Online tests of gia in chemistry. Online tests of GIA in chemistry Assignments of exam in chemistry 20 06

Specification
control measuring materials
to conduct a unified state exam in 2016
in chemistry

1. Appointment of KIM USE

The Unified State Exam (hereinafter - the Unified State Exam) is a form of objective assessment of the quality of training of persons who have mastered educational programs of secondary general education, using tasks of a standardized form (control measuring materials).

The Unified State Exam is conducted in accordance with the Federal Law of December 29, 2012 No. 273-FZ "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 specialized 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 2016 in chemistry was formed by those general methodological guidelines that were determined during the formation of 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 the current 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 graduates of the KIM Unified State Exam, they check the mastering of the 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 identify 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 a variety of actions while performing work is considered as an indicator of the assimilation of 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 state final certification of 2019 in chemistry for graduates of the 9th grade of educational institutions is carried out in order to assess the level of general education of graduates in this discipline. The tasks test the knowledge of the following sections of chemistry:

The structure of the atom.
Periodic Law and Periodic Table of Chemical Elements D.I. Mendeleev.
Molecular structure. Chemical bond: covalent (polar and non-polar), ionic, metallic.
Valence of chemical elements. Oxidation state of chemical elements.
Simple and complex substances.
Chemical reaction. Conditions and signs of chemical reactions. Chemical equations.
Electrolytes and non-electrolytes. Cations and anions. Electrolytic dissociation of acids, alkalis and salts (medium).
Ion exchange reactions and conditions for their implementation.
Chemical properties of simple substances: metals and non-metals.
Chemical properties of oxides: basic, amphoteric, acidic.
Chemical properties of bases. Chemical properties of acids.
Chemical properties of salts (medium).
Pure substances and mixtures. Rules for safe work in the school laboratory. Chemical pollution of the environment and its consequences.
Oxidation state of chemical elements. Oxidizing and reducing agent. Redox reactions.
Calculation of the mass fraction of a chemical element in a substance.
Periodic law of D.I. Mendeleev.
Initial information about organic substances. Biologically important substances: proteins, fats, carbohydrates.
Determination of the nature of the medium of the solution of acids and alkalis using indicators. Qualitative reactions to ions in the solution (chloride, sulfate, carbonation, ammonium ion). Qualitative reactions to gaseous substances (oxygen, hydrogen, carbon dioxide, ammonia).
Chemical properties of simple substances. Chemical properties of complex substances.

Date of passing the OGE in chemistry 2019:
June 4 (Tuesday).

There are no changes in the structure and content of the 2019 examination paper compared to 2018.

In this section, you will find online tests that will help you prepare for passing the OGE (GIA) in chemistry. We wish you every success!

The standard test of the OGE (GIA-9) of the 2019 format in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. Therefore, in this test, only the first part is presented (i.e. the first 19 tasks). According to the current structure of the exam, among these tasks, answer options are offered only in 15. However, for the convenience of passing the tests, the site administration decided to offer answer options in all tasks. But for tasks in which answer options are not provided by the compilers of real control and measuring materials (CMMs), the number of answer options has been significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

The standard OGE test (GIA-9) of the 2018 format in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. Therefore, in this test, only the first part is presented (i.e. the first 19 tasks). According to the current structure of the exam, among these tasks, answer options are offered only in 15. However, for the convenience of passing the tests, the site administration decided to offer answer options in all tasks. But for tasks in which answer options are not provided by the compilers of real control and measuring materials (CMMs), the number of answer options has been significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

The OGE standard test (GIA-9) of the 2017 format in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. Therefore, in this test, only the first part is presented (i.e. the first 19 tasks). According to the current structure of the exam, among these tasks, answer options are offered only in 15. However, for the convenience of passing the tests, the site administration decided to offer answer options in all tasks. But for tasks in which answer options are not provided by the compilers of real control and measuring materials (CMMs), the number of answer options has been significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

The standard test of the OGE (GIA-9) of the 2016 format in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. Therefore, in this test, only the first part is presented (i.e. the first 19 tasks). According to the current structure of the exam, among these tasks, answer options are offered only in 15. However, for the convenience of passing the tests, the site administration decided to offer answer options in all tasks. But for tasks in which answer options are not provided by the compilers of real control and measuring materials (CMMs), the number of answer options has been significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

The standard test of the OGE (GIA-9) of the 2015 format in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. Therefore, in this test, only the first part is presented (i.e. the first 19 tasks). According to the current structure of the exam, among these tasks, answer options are offered only in 15. However, for the convenience of passing the tests, the site administration decided to offer answer options in all tasks. But for tasks in which answer options are not provided by the compilers of real control and measuring materials (CMMs), the number of answer options has been significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

When performing tasks A1-A19, select only one correct option.
For tasks B1-B3, select two correct options.

When performing tasks A1-A15, select only one correct option.

When performing tasks A1-A15, choose only one correct option.

In 2-3 months it is impossible to learn (repeat, tighten up) such a complex discipline as chemistry.

There are no changes in the KIM USE 2020 in chemistry.

Don't postpone your preparation until later.

When starting to analyze the tasks, first study theory... The theory on the site is presented for each task in the form of recommendations that you need to know when completing the task. will guide you in the study of the main topics and determine what knowledge and skills will be required when completing the USE tasks in chemistry. For the successful completion of the exam in chemistry, theory is most important.
Theory needs to be backed up practice constantly solving tasks. Since most of the errors are due to the fact that I read the exercise incorrectly, I did not understand what is required in the task. The more often you solve thematic tests, the faster you will understand the structure of the exam. Training tasks developed on the basis of demos from FIPI give such an opportunity to decide and find out the answers. But don't rush to pry. First, decide for yourself and see how many points you scored.

Points for each chemistry task

1 point - for tasks 1-6, 11-15, 19-21, 26-28.
2 points - 7-10, 16-18, 22-25, 30, 31.
3 points - 35.
4 points - 32, 34.
5 points - 33.

Total: 60 points.

The structure of the examination paper consists of two blocks:

Questions that involve a short answer (in the form of a number or a word) - tasks 1-29.
Problems with detailed answers - tasks 30-35.

3.5 hours (210 minutes) are allotted for the performance of the examination work in chemistry.

There will be three cheat sheets on the exam. And you need to understand them

This is 70% of the information that will help you successfully pass the chemistry exam. The remaining 30% is the ability to use the presented cheat sheets.

If you want to get more than 90 points, you need to spend a lot of time on chemistry.
To successfully pass the exam in chemistry, you need to solve a lot:, training tasks, even if they seem easy and of the same type.
Distribute your strength correctly and do not forget about rest.

Dare, try and you will succeed!

To solve problems of this type, it is necessary to know the general formulas of the classes of organic substances and the general formulas for calculating the molar mass of substances of these classes:

Majority decision algorithm problems of finding a molecular formula includes the following actions:

- writing the reaction equations in general form;

- finding the amount of substance n for which the mass or volume is given, or the mass or volume of which can be calculated according to the condition of the problem;

- finding the molar mass of a substance M = m / n, the formula of which must be established;

- finding the number of carbon atoms in a molecule and drawing up the molecular formula of a substance.

Examples of solving the 35 USE problem in chemistry to find the molecular formula of organic matter by combustion products with an explanation

When 11.6 g of organic matter is burned, 13.44 liters of carbon dioxide and 10.8 g of water are formed. The vapor density of this substance in air is 2. It was found that this substance interacts with an ammoniacal solution of silver oxide, is catalytically reduced by hydrogen to form a primary alcohol, and is capable of being oxidized by an acidified solution of potassium permanganate to a carboxylic acid. Based on this data:
1) establish the simplest formula of the starting substance,
2) make up its structural formula,
3) give the equation for the reaction of its interaction with hydrogen.

Solution: general formula of organic matter СxHyOz.

Let's translate the volume of carbon dioxide and the mass of water in moles according to the formulas:

n = m/ M and n = V/ Vm,

Molar volume Vm = 22.4 l / mol

n (CO 2) = 13.44 / 22.4 = 0.6 mol, => the starting material contained n (C) = 0.6 mol,

n (H 2 O) = 10.8 / 18 = 0.6 mol, => the original substance contained twice as much n (H) = 1.2 mol,

This means that the desired compound contains oxygen in the amount:

n (O) = 3.2 / 16 = 0.2 mol

Let's look at the ratio of C, H and O atoms that make up the original organic matter:

n (C): n (H): n (O) = x: y: z = 0.6: 1.2: 0.2 = 3: 6: 1

Found the simplest formula: C 3 H 6 O

To find out the true formula, we find the molar mass of an organic compound using the formula:

M (CxHyOz) = Dair (CxHyOz) * M (air)

M East (СxHyOz) = 29 * 2 = 58 g / mol

Let's check if the true molar mass corresponds to the molar mass of the simplest formula:

M (C 3 H 6 O) = 12 * 3 + 6 + 16 = 58 g / mol - corresponds, => the true formula coincides with the simplest one.

Molecular Formula: C 3 H 6 O

From the problem data: "this substance interacts with an ammoniacal solution of silver oxide, is catalytically reduced by hydrogen to form a primary alcohol and is capable of being oxidized by an acidified solution of potassium permanganate to a carboxylic acid" we conclude that it is an aldehyde.

2) During the interaction of 18.5 g of saturated monobasic carboxylic acid with an excess of sodium bicarbonate solution, 5.6 liters of gas were released. Determine the molecular formula of the acid.

3) Some saturated carboxylic monobasic acid with a mass of 6 g requires the same mass of alcohol for complete esterification. This gives 10.2 g of an ester. Establish the molecular formula of the acid.

4) Determine the molecular formula of the acetylene hydrocarbon if the molar mass of the product of its reaction with an excess of hydrogen bromide is 4 times greater than the molar mass of the starting hydrocarbon

5) During the combustion of organic matter with a mass of 3.9 g, carbon monoxide (IV) with a mass of 13.2 g and water with a mass of 2.7 g were formed. Derive the formula of the substance, knowing that the vapor density of this substance in terms of hydrogen is 39.

6) During the combustion of organic matter weighing 15 g, carbon monoxide (IV) with a volume of 16.8 liters and water weighing 18 g were formed. Derive the formula of the substance, knowing that the vapor density of this substance in terms of hydrogen fluoride is 3.

7) During the combustion of 0.45 g of gaseous organic matter, 0.448 l (standard) of carbon dioxide, 0.63 g of water and 0.112 l (standard) of nitrogen were released. The density of the starting gaseous substance in terms of nitrogen is 1.607. Establish the molecular formula of this substance.

8) During the combustion of anoxic organic matter, 4.48 l (n.u.) of carbon dioxide, 3.6 g of water and 3.65 g of hydrogen chloride were formed. Determine the molecular formula of the burnt compound.

9) During the combustion of organic matter weighing 9.2 g, carbon monoxide (IV) with a volume of 6.72 liters (NU) and water weighing 7.2 g were formed. Set the molecular formula of the substance.

10) During the combustion of organic matter with a mass of 3 g, carbon monoxide (IV) with a volume of 2.24 l (n.u.) and water with a mass of 1.8 g were formed. It is known that this substance reacts with zinc.
Based on the given conditions of the assignment:
1) make the calculations necessary to establish the molecular formula of organic matter;
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 bonds of atoms in its molecule;
4) write the equation for the reaction of this substance with zinc.

Task number 1

An electron configuration corresponds to an excited state of an atom.

1.1s 2 2s 2 2p 6 3s 1
2.1s 2 2s 2 2p 6 3s 2 3p 6
3.1s 2 2s 2 2p 6 3s 1 3p 2
4.1s 2 2s 2 2p 6 3s 2 3p 6 3d 1 4s 2

Answer: 3

Explanation:

The energy of the 3s-sublevel is lower than the energy of the 3p-sublevel, but the 3s-sublevel, which should contain 2 electrons, is not completely filled. Consequently, such an electronic configuration corresponds to an excited state of an atom (aluminum).

The fourth option is not an answer due to the fact that, although the 3d level is not filled, its energy is higher than the 4s sublevel, i.e. in this case, it is filled in last.

Task number 2

In which row are the chemical elements in the order of decreasing their atomic radius?

1. Rb → K → Na
2. Mg → Ca → Sr
3. Si → Al → Mg
4. In → B → Al

Answer: 1

Explanation:

The atomic radius of the elements decreases with a decrease in the number of electron shells (the number of electron shells corresponds to the number of the period of the Periodic Table of Chemical Elements) and when passing to non-metals (i.e., with an increase in the number of electrons at the outer level). Consequently, in the table of chemical elements, the atomic radius of the elements decreases from bottom to top and from left to right.

Task number 3

A chemical bond is formed between atoms with the same relative electronegativity

2) covalent polar

3) covalent non-polar

4) hydrogen

Answer: 3

Explanation:

A covalent non-polar bond is formed between atoms with the same relative electronegativity, since there is no shift in the electron density.

Task number 4

The oxidation states of sulfur and nitrogen in (NH 4) 2 SO 3 are, respectively, equal

1. +4 and -3
2. -2 and +5
3. +6 and +3
4. -2 and +4

Answer: 1

Explanation:

(NH 4) 2 SO 3 (ammonium sulfite) is a salt formed by sulfurous acid and ammonia, therefore, the oxidation states of sulfur and nitrogen are +4 and -3, respectively (the oxidation state of sulfur in sulfurous acid is +4, the oxidation state of nitrogen in ammonia is 3).

Task number 5

The atomic crystal lattice has

1) white phosphorus

3) silicon

4) rhombic sulfur

Answer: 3

Explanation:

White phosphorus has a molecular crystal lattice, the formula of the white phosphorus molecule is P 4.

Both allotropic sulfur modifications (rhombic and monoclinic) have molecular crystal lattices with cyclic crown-shaped S 8 molecules at their sites.

Lead is a metal and has a metallic crystal lattice.

Silicon has a diamond-type crystal lattice, however, due to the longer length of the Si-Si bond compared to C-C, it is inferior to diamond in hardness.

Task number 6

From the listed substances, select three substances that belong to amphoteric hydroxides.

1. Sr (OH) 2
2. Fe (OH) 3
3. Al (OH) 2 Br
4. Be (OH) 2
5. Zn (OH) 2
6. Mg (OH) 2

Answer: 245

Explanation:

Amphoteric metals include Be, Zn, Al (you can remember "BeZnAl"), as well as Fe III and Cr III. Therefore, of the proposed answer options, amphoteric hydroxides include Be (OH) 2, Zn (OH) 2, Fe (OH) 3.

The compound Al (OH) 2 Br is a basic salt.

Task number 7

Are the following judgments about the properties of nitrogen correct?

A. Under normal conditions, nitrogen reacts with silver.

B. Nitrogen under normal conditions in the absence of a catalyst does not react with hydrogen.

1) only A is true

2) only B is true

3) both statements are true

4) both judgments are wrong.

Answer: 2

Explanation:

Nitrogen is a very inert gas and does not react with metals other than lithium under normal conditions.

The interaction of nitrogen with hydrogen refers to the industrial production of ammonia. The process is exothermic and reversible and only takes place in the presence of catalysts.

Task number 8

Carbon monoxide (IV) reacts with each of two substances:

1) oxygen and water

2) water and calcium oxide

3) potassium sulfate and sodium hydroxide

4) silicon oxide (IV) and hydrogen

Answer: 2

Explanation:

Carbon monoxide (IV) (carbon dioxide) is an acidic oxide, therefore, interacts with water to form unstable carbonic acid, alkalis and oxides of alkali and alkaline earth metals to form salts:

CO 2 + H 2 O ↔ H 2 CO 3

CO 2 + CaO → CaCO 3

Task number 9

Each of two substances reacts with sodium hydroxide solution:

1. KOH CO 2
2. KCl and SO 3
3.H 2 O and P 2 O 5
4. SO 2 and Al (OH) 3

Answer: 4

Explanation:

NaOH is an alkali (has basic properties), therefore, it can interact with acidic oxide - SO 2 and amphoteric metal hydroxide - Al (OH) 3:

2NaOH + SO 2 → Na 2 SO 3 + H 2 O or NaOH + SO 2 → NaHSO 3

NaOH + Al (OH) 3 → Na

Task number 10

Calcium carbonate interacts with the solution

1) sodium hydroxide

2) hydrogen chloride

3) barium chloride

4) ammonia

Answer: 2

Explanation:

Calcium carbonate is a water-insoluble salt and therefore does not interact with salts or bases. Calcium carbonate dissolves in strong acids with the formation of salts and the release of carbon dioxide:

CaCO 3 + 2HCl → CaCl 2 + CO 2 + H 2 O

Task number 11

In the transformation scheme

1) iron (II) oxide

2) iron (III) hydroxide

3) iron (II) hydroxide

4) iron (II) chloride

5) iron (III) chloride

Answer: X-5; Y-2

Explanation:

Chlorine is a strong oxidizing agent (the oxidizing ability of halogens increases from I 2 to F 2), it oxidizes iron to Fe +3:

2Fe + 3Cl 2 → 2FeCl 3

Iron (III) chloride is a soluble salt and enters into exchange reactions with alkalis with the formation of a precipitate - iron (III) hydroxide:

FeCl 3 + 3NaOH → Fe (OH) 3 ↓ + NaCl

Task number 12

Homologues are

1) glycerin and ethylene glycol

2) methanol and butanol-1

3) propyne and ethylene

4) propanone and propanal

Answer: 2

Explanation:

Homologues are substances belonging to the same class of organic compounds and differing by one or more CH 2 -groups.

Glycerin and ethylene glycol are trihydric and dihydric alcohols, respectively, differing in the number of oxygen atoms, therefore, they are neither isomers nor homologues.

Methanol and butanol-1 are primary alcohols with an unbranched skeleton, differ by two CH 2 -groups, therefore, are homologous.

Propyne and ethylene belong to the classes of alkynes and alkenes, respectively, contain different numbers of carbon and hydrogen atoms, therefore, they are neither homologues nor isomers.

Propanone and propanal belong to different classes of organic compounds, but contain 3 carbon atoms, 6 hydrogen atoms and 1 oxygen atom, therefore, are isomers in terms of the functional group.

Task number 13

For butene-2 impossible reaction

1) dehydration

2) polymerization

3) halogenation

4) hydrogenation

Answer: 1

Explanation:

Butene-2 belongs to the class of alkenes, enters into addition reactions with halogens, hydrogen halides, water and hydrogen. In addition, unsaturated hydrocarbons polymerize.

A dehydration reaction is a reaction that occurs with the elimination of a water molecule. Since butene-2 is a hydrocarbon, i.e. does not contain heteroatoms, water elimination is impossible.

Task number 14

Phenol does not interact with

1) nitric acid

2) sodium hydroxide

3) bromine water

Answer: 4

Explanation:

With phenol, nitric acid and bromic water enter into the reaction of electrophilic substitution on the benzene ring, resulting in the formation of nitrophenol and bromophenol, respectively.

Phenol, which has weak acidic properties, reacts with alkalis to form phenolates. In this case, sodium phenolate is formed.

Alkanes do not react with phenol.

Task number 15

Acetic acid methyl ester reacts with

1. NaCl
2. Br 2 (solution)
3. Cu (OH) 2
4. NaOH (solution)

Answer: 4

Explanation:

Acetic acid methyl ester (methyl acetate) belongs to the class of esters and undergoes acid and alkaline hydrolysis. Under conditions of acid hydrolysis, methyl acetate is converted into acetic acid and methanol, under conditions of alkaline hydrolysis with sodium hydroxide - sodium acetate and methanol.

Task number 16

Butene-2 can be obtained by dehydration

1) butanone

2) butanol-1

3) butanol-2

4) butanal

Answer: 3

Explanation:

One of the methods for obtaining alkenes is the reaction of intramolecular dehydration of primary and secondary alcohols, which occurs in the presence of anhydrous sulfuric acid and at temperatures above 140 o C. The elimination of a water molecule from an alcohol molecule proceeds according to Zaitsev's rule: a hydrogen atom and a hydroxyl group are split off from adjacent carbon atoms, moreover, hydrogen is split off from the carbon atom at which there is the smallest number of hydrogen atoms. Thus, the intramolecular dehydration of the primary alcohol, butanol-1, leads to the formation of butene-1, and the intramolecular dehydration of the secondary alcohol, butanol-2, to the formation of butene-2.

Task number 17

Methylamine can react with (c)

1) alkalis and alcohols

2) alkalis and acids

3) oxygen and alkalis

4) acids and oxygen

Answer: 4

Explanation:

Methylamine belongs to the class of amines and, due to the presence of a lone electron pair on the nitrogen atom, has basic properties. In addition, the main properties of methylamine are more pronounced than that of ammonia, due to the presence of a methyl group, which has a positive inductive effect. Thus, possessing basic properties, methylamine interacts with acids to form salts. In an oxygen atmosphere, methylamine burns to carbon dioxide, nitrogen and water.

Task number 18

In a given scheme of transformations

substances X and Y, respectively, are

1) ethanediol-1,2

3) acetylene

4) diethyl ether

Answer: X-2; Y-5

Explanation:

Bromoethane in an aqueous alkali solution enters into a nucleophilic substitution reaction with the formation of ethanol:

CH 3 -CH 2 -Br + NaOH (aq.) → CH 3 -CH 2 -OH + NaBr

Under the conditions of concentrated sulfuric acid at temperatures above 140 ° C, intramolecular dehydration occurs with the formation of ethylene and water:

All alkenes easily react with bromine:

CH 2 = CH 2 + Br 2 → CH 2 Br-CH 2 Br

Task number 19

Substitution reactions include the interaction

1) acetylene and hydrogen bromide

2) propane and chlorine

3) ethene and chlorine

4) ethylene and hydrogen chloride

Answer: 2

Explanation:

Addition reactions include the interaction of unsaturated hydrocarbons (alkenes, alkynes, alkadienes) with halogens, hydrogen halides, hydrogen and water. Acetylene (ethyne) and ethylene belong to the classes of alkynes and alkenes, respectively; therefore, they enter into addition reactions with hydrogen bromide, hydrogen chloride and chlorine.

Alkanes enter into substitution reactions with halogens in the light or at elevated temperatures. The reaction proceeds according to a chain mechanism with the participation of free radicals - particles with one unpaired electron:

Task number 20

On the speed of a chemical reaction

HCOOCH 3 (l) + H 2 O (l) → HCOOH (l) + CH 3 OH (l)

does not render influence

1) pressure increase

2) temperature rise

3) change in the concentration of HCOOCH 3

4) using a catalyst

Answer: 1

Explanation:

The reaction rate is influenced by changes in temperature and concentrations of the starting 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.

The use of a catalyst also speeds up the reactions without having a catalyst in the products.

The starting materials and reaction products are in the liquid phase; therefore, a change in pressure does not affect the rate of this reaction.

Task number 21

Abbreviated ionic equation

Fe +3 + 3OH - = Fe (OH) 3 ↓

corresponds to the molecular reaction equation

1. FeCl 3 + 3NaOH = Fe (OH) 3 ↓ + 3NaCl
2.4Fe (OH) 2 + O 2 + 2H 2 O = 4Fe (OH) 3 ↓
3. FeCl 3 + 3NaHCO 3 = Fe (OH) 3 ↓ + 3CO 2 + 3NaCl
4.4Fe + 3O 2 + 6H 2 O = 4Fe (OH) 3 ↓

Answer: 1

Explanation:

In an aqueous solution, soluble salts, alkalis and strong acids dissociate into ions, insoluble bases, insoluble salts, weak acids, gases, and simple substances are recorded in molecular form.

The condition for the solubility of salts and bases corresponds to the first equation, in which the salt enters into an exchange reaction with an alkali to form an insoluble base and another soluble salt.

The complete ionic equation is written as follows:

Fe +3 + 3Cl - + 3Na + + 3OH - = Fe (OH) 3 ↓ + 3Cl - + 3Na +

Task number 22

Which of the following gases is toxic and has a strong odor?

1) hydrogen

2) carbon monoxide (II)

4) carbon monoxide (IV)

Answer: 3

Explanation:

Hydrogen and carbon dioxide are odorless, non-toxic gases. Carbon monoxide and chlorine are toxic, but unlike CO, chlorine has a pungent odor.

Task number 23

The polymerization reaction enters

Answer: 4

Explanation:

All substances from the proposed options are aromatic hydrocarbons, but polymerization reactions are not typical for aromatic systems. The styrene molecule contains a vinyl radical, which is a fragment of the ethylene molecule, which is characterized by polymerization reactions. Thus, styrene polymerizes to form polystyrene.

Task number 24

To 240 g of a solution with a mass fraction of 10% salt was added 160 ml of water. Determine the mass fraction of salt in the resulting solution. (Write the number down to whole integers.)

The mass fraction of salt in the solution is calculated by the formula:

Based on this formula, we calculate the mass of salt in the original solution:

m (in-va) = ω (in-va in the outgoing solution). m (original solution) / 100% = 10%. 240 g / 100% = 24 g

When water is added to the solution, the mass of the resulting solution will be 160 g + 240 g = 400 g (water density is 1 g / ml).

The mass fraction of salt in the resulting solution will be:

Task number 25

Calculate how much nitrogen (n.o.) is formed with the complete combustion of 67.2 l (n.o.) ammonia. (Write the number down to tenths.)

Answer: 33.6 L

Explanation:

The complete combustion of ammonia in oxygen is described by the equation:

4NH 3 + 3O 2 → 2N 2 + 6H 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 (N 2) = 1 / 2ν (NH 3),

therefore, the volumes of ammonia and nitrogen are related to each other in the same way:

V (N 2) = 1 / 2V (NH 3)

V (N 2) = 1 / 2V (NH 3) = 67.2 l / 2 = 33.6 l

Task number 26

What is the volume (in liters at normal level) of oxygen formed during the decomposition of 4 mol of hydrogen peroxide? (Write the number down to tenths).

Answer: 44.8 L

Explanation:

In the presence of a catalyst - manganese dioxide, peroxide decomposes with the formation of oxygen and water:

2H 2 O 2 → 2H 2 O + O 2

According to the reaction equation, the amount of formed oxygen is half the amount of hydrogen peroxide:

ν (O 2) = 1/2 ν (H 2 O 2), therefore, ν (O 2) = 4 mol / 2 = 2 mol.

The volume of gases is calculated by the formula:

V = V m ν , where V m is the molar volume of gases at normal conditions, equal to 22.4 l / mol

The volume of oxygen generated during the decomposition of peroxide is equal to:

V (O 2) = V m ν (O 2) = 22.4 L / mol 2 mol = 44.8 L

Task number 27

Establish a correspondence between the classes of compounds and the trivial name of the substance, which is its representative.

Answer: A-3; B-2; IN 1; G-5

Explanation:

Alcohols are organic substances containing one or more hydroxyl groups (-OH) directly bonded to a saturated carbon atom. Ethylene glycol is a dihydric alcohol containing two hydroxyl groups: CH 2 (OH) -CH 2 OH.

Carbohydrates are organic substances containing carbonyl and several hydroxyl groups, the general formula of carbohydrates is written as C n (H 2 O) m (where m, n> 3). Of the proposed options, carbohydrates include starch - a polysaccharide, a high molecular weight carbohydrate consisting of a large number of monosaccharide residues, the formula of which is written in the form (C 6 H 10 O 5) n.

Hydrocarbons are organic substances that contain only two elements - carbon and hydrogen. The hydrocarbons from the proposed options include toluene - an aromatic compound consisting only of carbon and hydrogen atoms and does not contain functional groups with heteroatoms.

Carboxylic acids are organic substances, the molecules of which contain a carboxyl group, consisting of interconnected carbonyl and hydroxyl groups. The class of carboxylic acids includes butyric (butanoic) acid - C 3 H 7 COOH.

Task number 28

Establish a correspondence between the reaction equation and the change in the oxidation state of the oxidizing agent in it.

EQUATION OF REACTION

A) 4NH 3 + 5O 2 = 4NO + 6H 2 O

B) 2Cu (NO 3) 2 = 2CuO + 4NO 2 + O 2

B) 4Zn + 10HNO 3 = NH 4 NO 3 + 4Zn (NO 3) 2 + 3H 2 O

D) 3NO 2 + H 2 O = 2HNO 3 + NO

CHANGING THE DEGREE OF OXIDATION OF THE OXIDANT

Answer: A-1; B-4; AT 6; G-3

Explanation:

An oxidizing agent is a substance that contains atoms that can attach electrons during a chemical reaction and thus reduce the oxidation state.

Reducing agent is a substance that contains atoms that are capable of donating electrons during a chemical reaction and thus increase the oxidation state.

A) Oxidation of ammonia with oxygen in the presence of a catalyst leads to the formation of nitrogen monoxide and water. The oxidizing agent is molecular oxygen, initially having an oxidation state of 0, which, by attaching electrons, is reduced to an oxidation state of -2 in NO and H 2 O compounds.

B) Copper nitrate Cu (NO 3) 2 is a salt containing an acid residue with nitric acid. The oxidation states of nitrogen and oxygen in the nitrate anion are +5 and -2, respectively. In the course of the reaction, the nitrate anion is converted into nitrogen dioxide NO 2 (with an oxidation state of nitrogen +4) and oxygen O 2 (with an oxidation state of 0). Therefore, nitrogen is an oxidizing agent, since it lowers the oxidation state from +5 in nitrate ion to +4 in nitrogen dioxide.

C) In this redox reaction, the oxidizing agent is nitric acid, which, converting into ammonium nitrate, lowers the oxidation state of nitrogen from +5 (in nitric acid) to -3 (in the ammonium cation). The oxidation state of nitrogen in acidic residues of ammonium nitrate and zinc nitrate remains unchanged, i.e. the same as nitrogen in HNO 3.

D) In this reaction, nitrogen in the dioxide disproportionates, i.e. both increases (from N +4 in NO 2 to N +5 in HNO 3) and decreases (from N +4 in NO 2 to N +2 in NO) its oxidation state.

Task number 29

Establish a correspondence between the formula of a substance and the products of electrolysis of its aqueous solution, which were released on inert electrodes.

Answer: A-4; B-3; IN 2; G-5

Explanation:

Electrolysis is a redox process that occurs 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 evolved at the anode)

halide ions (except for F -) oxidation process 2Hal - - 2e → Hal 2 (free halogens are released)

organic acids oxidation process:

2RCOO - - 2e → R-R + 2CO 2

Total electrolysis equation:

A) Na 2 CO 3 solution:

2H 2 O → 2H 2 (at the cathode) + O 2 (at the anode)

B) Cu (NO 3) 2 solution:

2Cu (NO 3) 2 + 2H 2 O → 2Cu (at the cathode) + 4HNO 3 + O 2 (at the anode)

B) AuCl 3 solution:

2AuCl 3 → 2Au (at the cathode) + 3Cl 2 (at the anode)

D) BaCl 2 solution:

BaCl 2 + 2H 2 O → H 2 (at the cathode) + Ba (OH) 2 + Cl 2 (at the anode)

Task number 30

Establish a correspondence between the name of the salt and the ratio of this salt to hydrolysis.

Answer: A-2; B-3; IN 2; G-1

Explanation:

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) Sodium stearate is a salt formed by stearic acid (a weak monobasic carboxylic acid of the aliphatic series) and sodium hydroxide (an alkali - a strong base), therefore, undergoes hydrolysis at the anion.

C 17 H 35 COONa → Na + + C 17 H 35 COO -

C 17 H 35 COO - + H 2 O ↔ C 17 H 35 COOH + OH - (formation of weakly dissociating carboxylic acid)

Solution medium alkaline (pH> 7):

C 17 H 35 COONa + H 2 O ↔ C 17 H 35 COOH + NaOH

B) Ammonium phosphate is a salt formed by weak orthophosphoric acid and ammonia (weak base), therefore, it undergoes hydrolysis both by the cation and by the anion.

(NH 4) 3 PO 4 → 3NH 4 + + PO 4 3-

PO 4 3- + H 2 O ↔ HPO 4 2- + OH - (formation of weakly dissociating hydrogen phosphate ion)

NH 4 + + H 2 O ↔ NH 3 · H 2 O + H + (formation of ammonia dissolved in water)

The solution medium is close to neutral (pH ~ 7).

C) Sodium sulfide is a salt formed by weak hydrosulfuric acid and sodium hydroxide (alkali is a strong base), therefore, it undergoes hydrolysis at the anion.

Na 2 S → 2Na + + S 2-

S 2- + H 2 O ↔ HS - + OH - (formation of weakly dissociating hydrosulfide ion)

Solution medium alkaline (pH> 7):

Na 2 S + H 2 O ↔ NaHS + NaOH

D) Beryllium sulfate is a salt formed by strong sulfuric acid and beryllium hydroxide (weak base), therefore, it undergoes cation hydrolysis.

BeSO 4 → Be 2+ + SO 4 2-

Be 2+ + H 2 O ↔ Be (OH) + + H + (formation of weakly dissociating cation Be (OH) +)

The solution medium is acidic (pH< 7):

2BeSO 4 + 2H 2 O ↔ (BeOH) 2 SO 4 + H 2 SO 4

Task number 31

Establish a correspondence between the way of influencing the equilibrium system

MgO (solid) + CO 2 (g) ↔ MgCO 3 (solid) + Q

and a shift in chemical equilibrium as a result of this effect

Answer: A-1; B-2; IN 2; G-3Explanation:

This reaction is in chemical equilibrium, i.e. in such a state when the speed of the forward reaction is equal to the speed of the reverse. Displacement of equilibrium in the desired direction is achieved by changing the reaction conditions.

Le Chatelier's principle: if an equilibrium system is influenced from the outside, changing any of the factors that determine the equilibrium position, then the direction of the process that weakens this effect will increase in the system.

Factors determining 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 materials 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.

Thus,

A) since the reaction for obtaining magnesium carbonate is exothermic, a decrease in temperature will shift the equilibrium towards the direct reaction;

B) carbon dioxide is the initial substance in the production of magnesium carbonate, therefore, a decrease in its concentration will lead to a shift in equilibrium towards the initial substances, because in the direction of the reverse reaction;

C) Magnesium oxide and magnesium carbonate are solids, only CO 2 is a gas, so its concentration will affect the pressure in the system. With a decrease in the concentration of carbon dioxide, the pressure decreases, therefore, the equilibrium of the reaction shifts towards the starting substances (reverse reaction).

D) the introduction of the catalyst does not affect the displacement of the equilibrium.

Task number 32

Establish a correspondence between the formula of a substance and reagents, with each of which this substance can interact.

FORMULA OF SUBSTANCE

REAGENTS

1) H 2 O, NaOH, HCl

2) Fe, HCl, NaOH

3) HCl, HCHO, H 2 SO 4

4) O 2, NaOH, HNO 3

5) H 2 O, CO 2, HCl

Answer: A-4; B-4; IN 2; G-3

Explanation:

A) Sulfur is a simple substance that can burn in oxygen to form sulfur dioxide:

S + O 2 → SO 2

Sulfur (like halogens) in alkaline solutions disproportionates, resulting in the formation of sulfides and sulfites:

3S + 6NaOH → 2Na 2 S + Na 2 SO 3 + 3H 2 O

Concentrated nitric acid oxidizes sulfur to S +6, reducing to nitrogen dioxide:

S + 6HNO 3 (conc.) → H 2 SO 4 + 6NO 2 + 2H 2 O

B) Forfor (III) oxide is an acidic oxide, therefore, interacts with alkalis to form phosphites:

P 2 O 3 + 4NaOH → 2Na 2 HPO 3 + H 2 O

In addition, phosphorus (III) oxide is oxidized by atmospheric oxygen and nitric acid:

P 2 O 3 + O 2 → P 2 O 5

3P 2 O 3 + 4HNO 3 + 7H 2 O → 6H 3 PO 4 + 4NO

C) Iron (III) oxide is an amphoteric oxide, because exhibits both acidic and basic properties (reacts with acids and alkalis):

Fe 2 O 3 + 6HCl → 2FeCl 3 + 3H 2 O

Fe 2 O 3 + 2NaOH → 2NaFeO 2 + H 2 O (fusion)

Fe 2 O 3 + 2NaOH + 3H 2 O → 2Na 2 (dissolution)

Fe 2 O 3 enters into a proportional reaction with iron to form iron (II) oxide:

Fe 2 O 3 + Fe → 3FeO

D) Cu (OH) 2 - a base insoluble in water, dissolves with strong acids, turning into the corresponding salts:

Cu (OH) 2 + 2HCl → CuCl 2 + 2H 2 O

Cu (OH) 2 + H 2 SO 4 → CuSO 4 + 2H 2 O

Cu (OH) 2 oxidizes aldehydes to carboxylic acids (similar to the "silver mirror" reaction):

HCHO + 4Cu (OH) 2 → CO 2 + 2Cu 2 O ↓ + 5H 2 O

Task number 33

Establish a correspondence between substances and a reagent with which they can be distinguished from each other.

Answer: A-3; B-1; AT 3; G-5

Explanation:

A) The two soluble salts CaCl 2 and KCl can be distinguished using a potassium carbonate solution. Calcium chloride enters into an exchange reaction with it, as a result of which calcium carbonate precipitates:

CaCl 2 + K 2 CO 3 → CaCO 3 ↓ + 2KCl

B) Solutions of sulfite and sodium sulfate can be distinguished by an indicator - phenolphthalein.

Sodium sulfite is a salt formed by a weak unstable sulfurous acid and sodium hydroxide (alkali is a strong base), therefore, it undergoes hydrolysis at the anion.

Na 2 SO 3 → 2Na + + SO 3 2-

SO 3 2- + H 2 O ↔ HSO 3 - + OH - (formation of low-dissociating hydrosulfite ion)

The medium of the solution is alkaline (pH> 7), the color of the phenolphthalein indicator in the alkaline medium is crimson.

Sodium sulfate - a salt formed by strong sulfuric acid and sodium hydroxide (alkali - strong base), does not hydrolyze. The medium of the solution is neutral (pH = 7), the color of the phenolphthalein indicator in a neutral medium is pale pink.

B) The salts of Na 2 SO 4 and ZnSO 4 can also be distinguished with a potassium carbonate solution. Zinc sulfate enters into an exchange reaction with potassium carbonate, as a result of which zinc carbonate precipitates:

ZnSO 4 + K 2 CO 3 → ZnCO 3 ↓ + K 2 SO 4

D) Salts FeCl 2 and Zn (NO 3) 2 can be distinguished by a solution of lead nitrate. When it interacts with iron chloride, a poorly soluble substance PbCl 2 is formed:

FeCl 2 + Pb (NO 3) 2 → PbCl 2 ↓ + Fe (NO 3) 2

Task number 34

Establish a correspondence between the reactants and the carbon-containing products of their interaction.

REACTIVE SUBSTANCES

A) CH 3 -C≡CH + H 2 (Pt) →

B) CH 3 -C≡CH + H 2 O (Hg 2+) →

B) CH 3 -C≡CH + KMnO 4 (H +) →

D) CH 3 -C≡CH + Ag 2 O (NH 3) →

INTERACTION PRODUCT

1) CH 3 -CH 2 -CHO

2) CH 3 -CO-CH 3

3) CH 3 -CH 2 -CH 3

4) CH 3 -COOH and CO 2

5) CH 3 -CH 2 -COOAg

6) CH 3 -C≡CAg

Answer: A-3; B-2; AT 4; G-6

Explanation:

A) Propine adds hydrogen, converting in its excess into propane:

CH 3 -C≡CH + 2H 2 → CH 3 -CH 2 -CH 3

B) The addition of water (hydration) to alkynes in the presence of divalent mercury salts, resulting in the formation of carbonyl compounds, is a reaction of M.G. Kucherov. Hydration of propyne leads to the formation of acetone:

CH 3 -C≡CH + H 2 O → CH 3 -CO-CH 3

C) Oxidation of propyne with potassium permanganate in an acidic medium leads to the rupture of the triple bond in the alkyn, resulting in the formation of acetic acid and carbon dioxide:

5CH 3 -C≡CH + 8KMnO 4 + 12H 2 SO 4 → 5CH 3 -COOH + 5CO 2 + 8MnSO 4 + 4K 2 SO 4 + 12H 2 O

D) Silver propynide is formed and precipitated when propyne is passed through an ammonia solution of silver oxide. This reaction serves to detect alkyne with a triple bond at the end of the chain.

2CH 3 -C≡CH + Ag 2 O → 2CH 3 -C≡CAg ↓ + H 2 O

Task number 35

Establish a correspondence between the reactants and the organic matter, which is the reaction product.

INTERACTION PRODUCT

5) (CH 3 COO) 2 Cu

Answer: A-4; B-6; IN 1; G-6

Explanation:

A) When ethanol is oxidized with copper (II) oxide, acetaldehyde is formed, and the oxide is reduced to metal:

B) When concentrated sulfuric acid acts on alcohol at temperatures above 140 ° C, an intramolecular dehydration reaction occurs - the elimination of a water molecule, which leads to the formation of ethylene:

C) Alcohols react violently with alkali and alkaline earth metals. The active metal replaces hydrogen in the hydroxyl group of the alcohol:

2CH 3 CH 2 OH + 2K → 2CH 3 CH 2 OK + H 2

D) In an alcoholic solution of alkali, alcohols undergo an elimination (cleavage) reaction. In the case of ethanol, ethylene is formed:

CH 3 CH 2 Cl + KOH (alcohol) → CH 2 = CH 2 + KCl + H 2 O

Task number 36

Using the electronic balance method, write the reaction equation:

P 2 O 3 + HClO 3 +… → HCl +…

In this reaction, chloric acid is an oxidizing agent, since the chlorine contained in it lowers the oxidation state from +5 to -1 in HCl. Therefore, the reducing agent is acidic phosphorus (III) oxide, where phosphorus increases the oxidation state from +3 to a maximum +5, converting into orthophosphoric acid.

Let's compose half-reactions of oxidation and reduction:

Cl +5 + 6e → Cl −1 | 2

2P +3 - 4e → 2P +5 | 3

We write the equation of the redox reaction in the form:

3P 2 O 3 + 2HClO 3 + 9H 2 O → 2HCl + 6H 3 PO 4

Task number 37

Copper was dissolved in concentrated nitric acid. The evolved gas was passed over the heated zinc powder. The resulting solid was added to the sodium hydroxide solution. An excess of carbon dioxide was passed through the resulting solution, while the formation of a precipitate was observed. Write the equations for the four reactions described.

1) When copper is dissolved in concentrated nitric acid, copper is oxidized to Cu +2, while a brown gas is released:

Cu + 4HNO 3 (conc.) → Cu (NO 3) 2 + 2NO 2 + 2H 2 O

2) When brown gas is passed over a heated zinc powder, zinc is oxidized, and nitrogen dioxide is reduced to molecular nitrogen (it is assumed by many, with reference to Wikipedia, that zinc nitrate does not form when heated, since it is thermally unstable):

4Zn + 2NO 2 → 4ZnO + N 2

3) ZnO - amphoteric oxide, dissolves in an alkali solution, turning into tetrahydroxozincate:

ZnO + 2NaOH + H 2 O → Na 2

4) When an excess of carbon dioxide is passed through the sodium tetrahydroxozincate solution, an acidic salt is formed - sodium bicarbonate, zinc hydroxide precipitates:

Na 2 + 2CO 2 → Zn (OH) 2 ↓ + 2NaHCO 3

Task number 38

Write down the reaction equations with which you can carry out the following transformations:

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

1) The most typical reactions for alkanes are free radical substitution reactions, during which a hydrogen atom is replaced by a halogen atom. In the reaction of butane with bromine, the substitution of the hydrogen atom at the secondary carbon atom occurs predominantly, as a result of which 2-bromobutane is formed. This is due to the fact that the radical with an unpaired electron at the secondary carbon atom is more stable than the free radical with an unpaired electron at the primary carbon atom:

2) When 2-bromobutane interacts with alkali in an alcoholic solution, a double bond is formed as a result of the elimination of a hydrogen bromide molecule (Zaitsev's rule: when a hydrogen halide is removed from secondary and tertiary haloalkanes, a hydrogen atom is split off from the least hydrogenated carbon atom):

3) The interaction of butene-2 with bromine water or a solution of bromine in an organic solvent leads to rapid discoloration of these solutions as a result of the addition of a bromine molecule to butene-2 and the formation of 2,3-dibromobutane:

CH 3 -CH = CH-CH 3 + Br 2 → CH 3 -CHBr-CHBr-CH 3

4) When interacting with a dibromo derivative, in which halogen atoms are located at adjacent carbon atoms (or at the same atom), an alcoholic alkali solution, two molecules of hydrogen halide are split off (dehydrohalogenation) and a triple bond is formed:

5) In the presence of divalent mercury salts, alkynes add water (hydration) to form carbonyl compounds:

Task number 39

A mixture of iron and zinc powders reacts with 153 ml of 10% hydrochloric acid solution (ρ = 1.05 g / ml). Interaction with the same mass of the mixture requires 40 ml of a 20% sodium hydroxide solution (ρ = 1.10 g / ml). Determine the mass fraction of iron in the mixture.

In the answer, write down the reaction equations that are indicated in the condition of the problem, and provide all the necessary calculations.

Answer: 46.28%

Task number 40

When 2.65 g of organic matter was burned, 4.48 liters of carbon dioxide (NU) and 2.25 g of water were obtained.

It is known that when this substance is oxidized with a potassium permanganate sulfate solution, a monobasic acid is formed and carbon dioxide is released.

Based on the given conditions of the assignment:

1) make the calculations necessary to establish the molecular formula of organic matter;

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 bonds of atoms in its molecule;

4) write the equation of the oxidation reaction of this substance with potassium permanganate sulfate solution.

Answer:

1) C x H y; x = 8, y = 10

2) C 8 H 10

3) C 6 H 5 -CH 2 -CH 3 - ethylbenzene

4) 5C 6 H 5 -CH 2 -CH 3 + 12KMnO 4 + 18H 2 SO 4 → 5C 6 H 5 -COOH + 5CO 2 + 12MnSO 4 + 6K 2 SO 4 + 28H 2 O

Online tests of gia in chemistry. Online tests of GIA in chemistry Assignments of exam in chemistry 20 06

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