Demo exam for chemistry year. Changes in the demo versions of the exam in chemistry. Examination procedure
Specification
control measuring materials
for the unified state examination in 2018
in chemistry
1. Purpose of KIM USE
The Unified State Exam (hereinafter - the USE) is a form of objective assessment of the quality of training of persons who have mastered educational programs of secondary general education, using standardized tasks (control measuring materials).
The Unified State Exam is conducted in accordance with Federal Law No. 273-FZ of December 29, 2012 "On Education in the Russian Federation".
Control measuring materials allow to establish the level of mastering by graduates of the Federal component of the state standard of secondary (complete) general education in chemistry, basic and profile levels.
The results of the unified state exam in chemistry are recognized by educational organizations of secondary vocational education and educational organizations of higher vocational education as the results of entrance examinations in chemistry.
2. Documents defining the content of the KIM USE
3. Approaches to the selection of content, the development of the structure of the KIM USE
The basis of the approaches to the development of the CIM USE 2018 in chemistry was formed by those general methodological guidelines that were determined during the formation of the examination models of previous years. The essence of these settings is as follows.
- KIM are focused on testing the assimilation of the knowledge system, which is considered as the invariant core of the content of existing chemistry programs for general education organizations. In the standard, this knowledge system is presented in the form of requirements for the preparation of graduates. These requirements correlate with the level of presentation in the CMM of the checked content elements.
- In order to ensure the possibility of a differentiated assessment of the educational achievements of KIM USE graduates, they check the mastering of basic educational programs in chemistry at three levels of complexity: basic, advanced and high. The educational material on the basis of which the assignments are built is selected on the basis of its importance for the general education of secondary school graduates.
- The fulfillment of the tasks of the examination work involves the implementation of a certain set of actions. Among them, the most indicative are, for example, such as: to reveal the classification signs of substances and reactions; determine the oxidation state of chemical elements by the formulas of their compounds; explain the essence of a particular process, the relationship of the composition, structure and properties of substances. The examinee's ability to carry out various actions while performing work is considered as an indicator of mastering the studied material with the required depth of understanding.
- The equivalence of all variants of the examination work is ensured by observing the same ratio of the number of tasks that check the assimilation of the main elements of the content of the key sections of the chemistry course.
4. The structure of the KIM USE
Each version of the examination work is built according to a single plan: the work consists of two parts, including 40 tasks. Part 1 contains 35 tasks with a short answer, including 26 tasks of the basic level of difficulty (ordinal numbers of these tasks: 1, 2, 3, 4, ... 26) and 9 tasks of an increased level of difficulty (ordinal numbers of these tasks: 27, 28, 29, ... 35).
Part 2 contains 5 tasks of a high level of complexity, with a detailed answer (ordinal numbers of these tasks: 36, 37, 38, 39, 40).
The Unified State Exam in Chemistry is a variable component of the federal examination. It is taken only by those schoolchildren who are going to continue their studies at universities in such specialties as medicine, chemistry and chemical technology, construction, biotechnology or food industry.
This cannot be called easy - it will not work here with simple knowledge of the terms, because in recent years, tests with a choice of one answer from the proposed options have been excluded from CMMs. In addition, it will not be superfluous to learn everything about the procedure, timing and features of this exam, as well as prepare in advance for possible changes in the CMMs of 2018!
Demonstration version of the exam-2018
Dates of the exam in chemistry
The exact dates set aside for writing the exam in chemistry will be known in January, when the schedule for all examination tests will be posted on the Rosobrnadzor website. Fortunately, today we already have information about the approximate periods allocated for the examination of schoolchildren in the 2017/2018 academic year:
- The early stage of the exam starts on March 22, 2018. It will last until April 15. Writing the USE ahead of schedule is the prerogative of several categories of students. These include children who graduated from school earlier than the 2017/2018 academic year, but did not take the Unified State Exam for any reason; school graduates who previously received only a certificate, and not a matriculation certificate; evening school students; high school students who leave to live or study abroad; schoolchildren who received secondary education in other states, but entering the. Also, pupils representing the Russian Federation at international competitions and competitions, and schoolchildren who take part in all-Russian events, use early delivery. If you are indicated for medical intervention or rehabilitation, which in terms of time coincides with the main period for taking the exam, you can also take the exam ahead of time. An important point: any reason must be confirmed by appropriate documents;
- On May 28, 2018, the main dates of the exam will start. According to the preliminary plans of Rosobrnadzor, the examination period will end by June 10;
- On September 4, 2018, an additional period for passing the exam will begin.
Some statistics
Recently, an increasing number of schoolchildren have chosen this exam - in 2017, about 74 thousand people took it (12 thousand more than in 2016). In addition, the success rate has noticeably improved - the number of unsuccessful students (those who have not reached the threshold minimum points) has decreased by 1.1%. The average score in this subject ranges from 67.8-56.3 points, which corresponds to the level of the school "four". So, on the whole, despite its complexity, schoolchildren pass this subject quite well.
Examination procedure
When writing this USE, students are allowed to use the periodic system, a table with data on the solubility of salts, acids and bases, as well as reference materials of the electrochemical series of metal voltages. There is no need to take these materials with you - all permitted reference materials will be provided to students in one set with the examination card. In addition, an eleventh grader can take a calculator for the exam that does not have a programming function.
We remind you that the procedure for the USE strictly regulates any actions of students. Remember that you can easily lose your chance to enter a university if you suddenly want to discuss the solution to a problem with a friend, try to spy on the answer in a smartphone or a solver, or decide to call someone from the toilet room. By the way, you can go to the toilet or first-aid post, but only with permission and accompanied by a member of the examination committee.
In 2018, the exam in chemistry was expanded to 35 tasks, allocating 3.5 hours for them Innovations in the exam in chemistry
FIPI employees inform about the following changes in new type CMMs.
- In 2018, the number of complex tasks with a detailed answer will be increased. Introduced one new task, number 30, concerning redox reactions. Now the students have a total of 35 tasks to solve.
- You can still earn 60 primary points for all work. The balance was achieved by reducing the points that are awarded for completing simple tasks from the first part of the ticket.
What is included in the structure and content of the ticket?
In the exam, students will have to demonstrate how well they know topics from the course in inorganic, general and organic chemistry. The assignments will test the depth of your knowledge of chemical elements and substances, skills in conducting chemical reactions, knowledge of the basic laws and theoretical provisions of chemistry. In addition, it will become clear how well schoolchildren understand the systemic nature and causality of chemical phenomena, and how much they know about the genesis of substances and the methods of their cognition.
Structurally, the ticket is represented by 35 tasks, divided into two parts:
- part 1 - 29 tasks with a short answer. These tasks are devoted to the theoretical foundations of chemistry, inorganic and organic chemistry, methods of cognition and the use of chemistry in life. For this part of the KIM, you can score 40 points (66.7% of all points for the ticket);
- part 2 - 6 tasks of a high level of complexity, in which a detailed answer is provided. You have to solve problems with non-standard situations. All tasks are devoted to redox reactions, ion exchange reactions, transformations of inorganic and organic substances, or complex calculations. For this part of the KIM, you can score 20 points (33.3% of all points for the ticket).
In total, you can earn up to 60 primary points per ticket. 210 minutes will be allocated to solve it, which you should distribute as follows:
- for basic tasks from the first part - 2-3 minutes each;
- for tasks with an increased level of difficulty from the first part - from 5 to 7 minutes;
- for tasks with a high level of difficulty from the second part - from 10 to 15 minutes.
How are exam scores converted to grades?
Work scores affect the maturity certificate, so for several years in a row they have been translated into the grading system familiar to schoolchildren. The scores are first divided into certain intervals and then converted into grades:
- 0-35 points are identical to "two";
- 36-55 points show a satisfactory degree of preparation for the exam and are equal to the "three";
- 56-72 points - this is an opportunity to get a "four" in the certificate;
- 73 points and above is an indicator that the student knows the subject perfectly.
High-quality preparation for the chemistry exam will allow you not only to enter the chosen university, but also to improve your mark in the certificate! In order not to fill up the exam in chemistry, you will have to score at least 36 points. However, it is worth remembering that to enter a more or less prestigious university, you need to score at least 60-65 points. Top educational institutions accept only those who scored 85-90 points and above on the budget.
How to prepare for the exam in chemistry?
You can't pass a federal exam simply by relying on leftover knowledge from a school chemistry course. To fill in the gaps, it is worth getting down to textbooks and solution books already at the beginning of autumn! It is possible that some topic that you studied in grade 9 or 10 simply did not stick in your memory. In addition, competent preparation includes the development of demonstration tickets - KIMs, specially developed by the FIPI commission.
11/14/2016 on the FIPI website, approved demonstration options, codifiers and specifications of control measuring materials of the unified state exam and the main state exam of 2017, including in chemistry, were published.
Demo version of the exam in chemistry 2017 with answers
| Option of tasks + answers | Download demo |
| Specification | demo variant himiya ege |
| Codifier | kodifikator |
Demo versions of the exam in chemistry 2016-2015
| Chemistry | Download demo + answers |
| 2016 | ege 2016 |
| 2015 | ege 2015 |
There were significant changes in KIM in chemistry in 2017, therefore, demos of previous years are provided for your reference.
Chemistry - significant changes: The structure of the examination work has been optimized:
1. The structure of part 1 of the CMM has been fundamentally changed: tasks with a choice of one answer are excluded; tasks are grouped into separate thematic blocks, each of which has tasks of both basic and advanced levels of difficulty.
2. The total number of tasks has been reduced from 40 (in 2016) to 34.
3. Changed the assessment scale (from 1 to 2 points) for performing tasks of the basic level of complexity, which check the assimilation of knowledge about the genetic relationship of inorganic and organic substances (9 and 17).
4. The maximum primary score for the performance of the work as a whole will be 60 points (instead of 64 points in 2016).
Duration of the exam in chemistry
The total duration of the examination work is 3.5 hours (210 minutes).
The approximate time allotted for the completion of individual tasks is:
1) for each task of the basic level of complexity of part 1 - 2-3 minutes;
2) for each task of an increased level of complexity of part 1 - 5–7 minutes;
3) for each task of a high level of complexity of part 2 - 10-15 minutes.
For tasks 1-3, use the following row of chemical elements. The answer in tasks 1–3 is a sequence of numbers, under which the chemical elements in this row are indicated.
1) Na 2) K 3) Si 4) Mg 5) C
Task number 1
Determine which atoms of the elements indicated in the series have four electrons on the external energy level.
Answer: 3; five
The number of electrons on the external energy level (electronic layer) of the elements of the main subgroups is equal to the group number.
Thus, silicon and carbon are suitable from the options presented. they are in the main subgroup of the fourth group of the table D.I. Mendeleev (IVA group), i.e. Answers 3 and 5 are correct.
Task number 2
From the listed chemical elements, select three elements that are in the Periodic Table of Chemical Elements of D.I. Mendeleev are in the same period. Arrange the selected elements in ascending order of their metallic properties.
Write down the numbers of the selected elements in the required sequence in the answer field.
Answer: 3; 4; 1
Three of the elements presented in one period are sodium Na, silicon Si and magnesium Mg.
When moving within one period of the Periodic Table D.I. Mendeleev (horizontal lines) from right to left facilitates the return of electrons located on the outer layer, i.e. the metallic properties of the elements are enhanced. Thus, the metallic properties of sodium, silicon and magnesium are enhanced in the Si Task number 3 From among the elements listed in the row, select two elements that exhibit the lowest oxidation state of –4. Write down the numbers of the selected elements in the answer field. Answer: 3; five According to the octet rule, atoms of chemical elements tend to have 8 electrons at their outer electronic level, like noble gases. This can be achieved either by giving up electrons of the last level, then the previous one, containing 8 electrons, becomes external, or, conversely, by attaching additional electrons up to eight. Sodium and potassium are alkali metals and are in the main subgroup of the first group (IA). This means that there is one electron each on the outer electron layer of their atoms. In this regard, the loss of a single electron is energetically more favorable than the addition of seven more. With magnesium, the situation is similar, only it is in the main subgroup of the second group, that is, it has two electrons on the external electronic level. It should be noted that sodium, potassium and magnesium belong to metals, and for metals, in principle, a negative oxidation state is impossible. The minimum oxidation state of any metal is zero and is observed in simple substances. The chemical elements carbon C and silicon Si are non-metals and are in the main subgroup of the fourth group (IVA). This means that there are 4 electrons on their outer electron layer. For this reason, for these elements, both the release of these electrons and the addition of four more to a total of 8 are possible. The atoms of silicon and carbon cannot attach more than 4 electrons, so the minimum oxidation state for them is -4. Task number 4 From the proposed list, select two compounds in which an ionic chemical bond is present. Answer: 1; 3 In the overwhelming majority of cases, it is possible to determine the presence of an ionic type of bond in a compound by the fact that atoms of a typical metal and atoms of a non-metal are simultaneously included in its structural units. On this basis, we establish that there is an ionic bond in the compound under number 1 - Ca (ClO 2) 2, since in its formula you can see the atoms of a typical metal calcium and atoms of non-metals - oxygen and chlorine. However, there are no more compounds containing both metal and non-metal atoms in this list. In addition to the above feature, the presence of an ionic bond in a compound can be said if its structural unit contains an ammonium cation (NH 4 +) or its organic analogs - alkylammonium cations RNH 3 +, dialkylammonium R 2 NH 2 +, trialkylammonium R 3 NH + and tetraalkylammonium R 4 N +, where R is some hydrocarbon radical. For example, the ionic type of bond takes place in the compound (CH 3) 4 NCl between the cation (CH 3) 4 + and the chloride ion Cl -. Among the compounds indicated in the task there is ammonium chloride, in which the ionic bond is realized between the ammonium cation NH 4 + and the chloride ion Cl -. Task number 5 Establish a correspondence between the formula of a substance and the class / group to which this substance belongs: for each position indicated by a letter, select the corresponding position from the second column indicated by a number. Write down the numbers of the selected connections in the answer field. Answer: A-4; B-1; AT 3 Explanation: Acid salts are called salts resulting from incomplete replacement of mobile hydrogen atoms with a metal cation, ammonium or alkylammonium cation. In inorganic acids, which take place in the school curriculum, all hydrogen atoms are mobile, that is, they can be replaced by a metal. Examples of acidic inorganic salts among the presented list are ammonium bicarbonate NH 4 HCO 3 - the product of the replacement of one of two hydrogen atoms in carbonic acid with an ammonium cation. Basically, acidic salt is a cross between normal (medium) salt and acid. In the case of NH 4 HCO 3, the average between the normal salt (NH 4) 2 CO 3 and carbonic acid H 2 CO 3. In organic substances, only hydrogen atoms that are part of carboxyl groups (-COOH) or hydroxyl groups of phenols (Ar-OH) can be replaced by metal atoms. That is, for example, sodium acetate CH 3 COONa, despite the fact that in its molecule not all hydrogen atoms are replaced by metal cations, is a medium, not an acidic salt (!). Hydrogen atoms in organic substances, attached directly to a carbon atom, are almost never able to be replaced by metal atoms, with the exception of hydrogen atoms in the triple C≡C bond. Non-salt-forming oxides - oxides of non-metals that do not form salts with basic oxides or bases, that is, either do not react with them at all (most often), or give a different product (not salt) in reaction with them. It is often said that non-salt-forming oxides are nonmetal oxides that do not react with bases and basic oxides. However, this approach does not always work for the detection of non-salt-forming oxides. So, for example, CO, being a non-salt-forming oxide, reacts with the basic iron (II) oxide, but with the formation of not a salt, but a free metal: CO + FeO \u003d CO 2 + Fe Non-salt-forming oxides from the school chemistry course include oxides of non-metals in oxidation states +1 and +2. All of them are found in the exam 4 - these are CO, NO, N 2 O and SiO (the last SiO I personally never met in tasks). Task number 6 From the proposed list of substances, select two substances with each of which iron reacts without heating. Answer: 2; 4 Zinc chloride is a salt and iron is a metal. The metal reacts with the salt only if it is more active than that which is part of the salt. The relative activity of metals is determined by a number of metal activity (in another way, a number of metal stresses). Iron in the row of metal activity is located to the right of zinc, which means that it is less active and is not able to displace zinc from salt. That is, the reaction of iron with substance No. 1 does not go. Copper (II) sulfate CuSO 4 will react with iron, since iron is to the left of copper in the range of activity, that is, it is a more active metal. Concentrated nitric and concentrated sulfuric acids are not capable of reacting without heating with iron, aluminum and chromium in view of such a phenomenon as passivation: on the surface of these metals, under the action of these acids, a salt insoluble without heating is formed, which acts as a protective shell. However, when heated, this protective shell dissolves and the reaction becomes possible. Those. since it is indicated that there is no heating, the reaction of iron with conc. HNO 3 does not leak. Hydrochloric acid, regardless of concentration, belongs to non-oxidizing acids. Metals that are in the line of activity to the left of hydrogen react with non-oxidizing acids with the evolution of hydrogen. Iron belongs to such metals. Conclusion: the reaction of iron with hydrochloric acid proceeds. In the case of a metal and a metal oxide, the reaction, as in the case of a salt, is possible if the free metal is more active than that which is part of the oxide. Fe, according to the series of metal activities, is less active than Al. This means that Fe does not react with Al 2 O 3. Task number 7 From the list, select two oxides that react with hydrochloric acid solution, but do not react
with sodium hydroxide solution. Write down the numbers of the selected substances in the answer field. Answer: 3; 4 CO is a non-salt-forming oxide, does not react with an aqueous solution of alkali. (It should be remembered that, nevertheless, in harsh conditions - high pressure and temperature - it still reacts with solid alkali, forming formates - salts of formic acid.) SO 3 - sulfur oxide (VI) - acidic oxide, which corresponds to sulfuric acid. Acidic oxides do not react with acids and other acidic oxides. That is, SO 3 does not react with hydrochloric acid and reacts with a base - sodium hydroxide. Doesn't fit. CuO - copper (II) oxide - belongs to oxides with predominantly basic properties. Reacts with HCl and does not react with sodium hydroxide solution. Suitable MgO - magnesium oxide - belongs to the typical basic oxides. Reacts with HCl and does not react with sodium hydroxide solution. Suitable ZnO, an oxide with pronounced amphoteric properties, readily reacts with both strong bases and acids (as well as acidic and basic oxides). Doesn't fit. Task number 8 Answer: 4; 2 In the reaction between two salts of inorganic acids, gas is formed only when hot solutions of nitrites and ammonium salts are mixed due to the formation of thermally unstable ammonium nitrite. For instance, NH 4 Cl + KNO 2 \u003d t o \u003d\u003e N 2 + 2H 2 O + KCl However, both nitrites and ammonium salts are not listed. This means that one of the three salts (Cu (NO 3) 2, K 2 SO 3 and Na 2 SiO 3) reacts either with acid (HCl) or with alkali (NaOH). Among the salts of inorganic acids, only ammonium salts release gas when interacting with alkalis: NH 4 + + OH \u003d NH 3 + H 2 O Ammonium salts, as we said, are not on the list. There is only a variant of the interaction of salt with acid. The salts among the specified substances include Cu (NO 3) 2, K 2 SO 3 and Na 2 SiO 3. The reaction of copper nitrate with hydrochloric acid does not proceed, because neither gas, nor sediment, nor low-dissociating substance (water or weak acid) is formed. Sodium silicate reacts with hydrochloric acid, however, due to the release of a white gelatinous precipitate of silicic acid, and not gas: Na 2 SiO 3 + 2HCl \u003d 2NaCl + H 2 SiO 3 ↓ The last option remains - the interaction of potassium sulfite and hydrochloric acid. Indeed, as a result of the ion exchange reaction between sulfite and almost any acid, unstable sulfurous acid is formed, which instantly decomposes into colorless gaseous sulfur (IV) oxide and water. Task number 9 Write down the numbers of the selected substances in the table under the appropriate letters. Answer: 2; five CO 2 is an acidic oxide and must be treated with either a basic oxide or base to convert it to a salt. Those. to obtain potassium carbonate from CO 2, it must be acted upon with either potassium oxide or potassium hydroxide. Thus, substance X is potassium oxide: K 2 O + CO 2 \u003d K 2 CO 3 Potassium bicarbonate KHCO 3, like potassium carbonate, is a salt of carbonic acid, with the only difference that bicarbonate is a product of incomplete replacement of hydrogen atoms in carbonic acid. To obtain an acidic salt from a normal (average) salt, one must either act on it with the same acid that formed this salt, or else act with an acidic oxide corresponding to this acid in the presence of water. Thus, reactant Y is carbon dioxide. When it is passed through an aqueous solution of potassium carbonate, the latter passes into potassium bicarbonate: K 2 CO 3 + H 2 O + CO 2 \u003d 2KHCO 3 Task number 10 Establish a correspondence between the reaction equation and the property of the nitrogen element that it manifests in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number. Write down the numbers of the selected substances in the table under the appropriate letters. Answer: A-4; B-2; AT 2; G-1 A) NH 4 HCO 3 - salt, which contains the ammonium cation NH 4 +. In the ammonium cation, nitrogen always has an oxidation state of -3. As a result of the reaction, it is converted into ammonia NH 3. Hydrogen almost always (except for its compounds with metals) has an oxidation state of +1. Therefore, for the ammonia molecule to be electrically neutral, nitrogen must have an oxidation state of -3. Thus, no change in the oxidation state of nitrogen occurs; it does not exhibit redox properties. B) As already shown above, nitrogen in ammonia NH 3 has an oxidation state of -3. As a result of the reaction with CuO, ammonia is converted into a simple substance N 2. In any simple substance, the oxidation state of the element by which it is formed is zero. Thus, the nitrogen atom loses its negative charge, and since electrons are responsible for the negative charge, this means their loss by the nitrogen atom as a result of the reaction. An element that, as a result of the reaction, loses some of its electrons, is called a reducing agent. B) As a result of the reaction of NH 3 with a nitrogen oxidation state of -3, it turns into nitrogen oxide NO. Oxygen almost always has an oxidation state of -2. Therefore, in order for the nitrogen oxide molecule to be electrically neutral, the nitrogen atom must have an oxidation state of +2. This means that the nitrogen atom as a result of the reaction changed its oxidation state from -3 to +2. This indicates the loss of 5 electrons by the nitrogen atom. That is, nitrogen, as in case B, is a reducing agent. D) N 2 is a simple substance. In all simple substances, the element that forms them has an oxidation state of 0. As a result of the reaction, nitrogen is converted into lithium nitride Li3N. The only oxidation state of an alkali metal other than zero (any element has an oxidation state of 0) is +1. Thus, for the structural unit Li3N to be electrically neutral, nitrogen must have an oxidation state of -3. It turns out that as a result of the reaction, nitrogen acquired a negative charge, which means the addition of electrons. Nitrogen in this reaction is an oxidizing agent. Task number 11 Establish a correspondence between the formula of the substance and the reagents, with each of which this substance can interact: for each position indicated by a letter, select the corresponding position indicated by a number. D) ZnBr 2 (solution) 1) AgNO 3, Na 3 PO 4, Cl 2 2) BaO, H 2 O, KOH 3) H 2, Cl 2, O 2 4) HBr, LiOH, CH 3 COOH 5) H 3 PO 4, BaCl 2, CuO Write down the numbers of the selected substances in the table under the appropriate letters. Answer: A-3; B-2; AT 4; G-1 A) When gaseous hydrogen is passed through the sulfur melt, hydrogen sulfide H 2 S is formed: H 2 + S \u003d t o \u003d\u003e H 2 S When chlorine is passed over crushed sulfur at room temperature, sulfur dichloride is formed: S + Cl 2 \u003d SCl 2 To pass the exam, you do not need to know exactly how sulfur reacts with chlorine and, accordingly, you do not need to be able to write this equation. The main thing is to remember at a fundamental level that sulfur reacts with chlorine. Chlorine is a strong oxidizing agent, sulfur often exhibits a dual function - both oxidative and reductive. That is, if sulfur is acted upon by a strong oxidizing agent, which is molecular chlorine Cl 2, it will oxidize. Sulfur burns with a blue flame in oxygen with the formation of a gas with a pungent odor - sulfur dioxide SO 2: B) SO 3 - sulfur (VI) oxide has pronounced acidic properties. For such oxides, the most typical reactions are reactions with water, as well as with basic and amphoteric oxides and hydroxides. In the list at number 2, we just see water, and the basic oxide BaO, and hydroxide KOH. When an acidic oxide reacts with a basic oxide, a salt of the corresponding acid and the metal that is part of the basic oxide is formed. An acidic oxide corresponds to that acid in which the acid-forming element has the same oxidation state as in the oxide. Sulfuric acid H 2 SO 4 corresponds to SO 3 oxide (both there and there the oxidation state of sulfur is +6). Thus, when SO 3 interacts with metal oxides, sulfuric acid salts - sulfates containing sulfate ion SO 4 2- will be obtained: SO 3 + BaO \u003d BaSO 4 When interacting with water, the acidic oxide turns into the corresponding acid: SO 3 + H 2 O \u003d H 2 SO 4 And when acidic oxides react with metal hydroxides, a salt of the corresponding acid and water are formed: SO 3 + 2KOH \u003d K 2 SO 4 + H 2 O C) Zinc hydroxide Zn (OH) 2 has typical amphoteric properties, that is, it reacts with both acidic oxides and acids, and with basic oxides and alkalis. In list 4 we see both acids - hydrobromic HBr and acetic, and alkali - LiOH. Recall that alkalis are called water-soluble metal hydroxides: Zn (OH) 2 + 2HBr \u003d ZnBr 2 + 2H 2 O Zn (OH) 2 + 2CH 3 COOH \u003d Zn (CH 3 COO) 2 + 2H 2 O Zn (OH) 2 + 2LiOH \u003d Li 2 D) Zinc bromide ZnBr 2 is a salt, soluble in water. For soluble salts, ion exchange reactions are most common. A salt can be reacted with another salt, provided that both starting salts are soluble and a precipitate forms. Also ZnBr 2 contains the bromide ion Br-. For metal halides, it is characteristic that they are capable of reacting with the Hal 2 halogens located above in the periodic table. In this way? the described types of reactions proceed with all substances of list 1: ZnBr 2 + 2AgNO 3 \u003d 2AgBr + Zn (NO 3) 2 3ZnBr 2 + 2Na 3 PO 4 \u003d Zn 3 (PO 4) 2 + 6NaBr ZnBr 2 + Cl 2 \u003d ZnCl 2 + Br 2 Task number 12 Establish a correspondence between the name of a substance and the class / group to which this substance belongs: for each position indicated by a letter, select the corresponding position indicated by a number. Write down the numbers of the selected substances in the table under the appropriate letters. Answer: A-4; B-2; IN 1 Explanation: A) Methylbenzene, aka toluene, has the structural formula: As you can see, the molecules of this substance consist only of carbon and hydrogen, therefore methylbenzene (toluene) refers to hydrocarbons B) The structural formula of aniline (aminobenzene) is as follows: As can be seen from the structural formula, aniline molecule consists of an aromatic hydrocarbon radical (C 6 H 5 -) and an amino group (-NH 2), thus, aniline refers to aromatic amines, i.e. correct answer 2. C) 3-methylbutanal. The ending "al" indicates that the substance belongs to aldehydes. The structural formula of this substance: Task number 13 From the proposed list, select two substances that are structural isomers of butene-1. Write down the numbers of the selected substances in the answer field. Answer: 2; five Explanation: Isomers are substances that have the same molecular formula and different structural ones, i.e. substances that differ in the order of connection of atoms, but with the same composition of molecules. Task number 14 From the proposed list, select two substances, when interacting with a solution of potassium permanganate, a change in the color of the solution will be observed. Write down the numbers of the selected substances in the answer field. Answer: 3; five Explanation: Alkanes, as well as cycloalkanes with a ring size of 5 or more carbon atoms, are very inert and do not react with aqueous solutions of even strong oxidizing agents, such as, for example, potassium permanganate KMnO 4 and potassium dichromate K 2 Cr 2 O 7. Thus, options 1 and 4 disappear - when cyclohexane or propane is added to an aqueous solution of potassium permanganate, the color does not change. Among the hydrocarbons of the homologous series of benzene, only benzene is passive to the action of aqueous solutions of oxidizing agents, all other homologues are oxidized, depending on the medium, either to carboxylic acids or to their corresponding salts. Thus, option 2 (benzene) is eliminated. The correct answers are 3 (toluene) and 5 (propylene). Both substances discolor the purple potassium permanganate solution due to the reactions: CH 3 -CH \u003d CH 2 + 2KMnO 4 + 2H 2 O → CH 3 -CH (OH) –CH 2 OH + 2MnO 2 + 2KOH Task number 15 From the proposed list, select two substances with which formaldehyde reacts. Write down the numbers of the selected substances in the answer field. Answer: 3; 4 Explanation: Formaldehyde belongs to the class of aldehydes - oxygen-containing organic compounds with an aldehyde group at the end of the molecule: Typical reactions of aldehydes are oxidation and reduction reactions proceeding along the functional group. Among the list of answers for formaldehyde, reduction reactions are characteristic, where hydrogen is used as a reducing agent (cat. - Pt, Pd, Ni), and oxidation - in this case, the reaction of a silver mirror. When reduced with hydrogen on a nickel catalyst, formaldehyde is converted to methanol: The silver mirror reaction is the reduction of silver from an ammoniacal solution of silver oxide. When dissolved in an aqueous solution of ammonia, silver oxide is converted into a complex compound - diammine silver hydroxide (I) OH. After the addition of formaldehyde, a redox reaction occurs, in which silver is reduced: Task number 16 From the list provided, select two substances with which methylamine reacts. Write down the numbers of the selected substances in the answer field. Answer: 2; five Explanation: Methylamine is the simplest to represent organic compounds of the amine class. A characteristic feature of amines is the presence of a lone electron pair on the nitrogen atom, as a result of which amines exhibit the properties of bases and in reactions act as nucleophiles. Thus, in this regard, of the proposed answer options, methylamine as a base and nucleophile reacts with chloromethane and hydrochloric acid: CH 3 NH 2 + CH 3 Cl → (CH 3) 2 NH 2 + Cl - CH 3 NH 2 + HCl → CH 3 NH 3 + Cl - Task number 17 The following scheme of transformations of substances is given: Determine which of the specified substances are substances X and Y. Write down the numbers of the selected substances in the table under the appropriate letters. Answer: 4; 2 Explanation: One of the reactions for obtaining alcohols is the reaction of hydrolysis of halogenated alkanes. Thus, ethanol can be obtained from chloroethane by acting on the latter with an aqueous solution of alkali - in this case, NaOH. CH 3 CH 2 Cl + NaOH (aq) → CH 3 CH 2 OH + NaCl The next reaction is the oxidation reaction of ethyl alcohol. Oxidation of alcohols is carried out on a copper catalyst or using CuO: Task number 18 Establish a correspondence between the name of the substance and the product, which is predominantly formed by the interaction of this substance with bromine: for each position indicated by a letter, select the corresponding position indicated by a number. Answer: 5; 2; 3; 6 Explanation: For alkanes, the most typical reactions are free radical substitution reactions, during which a hydrogen atom is replaced by a halogen atom. Thus, brominating ethane, one can obtain bromoethane, and by brominating isobutane, 2-bromisobutane: Since the small cycles of the molecules of cyclopropane and cyclobutane are unstable, during bromination, the cycles of these molecules open, thus the addition reaction proceeds: Unlike the cyclopropane and cyclobutane cycles, the cyclohexane cycle is large, as a result of which the hydrogen atom is replaced by a bromine atom: Task number 19 Establish a correspondence between the reactants and the carbon-containing product that is formed during the interaction of these substances: for each position marked with a letter, select the corresponding position marked with a number. Write down the selected numbers in the table under the corresponding letters. Answer: 5; 4; 6; 2 Task number 20 From the proposed list of reaction types, select two types of reaction, which include the interaction of alkali metals with water. Write down the numbers of the selected reaction types in the answer field. Answer: 3; 4 Alkali metals (Li, Na, K, Rb, Cs, Fr) are located in the main subgroup of group I of the table of D.I. Mendeleev and are reducing agents, easily donating an electron located at the external level. If we designate an alkali metal with the letter M, then the reaction of an alkali metal with water will look like this: 2M + 2H 2 O → 2MOH + H 2 Alkali metals are very reactive towards water. The reaction proceeds violently with the release of a large amount of heat, is irreversible and does not require the use of a catalyst (non-catalytic) - a substance that accelerates the reaction and is not part of the reaction products. It should be noted that all highly exothermic reactions do not require the use of a catalyst and proceed irreversibly. Since metal and water are substances that are in different states of aggregation, this reaction occurs at the interface, therefore, it is heterogeneous. The type of this reaction is substitution. Reactions between inorganic substances are referred to as substitution reactions if a simple substance interacts with a complex one and, as a result, other simple and complex substances are formed. (The neutralization reaction takes place between the acid and the base, as a result of which these substances exchange their constituents and form a salt and a low-dissociating substance). As mentioned above, alkali metals are reducing agents, donating an electron from the outer layer, therefore, the reaction is redox. Task number 21 From the proposed list of external influences, select two influences that lead to a decrease in the rate of reaction of ethylene with hydrogen. Write down the numbers of the selected external influences in the answer field. Answer: 1; 4 The following factors affect the rate of a chemical reaction: changes in temperature and concentration of reagents, as well as the use of a catalyst. According to Van't Hoff's rule of thumb, for every 10 degrees, the rate constant of a homogeneous reaction increases by 2-4 times. Consequently, a decrease in temperature leads to a decrease in the reaction rate. The first answer is fine. As noted above, the reaction rate is also influenced by the change in the concentration of the reagents: if the concentration of ethylene is increased, the reaction rate will also increase, which does not correspond to the requirement of the problem. A decrease in the concentration of hydrogen - the initial component, on the contrary, decreases the reaction rate. Therefore, the second option is not suitable, and the fourth is suitable. A catalyst is a substance that accelerates the rate of a chemical reaction, but is not part of the products. The use of a catalyst accelerates the progress of the ethylene hydrogenation reaction, which also does not correspond to the condition of the problem, and therefore is not the correct answer. When ethylene interacts with hydrogen (on Ni, Pd, Pt catalysts), ethane is formed: CH 2 \u003d CH 2 (g) + H 2 (g) → CH 3 -CH 3 (g) All the components participating in the reaction and the product are gaseous substances, therefore, the pressure in the system will also affect the reaction rate. From two volumes of ethylene and hydrogen, one volume of ethane is formed, therefore, the reaction proceeds to decrease the pressure in the system. By increasing the pressure, we will speed up the reaction. The fifth answer doesn't fit. Task number 22 Establish a correspondence between the formula of the salt and the electrolysis products of an aqueous solution of this salt, which precipitated on the inert electrodes: to each position, SALT FORMULA ELECTROLYSIS PRODUCTS Write down the selected numbers in the table under the corresponding letters. Answer: 1; 4; 3; 2 Electrolysis is a redox process that takes place on the electrodes when a direct electric current passes through a solution or an electrolyte melt. At the cathode, the reduction of those cations that have the highest oxidative activity occurs predominantly. At the anode, first of all, those anions that have the highest reducing ability are oxidized. Electrolysis of aqueous solution 1) The process of electrolysis of aqueous solutions at the cathode does not depend on the material of the cathode, but depends on the position of the metal cation in the electrochemical series of voltages. For cations in a row Li + - Al 3+ reduction process: 2H 2 O + 2e → H 2 + 2OH - (H 2 is evolved at the cathode) Zn 2+ - Pb 2+ recovery process: Me n + + ne → Me 0 and 2H 2 O + 2e → H 2 + 2OH - (H 2 and Me are released at the cathode) Cu 2+ - Au 3+ reduction process Me n + + ne → Me 0 (Me is released at the cathode) 2) The process of electrolysis of aqueous solutions at the anode depends on the material of the anode and on the nature of the anion. If the anode is insoluble, i.e. is inert (platinum, gold, coal, graphite), the process will depend only on the nature of the anions. For anions F -, SO 4 2-, NO 3 -, PO 4 3-, OH - the oxidation process: 4OH - - 4e → O 2 + 2H 2 O or 2H 2 O - 4e → O 2 + 4H + (oxygen is liberated at the anode) halide ions (except for F-) oxidation process 2Hal - - 2e → Hal 2 (free halogens are liberated ) organic acids oxidation process: 2RCOO - - 2e → R-R + 2CO 2 Total electrolysis equation: A) Na 3 PO 4 solution 2H 2 O → 2H 2 (at the cathode) + O 2 (at the anode) B) KCl solution 2KCl + 2H 2 O → H 2 (at the cathode) + 2KOH + Cl 2 (at the anode) B) CuBr2 solution CuBr 2 → Cu (at the cathode) + Br 2 (at the anode) D) Cu (NO3) 2 solution 2Cu (NO 3) 2 + 2H 2 O → 2Cu (at the cathode) + 4HNO 3 + O 2 (at the anode) Task number 23 Establish a correspondence between the name of the salt and the ratio of this salt to hydrolysis: for each position marked with a letter, select the corresponding position marked with a number. Write down the selected numbers in the table under the corresponding letters. Answer: 1; 3; 2; 4 Hydrolysis of salts - the interaction of salts with water, leading to the addition of the hydrogen cation H + of the water molecule to the anion of the acid residue and (or) the hydroxyl group OH - of the water molecule to the metal cation. Salts formed by cations corresponding to weak bases and anions corresponding to weak acids undergo hydrolysis. A) Ammonium chloride (NH 4 Cl) - a salt formed by strong hydrochloric acid and ammonia (weak base), is hydrolyzed by the cation. NH 4 Cl → NH 4 + + Cl - NH 4 + + H 2 O → NH 3 · H 2 O + H + (formation of ammonia dissolved in water) The solution medium is acidic (pH< 7). B) Potassium sulfate (K 2 SO 4) - a salt formed by strong sulfuric acid and potassium hydroxide (alkali, i.e. a strong base), does not undergo hydrolysis. K 2 SO 4 → 2K + + SO 4 2- C) Sodium carbonate (Na 2 CO 3) - a salt formed by weak carbonic acid and sodium hydroxide (alkali, i.e. strong base), is hydrolyzed by anion. CO 3 2- + H 2 O → HCO 3 - + OH - (formation of weakly dissociating hydrocarbonate ion) The solution medium is alkaline (pH\u003e 7). D) Aluminum sulfide (Al 2 S 3) - a salt formed by weak hydrosulfuric acid and aluminum hydroxide (weak base), undergoes complete hydrolysis to form aluminum hydroxide and hydrogen sulfide: Al 2 S 3 + 6H 2 O → 2Al (OH) 3 + 3H 2 S The solution medium is close to neutral (pH ~ 7). Task number 24 Establish a correspondence between the chemical reaction equation and the direction of the displacement of the chemical equilibrium with increasing pressure in the system: for each position marked with a letter, select the corresponding position marked with a number. EQUATION OF REACTION A) N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) B) 2H 2 (d) + O 2 (d) ↔ 2H 2 O (d) C) H 2 (g) + Cl 2 (g) ↔ 2HCl (g) D) SO 2 (g) + Cl 2 (g) ↔ SO 2 Cl 2 (g) DIRECTION OF THE DISPLACEMENT OF THE CHEMICAL EQUILIBRIUM 1) shifts towards direct reaction 2) shifts towards the opposite reaction 3) there is no balance shift Write down the selected numbers in the table under the corresponding letters. Answer: A-1; B-1; AT 3; G-1 The reaction is in chemical equilibrium when the rate of the forward reaction is equal to the rate of the reverse. A shift in equilibrium in the desired direction is achieved by changing the reaction conditions. Factors that determine the position of equilibrium: - pressure: an increase in pressure shifts the equilibrium towards a reaction leading to a decrease in volume (conversely, a decrease in pressure shifts the equilibrium towards a reaction leading to an increase in volume) - temperature: an increase in temperature shifts the equilibrium towards an endothermic reaction (conversely, a decrease in temperature shifts the equilibrium towards an exothermic reaction) - concentration of starting substances and reaction products: an increase in the concentration of starting substances and the removal of products from the reaction sphere shift the equilibrium towards the direct reaction (on the contrary, a decrease in the concentration of starting substances and an increase in reaction products shift the equilibrium towards the opposite reaction) - catalysts do not affect the displacement of equilibrium, but only accelerate its achievement A) In the first case, the reaction proceeds with a decrease in volume, since V (N 2) + 3V (H 2)\u003e 2V (NH 3). By increasing the pressure in the system, the equilibrium will shift towards the side with a smaller volume of substances, therefore, in the forward direction (towards the direct reaction). B) In the second case, the reaction also proceeds with a decrease in volume, since 2V (H 2) + V (O 2)\u003e 2V (H 2 O). By increasing the pressure in the system, the equilibrium will also shift towards the direct reaction (towards the product). C) In the third case, the pressure does not change during the reaction, because V (H 2) + V (Cl 2) \u003d 2V (HCl), so the equilibrium does not shift. D) In \u200b\u200bthe fourth case, the reaction also proceeds with a decrease in volume, since V (SO 2) + V (Cl 2)\u003e V (SO 2 Cl 2). By increasing the pressure in the system, the equilibrium will shift towards the formation of the product (direct reaction). Task number 25 Establish a correspondence between the formulas of substances and a reagent with which you can distinguish between their aqueous solutions: for each position indicated by a letter, select the corresponding position indicated by a number. FORMULAS OF SUBSTANCES A) HNO 3 and H 2 O C) NaCl and BaCl 2 D) AlCl 3 and MgCl 2 Write down the selected numbers in the table under the corresponding letters. Answer: A-1; B-3; AT 3; G-2 A) Nitric acid and water can be distinguished using the salt - calcium carbonate CaCO 3. Calcium carbonate does not dissolve in water, and upon interaction with nitric acid forms a soluble salt - calcium nitrate Ca (NO 3) 2, while the reaction is accompanied by the release of colorless carbon dioxide: CaCO 3 + 2HNO 3 → Ca (NO 3) 2 + CO 2 + H 2 O B) Potassium chloride KCl and alkali NaOH can be distinguished by copper (II) sulfate solution. When copper (II) sulfate interacts with KCl, the exchange reaction does not occur, the solution contains ions K +, Cl -, Cu 2+ and SO 4 2-, which do not form with each other low-dissociating substances. When copper (II) sulfate interacts with NaOH, an exchange reaction occurs, as a result of which copper (II) hydroxide precipitates (blue base). C) Sodium chloride NaCl and barium chloride BaCl 2 are soluble salts that can also be distinguished by a solution of copper (II) sulfate. When copper (II) sulfate interacts with NaCl, the exchange reaction does not take place, the solution contains ions Na +, Cl -, Cu 2+ and SO 4 2-, which do not form low-dissociating substances with each other. When copper (II) sulfate interacts with BaCl 2, an exchange reaction takes place, as a result of which barium sulfate BaSO 4 precipitates. D) Aluminum chlorides AlCl 3 and magnesium MgCl 2 dissolve in water and behave differently when interacting with potassium hydroxide. Magnesium chloride with alkali forms a precipitate: MgCl 2 + 2KOH → Mg (OH) 2 ↓ + 2KCl When alkali interacts with aluminum chloride, a precipitate is first formed, which then dissolves to form a complex salt - potassium tetrahydroxoaluminate: AlCl 3 + 4KOH → K + 3KCl Task number 26 Establish a correspondence between the substance and its area of \u200b\u200bapplication: for each position indicated by a letter, select the corresponding position indicated by a number. Write down the selected numbers in the table under the corresponding letters. Answer: A-4; B-2; AT 3; G-5 A) Ammonia is the most important product of the chemical industry, its production is more than 130 million tons per year. Basically, ammonia is used in the production of nitrogen fertilizers (ammonium nitrate and sulfate, urea), drugs, explosives, nitric acid, soda. Among the proposed answers, the field of application of ammonia is the production of fertilizers (Fourth answer). B) Methane is the simplest hydrocarbon, the most thermally stable representative of a number of limiting compounds. It is widely used as a household and industrial fuel, as well as a raw material for industry (Second answer). Methane is 90-98% a constituent part of natural gas. C) Rubbers are materials that are obtained by polymerizing compounds with conjugated double bonds. Isoprene belongs to this type of compound and is used to obtain one of the types of rubbers: D) Low molecular weight alkenes are used to make plastics, in particular, ethylene is used to make a plastic called polyethylene: nCH 2 \u003d CH 2 → (-CH 2 -CH 2 -) n Task number 27 Calculate the mass of potassium nitrate (in grams), which should be dissolved in 150 g of a solution with a mass fraction of this salt of 10% to obtain a solution with a mass fraction of 12%. (Write the number down to tenths.) Answer: 3.4 g Explanation: Let x g be the mass of potassium nitrate, which is dissolved in 150 g of solution. We calculate the mass of potassium nitrate dissolved in 150 g of solution: m (KNO 3) \u003d 150 g 0.1 \u003d 15 g In order to make the salt mass fraction 12%, x g of potassium nitrate was added. The mass of the solution was (150 + x) g. The equation will be written in the form: (Write the number down to tenths.) Answer: 14.4 g Explanation: As a result of the complete combustion of hydrogen sulfide, sulfur dioxide and water are formed: 2H 2 S + 3O 2 → 2SO 2 + 2H 2 O A consequence of Avogadro's law is that the volumes of gases under the same conditions relate to each other in the same way as the number of moles of these gases. Thus, according to the reaction equation: ν (O 2) \u003d 3 / 2ν (H 2 S), therefore, the volumes of hydrogen sulfide and oxygen are related to each other in the same way: V (O 2) \u003d 3 / 2V (H 2 S), V (O 2) \u003d 3/2 6.72 L \u003d 10.08 L, hence V (O 2) \u003d 10.08 L / 22.4 L / mol \u003d 0.45 mol Let us calculate the mass of oxygen required for complete combustion of hydrogen sulfide: m (O 2) \u003d 0.45 mol 32 g / mol \u003d 14.4 g Task number 30 Using the electronic balance method, write the reaction equation: Na 2 SO 3 +… + KOH → K 2 MnO 4 +… + H 2 O Determine the oxidizing and reducing agent. Mn +7 + 1e → Mn +6 │2 reduction reaction S +4 - 2e → S +6 │1 oxidation reaction Mn +7 (KMnO 4) - oxidizing agent, S +4 (Na 2 SO 3) - reducing agent Na 2 SO 3 + 2KMnO 4 + 2KOH → 2K 2 MnO 4 + Na 2 SO 4 + H 2 O Task number 31 The iron was dissolved in hot concentrated sulfuric acid. The resulting salt was treated with an excess of sodium hydroxide solution. The resulting brown precipitate was filtered off and calcined. The resulting substance was heated with iron. Write the equations for the four reactions described. 1) Iron, like aluminum and chromium, does not react with concentrated sulfuric acid, becoming covered with a protective oxide film. The reaction occurs only when heated with the release of sulfur dioxide: 2Fe + 6H 2 SO 4 → Fe 2 (SO 4) 2 + 3SO 2 + 6H 2 O (when heated) 2) Iron (III) sulfate is a water-soluble salt that enters into an exchange reaction with alkali, as a result of which iron (III) hydroxide precipitates (a brown compound): Fe 2 (SO 4) 3 + 3NaOH → 2Fe (OH) 3 ↓ + 3Na 2 SO 4 3) Insoluble metal hydroxides decompose upon calcination to the corresponding oxides and water: 2Fe (OH) 3 → Fe 2 O 3 + 3H 2 O 4) When iron (III) oxide is heated with metallic iron, iron (II) oxide is formed (iron in the FeO compound has an intermediate oxidation state): Fe 2 O 3 + Fe → 3FeO (when heated) Task number 32 Write the reaction equations with which you can carry out the following transformations: When writing reaction equations, use the structural formulas of organic substances. 1) Intramolecular dehydration occurs at temperatures above 140 o C. This occurs as a result of the elimination of a hydrogen atom from the carbon atom of the alcohol, located through one to the alcoholic hydroxyl (in the β-position). CH 3 -CH 2 -CH 2 -OH → CH 2 \u003d CH-CH 3 + H 2 O (conditions - H 2 SO 4, 180 o C) Intermolecular dehydration occurs at temperatures below 140 o C under the action of sulfuric acid and ultimately comes down to the elimination of one water molecule from two alcohol molecules. 2) Propylene is an unsymmetrical alkenes. When hydrogen halides and water are added, a hydrogen atom is attached to a carbon atom at a multiple bond associated with a large number of hydrogen atoms: CH 2 \u003d CH-CH 3 + HCl → CH 3 -CHCl-CH 3 3) Acting with an aqueous solution of NaOH on 2-chloropropane, the halogen atom is replaced by a hydroxyl group: CH 3 -CHCl-CH 3 + NaOH (aq.) → CH 3 -CHOH-CH 3 + NaCl 4) Propylene can be obtained not only from propanol-1, but also from propanol-2 by the reaction of intramolecular dehydration at temperatures above 140 o C: CH 3 -CH (OH) -CH 3 → CH 2 \u003d CH-CH 3 + H 2 O (conditions H 2 SO 4, 180 o C) 5) In an alkaline medium, acting with a dilute aqueous solution of potassium permanganate, hydroxylation of alkenes occurs to form diols: 3CH 2 \u003d CH-CH 3 + 2KMnO 4 + 4H 2 O → 3HOCH 2 -CH (OH) -CH 3 + 2MnO 2 + 2KOH Task number 33 Determine the mass fractions (in%) of iron (II) sulfate and aluminum sulfide in the mixture if, when treating 25 g of this mixture with water, a gas was released, which completely reacted with 960 g of a 5% solution of copper (II) sulfate. In response, write down the reaction equations that are indicated in the condition of the problem, and provide all the necessary calculations (indicate the units of measurement of the desired physical quantities). Answer: ω (Al 2 S 3) \u003d 40%; ω (CuSO 4) \u003d 60% When a mixture of iron (II) sulfate and aluminum sulfide is treated with water, the sulfate simply dissolves, and the sulfide is hydrolyzed to form aluminum (III) hydroxide and hydrogen sulfide: Al 2 S 3 + 6H 2 O → 2Al (OH) 3 ↓ + 3H 2 S (I) When hydrogen sulfide is passed through a solution of copper (II) sulfate, copper (II) sulfide precipitates: CuSO 4 + H 2 S → CuS ↓ + H 2 SO 4 (II) We calculate the mass and amount of the substance of dissolved copper (II) sulfate: m (CuSO 4) \u003d m (solution) ω (CuSO 4) \u003d 960 g 0.05 \u003d 48 g; ν (CuSO 4) \u003d m (CuSO 4) / M (CuSO 4) \u003d 48 g / 160 g \u003d 0.3 mol According to the reaction equation (II) ν (CuSO 4) \u003d ν (H 2 S) \u003d 0.3 mol, and according to the reaction equation (III) ν (Al 2 S 3) \u003d 1 / 3ν (H 2 S) \u003d 0, 1 mole We calculate the masses of aluminum sulfide and copper (II) sulfate: m (Al 2 S 3) \u003d 0.1 mol * 150 g / mol \u003d 15 g; m (CuSO4) \u003d 25 g - 15 g \u003d 10 g ω (Al 2 S 3) \u003d 15 g / 25g · 100% \u003d 60%; ω (CuSO 4) \u003d 10 g / 25 g 100% \u003d 40% Task number 34 Burning a sample of some organic compound weighing 14.8 g yielded 35.2 g of carbon dioxide and 18.0 g of water. It is known that the relative density of vapors of this substance in terms of hydrogen is 37. During the study of the chemical properties of this substance, it was found that when this substance interacts with copper (II) oxide, ketone is formed. Based on the given conditions of the assignment: 1) make the calculations necessary to establish the molecular formula of organic matter (indicate the units of measurement of the sought physical quantities); 2) write down the molecular formula of the original organic matter; 3) make up the structural formula of this substance, which unambiguously reflects the order of the bonds of atoms in its molecule; 4) write the equation for the reaction of this substance with copper (II) oxide using the structural formula of the substance. Every year, demo versions of the USE of the current year are published on the official website of the FIPI. On August 21, 2017, draft documents regulating the structure and content of the KIM USE 2018 were presented (including the demo version of the USE in chemistry). There are documents that regulate the structure and content of CMM - codifier and specification. Total tasks - 35; of them by the level of difficulty: B - 21; P - 8; AT 6. Maximum primary score for work - 60. The total time to complete the work is 210 minutes. Changes in the KIM USE 2018 in chemistry of the year compared to 2017 The following changes were made in the examination paper in 2018 compared to the paper in 2017. 1. In order to more clearly distribute the tasks for individual thematic blocks and content lines, the order of the tasks of the basic and increased levels of difficulty in part 1 of the examination work has been slightly changed. 2. In the examination paper in 2018, the total number of tasks was increased from 34 (in 2017) to 35 by increasing the number of tasks in part 2 of the examination paper from 5 (in 2017) to 6 tasks. This is achieved through the introduction of assignments with a single context. In particular, in this format, tasks No. 30 and No. 31 are presented, which are focused on checking the assimilation of important elements of the content: "Redox reactions" and "Ion exchange reactions". 3. The grading scale for some tasks has been changed in connection with the clarification of the level of complexity of these tasks based on the results of their implementation in the examination paper in 2017: Task No. 9 of an increased level of complexity, focused on checking the assimilation of the content element "Characteristic chemical properties of inorganic substances" and presented in the format for establishing a correspondence between reactants and reaction products between these substances, will be evaluated with a maximum of 2 points; Task No. 21 of the basic level of complexity, focused on checking the assimilation of the content element "Redox reactions" and presented in the format for establishing a correspondence between the elements of two sets, will be evaluated by 1 point; Task number 26 of the basic level of complexity, focused on checking the assimilation of content lines "Experimental foundations of chemistry" and "General ideas about industrial methods of obtaining the most important substances" and presented in a format for establishing a correspondence between the elements of two sets, will be evaluated by 1 point; Task No. 30 of a high level of complexity with a detailed answer, focused on checking the assimilation of the content element "Redox reactions", will be evaluated with a maximum of 2 points; Task No. 31 of a high level of complexity with a detailed answer, focused on checking the assimilation of the content element "Ion exchange reactions", will be evaluated with a maximum of 2 points. In general, the adopted changes in the examination work in 2018 are aimed at increasing the objectivity of checking the formation of a number of important general educational skills, primarily such as: apply knowledge in the system, independently assess the correctness of the educational and educational-practical tasks, and also combine knowledge about chemical objects with an understanding of the mathematical relationship between various physical quantities. The structure of the KIM USE 2018 in chemistry Each variant of the examination work is built according to a single plan: the work consists of two parts, which include 35 tasks. Part 1 contains 29 tasks with a short answer, including 21 tasks of the basic level of difficulty (in the variant they are present under the numbers: 1-7, 10-15, 18-21, 26-29) and 8 tasks of an increased level of difficulty (their ordinal numbers: 8, 9, 16, 17, 22-25). Part 2 contains 6 tasks of a high level of difficulty, with a detailed answer. These are tasks numbered 30–35.FORMULA OF SUBSTANCE
REAGENTS
Unified State Exam in Chemistry 2018 - demo version with answers and criteria from FIPI
Demo version of the exam 2018 in chemistry
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demo variant ege
Codifier
kodifikator
