A Unified State Examination result in chemistry not lower than the minimum established number of points gives the right to admission to universities in a specialty where the list of entrance tests includes the subject of chemistry.
Universities do not have the right to set the minimum threshold for chemistry below 36 points. Prestigious universities tend to set their minimum threshold much higher. Because to study there, first-year students must have very good knowledge.
On the official website of FIPI, versions of the Unified State Examination in Chemistry are published every year: demonstration, early period. It is these options that give an idea of the structure of the future exam and the level of difficulty of the tasks and are sources of reliable information when preparing for the Unified State Exam.
Early version of the Unified State Exam in Chemistry 2017
| Year | Download early version |
| 2017 | variant po himii |
| 2016 | download |
Demo version of the Unified State Exam in Chemistry 2017 from FIPI
| Variant of tasks + answers | Download demo version |
| Specification | demo variant himiya ege |
| Codifier | codifier |
There are changes in the 2017 versions of the Unified State Exam in Chemistry compared to the previous 2016 KIM, so it is advisable to prepare according to the current version, and to use the versions of previous years for the diversified development of graduates.
Additional materials and equipment
The following materials are attached to each version of the Unified State Examination paper in chemistry:
− periodic table of chemical elements D.I. Mendeleev;
− table of solubility of salts, acids and bases in water;
− electrochemical series of metal voltages.
You are permitted to use a non-programmable calculator during the examination. The list of additional devices and materials, the use of which is permitted for the Unified State Examination, is approved by order of the Russian Ministry of Education and Science.
For those who want to continue their education at a university, the choice of subjects should depend on the list of entrance tests for the chosen specialty
(direction of training).
The list of entrance examinations at universities for all specialties (areas of training) is determined by order of the Russian Ministry of Education and Science. Each university selects from this list certain subjects that it indicates in its admission rules. You need to familiarize yourself with this information on the websites of the selected universities before applying for participation in the Unified State Exam with a list of selected subjects.
Early Unified State Exam in Chemistry 2017. Task 31
The gas obtained from the calcination of silver (I) nitrate was mixed with another gas obtained from the decomposition of potassium chlorate. The resulting mixture of gases was absorbed by water, and acid was formed. Magnesium phosphide was treated with hydrochloric acid, and gas was released. This gas was carefully passed through a hot concentrated solution of the resulting acid. Write equations for the five reactions described. In your answer, indicate the sum of the coefficients in all equations.
Early Unified State Exam in Chemistry 2017. Task 33
Potassium bicarbonate weighing 45 g was calcined to constant weight. The residue was dissolved in excess sulfuric acid. The resulting gas was passed through 200 g of a 5.6% potassium hydroxide solution. Determine the composition and mass of the formed salt, its mass fraction (%) in the solution. When solving, write down the reaction equations that are indicated in the problem statement and provide all the necessary calculations (indicate the units of measurement of the required physical quantities). In your answer, write down the sum of the molar mass (g/mol) of the salt formed, its mass (g) and its mass fraction (%, rounded to the nearest whole number) in the final solution. Neglect the solubility of gases in water.
Early Unified State Examination in Chemistry 2017. Task 34
Upon combustion of 12.24 g of organic matter of a non-cyclic structure, 20.16 l (n.s.) of carbon dioxide and 12.96 g of water were obtained. It is known that 1 mole of this organic substance adds only 1 mole of water and this substance does not react with an ammonia solution of silver oxide. Based on the data of the problem conditions: 1) make the calculations necessary to establish the molecular formula of an organic substance. 2) compose the molecular formula of an organic substance. 3) draw up a structural formula of an organic substance that unambiguously reflects the order of bonds of atoms in its molecule. 4) write an equation for the hydration reaction of organic matter. In your answer, write down the molar mass (g/mol) of the original organic substance.
Tasks 33 and 34 with solutions.
No. 33: A mixture of copper and copper (II) oxide was treated with concentrated sulfuric acid. The volume of gas released was 4.48 liters. This formed a solution weighing 300 g with a mass fraction of salt of 16%. Determine the mass fraction of copper (II) oxide in the initial mixture.
No. 34: A non-cyclic hydrocarbon was burned in oxygen, releasing carbon dioxide weighing 70.4 g and water weighing 21.6 g. When interacting with HCl, this hydrocarbon adds chlorine to the primary carbon atom. Based on the data of the problem conditions: 1) make the calculations necessary to establish the molecular formula of an organic substance; 2) write down the molecular formula of an organic substance; 3) draw up a structural formula of the original substance, which unambiguously reflects the order of bonds of atoms in its molecule; 4) write the equation for the reaction with HCl.
No. 34: When burning an organic substance of a non-cyclic structure weighing 16.2 g, 26.88 l (n.s.) of carbon dioxide and 16.2 g of water were released. 1 mole of a substance reacts completely with 1 mole of water. It is known that this substance does not react with OH. Based on the data of the problem conditions: 1) make the calculations necessary to establish the molecular formula of an organic substance; 2) write down the molecular formula of an organic substance; 3) draw up a structural formula of the original substance, which unambiguously reflects the order of bonds of atoms in its molecule; 4) write the equation for the reaction of this substance with water.
The national examination in chemistry is one of the subjects that a graduate can choose independently. This subject is necessary for those students who intend to continue their studies in the field of medicine, chemistry and chemical technology, construction, biotechnology, food industry and similar industries.
It is better to start preparing for this subject in advance, since in this case it will not be possible to leave by cramming. In addition, you need to clarify in advance possible changes and the dates of the exam in order to be able to correctly distribute your efforts during preparation. To make this task as easy as possible for you, we will analyze the features of the Unified State Exam in chemistry in 2017.
Demo version of the Unified State Exam-2017
Unified State Examination dates in chemistry
You can take the chemistry exam on the following dates:
- Early period. The early date for the exam will be 03/16/2017, and 05/3/2017 is declared as a reserve.
- Main stage. The main date for the examination is June 2, 2017.
- Reserve date. 06/19/2017 was selected as a reserve day.
Several categories of persons can take the Unified State Exam before the main deadline, which include:
- evening school students;
- students who are called up to serve in the ranks;
- schoolchildren leaving for a competition, competition or olympiad of federal or international significance,
- eleventh graders who go abroad due to a change of place of residence or to study at a foreign university;
- students who are prescribed preventive, health-improving treatment or rehabilitation procedures on the main date for passing the Unified State Exam;
- graduates of previous years;
- schoolchildren who studied abroad.
Let us remind you that an application for taking the exam early must be written and submitted before March 1, 2017.
Statistical information
The practice of conducting the Unified State Exam shows that chemistry is not particularly popular among graduates. This exam is not easy, so only one student out of ten takes it. The difficulty is also confirmed by the percentage of students who pass this subject with an unsatisfactory grade - in different years this figure ranges from 6.1 to 11% of the total number of schoolchildren taking exams in chemistry.
As for the average exam scores, recently they range from 67.8 (2013) to 56.3 (2015) points. On the one hand, you can notice a downward trend in this indicator, however, on the other hand, we will hasten to reassure students. These scores correspond to the school “B” level, so you shouldn’t be too afraid of chemistry.
Chemistry is considered one of the most difficult exams and requires serious preparation. What can you use on the Unified State Exam in chemistry?
In the chemistry exam, students can use the periodic table, a table containing information on the solubility of salts, acids and bases, as well as reference materials with data on the electrochemical voltage series of metals. All necessary materials will be given to students along with the ticket. A non-programmable calculator is also mentioned on the Unified State Exam in Chemistry.
Any other items such as smartphones, tablets, players, reference books and programmable computers are prohibited and are grounds for removing the student from the classroom. If you need to go to the first aid station or toilet, it is worth notifying the observer, who will escort you to the right place. Other activities (such as talking to neighbors or changing exam locations) are also prohibited.
Structure of the exam paper
The chemistry ticket consists of 34 tasks, divided into 2 parts:
- the first part includes 29 short-answer tasks;
- the second part consists of 5 tasks, the solution of which will require giving a detailed answer.
When completing chemistry assignments, students must complete the allotted 210 minutes.
The state exam in chemistry in 2017 will last 3.5 hours Changes in KIM-2017 in chemistry
The national exam in chemistry has undergone a lot of changes, which are reflected in the optimization of the ticket structure. The new KIM is aimed at increasing objectivity when assessing students’ knowledge and practical skills. It is worth paying attention to the following points:
- In the structure of the first part of the examination sheet, tasks involving the choice of one option from the proposed answers were excluded. New tasks provide a choice of several correct answers from the proposed ones (for example, 2 out of 5 or 3 out of 6), require students to be able to establish correspondence between individual positions from several sets, and also carry out calculations. In addition, the tasks were grouped into separate thematic blocks, each of which contains tasks related to the basic and advanced levels of complexity. In separate blocks, tasks are arranged in increasing complexity, that is, from one to the next the number of actions that need to be performed to obtain an answer will increase. According to a statement by FIPI representatives, these changes will bring the ticket in line with the school chemistry course program and will help students more effectively demonstrate knowledge of the terminology and patterns of chemical processes.
- In 2017, he reduced the total number of tasks - now there will be not 40, but only 34. Tasks involving similar types of activities have been removed from the ticket: for example, aimed at identifying knowledge about salts, acids and bases and their chemical properties. These changes are explained by the fact that the new ticket is practical-oriented, so even basic tasks will require students to systematically apply the acquired knowledge.
- Basic level tasks (numbers 9 and 17) test knowledge of the genetic relationships of substances of organic and inorganic nature. Now they are rated not at 1, but at 2 points.
- The primary score awarded for work has been changed - now it is not 64, but 60 points.
Grading system
Points for the Unified State Examination are assigned based on a maximum of one hundred. They were not transferred to the grading system familiar to schoolchildren until 2017, but this can be done independently.
To get an A, pay attention to discipline and demo options - If a student scores from 0 to 35 points, his level of knowledge is assessed as unsatisfactory and corresponds to a grade of “2”;
- Scores in the range from 36 to 55 are an indicator of a satisfactory level of knowledge and correspond to the mark “3”;
- Having scored from 56 to 72 points, you can count on a grade of “4”;
- With scores of 73 and above, the rating is considered excellent, that is, “5”.
You can view the final result on the Unified State Exam portal by identifying yourself using your passport data. Let us also recall that the minimum score you need to score for the Unified State Exam in chemistry is 36. It is also worth saying that, according to the latest news, the points for the Unified State Exam in chemistry will influence the grade on the certificate. You should definitely take this chance to correct the mark on your report card that you are not happy with.
To complete tasks 1–3, use the following series of chemical elements. The answer in tasks 1–3 is a sequence of numbers under which the chemical elements in a given row are indicated.
- 1. S
- 2. Na
- 3. Al
- 4. Si
- 5. Mg
Task No. 1
Determine which atoms of the elements indicated in the series contain one unpaired electron in the ground state.
Answer: 23
Explanation:
Let's write down the electronic formula for each of the indicated chemical elements and depict the electron-graphic formula of the last electronic level:
1) S: 1s 2 2s 2 2p 6 3s 2 3p 4
2) Na: 1s 2 2s 2 2p 6 3s 1
3) Al: 1s 2 2s 2 2p 6 3s 2 3p 1
4) Si: 1s 2 2s 2 2p 6 3s 2 3p 2
5) Mg: 1s 2 2s 2 2p 6 3s 2
Task No. 2
From the chemical elements indicated in the series, select three metal elements. Arrange the selected elements in order of increasing reducing properties.
Write down the numbers of the selected elements in the required sequence in the answer field.
Answer: 352
Explanation:
In the main subgroups of the periodic table, metals are located under the boron-astatine diagonal, as well as in secondary subgroups. Thus, the metals from this list include Na, Al and Mg.
The metallic and, therefore, reducing properties of the elements increase when moving to the left along the period and down the subgroup. Thus, the metallic properties of the metals listed above increase in the order Al, Mg, Na
Task No. 3
From among the elements indicated in the series, select two elements that, when combined with oxygen, exhibit an oxidation state of +4.
Write down the numbers of the selected elements in the answer field.
Answer: 14
Explanation:
The main oxidation states of elements from the presented list in complex substances:
Sulfur – “-2”, “+4” and “+6”
Sodium Na – “+1” (single)
Aluminum Al – “+3” (single)
Silicon Si – “-4”, “+4”
Magnesium Mg – “+2” (single)
Task No. 4
From the proposed list of substances, select two substances in which an ionic chemical bond is present.
- 1. KCl
- 2. KNO 3
- 3. H 3 BO 3
- 4.H2SO4
- 5.PCl 3
Answer: 12
Explanation:
In the vast majority of cases, the presence of an ionic type of bond in a compound can be determined by the fact that its structural units simultaneously include atoms of a typical metal and atoms of a non-metal.
Based on this criterion, the ionic type of bond occurs in the compounds KCl and KNO 3.
In addition to the above characteristic, the presence of an ionic bond in a compound can be said if its structural unit contains an ammonium cation (NH 4 +) or its organic analogues - alkylammonium cations RNH 3 +, dialkylammonium R 2 NH 2 +, trialkylammonium cations R 3 NH + and tetraalkylammonium R 4 N +, where R is some hydrocarbon radical. For example, the ionic type of bond occurs in the compound (CH 3) 4 NCl between the cation (CH 3) 4 + and the chloride ion Cl −.
Task No. 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 indicated by a number.
| A | B | IN |
Answer: 241
Explanation:
N 2 O 3 is a non-metal oxide. All non-metal oxides except N 2 O, NO, SiO and CO are acidic.
Al 2 O 3 is a metal oxide in the oxidation state +3. Metal oxides in the oxidation state +3, +4, as well as BeO, ZnO, SnO and PbO, are amphoteric.
HClO 4 is a typical representative of acids, because upon dissociation in an aqueous solution, only H + cations are formed from cations:
HClO 4 = H + + ClO 4 -
Task No. 6
From the proposed list of substances, select two substances, with each of which zinc interacts.
1) nitric acid (solution)
2) iron(II) hydroxide
3) magnesium sulfate (solution)
4) sodium hydroxide (solution)
5) aluminum chloride (solution)
Write down the numbers of the selected substances in the answer field.
Answer: 14
Explanation:
1) Nitric acid is a strong oxidizing agent and reacts with all metals except platinum and gold.
2) Iron hydroxide (ll) is an insoluble base. Metals do not react at all with insoluble hydroxides, and only three metals react with soluble (alkalis) - Be, Zn, Al.
3) Magnesium sulfate is a salt of a more active metal than zinc, and therefore the reaction does not proceed.
4) Sodium hydroxide - alkali (soluble metal hydroxide). Only Be, Zn, Al work with metal alkalis.
5) AlCl 3 – a salt of a metal more active than zinc, i.e. reaction is impossible.
Task No. 7
From the proposed list of substances, select two oxides that react with water.
- 1.BaO
- 2. CuO
- 3.NO
- 4. SO 3
- 5. PbO2
Write down the numbers of the selected substances in the answer field.
Answer: 14
Explanation:
Of the oxides, only oxides of alkali and alkaline earth metals, as well as all acidic oxides except SiO 2, react with water.
Thus, answer options 1 and 4 are suitable:
BaO + H 2 O = Ba(OH) 2
SO 3 + H 2 O = H 2 SO 4
Task No. 8
1) hydrogen bromide
3) sodium nitrate
4) sulfur oxide(IV)
5) aluminum chloride
Write down the selected numbers in the table under the corresponding letters.
Answer: 52
Explanation:
The only salts among these substances are sodium nitrate and aluminum chloride. All nitrates, like sodium salts, are soluble, and therefore sodium nitrate cannot form a precipitate in principle with any of the reagents. Therefore, salt X can only be aluminum chloride.
A common mistake among those taking the Unified State Exam in chemistry is not understanding that in an aqueous solution ammonia forms a weak base - ammonium hydroxide due to the reaction:
NH 3 + H 2 O<=>NH4OH
In this regard, an aqueous solution of ammonia gives a precipitate when mixed with solutions of metal salts that form insoluble hydroxides:
3NH 3 + 3H 2 O + AlCl 3 = Al(OH) 3 + 3NH 4 Cl
Task No. 9
In a given transformation scheme
Cu X> CuCl 2 Y> CuI
substances X and Y are:
- 1. AgI
- 2. I 2
- 3.Cl2
- 4.HCl
- 5. KI
Answer: 35
Explanation:
Copper is a metal located in the activity series to the right of hydrogen, i.e. does not react with acids (except H 2 SO 4 (conc.) and HNO 3). Thus, the formation of copper (ll) chloride is possible in our case only by reaction with chlorine:
Cu + Cl 2 = CuCl 2
Iodide ions (I -) cannot coexist in the same solution with divalent copper ions, because are oxidized by them:
Cu 2+ + 3I - = CuI + I 2
Task No. 10
Establish a correspondence between the reaction equation and the oxidizing substance in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.
Answer: 1433
Explanation:
An oxidizing agent in a reaction is a substance that contains an element that lowers its oxidation state
Task No. 11
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.
Answer: 1215
Explanation:
A) Cu(NO 3) 2 + NaOH and Cu(NO 3) 2 + Ba(OH) 2 – similar interactions. A salt reacts with a metal hydroxide if the starting substances are soluble, and the products contain a precipitate, gas or slightly dissociating substance. For both the first and second reactions, both requirements are met:
Cu(NO 3) 2 + 2NaOH = 2NaNO 3 + Cu(OH) 2 ↓
Cu(NO 3) 2 + Ba(OH) 2 = Na(NO 3) 2 + Cu(OH) 2 ↓
Cu(NO 3) 2 + Mg - a salt reacts with a metal if the free metal is more active than what is included in the salt. Magnesium in the activity series is located to the left of copper, which indicates its greater activity, therefore, the reaction proceeds:
Cu(NO 3) 2 + Mg = Mg(NO 3) 2 + Cu
B) Al(OH) 3 – metal hydroxide in the oxidation state +3. Metal hydroxides in the oxidation state +3, +4, as well as the hydroxides Be(OH) 2 and Zn(OH) 2 as exceptions, are classified as amphoteric.
By definition, amphoteric hydroxides are those that react with alkalis and almost all soluble acids. For this reason, we can immediately conclude that answer option 2 is appropriate:
Al(OH) 3 + 3HCl = AlCl 3 + 3H 2 O
Al(OH) 3 + LiOH (solution) = Li or Al(OH) 3 + LiOH(sol.) =to=> LiAlO 2 + 2H 2 O
2Al(OH) 3 + 3H 2 SO 4 = Al 2 (SO 4) 3 + 6H 2 O
C) ZnCl 2 + NaOH and ZnCl 2 + Ba(OH) 2 – interaction of the “salt + metal hydroxide” type. The explanation is given in paragraph A.
ZnCl 2 + 2NaOH = Zn(OH) 2 + 2NaCl
ZnCl 2 + Ba(OH) 2 = Zn(OH) 2 + BaCl 2
It should be noted that with an excess of NaOH and Ba(OH) 2:
ZnCl 2 + 4NaOH = Na 2 + 2NaCl
ZnCl 2 + 2Ba(OH) 2 = Ba + BaCl 2
D) Br 2, O 2 are strong oxidizing agents. The only metals that do not react are silver, platinum, and gold:
Cu + Br 2 t° > CuBr 2
2Cu + O2 t° >2CuO
HNO 3 is an acid with strong oxidizing properties, because oxidizes not with hydrogen cations, but with an acid-forming element - nitrogen N +5. Reacts with all metals except platinum and gold:
4HNO 3(conc.) + Cu = Cu(NO 3)2 + 2NO 2 + 2H 2 O
8HNO 3(dil.) + 3Cu = 3Cu(NO 3) 2 + 2NO + 4H 2 O
Task No. 12
Establish a correspondence between the general formula of a homologous series and the name of a substance belonging to this series: for each position indicated by a letter, select the corresponding position indicated by a number.
Write down the selected numbers in the table under the corresponding letters.
| A | B | IN | |
Answer: 231
Explanation:
Task No. 13
From the proposed list of substances, select two substances that are isomers of cyclopentane.
1) 2-methylbutane
2) 1,2-dimethylcyclopropane
3) penten-2
4) hexene-2
5) cyclopentene
Write down the numbers of the selected substances in the answer field.
Answer: 23
Explanation:
Cyclopentane has the molecular formula C5H10. Let's write the structural and molecular formulas of the substances listed in the condition
Substance name |
Structural formula |
Molecular formula |
cyclopentane |
C5H10 |
|
2-methylbutane |
||
1,2-dimethylcyclopropane |
C5H10 |
|
C5H10 |
||
cyclopentene |
Task No. 14
From the proposed list of substances, select two substances, each of which reacts with a solution of potassium permanganate.
1) methylbenzene
2) cyclohexane
3) methylpropane
Write down the numbers of the selected substances in the answer field.
Answer: 15
Explanation:
Of the hydrocarbons that react with an aqueous solution of potassium permanganate are those that contain C=C or C≡C bonds in their structural formula, as well as homologs of benzene (except benzene itself).
Methylbenzene and styrene are suitable in this way.
Task No. 15
From the proposed list of substances, select two substances with which phenol interacts.
1) hydrochloric acid
2) sodium hydroxide
4) nitric acid
5) sodium sulfate
Write down the numbers of the selected substances in the answer field.
Answer: 24
Explanation:
Phenol has weak acidic properties, more pronounced than alcohols. For this reason, phenols, unlike alcohols, react with alkalis:
C 6 H 5 OH + NaOH = C 6 H 5 ONa + H 2 O
Phenol contains in its molecule a hydroxyl group directly attached to the benzene ring. The hydroxy group is an orienting agent of the first kind, that is, it facilitates substitution reactions in the ortho and para positions:
Task No. 16
From the proposed list of substances, select two substances that undergo hydrolysis.
1) glucose
2) sucrose
3) fructose
5) starch
Write down the numbers of the selected substances in the answer field.
Answer: 25
Explanation:
All of the substances listed are carbohydrates. Of carbohydrates, monosaccharides do not undergo hydrolysis. Glucose, fructose and ribose are monosaccharides, sucrose is a disaccharide, and starch is a polysaccharide. Therefore, sucrose and starch from the above list are subject to hydrolysis.
Task No. 17
The following scheme of substance transformations is specified:
1,2-dibromoethane → X → bromoethane → Y → ethyl formate
Determine which of the indicated substances are substances X and Y.
2) ethanal
4) chloroethane
5) acetylene
Write down the numbers of the selected substances under the corresponding letters in the table.
Task No. 18
Establish a correspondence between the name of the starting substance and the product, which is mainly formed when this substance reacts with bromine: for each position indicated by a letter, select the corresponding position indicated by a number.
Write down the selected numbers in the table under the corresponding letters.
| A | B | IN | G |
Answer: 2134
Explanation:
Substitution at the secondary carbon atom occurs to a greater extent than at the primary one. Thus, the main product of propane bromination is 2-bromopropane, not 1-bromopropane:
Cyclohexane is a cycloalkane with a ring size of more than 4 carbon atoms. Cycloalkanes with a ring size of more than 4 carbon atoms, when interacting with halogens, enter into a substitution reaction with preservation of the cycle:
Cyclopropane and cyclobutane - cycloalkanes with a minimum ring size preferentially undergo addition reactions accompanied by ring rupture:
The replacement of hydrogen atoms at the tertiary carbon atom occurs to a greater extent than at the secondary and primary ones. Thus, the bromination of isobutane proceeds mainly as follows:
Task No. 19
Establish a correspondence between the reaction scheme and the organic substance that is the product of this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.
Write down the selected numbers in the table under the corresponding letters.
| A | B | IN | G |
Answer: 6134
Explanation:
Heating aldehydes with freshly precipitated copper hydroxide leads to the oxidation of the aldehyde group to a carboxyl group:
Aldehydes and ketones are reduced by hydrogen in the presence of nickel, platinum or palladium to alcohols:
Primary and secondary alcohols are oxidized by hot CuO to aldehydes and ketones, respectively:
When concentrated sulfuric acid reacts with ethanol upon heating, two different products may form. When heated to a temperature below 140 °C, intermolecular dehydration occurs predominantly with the formation of diethyl ether, and when heated above 140 °C, intramolecular dehydration occurs, as a result of which ethylene is formed:
Task No. 20
From the proposed list of substances, select two substances whose thermal decomposition reaction is redox.
1) aluminum nitrate
2) potassium bicarbonate
3) aluminum hydroxide
4) ammonium carbonate
5) ammonium nitrate
Write down the numbers of the selected substances in the answer field.
Answer: 15
Explanation:
Redox reactions are those reactions in which one or more chemical elements change their oxidation state.
The decomposition reactions of absolutely all nitrates are redox reactions. Metal nitrates from Mg to Cu inclusive decompose to metal oxide, nitrogen dioxide and molecular oxygen:
All metal bicarbonates decompose even with slight heating (60 o C) to metal carbonate, carbon dioxide and water. In this case, no change in oxidation states occurs:
Insoluble oxides decompose when heated. The reaction is not redox because Not a single chemical element changes its oxidation state as a result:
Ammonium carbonate decomposes when heated into carbon dioxide, water and ammonia. The reaction is not redox:
Ammonium nitrate decomposes into nitric oxide (I) and water. The reaction relates to OVR:
Task No. 21
From the proposed list, select two external influences that lead to an increase in the rate of reaction of nitrogen with hydrogen.
1) decrease in temperature
2) increase in pressure in the system
5) use of an inhibitor
Write down the numbers of the selected external influences in the answer field.
Answer: 24
Explanation:
1) temperature decrease:
The rate of any reaction decreases as the temperature decreases
2) increase in pressure in the system:
Increasing pressure increases the rate of any reaction in which at least one gaseous substance takes part.
3) decrease in hydrogen concentration
Decreasing the concentration always reduces the reaction rate
4) increase in nitrogen concentration
Increasing the concentration of reagents always increases the reaction rate
5) use of an inhibitor
Inhibitors are substances that slow down the rate of a reaction.
Task No. 22
Establish a correspondence between the formula of a substance and the products of electrolysis of an aqueous solution of this substance on inert electrodes: for each position indicated by a letter, select the corresponding position indicated by a number.
Write down the selected numbers in the table under the corresponding letters.
| A | B | IN | G |
Answer: 5251
Explanation:
A) NaBr → Na + + Br -
Na+ cations and water molecules compete with each other for the cathode.
2H 2 O + 2e - → H 2 + 2OH -
2Cl - -2e → Cl 2
B) Mg(NO 3) 2 → Mg 2+ + 2NO 3 -
Mg 2+ cations and water molecules compete with each other for the cathode.
Alkali metal cations, as well as magnesium and aluminum, are not able to be reduced in an aqueous solution due to their high activity. For this reason, water molecules are reduced instead according to the equation:
2H 2 O + 2e - → H 2 + 2OH -
NO 3 - anions and water molecules compete with each other for the anode.
2H 2 O - 4e - → O 2 + 4H +
So answer 2 (hydrogen and oxygen) is appropriate.
B) AlCl 3 → Al 3+ + 3Cl -
Alkali metal cations, as well as magnesium and aluminum, are not able to be reduced in an aqueous solution due to their high activity. For this reason, water molecules are reduced instead according to the equation:
2H 2 O + 2e - → H 2 + 2OH -
Cl - anions and water molecules compete with each other for the anode.
Anions consisting of one chemical element (except F -) outperform water molecules for oxidation at the anode:
2Cl - -2e → Cl 2
Therefore, answer option 5 (hydrogen and halogen) is appropriate.
D) CuSO 4 → Cu 2+ + SO 4 2-
Metal cations to the right of hydrogen in the activity series are easily reduced under aqueous solution conditions:
Cu 2+ + 2e → Cu 0
Acidic residues containing an acid-forming element in the highest oxidation state lose competition to water molecules for oxidation at the anode:
2H 2 O - 4e - → O 2 + 4H +
Thus, answer option 1 (oxygen and metal) is appropriate.
Task No. 23
Establish a correspondence between the name of the salt and the medium of the aqueous solution of this salt: for each position indicated by a letter, select the corresponding position indicated by a number.
Write down the selected numbers in the table under the corresponding letters.
| A | B | IN | G |
Answer: 3312
Explanation:
A) iron(III) sulfate - Fe 2 (SO 4) 3
formed by a weak “base” Fe(OH) 3 and a strong acid H 2 SO 4. Conclusion - the environment is acidic
B) chromium(III) chloride - CrCl 3
formed by the weak “base” Cr(OH) 3 and the strong acid HCl. Conclusion - the environment is acidic
B) sodium sulfate - Na 2 SO 4
Formed by the strong base NaOH and the strong acid H 2 SO 4. Conclusion - the environment is neutral
D) sodium sulfide - Na 2 S
Formed by the strong base NaOH and the weak acid H2S. Conclusion - the environment is alkaline.
Task No. 24
Establish a correspondence between the method of influencing the equilibrium system
CO (g) + Cl 2 (g) COCl 2 (g) + Q
and the direction of the shift in chemical equilibrium as a result of this effect: for each position indicated by a letter, select the corresponding position indicated by a number.
Write down the selected numbers in the table under the corresponding letters.
| A | B | IN | G |
Answer: 3113
Explanation:
The equilibrium shift under external influence on the system occurs in such a way as to minimize the effect of this external influence (Le Chatelier's principle).
A) An increase in the concentration of CO causes the equilibrium to shift toward the forward reaction because it results in a decrease in the amount of CO.
B) An increase in temperature will shift the equilibrium towards an endothermic reaction. Since the forward reaction is exothermic (+Q), the equilibrium will shift towards the reverse reaction.
C) A decrease in pressure will shift the equilibrium towards the reaction that results in an increase in the amount of gases. As a result of the reverse reaction, more gases are formed than as a result of the direct reaction. Thus, the equilibrium will shift towards the opposite reaction.
D) An increase in the concentration of chlorine leads to a shift in the equilibrium towards the direct reaction, since as a result it reduces the amount of chlorine.
Task No. 25
Establish a correspondence between two substances and a reagent that can be used to distinguish these substances: for each position indicated by a letter, select the corresponding position indicated by a number.
Answer: 3454
Explanation:
It is possible to distinguish two substances with the help of a third one only if these two substances interact with it differently, and, most importantly, these differences are externally distinguishable.
A) Solutions of FeSO 4 and FeCl 2 can be distinguished using a solution of barium nitrate. In the case of FeSO 4, a white precipitate of barium sulfate forms:
FeSO 4 + BaCl 2 = BaSO 4 ↓ + FeCl 2
In the case of FeCl 2 there are no visible signs of interaction, since the reaction does not occur.
B) Solutions of Na 3 PO 4 and Na 2 SO 4 can be distinguished using a solution of MgCl 2. The Na 2 SO 4 solution does not react, and in the case of Na 3 PO 4 a white precipitate of magnesium phosphate precipitates:
2Na 3 PO 4 + 3MgCl 2 = Mg 3 (PO 4) 2 ↓ + 6NaCl
C) Solutions of KOH and Ca(OH) 2 can be distinguished using a solution of Na 2 CO 3. KOH does not react with Na 2 CO 3, but Ca(OH) 2 gives a white precipitate of calcium carbonate with Na 2 CO 3:
Ca(OH) 2 + Na 2 CO 3 = CaCO 3 ↓ + 2NaOH
D) Solutions of KOH and KCl can be distinguished using a solution of MgCl 2. KCl does not react with MgCl 2, and mixing solutions of KOH and MgCl 2 leads to the formation of a white precipitate of magnesium hydroxide:
MgCl 2 + 2KOH = Mg(OH) 2 ↓ + 2KCl
Task No. 26
Establish a correspondence between the substance and its area of application: for each position indicated by a letter, select the corresponding position indicated by a number.
Write down the selected numbers in the table under the corresponding letters.
| A | B | IN | G |
Answer: 2331
Explanation:
Ammonia - used in the production of nitrogenous fertilizers. In particular, ammonia is a raw material for the production of nitric acid, from which, in turn, fertilizers are produced - sodium, potassium and ammonium nitrate (NaNO 3, KNO 3, NH 4 NO 3).
Carbon tetrachloride and acetone are used as solvents.
Ethylene is used to produce high molecular weight compounds (polymers), namely polyethylene.
The answer to tasks 27–29 is a number. Write this number in the answer field in the text of the work, while maintaining the specified degree of accuracy. Then transfer this number to ANSWER FORM No. 1 to the right of the number of the corresponding task, starting from the first cell. Write each character in a separate box in accordance with the samples given in the form. There is no need to write units of measurement of physical quantities.
Task No. 27
What mass of potassium hydroxide must be dissolved in 150 g of water to obtain a solution with a mass fraction of alkali of 25%? (Write the number to the nearest whole number.)
Answer: 50
Explanation:
Let the mass of potassium hydroxide that needs to be dissolved in 150 g of water be equal to x g. Then the mass of the resulting solution will be (150+x) g, and the mass fraction of alkali in such a solution can be expressed as x/(150+x). From the condition we know that the mass fraction of potassium hydroxide is 0.25 (or 25%). Thus, the equation is valid:
x/(150+x) = 0.25
Thus, the mass that must be dissolved in 150 g of water to obtain a solution with a mass fraction of alkali of 25% is 50 g.
Task No. 28
In a reaction whose thermochemical equation is
MgO (tv.) + CO 2 (g) → MgCO 3 (tv.) + 102 kJ,
88 g of carbon dioxide entered. How much heat will be released in this case? (Write the number to the nearest whole number.)
Answer: ___________________________ kJ.
Answer: 204
Explanation:
Let's calculate the amount of carbon dioxide:
n(CO 2) = n(CO 2)/ M(CO 2) = 88/44 = 2 mol,
According to the reaction equation, when 1 mole of CO 2 reacts with magnesium oxide, 102 kJ is released. In our case, the amount of carbon dioxide is 2 mol. Designating the amount of heat released as x kJ, we can write the following proportion:
1 mol CO 2 – 102 kJ
2 mol CO 2 – x kJ
Therefore, the equation is valid:
1 ∙ x = 2 ∙ 102
Thus, the amount of heat that will be released when 88 g of carbon dioxide participates in the reaction with magnesium oxide is 204 kJ.
Task No. 29
Determine the mass of zinc that reacts with hydrochloric acid to produce 2.24 L (N.S.) of hydrogen. (Write the number to the nearest tenth.)
Answer: ___________________________ g.
Answer: 6.5
Explanation:
Let's write the reaction equation:
Zn + 2HCl = ZnCl 2 + H 2
Let's calculate the amount of hydrogen substance:
n(H 2) = V(H 2)/V m = 2.24/22.4 = 0.1 mol.
Since in the reaction equation there are equal coefficients in front of zinc and hydrogen, this means that the amounts of zinc substances that entered into the reaction and the hydrogen formed as a result of it are also equal, i.e.
n(Zn) = n(H 2) = 0.1 mol, therefore:
m(Zn) = n(Zn) ∙ M(Zn) = 0.1 ∙ 65 = 6.5 g.
Do not forget to transfer all answers to answer form No. 1 in accordance with the instructions for completing the work.
Task No. 33
Sodium bicarbonate weighing 43.34 g was calcined to constant weight. The residue was dissolved in excess hydrochloric acid. The resulting gas was passed through 100 g of a 10% sodium hydroxide solution. Determine the composition and mass of the formed salt, its mass fraction in the solution. In your answer, write down the reaction equations that are indicated in the problem statement and provide all the necessary calculations (indicate the units of measurement of the required physical quantities).
Answer:
Explanation:
Sodium bicarbonate decomposes when heated according to the equation:
2NaHCO 3 → Na 2 CO 3 + CO 2 + H 2 O (I)
The resulting solid residue apparently consists only of sodium carbonate. When sodium carbonate is dissolved in hydrochloric acid, the following reaction occurs:
Na 2 CO 3 + 2HCl → 2NaCl + CO 2 + H 2 O (II)
Calculate the amount of sodium bicarbonate and sodium carbonate:
n(NaHCO 3) = m(NaHCO 3)/M(NaHCO 3) = 43.34 g/84 g/mol ≈ 0.516 mol,
hence,
n(Na 2 CO 3) = 0.516 mol/2 = 0.258 mol.
Let's calculate the amount of carbon dioxide formed by reaction (II):
n(CO 2) = n(Na 2 CO 3) = 0.258 mol.
Let's calculate the mass of pure sodium hydroxide and its amount of substance:
m(NaOH) = m solution (NaOH) ∙ ω(NaOH)/100% = 100 g ∙ 10%/100% = 10 g;
n(NaOH) = m(NaOH)/ M(NaOH) = 10/40 = 0.25 mol.
The interaction of carbon dioxide with sodium hydroxide, depending on their proportions, can proceed in accordance with two different equations:
2NaOH + CO 2 = Na 2 CO 3 + H 2 O (with excess alkali)
NaOH + CO 2 = NaHCO 3 (with excess carbon dioxide)
From the presented equations it follows that only average salt is obtained at the ratio n(NaOH)/n(CO 2) ≥2, and only acidic salt at the ratio n(NaOH)/n(CO 2) ≤ 1.
According to calculations, ν(CO 2) > ν(NaOH), therefore:
n(NaOH)/n(CO2) ≤ 1
Those. the interaction of carbon dioxide with sodium hydroxide occurs exclusively with the formation of an acid salt, i.e. according to the equation:
NaOH + CO 2 = NaHCO 3 (III)
We carry out the calculation based on the lack of alkali. According to reaction equation (III):
n(NaHCO 3) = n(NaOH) = 0.25 mol, therefore:
m(NaHCO 3) = 0.25 mol ∙ 84 g/mol = 21 g.
The mass of the resulting solution will be the sum of the mass of the alkali solution and the mass of carbon dioxide absorbed by it.
From the reaction equation it follows that it reacted, i.e. only 0.25 mol of CO 2 was absorbed out of 0.258 mol. Then the mass of absorbed CO 2 is:
m(CO 2) = 0.25 mol ∙ 44 g/mol = 11 g.
Then, the mass of the solution is equal to:
m(solution) = m(NaOH solution) + m(CO 2) = 100 g + 11 g = 111 g,
and the mass fraction of sodium bicarbonate in the solution will thus be equal to:
ω(NaHCO 3) = 21 g/111 g ∙ 100% ≈ 18.92%.
Task No. 34
Upon combustion of 16.2 g of organic matter of a non-cyclic structure, 26.88 l (n.s.) of carbon dioxide and 16.2 g of water were obtained. It is known that 1 mole of this organic substance in the presence of a catalyst adds only 1 mole of water and this substance does not react with an ammonia solution of silver oxide.
Based on the data of the problem conditions:
1) make the calculations necessary to establish the molecular formula of an organic substance;
2) write down the molecular formula of an organic substance;
3) draw up a structural formula of an organic substance that unambiguously reflects the order of bonds of atoms in its molecule;
4) write the equation for the hydration reaction of organic matter.
Answer:
Explanation:
1) To determine the elemental composition, let’s calculate the amounts of substances carbon dioxide, water and then the masses of the elements included in them:
n(CO 2) = 26.88 l/22.4 l/mol = 1.2 mol;
n(CO 2) = n(C) = 1.2 mol; m(C) = 1.2 mol ∙ 12 g/mol = 14.4 g.
n(H 2 O) = 16.2 g/18 g/mol = 0.9 mol; n(H) = 0.9 mol ∙ 2 = 1.8 mol; m(H) = 1.8 g.
m(org. substances) = m(C) + m(H) = 16.2 g, therefore, there is no oxygen in organic matter.
The general formula of an organic compound is C x H y.
x: y = ν(C) : ν(H) = 1.2: 1.8 = 1: 1.5 = 2: 3 = 4: 6
Thus, the simplest formula of the substance is C 4 H 6. The true formula of a substance may coincide with the simplest one, or it may differ from it by an integer number of times. Those. be, for example, C 8 H 12, C 12 H 18, etc.
The condition states that the hydrocarbon is non-cyclic and one molecule of it can attach only one molecule of water. This is possible if there is only one multiple bond (double or triple) in the structural formula of the substance. Since the desired hydrocarbon is non-cyclic, it is obvious that one multiple bond can only exist for a substance with the formula C 4 H 6. In the case of other hydrocarbons with a higher molecular weight, the number of multiple bonds is always more than one. Thus, the molecular formula of the substance C 4 H 6 coincides with the simplest one.
2) The molecular formula of an organic substance is C 4 H 6.
3) Of the hydrocarbons, alkynes in which the triple bond is located at the end of the molecule interact with an ammonia solution of silver oxide. In order to avoid interaction with an ammonia solution of silver oxide, the alkyne composition C 4 H 6 must have the following structure:
CH 3 -C≡C-CH 3
4) Hydration of alkynes occurs in the presence of divalent mercury salts.
