USE 2017 Chemistry Typical test tasks Medvedev

M .: 2017 .-- 120 p.

Typical test tasks in chemistry contain 10 options for sets of tasks, compiled taking into account all the features and requirements of the Unified State Exam in 2017. The purpose of the manual is to provide readers with information about the structure and content of KIM 2017 in chemistry, the degree of difficulty of the tasks. The collection provides answers to all test options and provides solutions to all tasks of one of the options. In addition, there are sample forms used on the exam to record answers and decisions. The author of the assignments is a leading scientist, teacher and methodologist who is directly involved in the development of control measuring materials for the exam. The manual is intended for teachers to prepare students for the exam in chemistry, as well as high school students and graduates - for self-preparation and self-control.

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CONTENT
Foreword 4
Work instructions 5
OPTION 1 8
Part 1 8
Part 2, 15
OPTION 2 17
Part 1 17
Part 2 24
OPTION 3 26
Part 1 26
Part 2 33
OPTION 4 35
Part 1 35
Part 2 41
OPTION 5 43
Part 1 43
Part 2 49
OPTION 6 51
Part 1 51
Part 2 57
OPTION 7 59
Part 1 59
Part 2 65
OPTION 8 67
Part 1 67
Part 2 73
OPTION 9 75
Part 1 75
Part 2 81
OPTION 10 83
Part 1 83
Part 2 89
ANSWERS AND SOLUTIONS 91
Answers to the tasks of part 1 91
Solutions and answers to the tasks of part 2 93
Solution of tasks of variant 10 99
Part 1 99
Part 2 113

This study guide is a collection of tasks for preparing for the delivery of the Unified State Examination (USE) in chemistry, which is both the final exam for a high school course and the entrance exam to the university. The structure of the manual reflects modern requirements for the procedure for passing the exam in chemistry, which will allow you to better prepare for the new forms of graduation certification and for admission to universities.
The manual consists of 10 variants of tasks, which in form and content are close to the demo version of the USE and do not go beyond the content of the chemistry course, which is normatively determined by the Federal component of the state standard of general education. Chemistry (order of the Ministry of Education No. 1089 dated 05.03.2004).
The level of presentation of the content of educational material in the tasks is correlated with the requirements of the state standard for the preparation of secondary (complete) school graduates in chemistry.
In the control measuring materials of the Unified State Exam, tasks of three types are used:
- assignments of the basic level of difficulty with a short answer,
- tasks of an increased level of complexity with a short answer,
- tasks of a high level of complexity with a detailed answer.
Each version of the examination paper is built according to a single plan. The work consists of two parts, including a total of 34 tasks. Part 1 contains 29 tasks with a short answer, including 20 tasks of a basic level of difficulty and 9 tasks of an increased level of difficulty. Part 2 contains 5 tasks of a high level of complexity, with a detailed answer (tasks numbered 30-34).
In tasks of a high level of complexity, the text of the solution is written on a special form. Assignments of this particular type make up the bulk of the written work in chemistry for university entrance exams.

Sodium nitride weighing 8.3 g reacted with sulfuric acid with a mass fraction of 20% and a mass of 490 g. Then crystalline soda weighing 57.2 g was added to the resulting solution. Find the mass fraction (%) of acid in the final solution. Write down the reaction equations that are indicated in the condition of the problem, bring all the necessary calculations (indicate the units of measurement of the desired physical quantities). Round up the answer for the site to the nearest whole number.

Real exam 2017. Task 34.

Cyclic substance A (does not contain oxygen and substituents) is oxidized with a break in the cycle to substance B weighing 20.8 g, the combustion products of which are carbon dioxide with a volume of 13.44 liters and water weighing 7.2 g. Based on the given conditions of the task: 1) make the calculations necessary to establish the molecular formula of organic matter B; 2) write down the molecular formulas of organic substances A and B; 3) make up the structural formulas of organic substances A and B, which unambiguously reflect the order of the bonds of atoms in the molecule; 4) write the equation of the oxidation reaction of substance A with potassium permanganate sulfate solution to form substance B. In the answer for the site, indicate the sum of all atoms in one molecule of the original organic substance A.

The following changes will be made in the CMMs of the USE 2017:

1. The approach to structuring part 1 of the examination paper will be fundamentally changed. It is assumed that, in contrast to the examination model of previous years, the structure of part 1 of the work will include several thematic blocks, each of which will present tasks of both basic and increased levels of complexity. Within each thematic block, tasks will be arranged in ascending order of the number of actions required to complete them. Thus, the structure of part 1 of the examination paper will be more consistent with the structure of the chemistry course itself. Such structuring of part 1 of the CMM will help the examinees during their work to more effectively focus their attention on the use of which knowledge, concepts and laws of chemistry and in what relationship will require the completion of tasks that check the assimilation of the educational material of a certain section of the chemistry course.

2. There will be noticeable changes in approaches to the design of tasks of the basic level of complexity. These can be tasks with a single context, with the choice of two correct answers out of five, three out of six, tasks "to establish a correspondence between the positions of two sets", as well as calculation tasks.

3. An increase in the differentiating ability of tasks makes it objective to formulate the question of reducing the total number of tasks in the examination work. It is assumed that the total number of tasks in the examination work will decrease from 40 to 34. This will be done mainly by streamlining the optimal number of those tasks, the implementation of which involved the use of similar activities. An example of such tasks, in particular, are tasks focused on checking the chemical properties of salts, acids, bases, conditions of ion exchange reactions.

4. A change in the format of tasks and their number will inevitably be associated with an adjustment of the grading scale for some tasks, which, in turn, will cause a change in the primary total score for the performance of the work as a whole, presumably in the range from 58 to 60 (instead of the previous 64 points).

The consequence of the planned changes in the examination model as a whole should be an increase in the objectivity of checking the formation of a number of subject and metasubject skills, which are an important indicator of the success of mastering the subject. We are talking, in particular, about such skills as: apply knowledge in the system, combine knowledge about chemical processes with an understanding of the mathematical relationship between various physical quantities, independently assess the correctness of the implementation of educational and educational-practical tasks, etc.

For tasks 1-3, use the following row of chemical elements. The answer in tasks 1–3 is a sequence of numbers, under which the chemical elements in this row are indicated.

• 1.S
• 2. Na
• 3. Al
• 4.Si
• 5. Mg

Task number 1

Determine which atoms of the elements indicated in the series contain one unpaired electron in the ground state.

Answer: 23

Explanation:

Let us 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 number 2

Select three metal elements from the listed chemical elements. Arrange the selected elements in ascending order of restorative 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 side subgroups. Thus, metals from the specified list include Na, Al and Mg.

The metallic and, consequently, the 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 series Al, Mg, Na

Task number 3

From among the elements indicated in the row, 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 number 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. H 2 SO 4
• 5.PCl 3

Answer: 12

Explanation:

In the overwhelming majority of cases, it is possible to determine the presence of an ionic type of bond in a compound by the fact that atoms of a typical metal and atoms of a non-metal are simultaneously included in its structural units.

Based on this criterion, the ionic type of bond takes place in the compounds KCl and KNO 3.

In addition to the above sign, the presence of an ionic bond in a compound can be said if its structural unit contains an ammonium cation (NH 4 +) or its organic analogs - alkylammonium cations RNH 3 +, dialkylammonium R 2 NH 2 +, trialkylammonium R 3 NH + and tetraalkylammonium R 4 N +, where R is some hydrocarbon radical. For example, the ionic type of bond takes place in the compound (CH 3) 4 NCl between the cation (CH 3) 4 + and the chloride ion Cl -.

Task number 5

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

AND	B	IN

Answer: 241

Explanation:

N 2 O 3 is a non-metal oxide. All oxides of non-metals, except for 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 acid, because upon dissociation in an aqueous solution, only H + cations are formed from cations:

HClO 4 \u003d H + + ClO 4 -

Task number 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 with insoluble hydroxides at all, 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 is a salt of a metal more active than zinc, i.e. reaction is impossible.

Task number 7

From the proposed list of substances, select two oxides that react with water.

• 1. BaO
• 2. CuO
• 3. NO
• 4. SO 3
• 5.PbO 2

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 react with water, as well as all acidic oxides except SiO 2.

Thus, answer options 1 and 4 are suitable:

BaO + H 2 O \u003d Ba (OH) 2

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

Task number 8

1) hydrogen bromide

3) sodium nitrate

4) sulfur (IV) oxide

5) aluminum chloride

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

Answer: 52

Explanation:

Salts among these substances are only sodium nitrate and aluminum chloride. All nitrates, as well as sodium salts, are soluble, and therefore sodium nitrate precipitate cannot, in principle, be produced with any of the reagents. Therefore, the salt X can only be aluminum chloride.

A common mistake among those who pass the exam in chemistry is a lack of understanding that in an aqueous solution ammonia forms a weak base - ammonium hydroxide due to the course of the reaction:

NH 3 + H 2 O<=> NH 4 OH

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 \u003d Al (OH) 3 + 3NH 4 Cl

Task number 9

In a given transformation scheme

Cu X\u003e CuCl 2 Y\u003e CuI

substances X and Y are:

• 1. AgI
• 2. I 2
• 3. Cl 2
• 4. HCl
• 5. KI

Answer: 35

Explanation:

Copper is a metal located in the line of activity to the right of hydrogen, i.e. does not react with acids (except for H 2 SO 4 (conc.) and HNO 3). Thus, the formation of copper chloride (ll) is possible in our case only when reacting with chlorine:

Cu + Cl 2 \u003d CuCl 2

Iodide ions (I -) cannot coexist in the same solution with bivalent copper ions, because oxidized by them:

Cu 2+ + 3I - \u003d CuI + I 2

Task number 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:

The oxidizing agent in the reaction is the substance that contains an element that lowers its oxidation state

Task number 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. The salt reacts with the metal hydroxide if the initial substances are soluble, and the products contain a precipitate, gas or a low-dissociating substance. For both the first and second reactions, both requirements are met:

Cu (NO 3) 2 + 2NaOH \u003d 2NaNO 3 + Cu (OH) 2 ↓

Cu (NO 3) 2 + Ba (OH) 2 \u003d Na (NO 3) 2 + Cu (OH) 2 ↓

Cu (NO 3) 2 + Mg - salt reacts with metal if the free metal is more active than that which is part of the salt. Magnesium in the row of activity is located to the left of copper, which indicates its greater activity, therefore, the reaction proceeds:

Cu (NO 3) 2 + Mg \u003d 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, as exceptions, the hydroxides Be (OH) 2 and Zn (OH) 2, are 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 suitable:

Al (OH) 3 + 3HCl \u003d AlCl 3 + 3H 2 O

Al (OH) 3 + LiOH (solution) \u003d Li or Al (OH) 3 + LiOH (tv.) \u003d To \u003d\u003e LiAlO 2 + 2H 2 O

2Al (OH) 3 + 3H 2 SO 4 \u003d 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. An explanation is given in A.

ZnCl 2 + 2NaOH \u003d Zn (OH) 2 + 2NaCl

ZnCl 2 + Ba (OH) 2 \u003d Zn (OH) 2 + BaCl 2

It should be noted that with an excess of NaOH and Ba (OH) 2:

ZnCl 2 + 4NaOH \u003d Na 2 + 2NaCl

ZnCl 2 + 2Ba (OH) 2 \u003d Ba + BaCl 2

D) Br 2, O 2 - strong oxidants. Of metals, they do not react only with silver, platinum, gold:

Cu + Br 2 t ° \u003e CuBr 2

2Cu + O 2 t ° \u003e 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 \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O

8HNO 3 (dil.) + 3Cu \u003d 3Cu (NO 3) 2 + 2NO + 4H 2 O

Task number 12

Establish a correspondence between the general formula of the 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.

AND	B	IN

Answer: 231

Explanation:

Task number 13

From the proposed list of substances, select two substances that are isomers of cyclopentane.

1) 2-methylbutane

2) 1,2-dimethylcyclopropane

3) pentene-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 C 5 H 10. Let's write the structural and molecular formulas of the substances listed in the condition

Name of substance	Structural formula	Molecular Formula
cyclopentane		C 5 H 10
2-methylbutane
1,2-dimethylcyclopropane		C 5 H 10
		C 5 H 10
cyclopentene

Task number 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 hydrocarbons with an aqueous solution of potassium permanganate, those that contain C \u003d C or C≡C bonds in their structural formula, as well as benzene homologues (except for benzene itself), react.

Thus methylbenzene and styrene are suitable.

Task number 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 mild acidic properties, more pronounced than that of alcohols. For this reason, phenols, unlike alcohols, react with alkalis:

C 6 H 5 OH + NaOH \u003d 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 orientant of the first kind, that is, it facilitates substitution reactions in the ortho and para positions:

Task number 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 these substances are carbohydrates. Of carbohydrates, monosaccharides are not hydrolyzed. Glucose, fructose and ribose are monosaccharides, sucrose is a disaccharide, and starch is a polysaccharide. Consequently, sucrose and starch from the specified list undergo hydrolysis.

Task number 17

The following scheme of transformations of substances is given:

1,2-dibromoethane → X → bromoethane → Y → ethyl formate

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

2) ethanal

4) chloroethane

5) acetylene

Write down the numbers of the selected substances in the table under the appropriate letters.

Task number 18

Establish a correspondence between the name of the starting substance and the product, which is predominantly formed by the interaction of this substance with bromine: for each position marked with a letter, select the corresponding position marked with a number.

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

AND	B	IN	D

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 cycle size of more than 4 carbon atoms. Cycloalkanes with a cycle size of more than 4 carbon atoms, when interacting with halogens, enter into a substitution reaction while maintaining the cycle:

Cyclopropane and cyclobutane - cycloalkanes with a minimum ring size predominantly enter into 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. Thus, the bromination of isobutane proceeds mainly as follows:

Task number 19

Establish a correspondence between the reaction scheme and the organic substance, which is the product of this reaction: for each position marked with a letter, select the corresponding position indicated by a number.

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

AND	B	IN	D

Answer: 6134

Explanation:

Heating aldehydes with freshly precipitated copper hydroxide leads to the oxidation of the aldehyde group to the carboxyl group:

Aldehydes and ketones are reduced with hydrogen in the presence of nickel, platinum or palladium to alcohols:

Primary and secondary alcohols are oxidized by incandescent CuO to aldehydes and ketones, respectively:

When concentrated sulfuric acid acts on ethanol when heated, two different products can form. When heated to temperatures below 140 ° C, intermolecular dehydration occurs predominantly with the formation of diethyl ether, and when heated to more than 140 ° C, intramolecular dehydration occurs, as a result of which ethylene is formed:

Task number 20

From the proposed list of substances, select two substances, the thermal decomposition reaction of which 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 as a result of which chemical 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 ° C) to metal carbonate, carbon dioxide and water. In this case, there is no change in oxidation states:

Insoluble oxides decompose when heated. In this case, the reaction is not redox because not a single chemical element changes the oxidation state as a result:

Ammonium carbonate decomposes when heated to carbon dioxide, water and ammonia. The reaction is not redox:

Ammonium nitrate decomposes into nitric oxide (I) and water. The reaction refers to OVR:

Task number 21

From the list provided, select two external influences that lead to an increase in the rate of reaction of nitrogen with hydrogen.

1) lowering the temperature

2) pressure increase in the system

5) using an inhibitor

Write down the numbers of the selected external influences in the answer field.

Answer: 24

Explanation:

1) lowering the temperature:

The rate of any reaction decreases with decreasing temperature.

2) pressure increase in the system:

Increasing pressure increases the speed of any reaction in which at least one gaseous substance is involved.

3) decrease in hydrogen concentration

A decrease in concentration always slows down the reaction rate

4) increase in nitrogen concentration

Increasing the concentration of reagents always increases the reaction rate

5) using an inhibitor

Inhibitors are substances that slow down the rate of reaction.

Task number 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 marked with a letter, select the corresponding position marked with a number.

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

AND	B	IN	D

Answer: 5251

Explanation:

A) NaBr → Na + + Br -

Na + cations and water molecules compete for the cathode.

2H 2 O + 2e - → H 2 + 2OH -

2Cl - -2e → Cl 2

B) Mg (NO 3) 2 → Mg 2+ + 2NO 3 -

For the cathode, Mg 2+ cations and water molecules compete with each other.

Cations of alkali metals, as well as magnesium and aluminum, are not able to be reduced in an aqueous solution due to their high activity. For this reason, instead of them, water molecules are restored in accordance with the equation:

2H 2 O + 2e - → H 2 + 2OH -

For the anode, NO 3 - anions and water molecules compete with each other.

2H 2 O - 4e - → O 2 + 4H +

So answer 2 (hydrogen and oxygen) is appropriate.

B) AlCl 3 → Al 3+ + 3Cl -

Cations of alkali metals, as well as magnesium and aluminum, are not able to be reduced in an aqueous solution due to their high activity. For this reason, instead of them, water molecules are restored in accordance with the equation:

2H 2 O + 2e - → H 2 + 2OH -

Cl - anions and water molecules compete for the anode.

Anions consisting of one chemical element (except for F -) outperform water molecules for oxidation at the anode:

2Cl - -2e → Cl 2

Thus 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 series of activity are easily reduced under the conditions of an aqueous solution:

Cu 2+ + 2e → Cu 0

Acid residues containing an acid-forming element in the highest oxidation state lose the competition to water molecules for oxidation at the anode:

2H 2 O - 4e - → O 2 + 4H +

So answer 1 (oxygen and metal) is appropriate.

Task number 23

Establish a correspondence between the name of the salt and the medium of an aqueous solution of this salt: for each position marked with a letter, select the corresponding position marked with a number.

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

AND	B	IN	D

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 - acidic environment

B) chromium (III) chloride - CrCl 3

formed by a weak base Cr (OH) 3 and a strong acid HCl. Conclusion - acidic environment

C) sodium sulfate - Na 2 SO 4

Formed by strong base NaOH and strong acid H 2 SO 4. Conclusion - neutral environment

D) sodium sulfide - Na 2 S

Formed by strong base NaOH and weak acid H 2 S. Conclusion - the medium is alkaline.

Task number 24

Establish a correspondence between the way of influencing the equilibrium system

СO (g) + Cl 2 (g) СOCl 2 (g) + Q

and the direction of the displacement of the chemical equilibrium as a result of this action: 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.

AND	B	IN	D

Answer: 3113

Explanation:

The balance 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 CO concentration leads to a shift in equilibrium towards a direct reaction, since as a result of it, the amount of CO decreases.

B) An increase in temperature will shift the equilibrium towards the endothermic reaction. Since the direct reaction is exothermic (+ Q), the equilibrium will shift towards the reverse reaction.

C) A decrease in pressure will shift the equilibrium towards the reaction resulting in an increase in the amount of gases. The reverse reaction produces more gases than the direct one. Thus, the balance will shift towards the opposite reaction.

D) An increase in the concentration of chlorine leads to a shift in equilibrium towards the direct reaction, since as a result of it the amount of chlorine decreases.

Task number 25

Establish a correspondence between the two substances and the reagent with which you can distinguish between 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 the third one only if these two substances interact with it in different ways, and, most importantly, these differences are outwardly distinguishable.

A) FeSO 4 and FeCl 2 solutions can be distinguished with barium nitrate solution. In the case of FeSO 4, a white precipitate of barium sulfate forms:

FeSO 4 + BaCl 2 \u003d BaSO 4 ↓ + FeCl 2

In the case of FeCl 2, there are no visible signs of interaction, since the reaction does not proceed.

B) Solutions of Na 3 PO 4 and Na 2 SO 4 can be distinguished using a solution of MgCl 2. The solution of Na 2 SO 4 does not enter into the reaction, and in the case of Na 3 PO 4 a white precipitate of magnesium phosphate precipitates:

2Na 3 PO 4 + 3MgCl 2 \u003d Mg 3 (PO 4) 2 ↓ + 6NaCl

C) KOH and Ca (OH) 2 solutions can be distinguished with Na 2 CO 3 solution. KOH does not react with Na 2 CO 3, and Ca (OH) 2 gives a white precipitate of calcium carbonate with Na 2 CO 3:

Ca (OH) 2 + Na 2 CO 3 \u003d CaCO 3 ↓ + 2NaOH

D) KOH and KCl solutions can be distinguished using MgCl 2 solution. KCl does not react with MgCl 2, and mixing of KOH and MgCl 2 solutions leads to the formation of a white precipitate of magnesium hydroxide:

MgCl 2 + 2KOH \u003d Mg (OH) 2 ↓ + 2KCl

Task number 26

Establish a correspondence between the substance and its area of \u200b\u200bapplication: for each position indicated by a letter, select the corresponding position indicated by a number.

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

AND	B	IN	D

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 fertilizers, in turn, are obtained - 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 number. Write this number in the answer field in the text of the work, observing the specified degree of accuracy. Then transfer this number to ANSWER FORM № 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. It is not necessary to write the units of measurement of physical quantities.

Task number 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 down to whole integers.)

Answer: 50

Explanation:

Let the mass of potassium hydroxide, which must 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) \u003d 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 number 28

Into the reaction, the thermochemical equation of which

MgO (solid) + CO 2 (g) → MgCO 3 (solid) + 102 kJ,

88 g of carbon dioxide entered. How much heat will be released in this case? (Write the number down to whole integers.)

Answer: ___________________________ kJ.

Answer: 204

Explanation:

Let's calculate the amount of carbon dioxide substance:

n (CO 2) \u003d n (CO 2) / M (CO 2) \u003d 88/44 \u003d 2 mol,

According to the reaction equation, when 1 mol of CO 2 interacts with magnesium oxide, 102 kJ is released. In our case, the amount of carbon dioxide is 2 mol. Denoting the amount of heat released in this case 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 true:

1 ∙ x \u003d 2 ∙ 102

Thus, the amount of heat that is released when 88 g of carbon dioxide participates in the reaction with magnesium oxide is 204 kJ.

Task number 29

Determine the mass of zinc, which reacts with hydrochloric acid to produce 2.24 L (NL) hydrogen. (Write the number down to tenths.)

Answer: ___________________________

Answer: 6.5

Explanation:

Let's write the reaction equation:

Zn + 2HCl \u003d ZnCl 2 + H 2

Let's calculate the amount of hydrogen substance:

n (H 2) \u003d V (H 2) / V m \u003d 2.24 / 22.4 \u003d 0.1 mol.

Since in the reaction equation before zinc and hydrogen there are equal coefficients, this means that the amounts of the substances of zinc that have entered into the reaction and hydrogen formed as a result of it are also equal, i.e.

n (Zn) \u003d n (H 2) \u003d 0.1 mol, therefore:

m (Zn) \u003d n (Zn) ∙ M (Zn) \u003d 0.1 ∙ 65 \u003d 6.5 g.

Do not forget to transfer all answers to answer form # 1 in accordance with the instructions for the work.

Task number 33

Sodium hydrogen carbonate 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 the answer, write down the reaction equations that are indicated in the condition of the problem, and provide all the necessary calculations (indicate the units of measurement of the desired physical quantities).

Answer:

Explanation:

Sodium bicarbonate decomposes when heated in accordance with 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 substance:

n (NaHCO 3) \u003d m (NaHCO 3) / M (NaHCO 3) \u003d 43.34 g / 84 g / mol ≈ 0.516 mol,

consequently,

n (Na 2 CO 3) \u003d 0.516 mol / 2 \u003d 0.258 mol.

Let's calculate the amount of carbon dioxide formed by reaction (II):

n (CO 2) \u003d n (Na 2 CO 3) \u003d 0.258 mol.

We calculate the mass of pure sodium hydroxide and its amount of substance:

m (NaOH) \u003d m solution (NaOH) ∙ ω (NaOH) / 100% \u003d 100 g ∙ 10% / 100% \u003d 10 g;

n (NaOH) \u003d m (NaOH) / M (NaOH) \u003d 10/40 \u003d 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 \u003d Na 2 CO 3 + H 2 O (with an excess of alkali)

NaOH + CO 2 \u003d NaHCO 3 (with excess carbon dioxide)

From the presented equations it follows that only the middle salt is obtained with the ratio n (NaOH) / n (CO 2) ≥2, and only acidic, with the ratio n (NaOH) / n (CO 2) ≤ 1.

According to calculations ν (CO 2)\u003e ν (NaOH), therefore:

n (NaOH) / n (CO 2) ≤ 1

Those. the interaction of carbon dioxide with sodium hydroxide occurs exclusively with the formation of an acidic salt, i.e. according to the equation:

NaOH + CO 2 \u003d NaHCO 3 (III)

The calculation is carried out for the lack of alkali. According to the reaction equation (III):

n (NaHCO 3) \u003d n (NaOH) \u003d 0.25 mol, therefore:

m (NaHCO 3) \u003d 0.25 mol ∙ 84 g / mol \u003d 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.

It follows from the reaction equation that it reacted, i.e. only 0.25 mol of CO 2 was absorbed from 0.258 mol. Then the mass of absorbed CO 2 is:

m (CO 2) \u003d 0.25 mol ∙ 44 g / mol \u003d 11 g.

Then, the mass of the solution is:

m (solution) \u003d m (solution of NaOH) + m (CO 2) \u003d 100 g + 11 g \u003d 111 g,

and the mass fraction of sodium bicarbonate in the solution will thus be equal to:

ω (NaHCO 3) \u003d 21 g / 111 g ∙ 100% ≈ 18.92%.

Task number 34

On combustion of 16.2 g of non-cyclic organic matter, 26.88 L (NU) of carbon dioxide and 16.2 g of water were obtained. It is known that 1 mol of this organic substance in the presence of a catalyst adds only 1 mol of water and this substance does not react with an ammoniacal solution of silver oxide.

Based on the given problem conditions:

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

2) write down the molecular formula of organic matter;

3) make up the structural formula of organic matter, which unambiguously reflects the order of bonds of atoms in its molecule;

4) write the reaction equation for the hydration of organic matter.

Answer:

Explanation:

1) To determine the elemental composition, we calculate the amount of substances carbon dioxide, water and then the masses of the elements included in them:

n (CO 2) \u003d 26.88 L / 22.4 L / mol \u003d 1.2 mol;

n (CO 2) \u003d n (C) \u003d 1.2 mol; m (C) \u003d 1.2 mol ∙ 12 g / mol \u003d 14.4 g.

n (H 2 O) \u003d 16.2 g / 18 g / mol \u003d 0.9 mol; n (H) \u003d 0.9 mol * 2 \u003d 1.8 mol; m (H) \u003d 1.8 g.

m (org. substances) \u003d m (C) + m (H) \u003d 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 \u003d ν (C): ν (H) \u003d 1.2: 1.8 \u003d 1: 1.5 \u003d 2: 3 \u003d 4: 6

Thus, the simplest formula of the substance is C 4 H 6. The true formula of a substance can coincide with the simplest one, or it can differ from it by an integer number of times. Those. be, for example, C 8 H 12, C 12 H 18, etc.

The condition says that the hydrocarbon is non-cyclic and one of its molecules can attach only one water molecule. This is possible if there is only one multiple bond (double or triple) in the structural formula of a substance. Since the desired hydrocarbon is non-cyclic, it is obvious that one multiple bond can only be 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 everywhere more than one. Thus, the molecular formula of the substance C 4 H 6 coincides with the simplest one.

2) The molecular formula of organic matter is C 4 H 6.

3) Of hydrocarbons, alkynes, in which a triple bond is located at the end of the molecule, interact with an ammonia solution of silver oxide. In order for there to be no interaction with the ammonia solution of silver oxide, the alkyne of the composition C 4 H 6 must have the following structure:

CH 3 -C≡C-CH 3

4) Hydration of alkynes takes place in the presence of divalent mercury salts.

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 put off your preparation until later.

1. 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 in the study of the main topics and determine what knowledge and skills will be required when completing the tasks of the exam in chemistry. For the successful completion of the exam in chemistry, theory is most important.
2. The theory needs to be supported practiceconstantly solving tasks. Since most of the mistakes 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 paperconsists of two blocks:

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

3.5 hours (210 minutes) are allocated for 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 pass successfully the USE in chemistry, you need to solve a lot:, training tasks, even if they seem easy and of the same type.
• Correctly distribute your strength and do not forget about rest.

Dare, try and you will succeed!