The Unified State Examination in Chemistry is a variable component of the federal examination. It is taken only by those schoolchildren who are planning to continue their studies at universities in such specialties as medicine, chemistry and chemical technology, construction, biotechnology or food industry.

This cannot be called easy - you won’t be able to get through it simply by knowing the terms, because last years Tests with a choice of one answer from the proposed options were excluded from the KIMs. In addition, it will not be amiss to learn everything about the procedure, timing and features of this exam, as well as prepare in advance for possible changes in the 2018 KIMs!

Demo version of the Unified State Exam-2018

Unified State Examination dates in chemistry

The exact dates allotted for writing the Unified State Exam in chemistry will be known in January, when the schedule for all examination tests will be posted on the Rosobrnadzor website. Fortunately, today we already have information about the approximate periods allocated for the examination of schoolchildren in the 2017/2018 academic year:

• The early stage of the examination starts on March 22, 2018. It will last until April 15. Writing the Unified State Examination ahead of schedule is the prerogative of several categories of students. These include children who graduated from school before 2017/2018 school year, but did not pass the Unified State Exam for any reason; school graduates who previously received only a certificate and not a matriculation certificate; evening school students; high school students who go to live or study abroad; schoolchildren who received secondary education in other states, but enter. Also, students representing the Russian Federation at international competitions and competitions, and schoolchildren who take part in all-Russian events, use early delivery. If you are indicated for medical intervention or rehabilitation, which coincides with the main period of passing the Unified State Exam, you can also take the exam ahead of schedule. Important point: any reason must be confirmed by relevant documents;
• On May 28, 2018 the main Unified State Exam dates. According to preliminary plans of Rosobrnadzor, the examination period will end before June 10;
• On September 4, 2018, an additional period for taking the Unified State Exam will begin.

Some statistics

Recently, an increasing number of schoolchildren are choosing this exam - in 2017, about 74 thousand people took it (12 thousand more than in 2016). In addition, the success rate has noticeably improved - the number of unsuccessful students (those who did not reach the minimum threshold score) decreased by 1.1%. Average score in this subject ranges from 67.8-56.3 points, which corresponds to the school “four” level. So, in general, despite its complexity, students pass this subject quite well.

Examination procedure

When writing this Unified State Examination, students are allowed to use the periodic system, a table with data on the solubility of salts, acids and bases, as well as reference materials electrochemical voltage series of metals. There is no need to take these materials with you - all permitted reference materials will be provided to schoolchildren in one set with the exam card. In addition, an eleventh grader can take a calculator for the exam that does not have a programming function.

We remind you that the procedure for conducting the Unified State Exam strictly regulates any actions of students. Remember that you can easily lose your chance to enter a university if you suddenly want to discuss the solution to a problem with a friend, try to peek the answer in a smartphone or a workbook, or decide to call someone from the restroom. By the way, you can go to the toilet or first aid station, but only with permission and accompanied by a member of the examination committee.

In 2018 year of the Unified State Exam in chemistry expanded to 35 tasks, allocating 3.5 hours to them

Innovations in the Unified State Examination in Chemistry

FIPI employees report the following changes in the new model CMMs.

1. In 2018, the number of complex tasks with detailed answers will be increased. One new task has been introduced, number 30, concerning redox reactions. Now students have to solve a total of 35 tasks.
2. You can still get 60 for all the work primary points. The balance was achieved by reducing the points awarded for completing simple tasks from the first part of the ticket.

What is included in the structure and content of the ticket?

In the exam, students will have to demonstrate how well they know the topics from the course of inorganic, general and organic chemistry. The tasks will test the depth of your knowledge about chemical elements and substances, skills in conducting chemical reactions, knowledge of the basic laws and theoretical principles of chemistry. In addition, it will become clear how well schoolchildren understand the systematicity and causality of chemical phenomena, and how much they know about the genesis of substances and methods of knowing them.

Structurally, the ticket is represented by 35 tasks, divided into two parts:

• Part 1 – 29 short answer tasks. These assignments are devoted to the theoretical foundations of chemistry, inorganic and organic chemistry, methods of knowledge and the use of chemistry in life. For this part of the KIM you can score 40 points (66.7% of all points for the ticket);
• Part 2 – 6 tasks of a high level of complexity, which provide a detailed answer. You will have to solve problems with non-standard situations. All tasks are devoted to redox reactions, ion exchange reactions, transformations of inorganic and organic substances, or complex calculations. For this part of the KIM you can score 20 points (33.3% of all points for the ticket).

In total, you can earn up to 60 primary points per ticket. You will be allocated 210 minutes to solve it, which you should distribute as follows:

• for basic tasks from the first part - 2-3 minutes;
• for tasks with an increased level of complexity from the first part - from 5 to 7 minutes;
• for tasks with a high level of complexity from the second part - from 10 to 15 minutes.

How do exam scores translate into grades?

Points for work affect the matriculation certificate, so for several years in a row they have been transferred to the marking system familiar to schoolchildren. First, the scores are divided into certain intervals and then converted into grades:

• 0-35 points are identical to “two”;
• 36-55 points indicate a satisfactory degree of preparation for the Unified State Exam and are equal to a “three”;
• 56-72 points is an opportunity to get a “B” in the certificate;
• 73 points and above is an indicator that the student knows the subject “excellent”.

High-quality preparation for the chemistry exam will allow you not only to enter the university of your choice, but to improve your grade on the certificate!

In order not to fail the Unified State Exam in Chemistry, you will have to score at least 36 points. However, it is worth remembering that for admission to more or less prestigious university you need to score at least 60-65 points. Top educational establishments and only those who score 85-90 points and above are accepted for the budget.

How to prepare for the Unified State Exam in Chemistry?

It is impossible to pass a federal level exam simply relying on residual knowledge from a school chemistry course. To fill in the gaps, it’s worth sitting down to textbooks and workbooks in early autumn! It is possible that some topic that you studied in the 9th or 10th grade simply did not stick in your memory. In addition, competent preparation includes the development of demonstration tickets - CIMs, specially developed by the FIPI commission.

It is advisable to start preparing for the Unified State Exam 2018 in chemistry for 11th grade graduates by familiarizing yourself with the demo versions of CMMs published on the official FIPI website. Also open bank FIPI assignments contains examples of real options included in exam tests.

Demo version of the Unified State Exam in Chemistry FIPI 2018, assignments with answers

Unified State Exam 2018 chemistry demo	Download demo version 2018
Specification	demo variant
Codifier	codifier

Total tasks – 35

Maximum Primary Score for work - 60.

The total time to complete the work is 210 minutes.

System for assessing the performance of individual tasks and exams Unified State Examination work 2018 in chemistry in general

Answers to tasks in Part 1 are automatically processed after scanning answer forms No. 1.

Answers to tasks in Part 2 are checked by a subject commission. For the correct answer to each of tasks 1–6, 11–15, 19–21, 26–29, 1 point is given.

The task is considered completed correctly if the examinee gave the correct answer in the form of a sequence of numbers or a number with a given degree of accuracy. Tasks 7–10, 16–18, 22–25 are considered completed correctly if the sequence of numbers is indicated correctly.

For a complete correct answer in tasks 7–10, 16–18, 22–25, 2 points are given; if one mistake is made - 1 point; for an incorrect answer (more than one error) or lack thereof – 0 points.

Part 2 tasks (with a detailed answer) involve checking from two to five answer elements.

Assignments with extended answers can be completed by graduates different ways. The presence of each required element of the answer is assessed by 1 point, so the maximum score for a correctly completed task is from 2 to 5 points, depending on the degree of complexity of the task: tasks 30 and 31 - 2 points; 32 – 4 points; 33 – 5 points; 34 – 4 points; 35 – 3 points.

Testing of assignments in Part 2 is carried out on the basis of an element-by-element analysis of the graduate’s response in accordance with the assignment assessment criteria.

We have developed practice tests in chemistry for the Unified State Exam 2020 with answers and solutions.

In preparation, study 10 training options, based on the new demo versions.

Features of tasks in Unified State Exam tests in chemistry

Let's look at the typology and structure of some tasks in the first part:

• – the condition contains a number of chemical elements and questions regarding each of them, pay attention to the number of cells for the answer - there are two of them, therefore, there are two solution options;
• – correspondence between two sets: there will be two columns, one contains formulas of substances, and the second contains a group of substances; it will be necessary to find correspondences.
• In the first part there will also be problems that require the behavior of a “chemical thought experiment”, in which the student chooses formulas that allow him to find the correct answer to the exam question.
• The tasks of the second block are higher in complexity and require mastery of several content elements and several skills.

Clue: when solving a problem, it is important to determine the class, group of substance and properties.

Assignments with detailed answers are aimed at testing knowledge in the main courses:

• Atomic structure;
• Periodic laws;
• Inorganic chemistry;
• Organic chemistry;
• Calculations using formulas;
• Application of chemistry in life.

Preparation for the Unified State Exam in Chemistry - quickly and efficiently

Fast- means, no less than six months:

1. Improve your math.
2. Repeat the whole theory.
3. Solve online trial options in chemistry, watch video lessons.

Our website has provided such an opportunity - come in, train and get high scores in exams.

Every year, demo versions of the current year's Unified State Exam are published on the official website of the FIPI.

On August 21, 2017, draft documents were presented regulating the structure and content of the KIM Unified State Exam 2018 (including demo version of the Unified State Exam in chemistry).

There are documents that regulate the structure and content of CMMs - codifier and specification.

Unified State Exam in Chemistry 2018 - demo version with answers and criteria from FIPI

Demo version of the Unified State Exam 2018 in chemistry	Download demo version 2018
Specification	demo variant
Codifier	codifier

Total tasks - 35; of which by level of difficulty: B – 21; P – 8; AT 6.

Maximum Primary Score for work - 60.

The total time to complete the work is 210 minutes.

Changes in the KIM Unified State Exam 2018 in Chemistry of the Year compared to 2017

The following changes have been made in the 2018 exam paper compared to the 2017 paper.

1. In order to more clearly distribute tasks into individual thematic blocks and content lines, the order of tasks of basic and advanced levels of complexity in part 1 of the examination paper has been slightly changed.

2. In the 2018 examination paper, the total number of tasks was increased from 34 (in 2017) to 35 due to an increase in the number of tasks in part 2 of the examination paper from 5 (in 2017) to 6 tasks. This is achieved through the introduction of tasks with a single context. In particular, this format presents tasks No. 30 and No. 31, which are aimed at testing the assimilation of important content elements: “Redox reactions” and “Ion exchange reactions.”

3. The grading scale for some tasks has been changed due to the clarification of the level of difficulty of these tasks based on the results of their completion in the 2017 examination paper:

Task No. 9 of an increased level of complexity, aimed at testing the mastery of the content element “Characteristic Chemical properties inorganic substances"and presented in a format for establishing correspondence between reacting substances and reaction products between these substances, will be assessed with a maximum of 2 points;

Task No. 21 basic level complexity, aimed at testing the assimilation of the content element “Reactions” redox” and presented in a format for establishing correspondence between the elements of two sets, will be scored 1 point;

Task No. 26 of a basic level of complexity, aimed at testing the assimilation of the content lines “Experimental foundations of chemistry” and “ General views on industrial methods for obtaining essential substances” and presented in a format for establishing correspondence between the elements of two sets, will be assessed 1 point;

Task No. 30 of a high level of complexity with a detailed answer, aimed at testing the assimilation of the content element “Redox reactions”, will be assessed with a maximum of 2 points;

Task No. 31 of a high level of complexity with a detailed answer, aimed at testing the assimilation of the content element “Ion exchange reactions”, will be assessed with a maximum of 2 points.

In general, the adopted changes in the 2018 examination work are aimed at increasing the objectivity of testing the formation of a number of important general educational skills, primarily such as: applying knowledge in the system, independently assessing the correctness of completing an educational and educational-practical task, as well as combining knowledge about chemical objects with an understanding of the mathematical relationship between various physical quantities.

Structure of KIM Unified State Exam 2018 in Chemistry

Each version of the examination paper is built according to a single plan: the paper consists of two parts, including 35 tasks.

Part 1 contains 29 tasks with a short answer, including 21 tasks of a basic level of complexity (in the version they are numbered: 1–7, 10–15, 18–21, 26–29) and 8 tasks of an increased level of complexity (their serial numbers: 8, 9, 16, 17, 22–25).

Part 2 contains 6 tasks of a high level of difficulty, with detailed answers. These are tasks numbered 30–35.

Average general education

Preparing for the Unified State Exam 2018 in Chemistry: analysis of the demo version

We bring to your attention an analysis of the demo version of the 2018 Unified State Exam in chemistry. This article contains explanations and detailed algorithms for solving problems. To help you prepare for the Unified State Exam, we recommend our selection of reference books and manuals, as well as several articles on current topics published earlier.

Exercise 1

Determine which atoms of the elements indicated in the series in the ground state have four electrons in the outer 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, consisting of a main and secondary subgroup. If an element is in the main subgroup of a certain group, then the group number indicates the number of electrons in the last layer. Therefore, to answer this question, you need 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 indicated in the series

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

select three elements that are in Periodic table chemical elements of D.I. Mendeleev are in the same period.

Arrange the chemical elements in increasing order of their metallic properties.

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

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 series of chemical elements arranged in order of increasing electronegativity, which means decreasing metallic properties and increasing non-metallic ones. Each period (except the first) begins with an active metal, which is called an alkali, and ends with an inert element, i.e. 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 order of increasing metallic properties. From what was written above we determine if metallic properties decrease from left to right, which means they increase on the contrary, from right to left. Therefore, the correct answers will be 3; 4; 1.

Task 3

From the number of elements indicated in the row

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

select 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 it is located chemical element 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 in a compound can exhibit) minus the group number, with a minus sign. For example, the element is in group 5, the main subgroup; therefore, its highest oxidation state in compounds will be +5; the lowest oxidation state, respectively, is 8 – 5 = 3 with a minus sign, i.e. –3. For elements of period 4, the highest valence is +4, and the lowest is –4. Therefore, from the list of data elements in the task, we look for two elements located in group 4 of the main subgroup. This will be C and Si numbers of the correct answer 3; 5.

Task 4

From the list provided, select two compounds that contain 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 the interaction of atoms that binds them into molecules, ions, radicals, and crystals. There are four types chemical bonds: ionic, covalent, metallic and hydrogen.

Ionic bond - a bond that arises as a result of electrostatic attraction of oppositely charged ions (cations and anions), in other words, between a typical metal and a typical non-metal; those. elements that differ sharply from each other in electronegativity. (> 1.7 on the Pauling scale). The 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 group 7 of the main subgroup. The exception is ammonium salts; they do not contain a metal atom, instead an ion, but in ammonium salts the bond between the ammonium ion and the acid residue 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 down the selected numbers in the table 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 acidic ions. The exception is ammonium salts. These salts have ammonium ions instead of metal ions. Salts are medium, acidic, double, basic and complex. Medium salts are products of complete replacement of acid hydrogen with a metal or ammonium ion; For example:

H 2 SO 4 + 2Na = H 2 + Na 2 SO 4 .

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

2H 2 SO 4 + 2Na = H 2 + 2 NaHSO 4 .

This salt is acidic. Now let's look at our assignment. It contains two salts: NH 4 HCO 3 and KF. The first salt is acidic because it is the product of incomplete replacement of hydrogen in the acid. Therefore, in the sign with the answer under the letter “A” we will put the number 4; the other salt (KF) does not contain hydrogen between the metal and the acidic residue, so in the answer sheet under the letter “B” we will put the number 1. Oxides are a binary compound that contains 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. non-metal oxides P 2 O 5, SO 3 - they correspond to acids ; H 3 PO 4 and H 2 SO 4), amphoteric (oxides that, depending on the circumstances, may exhibit basic and acid properties– Al 2 O 3 , ZnO) and non-salt-forming. These are oxides of non-metals that exhibit neither basic, nor acidic, nor amphoteric properties; this is CO, N 2 O, NO. Consequently, NO oxide is a non-salt-forming oxide, so in the table with the answer under the letter “B” we will 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) diluted 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 metals from salt solutions that are in the electrochemical voltage series to the right of iron:

For example:

Fe + CuSO 4 = FeSO 4 + Cu

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

Fe + 2НCl = FeCl 2 + H 2

with nitric acid solution

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

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

Based on this, the correct answers will be: 2; 4.

Task 7

Strong acid X was added to water from a test tube with a precipitate of aluminum hydroxide, and a solution of substance Y was added to another. As a result, dissolution of the precipitate was observed in each test tube. 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 down the numbers of the selected substances under the corresponding letters in the table.

Answer: Aluminum hydroxide is an amphoteric base, so 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 aluminum hydroxide precipitate 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

FORMULA OF THE SUBSTANCE	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 (solution) 5) H 3 PO 4 (solution), BaCl 2, CuO

Answer: Below 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 by solutions of concentrated sulfuric and hydrochloric acids. Interacts with alkalis. Of all the reagents numbered 1–5, the ones most suitable for the properties described above are the simple substances numbered 3.

S + Cl 2 = 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 oxidation state +6. This is the highest degree of oxidation 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. In this case, its oxidation state can decrease 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 numbered 2, that is:

SO 3 + BaO = BaSO 4

SO 3 + H 2 O = H 2 SO 4

SO 3 + 2KOH = 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 presented reagents, you can safely choose the reagents numbered 4.

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

Zn(OH) 2 + LiOH = 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 oxygen-free 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. Reagents numbered 1 meet these criteria.

ZnBr 2 + 2AgNO 3 = 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 answer options are as follows:

The new directory contains all theoretical material in the chemistry course required to pass the Unified State Exam. It includes all elements of content, verified by test materials, and helps to generalize and systematize knowledge and skills for a secondary (high) school course. The theoretical material is presented in a concise and accessible form. Each topic is accompanied by examples test tasks. Practical tasks correspond to the Unified State Exam format. Answers to the tests are provided 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 that are in the electrochemical voltage series 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 = 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 number 4)

D) When complex substances burn with the participation of oxygen, oxides of all elements included in the composition are formed complex substance; For example:

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

So the general answer would be:

Determine which of the indicated 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, resulting in the formation of weak carbonic acid, which at the moment of formation decomposes into carbon dioxide and water:

K 2 CO 3 + 2HCl (excess) = 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 = KHCO 3

We write the answer in the table:

Answer: A) Methylbenzene belongs to the homologous series 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)

B) 3-methylbutanal belongs to the homologous series of aldehydes. Since aldehydes have the ending -al. Its formula:

Task 12

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

1) butane
2) cyclobutane
3) butine-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 composition, but the order of atomic bonding (chemical structure) differs. 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) butine-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 structural isomers of butene-1.

We write down the correct answers in the table:

Task 13

From the proposed list, select two substances whose interaction with a solution of potassium permanganate in the presence of sulfuric acid will result in a change in the color of the solution.

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, so we cross out hexane No. 1 and propane No. 4.

Cross out number 2 (benzene). In benzene homologues, the alkyl groups are easily 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 successful completion Unified State Examination in Chemistry: brief theoretical information on all topics, assignments different types and levels of difficulty, methodological comments, answers and evaluation criteria. Students will not have to search for additional information on the Internet and buy other textbooks. In this book they will find everything they need to independently and effectively prepare for the exam. The publication sets out in a concise form the basics of the subject in accordance with current educational standards and examines the most difficult examination questions of an increased level of complexity in as much detail as possible. In addition, training tasks are provided with which you can check the level of mastery of the material. The book appendix contains the necessary reference materials on the subject.

Task 15

From the list provided, 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 similar properties. 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 methyl ammonium chloride.

CH 3 –NH 2 + HCl =Cl.

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

CH 3 –NH 2 + CH 3 Cl = [(CH 3) 2 NH 2 ]Cl

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

Task 16

Match the name of the substance with the product that is predominantly formed when this substance reacts 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, so the hydrogen atom is replaced by bromine. And the result is bromoethane:

CH 3 –СH3 + Br 2 = 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 (combined with three hydrogen atoms), but also one primary carbon atom. And since the replacement of a hydrogen atom with a halogen occurs most easily 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	H 3	C	H 3
	│		│
CH 3 –	C	–CH 3 + Br 2 = CH 3 –	C	–CH 3 + HBr	(answer 2)
	│		│
	N		B	r

C) Cycloalkanes, which include cyclopropane, differ greatly in cycle stability: three-membered rings are the least stable and five- and six-membered rings are the most stable. When bromination of 3- and 4-membered rings occurs, 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 comes down to the reaction of replacing hydrogen with 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 in which complex substances exchange constituent parts.

CH 3 COOH + Na 2 S = CH 3 COONa + H 2 S.

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

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

HCOOH + NaOH = HCOONa + H2O.

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

B) Formic acid, unlike others carboxylic acids- an amazing substance. In addition to the functional carboxyl group –COOH, it also contains the aldehyde group СОН. 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 is formed.

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

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

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

2C 2 H 5 OH + 2Na = 2C 2 H 5 ONa + H 2 (answer 2)

Thus, the answers to this task will be:

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Task 18

The following scheme for the transformation of substances is specified:

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

Task 19

From the proposed list of reaction types, select two reaction types, 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 equation for the reaction, for example, sodium with water:

2Na + 2H 2 O = 2NaOH + H 2.

Sodium is a very active metal, so it will react vigorously with water, in some cases even with an explosion, so the reaction occurs without catalysts. Sodium is a metal 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 in the form of a gas. During the reaction, the oxidation states of sodium and hydrogen change,

therefore, the reaction is a redox reaction, 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 environment, and here an alkali is formed. From this we can conclude that the answers will be correct

Task 20

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

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

Answer: The rate of a chemical reaction is a value that shows how the concentrations of starting substances or reaction products change per unit 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 elevated temperatures, so lowering the temperature will reduce its rate. Answer No. 1. Next: if you increase the concentration of one of the reactants, the reaction will go faster. This doesn't suit us. A catalyst, a substance that increases the rate of a reaction, is also not suitable. Reducing the hydrogen concentration will slow down the reaction, which is what we need. This means that another correct answer is No. 4. To answer point 4 of the question, let’s write the equation for this reaction:

CH 2 = CH 2 + H 2 = CH 3 -CH 3.

From the equation of the reaction it is clear that it proceeds with a decrease in volume (2 volumes of substances entered the reaction - ethylene + hydrogen), but only one volume of the reaction product was formed. Therefore, as the pressure increases, the reaction rate should increase - this is also not suitable. Summarize. The correct answers were:

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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 oxidation states change in reactions:

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

in this reaction, nitrogen changes its oxidation state from 3– to 0, that is, it is oxidized. This means that he is a reducer. 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 formula of the salt and the products of electrolysis of an aqueous solution of this salt, which were released on the 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 passing through a constant electric current through a solution or molten electrolyte. At the cathode Always the recovery process is underway; at the anode Always the oxidation process is underway. If the metal is in the electrochemical voltage series of metals up to manganese, then water is reduced at the cathode; from manganese to hydrogen, the release of water and metal is possible; 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), the anions are oxidized:

In the case of oxygen-containing anions and fluorides, the process of water oxidation occurs, but 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 the acidic residue of an oxygen-containing acid.

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

2H 2 O = H 2 + 2OH – .

Hydrogen is released at the cathode.

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

2H 2 O – 4e = O 2 + 4H +

Oxygen is released at the anode. Thus, the overall reaction equation will be as follows:

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

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

2H 2 O = H 2 + 2OH – .

Hydrogen will be released as a reaction product. Cl – will be oxidized at the anode to a free state according to the following equation:

2CI – – 2e = Cl 2 .

The overall process on the electrodes is as follows:

2KCl + 2H 2 O = 2KOH + H 2 + Cl 2 (answer 4)

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

Cu 2+ + 2e = Cu 0 .

Bromine is oxidized at the anode:

The overall reaction equation will be as follows:

Correct answer 3.

D) 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 = O 2 + 4H +

Correct answer 2.

The general answer to this question is:

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Task 23

Establish a correspondence between the name of the salt and the relationship 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 salt ions with water molecules, leading to the formation of a weak electrolyte. Any salt can be thought of as the product of the interaction 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 by a weak base and strong acid.
3. Salts formed by a weak base and a weak acid.
4. Salts formed by a strong base and a strong acid.

Let's now look at this task from this point of view.

A) NH 4 Cl - a salt formed by the weak base NH 4 OH and the strong acid HCl - undergoes hydrolysis. The result is a weak base and a strong acid. This salt is hydrolyzed by 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 a weak electrolyte is not formed. Answer 3.

B) Sodium carbonate Na 2 CO 3 is a salt formed by the strong base NaOH and the weak carbonic acid H 2 CO 3 – undergoes hydrolysis. Since the salt is formed by a dibasic acid, hydrolysis can theoretically occur in two stages. As a result of the first stage, an alkali and an acidic 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) The aluminum sulfide salt Al 2 S 3 is formed by the weak base Al (OH) 3 and the weak acid H 2 S. Such salts undergo hydrolysis. The result is a weak base and a weak acid. Hydrolysis occurs along the cation and anion. The correct answer is 4.

Thus, the general answer to the task looks like:

Task 24

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

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

Answer: Reversible reactions are reactions that can simultaneously go in two opposite directions: towards the direct and reverse reactions, therefore in the equations of reversible reactions, instead of equality, the reversibility sign is put. Every reversible reaction ends in chemical equilibrium. This is a dynamic process. In order to remove a reaction from a 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 Le Chatelier’s principle: if a system in a state of chemical equilibrium is acted upon from the outside, by changing concentration, temperature or pressure, then the system tends to take a position that counteracts this action.

Let's look at this using examples from our assignment.

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

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

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

D) The reaction between sulfur oxide (IV) and chlorine SO 2 (g) + Cl 2 (g) = SO 2 Cl 2 (g) is a reaction that occurs with a decrease in the volume of substances (2 volumes of gases entered 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:

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Task 25

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

FORMULAS OF SUBSTANCES
A) HNO 3 and NaNO 3 B) KCI and NaOH B) NaCI and BaCI 2 D) AICI 3 and MgCI 2

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

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

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

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

CuSO 4 + 2NaOH = 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 it is the opposite - many barium salts are insoluble. Using the solubility table, we determine that barium sulfate is insoluble, so the reagent will be copper sulfate. Answer 5.

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

The general answer to this task looks like this:

Task 26

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

Answer: A) When burned, methane is released a large number of heat, so it can be used as fuel (answer 2).

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

C) Ethylene is an unsaturated hydrocarbon that undergoes 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 the number to the nearest tenth).

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 the substance is equal to: ω · m(solution) = 0.1 · 150 = 15 g.

2. Let the mass of added potassium nitrate be equal to 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) = 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, you need to add 3.4 g of KNO3.

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Task 28

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

2H 2 (g) + O 2 (g) = H 2 O (g) + 484 kJ,

1452 kJ of heat was released. Calculate the mass of water formed in this case (in grams).

This problem can be solved in one action.

According to the reaction equation, as a result, 36 grams of water were formed and 484 kJ of energy were released. And 1454 kJ of energy will be released when X g of water is formed.

Answer: When 1452 kJ of energy is released, 108 g of water are formed.

Task 29

Calculate the mass of oxygen (in grams) required to completely burn 6.72 liters (n.s.) of hydrogen sulfide.

To solve this problem, we will write the reaction equation for the combustion of hydrogen sulfide and calculate the masses of oxygen and hydrogen sulfide that entered into the reaction using 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, 0.3 mol of H 2 S will react with X mol of O 2.

Hence X = 0.45 mol.

3. Determine the mass of 0.45 mol of oxygen

m(O2) = n · M= 0.45 mol · 32 g/mol = 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 an oxidation-reduction reaction is possible. In your answer, write down the equation of 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; it 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 oxidation states: 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 oxidation state 4+, which can oxidize 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 = K 2 MnO 4 + Na 2 SO 4 + H 2 O

After arranging the coefficients, the formula takes on the following form:

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

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

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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, complete 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 = K 2 CO 3 + H 2 O

If, as a result of a reaction in electrolyte solutions, an insoluble or gaseous or slightly 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 poorly dissociating substance. Let's write down the complete ionic equation.

Since water is a poorly dissociating substance, it is written in the form of a molecule. Next, we create the abbreviated ionic equation. Those ions that moved 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 the abbreviated ionic equation.

This equation will be the answer to this task.

Task 32

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

Answer: Electrolysis is a redox process that takes place on electrodes when a direct electric current is passed through a solution or melt of an electrolyte. The task talks about the electrolysis of a copper nitrate solution. During the electrolysis of salt solutions, water can also take part in electrode processes. When salt is dissolved in water, it breaks down into ions:

Reduction processes occur at the cathode. Depending on the activity of the metal, metal, metal and water can be reduced. Since copper in the electrochemical voltage series 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 occur 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 = 3S + 2H 2 O.

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

S + KOH = 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 perform 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, so it should result in an unsaturated hydrocarbon.

The following reaction is interesting because it proceeds according to Markovnikov’s rule. According to this rule, when hydrogen halides combine with asymmetrically constructed alkenes, the halogen attaches to the less hydrogenated carbon atom at the double bond, and hydrogen, vice versa.

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Task 34

When a sample of calcium carbonate was heated, some of the substance decomposed. At the same time, 4.48 liters (n.s.) of carbon dioxide were released. The mass of the solid residue was 41.2 g. This residue was added to 465.5 g of a solution of hydrochloric acid 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 problem statement and provide all the necessary calculations (indicate the units of measurement of the required quantities).

Answer: Let us write down a brief condition for this problem.

After all the preparations have been made, we proceed to the solution.

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

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

2) Determine the amount of calcium oxide formed.

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

Hence: 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, a 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(CaCO 3) / M(CaCO 3) = 30 g / 100 g/mol = 0.3 mol

CaO + HCl = CaCl 2 + H 2 O

CaCO 3 + HCl = CaCl 2 + H 2 O + CO 2

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

The reaction involved 0.3 mol of CaCO 3 and 0.2 mol of CaO for a total of 0.5 mol.

Accordingly, 0.5 mol CaCl 2 is formed

6) Calculate the mass of 0.5 mol calcium chloride

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

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

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

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

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

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

9) And finally, we will answer the question of the task. Let's find the mass fraction in % of salt in the solution using the following magic triangle:

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

Answer: ω% (CaCI 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 interaction organic matter B with propanol-2. It is known that substance B has 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) Compose structural formula this substance, which will unambiguously show the order of bonds of atoms in its molecule;

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

Answer: Let's try to figure out 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 elements that make up the molecule, we can determine its molecular formula.

Let us take the mass of substance A as 100 g. Then the masses of all elements included in 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. Let's 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 compose the structural formula of substance A. We already know that carbon in organic chemistry is always tetravalent, hydrogen is monovalent, oxygen is divalent and nitrogen is trivalent. The problem statement also states that substance B is capable of interacting with both acids and alkalis, that is, it is amphoteric. From natural amphoteric substances, we know that amino acids have pronounced amphotericity. Therefore, it can be assumed that substance B refers to amino acids. And of course, we take into account that it is obtained by interaction with 2-propanol. Having counted the number of carbon atoms in propanol-2, we can make a bold conclusion that substance B is aminoacetic acid. After a certain number of attempts, the following formula was obtained:

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

For the first time, schoolchildren and applicants are invited to tutorial to prepare for the Unified State Exam in chemistry, which contains training tasks collected by topic. The book presents tasks of different types and levels of complexity on all tested topics in the chemistry course. Each section of the manual includes at least 50 tasks. The tasks correspond to modern educational standard and the regulations on holding a unified state exam in chemistry for graduates of secondary educational institutions. Completing the proposed training tasks on the topics will allow you to qualitatively prepare for passing the Unified State Exam in chemistry. The manual is addressed to high school students, applicants and teachers.