 0 (in a simple substance)!

2. Metals are not capable of exhibiting negative oxidation states. This is quite important, given that more than 70% of chemical elements belong specifically to metals.

And now the question: "Can Mn (+7) act as a reducing agent in chemical reactions?" Take your time, try to answer yourself.

The correct answer is: "No, it cannot!" This is very easy to explain. Take a look at the position of this element in the periodic table. Mn is in group VII, therefore, its HIGHER oxidation state is +7. If Mn (+7) were to act as a reducing agent, its oxidation state would increase (remember the definition of a reducing agent!), Which is impossible, since it already has a maximum value. Conclusion: Mn (+7) can only be an oxidizing agent.

For the same reason, ONLY OXIDIZING properties can exhibit S (+6), N (+5), Cr (+6), V (+5), Pb (+4), etc. Take a look at the position of these elements in the periodic system and see for yourself.

And another question: "Can Se (-2) act as an oxidizing agent in chemical reactions?"

Again, negative answer. You've probably already guessed what this is about. Selenium is in group VI, its LOWEST oxidation state is -2. Se (-2) cannot BUY electrons, that is, it cannot be an oxidizing agent. If Se (-2) participates in OVR, then only in the role of RESTORER.

For a similar reason, ONLY RESTORER can be N (-3), P (-3), S (-2), Te (-2), I (-1), Br (-1), etc.

The final conclusion: an element in the lowest oxidation state can act in the ORP only as a reducing agent, and an element with the highest oxidation state can only act as an oxidizing agent.

"What if the element has an intermediate oxidation state?" - you ask. Well, then both its oxidation and its reduction are possible. For example, sulfur in reaction with oxygen is oxidized, and in reaction with sodium it is reduced.

It is probably logical to assume that each element in the highest oxidation state will be a pronounced oxidizing agent, and in the lowest one, a strong reducing agent. In most cases, this is true. For example, all compounds Mn (+7), Cr (+6), N (+5) can be classified as strong oxidizing agents. But, for example, P (+5) and C (+4) are restored with difficulty. And it is almost impossible to force Ca (+2) or Na (+1) to act as an oxidizing agent, although, formally speaking, +2 and +1 are also the highest oxidation states.

On the contrary, many chlorine compounds (+1) are powerful oxidizing agents, although the oxidation state +1 in this case is far from the highest.

F (-1) and Cl (-1) are bad recovers, and their counterparts (Br (-1) and I (-1)) are good. Oxygen in the lowest oxidation state (-2) practically does not exhibit reducing properties, and Te (-2) is a powerful reducing agent.

We see that everything is not as obvious as we would like. In some cases, the ability to oxidation - reduction can be easily foreseen, in other cases - you just need to remember that substance X is, say, a good oxidizing agent.

We seem to have finally gotten to a list of typical oxidizing and reducing agents. I would like you not only to "memorize" these formulas (although it will be good too!), But also to be able to explain why this or that substance was included in the corresponding list.

Typical oxidants

1. Simple substances - non-metals: F 2, O 2, O 3, Cl 2, Br 2.
2. Concentrated sulfuric acid (H 2 SO 4), nitric acid (HNO 3) in any concentration, hypochlorous acid (HClO), perchloric acid (HClO 4).
3. Potassium permanganate and potassium manganate (KMnO 4 and K 2 MnO 4), chromates and dichromates (K 2 CrO 4 and K 2 Cr 2 O 7), bismuthates (eg NaBiO 3).
4. Oxides of chromium (VI), bismuth (V), lead (IV), manganese (IV).
5. Hypochlorites (NaClO), chlorates (NaClO 3) and perchlorates (NaClO 4); nitrates (KNO 3).
6. Peroxides, superoxides, ozonides, organic peroxides, peroxo acids, all other substances containing the -O-O- group (eg, hydrogen peroxide - H 2 O 2, sodium peroxide - Na 2 O 2, potassium superoxide - KO 2).
7. Metal ions located on the right side of the stress row: Au 3+, Ag +.

Typical reducing agents

1. Simple substances - metals: alkali and alkaline earth, Mg, Al, Zn, Sn.
2. Simple substances - non-metals: H 2, C.
3. Metal hydrides: LiH, CaH 2, lithium aluminum hydride (LiAlH 4), sodium borohydride (NaBH 4).
4. Hydrides of some non-metals: HI, HBr, H 2 S, H 2 Se, H 2 Te, PH 3, silanes and boranes.
5. Iodides, bromides, sulfides, selenides, phosphides, nitrides, carbides, nitrites, hypophosphites, sulfites.
6. Carbon monoxide (CO).

I would like to emphasize a few points:

1. I did not set myself the goal of listing all oxidizing and reducing agents. This is impossible and unnecessary.
2. One and the same substance can act as an oxidizing agent in one process, and as an agent in another.
3. No one can guarantee that you will definitely encounter one of these substances in exam problem C1, but the probability of this is very high.
4. It is not rote memorization of formulas that is important, but UNDERSTANDING. Try to test yourself: write down a mixture of substances from two lists, and then try to separate them yourself into typical oxidizing and reducing agents. Be guided by the considerations we discussed at the beginning of this article.

And now a little test work. I will offer you some incomplete equations, and you will try to find an oxidizing agent and a reducing agent. It is not necessary to add the right-hand sides of the equations yet.

Example 12... Determine the oxidizing and reducing agent in the ORP:

HNO 3 + Zn = ...

CrO 3 + C 3 H 6 + H 2 SO 4 = ...

Na 2 SO 3 + Na 2 Cr 2 O 7 + H 2 SO 4 = ...

O 3 + Fe (OH) 2 + H 2 O = ...

CaH 2 + F 2 = ...

KMnO 4 + KNO 2 + KOH = ...

H 2 O 2 + K 2 S + KOH = ...

I think you coped with this task without difficulty. If you have problems, read the beginning of this article again, work on a list of typical oxidants.

“All this is wonderful!” The impatient reader exclaims. “But where are the promised problems C1 with incomplete equations? Yes, in example 12 we were able to determine the oxidizer and g-tel, but this is not the main thing. The main thing is to be able to COMPLETE the reaction equation, but can a list of oxidants help us with this? "

Yes, it can, if you understand WHAT HAPPENS to typical oxidants under different conditions. This is exactly what we are going to do now.

Sixth step: transformations of some oxidants in different environments. "Destiny" of permanganates, chromates, nitric and sulfuric acids

So, we must not only be able to recognize typical oxidizing agents, but also understand what these substances are converted into during the OVR. Obviously, without this understanding, we will not be able to correctly solve Problem 30. The situation is complicated by the fact that the interaction products cannot be indicated UNIVERSAL. It makes no sense to ask: "What will the potassium permanganate turn into during the recovery process?" It all depends on many reasons. In the case of KMnO 4, the main one is the acidity (pH) of the medium. In principle, the nature of the recovery products may depend on:

1. the reducing agent used during the process,
2. acidity of the environment,
3. the concentration of the participants in the reaction,
4. process temperature.

We will not now talk about the effect of concentration and temperature (although inquisitive young chemists may remember that, for example, chlorine and bromine interact differently with an aqueous solution of alkali in the cold and when heated). Let's focus on the pH of the medium and the strength of the reducing agent.

The information below should simply be remembered. Don't try to analyze the causes, just REMEMBER the reaction products. I assure you, on the exam in chemistry, this may be useful to you.

Potassium permanganate reduction products (KMnO 4) in various media

Example 13... Supplement the equations of redox reactions:

KMnO 4 + H 2 SO 4 + K 2 SO 3 = ...
KMnO 4 + H 2 O + K 2 SO 3 = ...
KMnO 4 + KOH + K 2 SO 3 = ...

Solution... Guided by the list of typical oxidizing and reducing agents, we come to the conclusion that potassium permanganate is the oxidizing agent in all these reactions, and potassium sulfite is the reducing agent.

H 2 SO 4, H 2 O and KOH determine the nature of the solution. In the first case, the reaction takes place in an acidic medium, in the second - in a neutral one, in the third - in an alkaline one.

Conclusion: in the first case, permanganate will be reduced to the Mn (II) salt, in the second - to manganese dioxide, in the third - to potassium manganate. Let's supplement the reaction equations:

KMnO 4 + H 2 SO 4 + K 2 SO 3 = MnSO 4 + ...
KMnO 4 + H 2 O + K 2 SO 3 = MnO 2 + ...
KMnO 4 + KOH + K 2 SO 3 = K 2 MnO 4 + ...

And what will potassium sulfite turn into? Well, naturally, in sulfate. It is obvious that K in the composition of K 2 SO 3 is simply nowhere to be oxidized, oxygen oxidation is extremely unlikely (although, in principle, it is possible), but S (+4) easily turns into S (+6). The oxidation product is K 2 SO 4, you can add this formula to the equations:

KMnO 4 + H 2 SO 4 + K 2 SO 3 = MnSO 4 + K 2 SO 4 + ...
KMnO 4 + H 2 O + K 2 SO 3 = MnO 2 + K 2 SO 4 + ...
KMnO 4 + KOH + K 2 SO 3 = K 2 MnO 4 + K 2 SO 4 + ...

Our equations are almost ready. It remains to add substances that are not directly involved in the OVR and arrange the coefficients. By the way, if you start from the second point, it may be even easier. Let's build, for example, an electronic balance for the last reaction

We put the coefficient 2 in front of the formulas KMnO 4 and K 2 MnO 4; before the formulas of sulfite and potassium sulfate we mean coeff. 1:

2KMnO 4 + KOH + K 2 SO 3 = 2K 2 MnO 4 + K 2 SO 4 + ...

On the right we see 6 potassium atoms, on the left - so far only 5. It is necessary to correct the situation; we put the coefficient 2 in front of the KOH formula:

2KMnO 4 + 2KOH + K 2 SO 3 = 2K 2 MnO 4 + K 2 SO 4 + ...

The final touch: on the left we see hydrogen atoms, on the right they are not. Obviously, we urgently need to find some substance that contains hydrogen in the oxidation state +1. Let's get some water!

2KMnO 4 + 2KOH + K 2 SO 3 = 2K 2 MnO 4 + K 2 SO 4 + H 2 O

We check the equation again. Yes, everything is great!

“An interesting movie!” The alert young chemist remarks. “Why did you add water in the last step? And if I want to add hydrogen peroxide, or just H 2 or potassium hydride or H 2 S? NEEDED to add or did you just feel like it? "

Well, let's figure it out. Well, first of all, we naturally have no right to add substances to the reaction equation at our will. The reaction goes exactly the way it goes; as nature ordered. Our likes and dislikes are unable to influence the course of the process. We can try to change the reaction conditions (increase the temperature, add a catalyst, change the pressure), but if the reaction conditions are set, its result can no longer depend on our will. Thus, the formula for water in the equation of the last reaction is not my desire, but a fact.

Secondly, you can try to equalize the reaction in cases where the substances you listed are present instead of water. I assure you: in no case will you be able to do this.

Thirdly, options with H 2 O 2, H 2, KH or H 2 S are simply unacceptable in this case for one reason or another. For example, in the first case, the oxidation state of oxygen changes, in the second and third - hydrogen, and we agreed that the oxidation state will change only for Mn and S. In the fourth case, sulfur generally acted as an oxidizing agent, and we agreed that S - reducing agent. In addition, potassium hydride is unlikely to "survive" in an aqueous medium (and the reaction, recall, takes place in an aqueous solution), and H 2 S (even if this substance was formed) will inevitably enter into a solution with KOH. As you can see, knowledge of chemistry allows us to reject these issues.

"But why exactly water?" - you ask.

Yes, because, for example, in this process (as in many others), water acts as a solvent. Because, for example, if you analyze all the reactions that you have written in 4 years of studying chemistry, you will find that H 2 O occurs in almost half of the equations. Water is generally quite a "popular" compound in chemistry.

Understand, I am not saying that every time in Problem 30 you need to "send hydrogen somewhere" or "get oxygen from somewhere," you need to grab onto water. But this will probably be the first substance to think about.

Similar logic is used for the equations of reactions in acidic and neutral media. In the first case, it is necessary to add the formula of water to the right side, in the second - potassium hydroxide:

KMnO 4 + H 2 SO 4 + K 2 SO 3 = MnSO 4 + K 2 SO 4 + H 2 O,
KMnO 4 + H 2 O + K 2 SO 3 = MnO 2 + K 2 SO 4 + KOH.

The arrangement of the coefficients for highly experienced young chemists should not cause the slightest difficulty. Final answer:

2KMnO 4 + 3H 2 SO 4 + 5K 2 SO 3 = 2MnSO 4 + 6K 2 SO 4 + 3H 2 O,
2KMnO 4 + H 2 O + 3K 2 SO 3 = 2MnO 2 + 3K 2 SO 4 + 2KOH.

In the next part we will talk about the products of the reduction of chromates and dichromates, about nitric and sulfuric acids.

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

There are no changes in the KIM USE 2020 in chemistry.

Don't postpone your preparation until later.

1. When starting to analyze the 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. will guide you in the study of the main topics and determine what knowledge and skills will be required when completing the USE tasks in chemistry. For the successful completion of the exam in chemistry, theory is most important.
2. Theory needs to be backed up practice constantly solving tasks. 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 such an opportunity to decide and find out the answers. But don't rush to pry. First, decide for yourself and see how many points you scored.

Points for each chemistry task

• 1 point - for tasks 1-6, 11-15, 19-21, 26-28.
• 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 involve a short answer (in the form of a number or a word) - tasks 1-29.
2. Problems with detailed answers - tasks 30-35.

3.5 hours (210 minutes) are allotted for the performance of the examination work in chemistry.

There will be three cheat sheets on the exam. And you need to understand them

This is 70% of the information that will help you successfully pass the chemistry exam. The remaining 30% is the ability to use the presented 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.
• Distribute your strength correctly and do not forget about rest.

Dare, try and you will succeed!