The State Core Chemistry Examination is one of the non-mandatory exams that ninth grade students may take among other subjects of their choice. This is a rather difficult subject, so it is chosen by a small number of schoolchildren who plan to study in specialized classes or enter other educational institutions for specialties related to the medical, chemical, construction or food industries.

The Federal Institute of Pedagogical Measurements (FIPI) does not provide for any changes for the OGE in chemistry for 2018: control and measuring materials (KIM) will retain the content and structure similar to the materials of previous years. They are designed to test students' knowledge of the program completed in grades 8-9: methods for studying chemical phenomena and substances, basic concepts of organic and inorganic chemistry, tasks for calculating chemical reactions and other sections of the school curriculum.

For the OGE, two models of work No. 1 and No. 2 are provided, for the implementation of which the regulations allocated 120 and 180 minutes, respectively (for the model of the second type, 60 minutes were added to conduct and describe the chemical experiment). At the same time, it is recommended to distribute the total amount of time for completing tasks as follows:

• 3-8 minutes - for answers to each of the questions of the first part;
• for 12-17 minutes - for answers to each of the questions of the second part.

During the exam, students are allowed to use a calculator that is not equipped with a calculation programming function, and the following auxiliary materials:

• periodic system of chemical elements of Mendeleev;
• electrochemical series of metal voltages;
• table of solubility of acids, salts and bases in water.

Items, reference literature, telephones and other objects that are not on the list of permitted items are not allowed to be taken to the exam. If they are detected or other violations of the OGE regulations, the student will be removed from the class, and the examination paper will not be counted.

For the OGE in chemistry in 2018, Rosobrnadzor set the following dates:

• April 27 - the day of early delivery (with a reserve day on May 7);
• June 7 - the day of the main delivery (with a reserve day on June 22);
• September 12 is the day of additional delivery (with a reserve day of September 20).

Structure of KIM

As before, in 2018 OGE tickets in chemistry grade 9 include theoretical and practical parts. The theoretical part is intended to test the knowledge of ninth-graders of the basic formulas and definitions of organic and inorganic chemistry, which should be used later to solve tasks from the practical part. The latter is designed to test students' ability to conduct redox and ion exchange reactions, understanding the relationships between different classes of substances, molar mass and volume of substances.

Structurally, each ticket can be divided into two parts:

1. The first part includes 19 tasks, of which questions 1-15 are presented in the form of basic tests, and 16-19 belong to the category of questions of increased complexity. To answer these questions, you need to put a number, a sequence of numbers or a word in the examination form.
2. The second part consists of 3 or 4 tasks for the model of CMMs No. 1 and No. 2, respectively, which require a detailed answer, supported by reaction equations and chemical calculations. For model KIMs No. 2, it is required to conduct a practical experiment under the supervision of members of the examination committee and record its results.

Evaluation of work

To assess the knowledge of ninth-graders, depending on the level of preparation, the tasks in the tickets are grouped into three levels: simple, advanced and high complexity. The maximum score that you can get for the OGE in chemistry in grade 9 depends on the chosen exam model:

• 34 points for model No. 1, of which 15 (44.1%) - for solving the basic part, 8 (23.5%) - for questions of increased complexity and 11 - for the most difficult tasks;
• 38 points for model No. 2, of which 15 (39.5%) for basic tests, 8 (21%) for questions of increased complexity and 15 (39.5%) for the most difficult tasks.

Most of the students pass in the 9th grade the OGE in chemistry according to model No. 1, for which you can get a maximum of 34 points. For this option, the following system is provided for converting into grades on a five-point scale:

• 0-8 points - deuce;
• 9-17 points - triple;
• 18-26 points - four;
• 27-34 points - five.

The minimum threshold for passing the OGE in chemistry is 9 points, for which it is enough to answer the first nine questions of the ticket. But ninth-graders who will continue their education in a specialized class must receive at least 23 points.

Preparation for the OGE

To prepare for passing the OGE in chemistry, you can do self-training, additionally work with a tutor or attend relevant courses.

For students who choose the option of self-study, the following recommendations will be useful:

• Use educational and reference manuals that provide information within the framework of the school curriculum.
• You need to repeat the material starting with simple topics (studied in grade 8) and gradually move towards more complex material.
• Watch video tutorials to master chemical experiments and reactions. In this case, it is important to choose those options that describe in detail each stage of the process and the reactions obtained.
• Work out the basic chemical formulas to refresh your memory and, if necessary, fill in the gaps in knowledge.
• Make a note where to write down the material for preparing for the OGE in a form that is understandable to you.
• Pass online tests on the Internet, the complexity and structure of which is identical to real exams.
• Familiarize yourself with the demo versions of tests developed by FIPI in order to have an idea about the structure of the ticket, the level of complexity and wording of tasks, the requirements for the completeness of the explanation of questions that require writing detailed answers.

Note: it should be understood that the tasks presented in the demo versions will most likely not be on the exam itself, but there will be tasks of a similar topic or simply with other digital data.

• Pay attention to safety rules, since passing the OGE in chemistry according to model No. 2 provides for a real chemical experiment using laboratory equipment that you need to be able to use correctly.

Video example OGE tasks in chemistry:

OGE in chemistry is given only at the choice of the student, this test is not included in the list of mandatory. Chemistry is chosen by students who, after the 9th grade, plan to enter the profile 10th grade of a school or a specialized college, technical school. For admission to medical school, it is required to pass not only chemistry, but also biology. The exam implies orientation in theory, its successful application in practice. The subject needs to solve a lot of tasks of different levels of complexity from a wide range of topics. To decide which topics to pay attention to, check out the OGE Preparation Program in Chemistry.

The exam consists of tasks, which are divided into two logical blocks:

• The first part includes tasks for knowledge of the theory: here you need to give a short answer - a number, a sequence of numbers, a word.
• In the second part, there are several questions to which you need to give detailed, complete answers, conduct a laboratory experiment, write conclusions, and perform calculations. It is extremely important to be able to use special equipment, to use algorithms for solving problems of different levels of complexity.
  

In 2018, the minimum threshold was 9 points - this is the minimum that will allow you to get a minimum grade and a certificate.
In the exam, the subject has tips: tables of the solubility of salts, acids, bases in water, the periodic table of Mendeleev, tables of stresses of metals. With the ability to use these materials, you can solve many tasks without difficulty.

• The main advice that is relevant for every exam is to plan your study. Without a clear plan, you will not be able to achieve a high level of preparation. To make your planning as efficient as possible, check out- it indicates topics and sections that need special attention.
• Evaluate your strengths: the easiest way is online testing. Upon passing the test, you get the result, and you can evaluate which types of tasks and topics cause you the most difficulty.
• Once you have identified problematic topics, give them more attention than others. For training, take textbooks, reference books.
• Be sure to solve problems! The more problems you solve for preparation, the easier it will be on the exam.
• Ask questions: find a specialist who can help you in problem situations. It could be a tutor or a school teacher. Only a specialist can help you analyze your mistakes and not make them again.
• Learn to use hints - those tables that you can take with you to the exam.
• It is not enough to study theory, it is very important to train to pass tests. This form of knowledge testing causes difficulties for many, especially if it was not used in the lessons. Solve more test tasks of different types so that they do not cause fear and misunderstanding during the exam.
  
• "I will solve the OGE in chemistry" will help you prepare for the exam and successfully pass it, rationally using the allotted time, without stress.

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.

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 the Le Chatelier principle, if the reaction proceeds with a decrease in volume, then an increase in pressure shifts the chemical equilibrium towards the formation of a reaction product. Correct answer 1.

B) The reaction 2H 2 (g) + O 2 (g) \u003d 2H 2 O (g) is similar to the previous reaction, it also goes with a decrease in volume (3 volumes of gas entered, and 2 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 decision.

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

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

2) Determine the amount of calcium oxide formed.

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

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

3) Determine the mass of 0.2 mol CaO

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

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

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

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

CaO + HCl \u003d CaCl 2 + H 2 O

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

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

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

Accordingly, 0.5 mol CaCl 2 is formed

6) Calculate the mass of 0.5 mol of calcium chloride

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

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

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

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

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

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

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

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

Answer: ω% (СaCI 2) = 11.2%

Task 35

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

Based on these conditions, complete the tasks:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Dear teachers and 9th graders!

Basic general education is completed by the Basic State Examination of Graduates, during which the compliance of their knowledge with the requirements of the state educational standard is checked.

The examination of graduates of the 9th grade of general educational organizations is carried out in the form of testing.

The requirements for the level of training of graduates in chemistry, specified in the federal component of the state standard for general education, are the basis for the development of control measuring materials for the OGE.

According to these requirements, a certain system of knowledge about inorganic and organic substances, their composition, properties and application is mandatory for assimilation. This system of knowledge, which is based on the Periodic Law and the Periodic Table of Chemical Elements by D.I. Mendeleev, is the invariant core of all general educational programs in chemistry. In the proposed examination work, it was this content that was the basis for the development of control measuring materials.

The purpose of this manual is to acquaint teachers and students with the structure and content of the examination work, to enable the graduate to independently check his readiness for a new form of examination in chemistry - in the form of testing.

The options presented in the manual, comments on the solution of all tasks of one of the options and the answers given to the tasks of all options can be of great help in this.

The implementation of the presented tasks is one of the ways to consolidate, systematize and generalize the knowledge gained, as well as a way of self-control of the knowledge available to graduates.

Let's pay attention to one important point. The tasks contained in the variants of the examination paper are different in their form and require different types of answers for their implementation: when completing tasks of part 1 with a short answer (tasks 1-15), it is enough to write down the number of the selected correct answer, when completing tasks of part 1 with a short answer (tasks 16-19) you must answer in the form of a number or a set of numbers. The answer in part 2 suggests recording the necessary reaction equations or calculations made when solving the problem.

Tasks of part 1 (1-15) with a short answer meet the requirements of the basic level of training of graduates of the basic school in chemistry. They are formulated as a short statement, the end of which is the corresponding answer option. Each question has four possible answers, only one of which is correct.

Tasks of part 1 (16-19) with a short answer, unlike tasks 1-15, have an increased level of complexity and therefore contain more information that needs to be comprehended and understood. That is why the implementation of such tasks will require the implementation of a greater number of learning activities than in the case of choosing one correct answer. The answer should be a set of numbers.

The tasks of part 2 with a detailed answer correspond in their content to the most difficult tasks of traditional written work. They are designed to test the possession of skills that meet the highest requirements for the level of preparation of graduates of the basic school. To complete these tasks, you must be able to:

1) draw up equations of redox reactions and arrange coefficients in them using the electronic balance method;

2) carry out calculations of the mass fraction of a dissolved substance, the amount of a substance, mass or volume by the amount of a substance. The answer involves writing the necessary reaction equations or calculations made when solving the problem;

3) select the necessary reagents from the proposed list of substances to obtain the specified substance, draw up reaction equations, describe the signs of reactions, write down abbreviated ionic equations for ion exchange reactions.

Unlike previous years, starting from 2014, the content of the exam includes an experiment to obtain the substance named in task 22. During the experiment, the student must be able to:

2) prepare laboratory equipment for the experiment;

3) draw up a scheme of transformations, as a result of which the specified substance can be obtained;

4) carry out reactions in accordance with the prepared scheme of transformations and obtain a substance;

5) draw a conclusion about the chemical properties of the substances involved in the reaction, and signs of the classification of reactions.

In order to successfully complete each of the tasks in the process of independent work when performing one or another option, one should not only carefully consider the solution of tasks of options 9 and 18, but also analyze them.