Specification
control measuring materials
to be held in 2018
main state exam
in chemistry

1. Appointment of KIM for OGE- to evaluate the level of general education in chemistry of graduates of the ninth grade of general education organizations for the purpose of the state final certification of graduates. The results of the exam can be used when enrolling students in specialized secondary school classes.

The OGE is conducted in accordance with the Federal Law of the Russian Federation dated December 29, 2012 No. 273-FZ “On Education in the Russian Federation”.

2. Documents defining the content of KIM

3. Approaches to the selection of content, the development of the structure of KIM

The development of KIM for the OGE in chemistry was carried out taking into account the following general provisions.

• KIM are focused on testing the assimilation of the knowledge system, which is considered as an invariant core of the content of existing chemistry programs for the basic school. In the Federal component of the state educational standard in chemistry, this system of knowledge is presented in the form of requirements for the preparation of graduates.
• KIM are designed to provide an opportunity for a differentiated assessment of the training of graduates. For this purpose, the verification of mastering the basic elements of the content of the chemistry course in grades VIII-IX is carried out at three levels of complexity: basic, advanced and high.
• The educational material on the basis of which the tasks are built is selected on the basis of its significance for the general educational preparation of graduates of the basic school. At the same time, special attention is paid to those elements of the content that are developed in the course of chemistry of X-XI classes.

4. Connection of the examination model of the OGE with KIM USE

The most important principle taken into account when developing KIM for the OGE is their continuity with the KIM USE, which is due to unified approaches to assessing the educational achievements of students in chemistry in primary and secondary schools.

The implementation of this principle is ensured by: the unity of the requirements for the selection of content, checked by the tasks of the OGE; the similarity of the structures of the examination options for KIM for the OGE and the USE; the use of similar task models, as well as the identity of the assessment systems for tasks of similar types used both in the OGE and in the USE.

5. Characteristics of the structure and content of KIM 1

In 2018, the executive authorities of the constituent entities of the Russian Federation that manage in the field of education are offered two models of examination work, in terms of their structure and content of the tasks included in it, similar to the models of examination work in 2014.

Each version of the examination paper consists of two parts.

Part 1 contains 19 tasks with a short answer, including 15 tasks of a basic level of complexity (the serial numbers of these tasks: 1, 2, 3, 4, ... 15) and 4 tasks of an increased level of complexity (the serial numbers of these tasks: 16, 17, 18, 19). For all their differences, the tasks of this part are similar in that the answer to each of them is written briefly in the form of one digit or a sequence of digits (two or three). The sequence of numbers is written in the answer sheet without spaces and other additional characters.

Part 2 depending on the model, the CIM contains 3 or 4 tasks of a high level of complexity, with a detailed answer. The difference between examination models 1 and 2 is in the content and approaches to the implementation of the last tasks of the examination options:

• exam model 1 contains task 22, which provides for the implementation of a "thought experiment";
• exam model 2 contains tasks 22 and 23, which provide for the implementation of a real chemical experiment.

Tasks are arranged according to the principle of gradual increase in the level of their complexity. The share of tasks of basic, advanced and high levels of complexity in the work was 68, 18 and 14%, respectively.
Table 1 gives a general idea of ​​the number of tasks in each part of the examination paper of models 1 and 2.

..............................

1 Model 1 (M1) corresponds to demo #1; model 2 (M2) - demo version No. 2.

Training test to prepare for the OGE - 2018 in chemistry in grade 9

Work instructions

2 hours (120 minutes) are given to complete the work. The work consists of 2 parts, including 22 tasks. Part 1 contains 19 short answer tasks, part 2 contains 3 long answer tasks.

Answers to tasks 1-15 are written as one digit, which corresponds to the number of the correct answer.

Answers to tasks 16-19 are written as a sequence of numbers.

For tasks 20-22, a complete detailed answer should be given, including the necessary reaction equations and the solution of the problem.

When performing work, you can use the Periodic Table of Chemical Elements D.I. Mendeleev, a table of the solubility of salts, acids and bases in water, an electrochemical series of metal voltages and a non-programmable calculator.

Part 1

1. The chemical element of the 2nd period of the VIA group corresponds to the electron distribution scheme

1) Fig. one

2) Fig. 2

3) Fig. 3

4) Fig. 4

Answer:

2. The non-metallic properties of simple substances are enhanced in the series

1) phosphorus → silicon → aluminum

2) fluorine → chlorine → bromine

3) selenium → sulfur → oxygen

4) nitrogen → phosphorus → arsenic

Answer:

3. A covalent polar bond is realized in a substance

1) CuO

2) P4

3) SO2

4) MgCl 2

Answer:

4 . In which compound is the oxidation state of chlorine +7?

1)HCl

2) Cl2O

3) KClO 3

4) KClO 4

Answer:

5. Substances whose formulas are ZnO and Na 2 SO 4 , are respectively

1) basic oxide and acid

2) amphoteric hydroxide and salt

3) amphoteric oxide and salt

4) basic oxide and base

Answer:

6. The reaction whose equation is

2NaOH + CuCl 2 = Cu(OH) 2 + 2NaCl

refers to reactions

1) expansions

2) connections

3) substitution

4) exchange

Answer:

7. The smallest number of positive ions is formed during the dissociation of 1 mol

1) nitric acid

2) sodium carbonate

3) aluminum sulfate

4) potassium phosphate

Answer:

8. The irreversible course of the ion exchange reaction between solutions of barium hydroxide and potassium carbonate is due to the interaction of ions

1) K + and OH -

2) K + and CO 3 2―

3) Ba 2+ and CO 3 2–

4) Ba 2+ and OH -

Answer:

9. Copper reacts with solution

1) AgNO3

2) Al 2 (SO 4) 3

3) FeSO4

4) NaOH

Answer:

10 . Copper(II) oxide can react with each substance of the pair

1) HCl, O 2

2) Ag, SO 3

3) H2, SO4

4) Al, N 2

Answer:

11 . Determine the formula of the unknown substance in the reaction scheme:

KOH + ... → K 2 CO 3 + H2O

1) CO

2) CO2

3)CH4

4) C

Answer:

12. CaNO3 can be converted to CaSO3 using

1) hydrogen sulfide

2) barium sulfite

3) sodium sulfite

4) sour gas

Answer:

13. Are the judgments about the ways of separating mixtures correct?

A. Evaporation refers to the physical methods of separation of mixtures.

B. Separation of a mixture of water and ethanol is possible by filtration.

1) only A is true

2) only B is true

3) both statements are correct

4) both judgments are wrong

Answer:

14. In the reaction 3CuO + 2NH 3 \u003d 3Cu + N 2 + 3H 2 O

The change in the oxidation state of the oxidizer corresponds to the scheme

1) +2 → 0

2) −3 → 0

3) −2 → 0

4) 0 → +2

Answer:

15 . On which diagram is the distribution of mass fractions of elements

corresponds to NHNO 3

Part 2

16. When completing the task, from the proposed list of answers, select the two correct ones and write down the numbers under which they are indicated.

In the series of chemical elements Be- Mg-Ca

1) the atomic radius increases

2) the highest degree of oxidation increases

3) the value of electronegativity increases

4) the main properties of the formed hydroxides increase

5) the number of electrons in the outer level decreases

Answer:

18. Match the two substances with a reagent that can be used to distinguish between these substances.

SUBSTANCES		REAGENT
A) NaNO 3 and Ca (NO 3) 2 B) FeCl 2 and FeCl 3 C) H 2 SO 4 and HNO 3		1) BaCl2 2) Na 2 CO 3 3) HCl 4) NaOH

Write down the numbers in response, arranging them in the order corresponding to the letters:

19. Establish a correspondence between the substance and the reagents, with each of which it can react.

Answer:

20. Using the electronic balance method, arrange the coefficients in the reaction equation, the scheme of which

P + H 2 SO 4 → H 3 PO 4 + SO 2 + H 2 0

Determine the oxidizing agent and reducing agent

2 , H 2 SO 4 , CaCO 3

Chemistry Test Estimation System

Correct execution of each task parts 1 the basic level of difficulty (1–15) is estimated at 1 point.

Correct completion of each task parts 1 advanced level of difficulty (16–19) is rated with a maximum of 2 points. Tasks 16 and 17 are considered correctly completed if two answers are correctly selected in each of them. For an incomplete answer - one of the two answers is correctly named or three answers are named, of which two are correct - 1 point is given. The rest of the answers are considered incorrect and are evaluated at 0 points.

Tasks 18 and 19 are considered completed correctly if three matches are correctly established. Partially correct is the answer in which two out of three matches are established; it is worth 1 point. The remaining options are considered an incorrect answer and are evaluated at 0 points.

Part 1

Part 2

20. Using the electronic balance method, arrange the coefficients in the reaction equation, the scheme of which is:

HNO 3 + Zn \u003d Zn (NO 3) 2 + NO + H 2 O

Specify the oxidizing agent and reducing agent.

Response elements
1) Let's make an electronic balance: S +6 + 2ē = S +4 │2 │5 P 0 - 5ē \u003d P +5 │5 │2 2) We indicate that S +6 (H 2 SO 4 ) is an oxidizing agent, and P 0 (P) - reducing agent 3) Let's arrange the coefficients in the reaction equations: 2P + 5H 2 SO 4 →2H 3 PO 4 + 5SO 2 + 2H 2 0
Evaluation criteria	Points
The answer contains an error in only one of the elements
There are two errors in the response.
Maximum score

21. When an excess of potassium carbonate solution reacted with a 10% solution of barium nitrate, 3.94 g of a precipitate fell out. Determine the mass of the barium nitrate solution taken for the experiment.

Response elements (Other formulations of the answer are allowed that do not distort its meaning)
Explanation. The reaction equation is composed: K 2 CO 3 + Ba(NO 3 ) 2 = ↓ + 2KNO 3 2) The amount of barium carbonate substance and the mass of barium nitrate were calculated: N (BaCO 3 ) \u003d m (BaCO 3 ) / M (BaCO 3 ) \u003d 3.94: 197 \u003d 0.02 mol n (Ba (NO 3) 2) \u003d n (BaCO 3) \u003d 0.02 mol m (Ba (NO 3 ) 2 ) \u003d n (Ba (NO 3 ) 2 ) M (Ba (NO 3 ) 2 ) \u003d 0.02 261 \u003d 5.22 g. 3) The mass of the barium nitrate solution is determined: M (solution) = m(Ba(NO 3) 2 / ω (Ba (NO 3) 2 \u003d 5.22 / 0.1 \u003d 52.2 g Answer: 52.2 g.
Evaluation criteria	Points
The answer is correct and complete, includes all named elements
Correctly written 2 of the elements named above
Correctly written 1 element of the above (1st or 2nd)
All elements of the answer are written incorrectly
Maximum score

22. Given substances: CuO, NaCl, KOH, MnO 2 , H 2 SO 4 , CaCO 3

Using water and the necessary substances only from this list, get copper (II) chloride in two stages. Describe the signs of ongoing reactions. For the second reaction, write the abbreviated ionic reaction equation.

Response elements (Other formulations of the answer are allowed that do not distort its meaning)
Let's write 2 reaction equations: 2NaCl + H 2 SO 4 \u003d 2HCl + Na 2 SO 4 CuO + 2HCl \u003d CuCl 2 + H 2 O Let us indicate signs of reactions. The first reaction is gas evolution. For the dissolution reaction of CuO - color change, the formation of a blue solution. Let's make an abbreviated ionic equation for the first reaction: CuO + 2H + \u003d Cu 2+ + H 2 O
Evaluation criteria	Points
The answer is correct and complete, includes all named elements
The four elements of the answer are correctly written
Three elements of the answer are correctly written
Correctly written two elements of the answer
One element of the answer is correctly written
All elements of the answer are written incorrectly
Maximum score

2018 year.

The maximum number of points that an examinee can receive for completing the entire examination work (without a real experiment) is 34 points.

Table 4
The scale for converting the initial score for the performance of the examination paper into a mark on a five-point scale (work without a real experiment, demo version 1)

• 0-8 points - mark "2"
• 9-17 points - mark "3"
• 18-26 points - mark "4"
• 27-34 points - mark "5"

The mark “5” is recommended to be set if, out of the total score sufficient to obtain this mark, the graduate scored 5 or more points for completing the tasks of part 3. The results of the exam can be used when enrolling students in specialized secondary school classes. The benchmark for selection to profile classes can be an indicator, the lower limit of which corresponds to 23 points.

Part 1 contains 19 tasks with a short answer, including 15 tasks of a basic level of complexity (the serial numbers of these tasks: 1, 2, 3, 4, ... 15) and 4 tasks of an increased level of complexity (the serial numbers of these tasks: 16, 17, 18, 19). For all their differences, the tasks of this part are similar in that the answer to each of them is written briefly in the form of one digit or a sequence of digits (two or three). The sequence of numbers is written in the answer sheet without spaces and other additional characters.

Part 2, depending on the CMM model, contains 3 or 4 tasks of a high level of complexity, with a detailed answer. The difference between examination models 1 and 2 is in the content and approaches to the implementation of the last tasks of the examination options:

Exam Model 1 contains task 22, which involves performing a "thought experiment";

Exam model 2 contains tasks 22 and 23, which provide for the performance of laboratory work (a real chemical experiment).

Scale for converting points into grades:

"2"– from 0 to 8

"3"– from 9 to 17

"4"– from 18 to 26

"five"– from 27 to 34

The system for assessing the performance of individual tasks and the examination work as a whole

The correct performance of each of the tasks 1-15 is estimated at 1 point. The correct performance of each of the tasks 16-19 is estimated as a maximum of 2 points. Tasks 16 and 17 are considered completed correctly if two answers are correctly selected in each of them. For an incomplete answer - one of the two answers is correctly named or three answers are named, of which two are correct - 1 point is given. The rest of the answers are considered incorrect and are scored 0 points. Tasks 18 and 19 are considered completed correctly if three matches are correctly established. Partially correct is the answer in which two out of three matches are established; it is worth 1 point. The remaining options are considered incorrect answers and are scored 0 points.

Checking the tasks of part 2 (20–23) is carried out by the subject commission. The maximum score for a correctly completed task: for tasks 20 and 21 - 3 points each; in model 1 for task 22 - 5 points; in model 2 for task 22 - 4 points, for task 23 - 5 points.

To complete the examination work in accordance with model 1, 120 minutes are allotted; according to model 2 – 140 minutes

At the end of the 9th grade, students are required to pass the final state certification. The test consists of passing several mandatory exams.

GIA in chemistry 2018 is optional. Chemistry is a discipline that students have the right to choose for their own examination.

It is worth noting that state certification is necessary to obtain a certificate of incomplete school education. Also, the GIA is an intermediate stage of preparation for the final exams in grade 11. However, the results of the examination in the certificate are not reflected in any way.

Every year, certain adjustments are made to the process and form of state certification. This is necessary so that students can fully demonstrate their abilities in the exam.

The updated OGE procedure will include three possibilities correct negative exam results and lengthening the period for retakes.

Regarding the calculation of results, changes are also possible, namely, it is planned to develop a unified federal scale of points, according to which the results of the examination will be calculated. It should be noted that the data will not be tied to regional indicators.

Concerning exams in chemistry, no significant changes are expected.

The structure of tasks of the GIA 2018 in chemistry

The total number of tasks in KIMah in chemistry is 22 . All questions are divided into two parts. The first category is represented by questions as answers to which you need to indicate the correct answer. In this part of everything 19 tasks and for the correct answer, the examiner receives 1 point.

The second block consists of questions of increased complexity, to which you need to write a detailed answer or solution. There are only four questions in this part, and for each correct answer, the student receives from 3 to 5 points.

Categories of those who pass the GIA in chemistry

By relevant order Ministry of Education of the Russian Federation certain categories of students who have the right to register an application for state certification.

Students who have completed grade 9 and have grades in all subjects not lower than three can try their hand at the final certification. Those students who have an “unsatisfactory” mark in one discipline are also allowed to pass the state exam, but this discipline is included in the list of exams that the teenager will take.

Also, those students who in previous periods passed the examination unsatisfactorily or for results that do not suit them for a number of reasons have the right to pass the examination.

Final certification in chemistry in the early period in 2018 will take place April 27, in the main period - June 7. For those who wished to improve their results, the retake was carried out - 12-th of September.

How to become a member of the GIA 2018

The student is required not only to prepare for the exam, but also to register his written desire to become a member of the GIA.

To do this, the student must submit an application in a unified form to the school administration at the place of study. Such an application must be submitted before 1st of February.

In order for the document to be registered, and in the future, the student has the opportunity to take exams, the application must be submitted in person, provide a passport or other certifying document. In addition, the application must indicate all those disciplines in which the student plans to be examined.

It is possible to change the list of subjects in the future, however, for this, the student will need to submit an explanatory document with copies of supporting documents attached to it. Changes can be made no later than one month before the guest test.

Early delivery of the GIA in chemistry in 2018

Until recently, only those students who did not have the opportunity to attend state certification in a timely manner could pass early examinations. Such schoolchildren included those children who were involved in sports, undergoing medical treatment or going to study abroad.

This option has both advantages and disadvantages. The advantage of early state certification is that students have more time to prepare if they fail the exam. Also, children do not have to worry about the violation of the terms of the examination.

However, the disadvantages are also significant. Schoolchildren are subjected to additional psychological and physiological burdens. Students have to combine basic studies with preparation for exams, which, of course, is the reason for the violation of the regime and as a result of poor results at the GIA.

Demo version of GIA in Chemistry 2018

Quite often, students use demo version of FIPI GIA in chemistry 2018. Demo option- these are materials that include groups of tasks presented at state certifications of previous years. Materials are stored in An open bank of tasks on the official website of FIPI(fipi.ru).

The form of materials is in full compliance with CIMs that will be offered for state certification, with the exception of numerical indicators.

The manual is quite suitable so that the student can practice in completing assignments and decide on topics that require additional preparation. Also, a ninth grader will be able to draw up a plan for completing tasks for state certification.

Our website offers the most modern demo versions of GIA tasks in chemistry and, if desired, they can be downloaded and printed.

Additional information about the GIA 2018

The list of items allowed for the GIA is approved annually. It is allowed to take a black pen, gel or capillary pen and a passport to the examination in chemistry. Also, at the state certification in this discipline, you can use the periodic system of chemical elements of D.I. Mendeleev, the table of the solubility of salts, acids and bases in water, the electrochemical series of metal voltages and a non-programmable calculator.

You can not bring a mobile phone and other electronic computing devices for state certification. It is also not allowed to use other reference materials not allowed by the organizers.

In order for the exam to be counted, the student must score at least 9 points. This result corresponds to the mark "satisfactory". The rating "Excellent" corresponds to the result in 34 points.

From time to time, situations arise during the examination when students do not agree with the result of the state certification or believe that the examination procedure was violated by the organizers. In this case, the examiners have the right to protest those points with which they do not agree.

If the student believes that the examination process did not meet the mandatory standards, then, without leaving the classroom, he must apply in writing to the conflict commission and demand a retake. If the student does not triple the mark, then he has the right to demand a re-check of his work. It is noteworthy that all work will be checked completely and other experts are required to check. To do this, the student must write an application to the conflict commission within four-day deadline and substantiate your claims.

B may be denied if the complaint expresses claims regarding the content or form of tasks. Also, the conflict commission may refuse the student if the student himself caused a violation of the state certification procedure.

How to prepare for the GIA 2018 in chemistry

Preparation for the GIA in chemistry 2018 takes a lot of time, but it is very important to use the right aids correctly. Only the choice of the most optimal method of preparation can ensure good performance in state certification. The most optimal preparation option includes the integrated use of the GIA demo version in chemistry, online tests in chemistry, demonstration materials and various books and collections of problems.

The following brochures and textbooks can be used as aids:

All materials help the student to briefly repeat the previously studied theory and practice in practice.

The GIA demo and demonstration materials provide an opportunity to get acquainted with the structure of examination tasks, as well as identify those topics that need to be repeated. Online testing is used when training is required regarding the time spent on exam tasks.

In order to pass the test successfully, the student needs to take care of his mood. A positive mood during preparation and at the state certification itself guarantees good results in the exam.

It is extremely important to help your child and for parents. In order for a student to fully prepare for state certification, an atmosphere of peace and understanding is necessary.

GIA statistics in chemistry for the past years

Over the past few years, such an exam as the GIA in chemistry has not been popular with schoolchildren. Only a small percentage of students chose this subject. Statistics from previous years show that due to the fact that there was little time to complete the tasks, and the tasks were difficult, the children did not cope well with the chemistry exam. 100 points were absent, and the number of "losers" was large. Over time, the situation began to change, and today, 100-pointers have already appeared, and the percentage of “losers” has decreased by 0,6% .

Exam Schedule

Secondary general education

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

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

Exercise 1

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

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

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

Task 2

Of the chemical elements listed in the series

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

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

Arrange the elements in ascending order of their metallic properties.

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

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

Looking at the table of chemical elements, we note that from the data in the element task, Na, Mg and Si are located in the 3rd period. Next, you need to arrange these elements in ascending order of metallic properties. From the above, we determine if the metallic properties decrease from left to right, then they increase on the contrary, from right to left. Therefore, the correct answers will be 3; 4; one.

Task 3

From among the elements indicated in the row

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

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

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

Task 4

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

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

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

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

Task 5

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

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

Answer:

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

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

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

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

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

Answer:

Task 6

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

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

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

2Fe + 3Cl 2 = 2FeCl 3

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

For example:

Fe + CuSO 4 \u003d FeSO 4 + Cu

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

Fe + 2НCl \u003d FeCl 2 + H 2

with nitric acid solution

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

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

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

Task 7

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

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

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

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

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

In this case, the precipitate of aluminum hydroxide dissolves.

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

In this case, the aluminum hydroxide precipitate also dissolves.

Answer:

Task 8

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

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

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

S + Cl 2 \u003d SCl 2

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

SO 3 + BaO = BaSO 4

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

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

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

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

Zn (OH) 2 + LiOH \u003d Li 2

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

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

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

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

ZnBr 2 + Cl 2 = ZnCl 2 + Br 2

The response options are as follows:

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

Task 9

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

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

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

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

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

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

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

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

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

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

So the general answer would be:

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

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

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

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

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

CO 2 + KOH \u003d KHCO 3

We write the answer in the table:

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

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

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

Task 12

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

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

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

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

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

Write the correct answers in the table:

Task 13

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

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

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

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

Correct answer:

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

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

Task 15

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

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

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

CH 3 -NH 2 + HCl \u003d Cl.

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

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

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

Task 16

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

3) Br–CH 2 –CH 2 –CH 2 –Br

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

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

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

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

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

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

So the general answer would be:

Task 17

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

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

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

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

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

HCOOH + NaOH \u003d HCOONa + H 2 O.

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

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

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

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

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

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

So the answers to this question will be:

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

Task 18

The following scheme of transformation of substances is given:

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

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

The general answer of this number:

Task 19

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

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

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

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

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

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

Task 20

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

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

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

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

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

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

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

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

Task 21

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

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

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

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

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

General answer:

Task 22

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

SALT FORMULA	ELECTROLYSIS PRODUCTS

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

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

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

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

Now let's look at this task:

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

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

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

Hydrogen is released at the cathode.

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

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

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

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

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

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

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

2CI - - 2e \u003d Cl 2.

The overall process on the electrodes is as follows:

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

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

Cu 2+ + 2e = Cu 0 .

Bromine is oxidized at the anode:

The overall reaction equation will have the following form:

Correct answer 3.

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

Cu 2+ + 2e = Cu 0 .

Oxygen is released at the anode:

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

Correct answer 2.

General answer to this question:

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

Task 23

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

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

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

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

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

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

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

Na 2 CO 3 + H 2 O ↔NaHCO 3 + NaOH;

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

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

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

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

Thus, the general answer to the task is:

Task 24

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

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

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

Let's analyze this with examples of our task.

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

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

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

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

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

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

Task 25

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

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

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

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

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

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

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

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

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

The general answer to this question looks like this:

Task 26

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

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

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

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

Task 27

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

Let's solve this problem:

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

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

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

100% – (15 + x)/(150 + x) = 12%

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

15 + x = 18 + 0,12x

0,88x = 3

x = 3/0,88 = 3,4

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

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

Task 28

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

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

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

This task can be solved in one step.

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

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

Task 29

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

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

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

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

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

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

Hence X = 0.45 mol.

3. Determine the mass of 0.45 mol of oxygen

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

Answer: the mass of oxygen is 14.4 grams.

Task 30

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

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

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

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

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

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

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

Task 31

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

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

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

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

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

This equation will be the answer to this task.

Task 32

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

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

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

Cu 2+ + 2e = Cu 0 .

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

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

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

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

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

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

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

Task 33

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

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

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

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

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

Task 34

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

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

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

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

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

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

2) Determine the amount of calcium oxide formed.

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

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

3) Determine the mass of 0.2 mol CaO

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

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

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

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

CaO + HCl \u003d CaCl 2 + H 2 O

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

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

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

Accordingly, 0.5 mol CaCl 2 is formed

6) Calculate the mass of 0.5 mol of calcium chloride

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

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

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

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

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

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

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

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

Answer: ω% (СaCI 2) = 11.2%

Task 35

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

Based on these conditions, complete the tasks:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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