In our last article, we talked about the basic tasks in the Unified State Exam in Chemistry 2018. Now, we have to analyze in more detail the tasks of an increased (in the 2018 Unified State Exam codifier in chemistry - high level of complexity) level of complexity, previously called part C.

To tasks increased level There are only five (5) tasks of complexity - No. 30,31,32,33,34 and 35. Let's consider the topics of the tasks, how to prepare for them and how to solve complex tasks in the Unified State Exam in Chemistry 2018.

Example of task 30 in the Unified State Examination in Chemistry 2018

Aimed at testing the student's knowledge about oxidation-reduction reactions (ORR). The assignment always gives the equation chemical reaction with missing substances from either side of the reaction ( left-hand side- reagents, right side - products). A maximum of three (3) points may be awarded for this assignment. The first point is given for correctly filling in the gaps in the reaction and correct equalization of the reaction (arrangement of coefficients). The second point can be obtained by correctly describing the ORR balance, and the last point is given for correctly determining who is the oxidizing agent in the reaction and who is the reducing agent. Let's look at the solution to task No. 30 from the demo version of the Unified State Exam in Chemistry 2018:

Using the electron balance method, create an equation for the reaction

Na 2 SO 3 + … + KOH à K 2 MnO 4 + … + H 2 O

Identify the oxidizing agent and the reducing agent.

The first thing you need to do is arrange the charges of the atoms indicated in the equation, it turns out:

Na + 2 S +4 O 3 -2 + … + K + O -2 H + à K + 2 Mn +6 O 4 -2 + … + H + 2 O -2

Often after this action, we immediately see the first pair of elements that changed the oxidation state (CO), that is, from different sides of the reaction, for the same atom, varying degrees oxidation. In this particular task, we do not observe this. Therefore, it is necessary to take advantage of additional knowledge, namely, on the left side of the reaction, we see potassium hydroxide ( CON), the presence of which tells us that the reaction occurs in an alkaline environment. WITH right side, we see potassium manganate, and we know that in an alkaline reaction medium, potassium manganate is obtained from potassium permanganate, therefore, the gap on the left side of the reaction is potassium permanganate ( KMnO 4 ). It turns out that on the left we had manganese at CO +7, and on the right at CO +6, which means we can write the first part of the OVR balance:

Mn +7 +1 e — à Mn +6

Now, we can guess what else should happen in the reaction. If manganese receives electrons, then someone must have given them to it (we follow the law of conservation of mass). Let's consider all the elements on the left side of the reaction: hydrogen, sodium and potassium are already in CO +1, which is the maximum for them, oxygen will not give up its electrons to manganese, which means sulfur remains in CO +4. We conclude that sulfur gives up electrons and goes into the sulfur state with CO +6. Now we can write the second part of the balance sheet:

S +4 -2 e — à S +6

Looking at the equation, we see that on the right hand side, there is no sulfur or sodium anywhere, which means they must be in the gap, and the logical compound to fill it is sodium sulfate ( NaSO 4 ).

Now the OVR balance is written (we get the first point) and the equation takes the form:

Na 2 SO 3 + KMnO 4 + KOHà K 2 MnO 4 + NaSO 4 + H 2 O

It is important at this point to immediately write who is the oxidizing agent and who is the reducing agent, since students often concentrate on balancing the equation and simply forget to do this part of the task, thereby losing a point. By definition, an oxidizing agent is the particle that receives electrons (in our case, manganese), and a reducing agent is the particle that gives up electrons (in our case, sulfur), so we get:

Oxidizer: Mn +7 (KMnO 4 )

Reducing agent: S +4 (Na 2 SO 3 )

Here we must remember that we are indicating the state of the particles in which they were when they began to exhibit the properties of an oxidizing or reducing agent, and not the states to which they came as a result of redox reaction.

Now, in order to get the last point, you need to correctly equalize the equation (arrange the coefficients). Using the balance, we see that in order for it to be sulfur +4, to go into the +6 state, two manganese +7 must become manganese +6, and what matters is we put 2 in front of the manganese:

Na 2 SO 3 + 2KMnO 4 + KOHà 2K 2 MnO 4 + NaSO 4 + H 2 O

Now we see that we have 4 potassium on the right, and only three on the left, which means we need to put 2 in front of potassium hydroxide:

Na 2 SO 3 + 2KMnO 4 + 2KOHà 2K 2 MnO 4 + NaSO 4 + H 2 O

As a result, the correct answer to task No. 30 looks like this:

Na 2 SO 3 + 2KMnO 4 + 2KOHà 2K 2 MnO 4 + NaSO 4 + H 2 O

Oxidizer: Mn +7 (KMnO 4)

Reducing agent: S +4 (Na 2 SO 3 )

Solution to task 31 in the Unified State Exam in chemistry

This is a chain of inorganic transformations. To successfully complete this task, you must have a good understanding of the reactions characteristic of inorganic compounds. The task consists of four (4) reactions, for each of which you can get one (1) point, for a total of four (4) points for the task. It is important to remember the rules for completing the assignment: all equations must be equalized, even if a student wrote the equation correctly but did not equalize, he will not receive a point; it is not necessary to solve all the reactions, you can do one and get one (1) point, two reactions and get two (2) points, etc., and it is not necessary to complete the equations strictly in order, for example, a student can do reaction 1 and 3, which means you need to do this and get two (2) points, the main thing is to indicate that these are reactions 1 and 3. Let’s look at the solution to task No. 31 from the demo version of the Unified State Exam in Chemistry 2018:

Iron was dissolved in hot concentrated sulfuric acid. The resulting salt was treated with an excess of sodium hydroxide solution. The brown precipitate that formed was filtered and calcined. The resulting substance was heated with iron.
Write equations for the four reactions described.

To make the solution easier, you can draw up the following diagram in a draft:

To complete the task, of course, you need to know all the proposed reactions. However, there are always hidden clues in the condition (concentrated sulfuric acid, excess sodium hydroxide, brown precipitate, calcined, heated with iron). For example, a student does not remember what happens to iron when interacting with conc. sulfuric acid, but he remembers that the brown precipitate of iron after treatment with alkali is most likely iron hydroxide 3 ( Y = Fe(OH) 3 ). Now we have the opportunity, by substituting Y into the written diagram, to try to make equations 2 and 3. The subsequent steps are purely chemical, so we will not describe them in such detail. The student must remember that heating iron hydroxide 3 results in the formation of iron oxide 3 ( Z = Fe 2 O 3 ) and water, and heating iron oxide 3 with pure iron will lead them to the middle state - iron oxide 2 ( FeO). Substance X, which is a salt obtained after reaction with sulfuric acid, yielding iron hydroxide 3 after treatment with alkali, will be iron sulfate 3 ( X = Fe 2 (SO 4 ) 3 ). It is important to remember to balance the equations. As a result, the correct answer to task No. 31 looks like this:

Task 32 Unified State Exam in Chemistry

Very similar to task No. 31, only it contains a chain of organic transformations. The design requirements and solution logic are similar to task No. 31, the only difference is that in task No. 32 five (5) equations are given, which means you can score five (5) points in total. Due to its similarity to task No. 31, we will not consider it in detail.

Solution to task 33 in chemistry 2018

A calculation task, to complete it you need to know the basic calculation formulas, be able to use a calculator and draw logical parallels. Assignment 33 is worth four (4) points. Let's look at part of the solution to task No. 33 from the demo version of the Unified State Exam in Chemistry 2018:

Determine the mass fractions (in %) of iron (II) sulfate and aluminum sulfide in the mixture if, when treating 25 g of this mixture with water, a gas was released that completely reacted with 960 g of a 5% solution of copper sulfate. In your answer, write down the reaction equations that indicated in the problem statement, and provide all the necessary calculations (indicate the units of measurement of the required physical quantities).

We get the first (1) point for writing the reactions that occur in the problem. Obtaining this particular point depends on knowledge of chemistry, the remaining three (3) points can only be obtained through calculations, therefore, if a student has problems with mathematics, he must receive at least one (1) point for completing task No. 33:

Since further actions are purely mathematical, we will not go into detail here. You can watch a selection of the analysis on our YouTube channel (link to the video analysis of task No. 33).

Formulas that will be required to solve this task:

Chemistry assignment 34 2018

Calculation task, which differs from task No. 33 in the following:

• • • If in task No. 33 we know between which substances the interaction occurs, then in task No. 34 we must find what reacted;
    • In task No. 34 organic compounds are given, while in task No. 33 inorganic processes are most often given.

In fact, task No. 34 is the reverse of task No. 33, which means the logic of the task is reverse. For task No. 34 you can get four (4) points, and, as in task No. 33, only one of them (in 90% of cases) is obtained for knowledge of chemistry, the remaining 3 (less often 2) points are obtained for mathematical calculations. To successfully complete task No. 34 you must:

Know the general formulas of all main classes of organic compounds;

Know the basic reactions of organic compounds;

Be able to write an equation in general form.

Once again I would like to note that necessary for successful passing the Unified State Exam in chemistry in 2018, the theoretical bases have remained virtually unchanged, which means that all the knowledge that your child received at school will help him pass the chemistry exam in 2018. In our center for preparing for the Unified State Exam and the Unified State Examination Hodograph, your child will receive All necessary for preparation theoretical materials, and in the classroom will consolidate the acquired knowledge for successful implementation everyone exam assignments. The best teachers who have undergone very advanced training will work with him. big competition and difficult entrance tests. Classes are held in small groups, which allows the teacher to devote time to each child and formulate his individual implementation strategy exam paper.

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

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Unified State Exam 2017 Chemistry Typical test tasks Medvedev

M.: 2017. - 120 p.

Typical test tasks in chemistry contain 10 versions of sets of tasks, compiled taking into account all the features and requirements of the Unified State Exam in 2017. The purpose of the manual is to provide readers with information about the structure and content of the 2017 KIM in chemistry, the degree of difficulty of the tasks. The collection contains answers to all test options and provides solutions to all tasks of one of the options. In addition, samples of forms used in the Unified State Exam for recording answers and solutions are provided. The author of the assignments is a leading scientist, teacher and methodologist who is directly involved in the development of control measuring instruments Unified State Exam materials. The manual is intended for teachers to prepare students for the chemistry exam, as well as for high school students and graduates - for self-preparation and self-control.

Format: pdf

Size: 1.5 MB

Watch, download:drive.google

CONTENT
Preface 4
Instructions for performing work 5
OPTION 1 8
Part 1 8
Part 2, 15
OPTION 2 17
Part 1 17
Part 2 24
OPTION 3 26
Part 1 26
Part 2 33
OPTION 4 35
Part 1 35
Part 2 41
OPTION 5 43
Part 1 43
Part 2 49
OPTION 6 51
Part 1 51
Part 2 57
OPTION 7 59
Part 1 59
Part 2 65
OPTION 8 67
Part 1 67
Part 2 73
OPTION 9 75
Part 1 75
Part 2 81
OPTION 10 83
Part 1 83
Part 2 89
ANSWERS AND SOLUTIONS 91
Answers to the tasks of part 1 91
Solutions and answers to the tasks of part 2 93
Solving problems of option 10 99
Part 1 99
Part 2 113

The present tutorial is a collection of tasks for preparing for the Unified State Exam (USE) in chemistry, which is like the final exam for the course high school, and the entrance exam to the university. The structure of the manual reflects modern requirements for the procedure for passing the Unified State Exam in Chemistry, which will allow you to better prepare for new forms of final certification and for admission to universities.
The manual consists of 10 variants of tasks, which in form and content are close to the demo version of the Unified State Exam and do not go beyond the content of the chemistry course, normatively determined by the Federal component of the state standard of general education. Chemistry (Order of the Ministry of Education No. 1089 of 03/05/2004).
Content presentation level educational material in the assignments it is correlated with the requirements of the state standard for the preparation of secondary (full) school graduates in chemistry.
The control measurement materials of the Unified State Exam use tasks of three types:
- tasks basic level difficulty with a short answer,
- tasks of an increased level of complexity with a short answer,
- tasks high level difficulties with a detailed answer.
Each version of the examination paper is built according to a single plan. The work consists of two parts, including a total of 34 tasks. Part 1 contains 29 short answer questions, including 20 basic level tasks and 9 advanced level tasks. Part 2 contains 5 tasks of a high level of complexity, with detailed answers (tasks numbered 30-34).
In tasks of a high level of complexity, the text of the solution is written on a special form. Tasks of this type make up the bulk of the written work in chemistry in university entrance exams.

The Unified State Exam in Chemistry is an exam taken by graduates planning to enter a university for certain specialties related to this discipline. Chemistry is not included in the list of compulsory subjects; according to statistics, 1 out of 10 graduates takes chemistry.

• The graduate receives 3 hours of time to test and complete all tasks - planning and distributing time to work with all tasks is an important task for the test taker.
• Usually the exam includes 35-40 tasks, which are divided into 2 logical blocks.
• Like the rest of the Unified State Exam, the chemistry test is divided into 2 logical blocks: testing (selecting the correct option or options from those proposed) and questions that require detailed answers. It is the second block that usually takes longer, so the subject needs to manage time rationally.

• The main thing is to have reliable, deep theoretical knowledge that will help you successfully complete various tasks of the first and second blocks.
• You need to start preparing in advance in order to systematically work through all the topics - six months may not be enough. The best option– start preparing in the 10th grade.
• Identify the topics that give you the most problems so that when you ask a teacher or tutor for help, you know what to ask.
• Learn to perform tasks typical for the Unified State Exam in chemistry - it is not enough to master the theory, you need to develop the skills to complete tasks and various tasks to automaticity.
  

• Not always self-study effective, so it’s worth finding a specialist to whom you can turn for help. The best option is a professional tutor. Also, don’t be afraid to ask questions to your school teacher. Do not neglect your school education, carefully complete assignments in class!
• There are hints in the exam! The main thing is to learn how to use these sources of information. The student has the periodic table, tables of metal stress and solubility - this is about 70% of the data that will help to understand various tasks.

• Chemistry requires a thorough knowledge of mathematics - without this it will be difficult to solve problems. Be sure to repeat the work with percentages and proportions.
• Learn the formulas needed to solve chemistry problems.
• Study the theory: textbooks, reference books, collections of problems will be useful.
• The best way to consolidate theoretical assignments is to actively solve chemistry assignments. IN online mode you can solve in any quantity, improve your problem solving skills different types and level of difficulty.
• Controversial issues in assignments and errors are recommended to be sorted out and analyzed with the help of a teacher or tutor.

Determine which atoms of the elements indicated in the series contain one unpaired electron in the ground state.
Write down the numbers of the selected elements in the answer field.
Answer:

Answer: 23
Explanation:
Let's write down the electronic formula for each of the indicated chemical elements and depict the electron-graphic formula of the last electronic level:
1) S: 1s 2 2s 2 2p 6 3s 2 3p 4

2) Na: 1s 2 2s 2 2p 6 3s 1

3) Al: 1s 2 2s 2 2p 6 3s 2 3p 1

4) Si: 1s 2 2s 2 2p 6 3s 2 3p 2

5) Mg: 1s 2 2s 2 2p 6 3s 2

From the chemical elements indicated in the series, select three metal elements. Arrange the selected elements in order of increasing reducing properties.

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

Answer: 352
Explanation:
In the main subgroups of the periodic table, metals are located under the boron-astatine diagonal, as well as in secondary subgroups. Thus, the metals from this list include Na, Al and Mg.
The metallic and, therefore, reducing properties of the elements increase when moving to the left along the period and down the subgroup.
Thus, the metallic properties of the metals listed above increase in the order Al, Mg, Na

From among the elements indicated in the series, select two elements that, when combined with oxygen, exhibit an oxidation state of +4.

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

Answer: 14
Explanation:
The main oxidation states of elements from the presented list in complex substances:
Sulfur – “-2”, “+4” and “+6”
Sodium Na – “+1” (single)
Aluminum Al – “+3” (single)
Silicon Si – “-4”, “+4”
Magnesium Mg – “+2” (single)

From the proposed list of substances, select two substances in which an ionic chemical bond is present.

Answer: 12

Explanation:

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 atoms of a non-metal.

Based on this criterion, the ionic type of bond occurs in the compounds KCl and KNO 3.

In addition to the above characteristic, 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 analogues - alkylammonium cations RNH 3 + , dialkylamonium R 2NH2+ , trialkylammonium R 3NH+ and tetraalkylammonium R 4N+ , where R is some hydrocarbon radical. For example, the ionic type of bond occurs in the compound (CH 3 ) 4 NCl between the cation (CH 3 ) 4 + and chloride ion Cl − .

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 indicated by a number.

Answer: 241

Explanation:

N 2 O 3 is a non-metal oxide. All non-metal oxides except N 2 O, NO, SiO and CO are acidic.

Al 2 O 3 is a metal oxide in the oxidation state +3. Metal oxides in the oxidation state +3, +4, as well as BeO, ZnO, SnO and PbO, are amphoteric.

HClO 4 is a typical representative of acids, because upon dissociation in an aqueous solution, only H + cations are formed from cations:

HClO 4 = H + + ClO 4 -

From the proposed list of substances, select two substances, with each of which zinc interacts.

1) nitric acid (solution)

2) iron(II) hydroxide

3) magnesium sulfate (solution)

4) sodium hydroxide (solution)

5) aluminum chloride (solution)

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

Answer: 14

Explanation:

1) Nitric acid is a strong oxidizing agent and reacts with all metals except platinum and gold.

2) Iron hydroxide (ll) is an insoluble base. Metals do not react at all with insoluble hydroxides, and only three metals react with soluble (alkalis) - Be, Zn, Al.

3) Magnesium sulfate – salt more active metal, than zinc, and therefore the reaction does not proceed.

4) Sodium hydroxide - alkali (soluble metal hydroxide). Only Be, Zn, Al work with metal alkalis.

5) AlCl 3 – a salt of a metal more active than zinc, i.e. reaction is impossible.

From the proposed list of substances, select two oxides that react with water.

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

Answer: 14

Explanation:

Of the oxides, only oxides of alkali and alkaline earth metals, as well as all acidic oxides except SiO 2, react with water.

Thus, answer options 1 and 4 are suitable:

BaO + H 2 O = Ba(OH) 2

SO 3 + H 2 O = H 2 SO 4

1) hydrogen bromide

3) sodium nitrate

4) sulfur oxide(IV)

5) aluminum chloride

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

Answer: 52

Explanation:

The only salts among these substances are sodium nitrate and aluminum chloride. All nitrates, like sodium salts, are soluble, and therefore sodium nitrate cannot form a precipitate in principle with any of the reagents. Therefore, salt X can only be aluminum chloride.

A common mistake among those taking the Unified State Exam in chemistry is a failure to understand that in an aqueous solution ammonia forms a weak base - ammonium hydroxide due to the reaction occurring:

NH 3 + H 2 O<=>NH4OH

In this regard, an aqueous solution of ammonia gives a precipitate when mixed with solutions of metal salts, forming insoluble hydroxides:

3NH 3 + 3H 2 O + AlCl 3 = Al(OH) 3 + 3NH 4 Cl

In a given transformation scheme

Cu X > CuCl 2 Y > CuI

substances X and Y are:

Answer: 35

Explanation:

Copper is a metal located in the activity series to the right of hydrogen, i.e. does not react with acids (except for H 2 SO 4 (conc.) and HNO 3). Thus, the formation of copper (ll) chloride is possible in our case only by reaction with chlorine:

Cu + Cl 2 = CuCl 2

Iodide ions (I -) cannot coexist in the same solution with divalent copper ions, because are oxidized by them:

Cu 2+ + 3I - = CuI + I 2

Establish a correspondence between the reaction equation and the oxidizing substance in this reaction: 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.

Answer: 1433
Explanation:
An oxidizing agent in a reaction is a substance that contains an element that lowers its oxidation state

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.

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

Answer: 1215

Explanation:

A) Cu(NO 3) 2 + NaOH and Cu(NO 3) 2 + Ba(OH) 2 – similar interactions. A salt reacts with a metal hydroxide if the starting substances are soluble, and the products contain a precipitate, gas or slightly dissociating substance. For both the first and second reactions, both requirements are met:

Cu(NO 3) 2 + 2NaOH = 2NaNO 3 + Cu(OH) 2 ↓

Cu(NO 3) 2 + Ba(OH) 2 = Na(NO 3) 2 + Cu(OH) 2 ↓

Cu(NO 3) 2 + Mg - a salt reacts with a metal if the free metal is more active than what is included in the salt. Magnesium in the activity series is located to the left of copper, which indicates its greater activity, therefore, the reaction proceeds:

Cu(NO 3) 2 + Mg = Mg(NO 3) 2 + Cu

B) Al(OH) 3 – metal hydroxide in the oxidation state +3. Metal hydroxides in the oxidation state +3, +4, as well as the hydroxides Be(OH) 2 and Zn(OH) 2 as exceptions, are classified as amphoteric.

By definition, amphoteric hydroxides are those that react with alkalis and almost all soluble acids. For this reason, we can immediately conclude that answer option 2 is appropriate:

Al(OH) 3 + 3HCl = AlCl 3 + 3H 2 O

Al(OH) 3 + LiOH (solution) = Li or Al(OH) 3 + LiOH(sol.) =to=> LiAlO 2 + 2H 2 O

2Al(OH) 3 + 3H 2 SO 4 = Al 2 (SO 4) 3 + 6H 2 O

C) ZnCl 2 + NaOH and ZnCl 2 + Ba(OH) 2 – interaction of the “salt + metal hydroxide” type. The explanation is given in paragraph A.

ZnCl 2 + 2NaOH = Zn(OH) 2 + 2NaCl

ZnCl 2 + Ba(OH) 2 = Zn(OH) 2 + BaCl 2

It should be noted that with an excess of NaOH and Ba(OH) 2:

ZnCl 2 + 4NaOH = Na 2 + 2NaCl

ZnCl 2 + 2Ba(OH) 2 = Ba + BaCl 2

D) Br 2, O 2 are strong oxidizing agents. The only metals that do not react are silver, platinum, and gold:

Cu + Br 2 t° > CuBr 2

2Cu + O2 t° >2CuO

HNO 3 is an acid with strong oxidizing properties, because oxidizes not with hydrogen cations, but with an acid-forming element - nitrogen N +5. Reacts with all metals except platinum and gold:

4HNO 3(conc.) + Cu = Cu(NO 3)2 + 2NO 2 + 2H 2 O

8HNO 3(dil.) + 3Cu = 3Cu(NO 3) 2 + 2NO + 4H 2 O

Establish a correspondence between the general formula of a homologous series and the name of a substance belonging to this series: 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.

Answer: 231

Explanation:

From the proposed list of substances, select two substances that are isomers of cyclopentane.

1) 2-methylbutane

2) 1,2-dimethylcyclopropane

3) penten-2

4) hexene-2

5) cyclopentene

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

Answer: 23
Explanation:
Cyclopentane has the molecular formula C5H10. Let's write the structural and molecular formulas of the substances listed in the condition

From the proposed list of substances, select two substances, each of which reacts with a solution of potassium permanganate.

1) methylbenzene

2) cyclohexane

3) methylpropane

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

Answer: 15

Explanation:

Of the hydrocarbons that react with an aqueous solution of potassium permanganate, those that contain structural formula C=C or C≡C bonds, as well as homologues of benzene (except benzene itself).
Methylbenzene and styrene are suitable in this way.

From the proposed list of substances, select two substances with which phenol interacts.

1) hydrochloric acid

2) sodium hydroxide

4) nitric acid

5) sodium sulfate

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

Answer: 24

Explanation:

Phenol has weak acidic properties, more pronounced than alcohols. For this reason, phenols, unlike alcohols, react with alkalis:

C 6 H 5 OH + NaOH = C 6 H 5 ONa + H 2 O

Phenol contains in its molecule a hydroxyl group directly attached to the benzene ring. The hydroxy group is an orienting agent of the first kind, that is, it facilitates substitution reactions in the ortho and para positions:

From the proposed list of substances, select two substances that undergo hydrolysis.

1) glucose

2) sucrose

3) fructose

5) starch

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

Answer: 25

Explanation:

All of the substances listed are carbohydrates. Of carbohydrates, monosaccharides do not undergo hydrolysis. Glucose, fructose and ribose are monosaccharides, sucrose is a disaccharide, and starch is a polysaccharide. Therefore, sucrose and starch from the above list are subject to hydrolysis.

The following scheme of substance transformations is specified:

1,2-dibromoethane → X → bromoethane → Y → ethyl formate

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

2) ethanal

4) chloroethane

5) acetylene

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

Answer: 31

Explanation:

Establish a correspondence between the name of the starting substance and the product, which is mainly formed when this substance reacts with bromine: 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.

Answer: 2134

Explanation:

Substitution at the secondary carbon atom occurs in to a greater extent than with the primary one. Thus, the main product of propane bromination is 2-bromopropane, not 1-bromopropane:

Cyclohexane is a cycloalkane with a ring size of more than 4 carbon atoms. Cycloalkanes with a ring size of more than 4 carbon atoms, when interacting with halogens, enter into a substitution reaction with preservation of the cycle:

Cyclopropane and cyclobutane are cycloalkanes with minimum size cycles predominantly undergo addition reactions accompanied by ring rupture:

The replacement of hydrogen atoms at the tertiary carbon atom occurs to a greater extent than at the secondary and primary ones. Thus, the bromination of isobutane proceeds mainly as follows:

Establish a correspondence between the reaction scheme and the organic substance that is the product of this reaction: 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.

Answer: 6134

Explanation:

Heating aldehydes with freshly precipitated copper hydroxide leads to the oxidation of the aldehyde group to a carboxyl group:

Aldehydes and ketones are reduced by hydrogen in the presence of nickel, platinum or palladium to alcohols:

Primary and secondary alcohols are oxidized by hot CuO to aldehydes and ketones, respectively:

When concentrated sulfuric acid reacts with ethanol upon heating, two different products may form. When heated to a temperature below 140 °C, intermolecular dehydration occurs predominantly with the formation of diethyl ether, and when heated above 140 °C, intramolecular dehydration occurs, resulting in the formation of ethylene:

From the proposed list of substances, select two substances whose thermal decomposition reaction is redox.

1) aluminum nitrate

2) potassium bicarbonate

3) aluminum hydroxide

4) ammonium carbonate

5) ammonium nitrate

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

Answer: 15

Explanation:

Redox reactions are those reactions in which one or more chemical elements change their oxidation state.

The decomposition reactions of absolutely all nitrates are redox reactions. Metal nitrates from Mg to Cu inclusive decompose to metal oxide, nitrogen dioxide and molecular oxygen:

All metal bicarbonates decompose even with slight heating (60 o C) to metal carbonate, carbon dioxide and water. In this case, no change in oxidation states occurs:

Insoluble oxides decompose when heated. The reaction is not redox because Not a single chemical element changes its oxidation state as a result:

Ammonium carbonate decomposes when heated into carbon dioxide, water and ammonia. The reaction is not redox:

Ammonium nitrate decomposes into nitric oxide (I) and water. The reaction relates to OVR:

From the proposed list, select two external influences that lead to an increase in the rate of reaction of nitrogen with hydrogen.

1) decrease in temperature

2) increase in pressure in the system

5) use of an inhibitor

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

Answer: 24

Explanation:

1) temperature decrease:

The rate of any reaction decreases as the temperature decreases

2) increase in pressure in the system:

Increasing pressure increases the rate of any reaction in which at least one gaseous substance takes part.

3) decrease in hydrogen concentration

Decreasing the concentration always reduces the reaction rate

4) increase in nitrogen concentration

Increasing the concentration of reagents always increases the reaction rate

5) use of an inhibitor

Inhibitors are substances that slow down the rate of a reaction.

Establish a correspondence between the formula of a substance and the products of electrolysis of an aqueous solution of this substance on inert electrodes: 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.

Answer: 5251

Explanation:

A) NaBr → Na + + Br -

Na+ cations and water molecules compete with each other for the cathode.

2H 2 O + 2e — → H 2 + 2OH —

2Cl - -2e → Cl 2

B) Mg(NO 3) 2 → Mg 2+ + 2NO 3 —

Mg 2+ cations and water molecules compete with each other for the cathode.

Alkali metal cations, as well as magnesium and aluminum, are not able to be reduced in an aqueous solution due to their high activity. For this reason, water molecules are reduced instead according to the equation:

2H 2 O + 2e — → H 2 + 2OH —

NO3 anions and water molecules compete with each other for the anode.

2H 2 O - 4e - → O 2 + 4H +

So answer 2 (hydrogen and oxygen) is appropriate.

B) AlCl 3 → Al 3+ + 3Cl -

Alkali metal cations, as well as magnesium and aluminum, are not able to be reduced in an aqueous solution due to their high activity. For this reason, water molecules are reduced instead according to the equation:

2H 2 O + 2e — → H 2 + 2OH —

Cl anions and water molecules compete with each other for the anode.

Anions consisting of one chemical element(except F -) win competition from water molecules for oxidation at the anode:

2Cl - -2e → Cl 2

Therefore, answer option 5 (hydrogen and halogen) is appropriate.

D) CuSO 4 → Cu 2+ + SO 4 2-

Metal cations to the right of hydrogen in the activity series are easily reduced under aqueous solution conditions:

Cu 2+ + 2e → Cu 0

Acidic residues containing an acid-forming element in highest degree oxidation, lose competition to water molecules for oxidation at the anode:

2H 2 O - 4e - → O 2 + 4H +

Thus, answer option 1 (oxygen and metal) is appropriate.

Establish a correspondence between the name of the salt and the medium of the aqueous solution of this salt: 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.

Answer: 3312

Explanation:

A) iron(III) sulfate - Fe 2 (SO 4) 3

formed by a weak “base” Fe(OH) 3 and a strong acid H 2 SO 4. Conclusion - the environment is acidic

B) chromium(III) chloride - CrCl 3

formed by the weak “base” Cr(OH) 3 and the strong acid HCl. Conclusion - the environment is acidic

B) sodium sulfate - Na 2 SO 4

Formed by the strong base NaOH and the strong acid H 2 SO 4. Conclusion - the environment is neutral

D) sodium sulfide - Na 2 S

Formed by the strong base NaOH and the weak acid H2S. Conclusion - the environment is alkaline.

Establish a correspondence between the method of influencing the equilibrium system

CO (g) + Cl 2 (g) COCl 2 (g) + Q

and the direction of the shift in chemical equilibrium as a result of this effect: 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.

Answer: 3113

Explanation:

The equilibrium shift under external influence on the system occurs in such a way as to minimize the effect of this external influence (Le Chatelier's principle).

A) An increase in the concentration of CO causes the equilibrium to shift toward the forward reaction because it results in a decrease in the amount of CO.

B) An increase in temperature will shift the equilibrium towards an endothermic reaction. Since the forward reaction is exothermic (+Q), the equilibrium will shift towards the reverse reaction.

C) A decrease in pressure will shift the equilibrium towards the reaction that results in an increase in the amount of gases. As a result of the reverse reaction, more gases are formed than as a result of the direct reaction. Thus, the equilibrium will shift towards the opposite reaction.

D) An increase in the concentration of chlorine leads to a shift in equilibrium towards the direct reaction, since as a result it reduces the amount of chlorine.

Establish a correspondence between two substances and a reagent that can be used to distinguish these substances: 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.

Answer: 3454

Explanation:

It is possible to distinguish two substances with the help of a third one only if these two substances interact with it differently, and, most importantly, these differences are externally distinguishable.

A) Solutions of FeSO 4 and FeCl 2 can be distinguished using a solution of barium nitrate. In the case of FeSO 4, a white precipitate of barium sulfate forms:

FeSO 4 + BaCl 2 = BaSO 4 ↓ + FeCl 2

In the case of FeCl 2 there are no visible signs of interaction, since the reaction does not occur.

B) Solutions of Na 3 PO 4 and Na 2 SO 4 can be distinguished using a solution of MgCl 2. The Na 2 SO 4 solution does not react, and in the case of Na 3 PO 4 a white precipitate of magnesium phosphate precipitates:

2Na 3 PO 4 + 3MgCl 2 = Mg 3 (PO 4) 2 ↓ + 6NaCl

C) Solutions of KOH and Ca(OH) 2 can be distinguished using a solution of Na 2 CO 3. KOH does not react with Na 2 CO 3, but Ca(OH) 2 gives a white precipitate of calcium carbonate with Na 2 CO 3:

Ca(OH) 2 + Na 2 CO 3 = CaCO 3 ↓ + 2NaOH

D) Solutions of KOH and KCl can be distinguished using a solution of MgCl 2. KCl does not react with MgCl 2, and mixing solutions of KOH and MgCl 2 leads to the formation of a white precipitate of magnesium hydroxide:

MgCl 2 + 2KOH = Mg(OH) 2 ↓ + 2KCl

Establish a correspondence between the substance and its area of ​​application: 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.

Answer: 2331
Explanation:
Ammonia - used in the production of nitrogenous fertilizers. In particular, ammonia is a raw material for the production of nitric acid, from which, in turn, fertilizers are obtained - sodium, potassium and ammonium nitrate (NaNO 3, KNO 3, NH 4 NO 3).
Carbon tetrachloride and acetone are used as solvents.
Ethylene is used to produce high molecular weight compounds (polymers), namely polyethylene.

Tips for preparing for the Unified State Exam in chemistry on the website website

How to competently pass the Unified State Exam (and Unified State Exam) in chemistry? If you only have 2 months and you’re not ready yet? And don’t be friends with chemistry...

It offers tests with answers for each topic and task, by passing which you can study the basic principles, patterns and theory found in the Unified State Exam in chemistry. Our tests allow you to find answers to most questions encountered in the Unified State Exam in chemistry, and our tests allow you to consolidate the material, find weak spots, and work through the material.

All you need is the Internet, stationery, time and a website. It is best to have a separate notebook for formulas/solutions/notes and a dictionary of trivial names of compounds.

1. From the very beginning, you need to assess your current level and the number of points you need, for this it is worth going through. If everything is very bad and you need excellent performance, congratulations, even now all is not lost. You can train yourself to pass successfully without the help of a tutor.
   Decide on minimum quantity points you want to score, this will allow you to understand how many tasks you must solve accurately to get the score you need.
   Naturally, take into account that everything may not go so smoothly and solve it as best you can. larger number tasks, or better yet, everything. The minimum that you have determined for yourself - you must decide ideally.
2. Let's move on to the practical part - training for the solution.
   Most effective method- next. Select only the exam you are interested in and solve the corresponding test. About 20 solved tasks guarantee that you will meet all types of problems. As soon as you begin to feel that you know how to solve every task you see from beginning to end, proceed to the next task. If you don’t know how to solve a task, use the search on our website. There is almost always a solution on our website, otherwise just write to the tutor by clicking on the icon in the lower left corner - it’s free.
3. At the same time, we repeat the third point for everyone on our website, starting with.
4. When the first part is given to you at least at an average level, you begin to decide. If one of the tasks is difficult, and you made a mistake in completing it, then return to the tests on this task or the corresponding topic with tests.
5. Part 2. If you have a tutor, focus on studying this part with him. (provided that you are able to solve the rest at least 70%). If you started part 2, then you should score a passing grade without any problems 100% of the time. If this does not happen, it is better to stay on the first part for now. When you are ready for part 2, we recommend that you get a separate notebook where you will write down only the solutions to part 2. The key to success is solving as many tasks as possible, just like in part 1.