Difficult exam options in chemistry. Ege in chemistry
However, it is often chosen by students who want to enter universities of the corresponding direction. This test is necessary for those who want to further study chemistry, chemical engineering and medicine, or will specialize in biotechnology. It is inconvenient that the date of the exam coincides with the exam in history and literature.
However, these subjects are rarely taken together - they are too different in focus for universities to require USE results in such a set. This exam quite difficult - the percentage of those who do not cope with it ranges from 6 to 11%, and the average test score is about 57. All this does not contribute to the popularity of this subject - chemistry is only seventh in the popularity rating among graduates of past years.
The exam in chemistry is important for future doctors, chemists and biotechnologists
Demo version of the USE-2016
USE dates in chemistry
early period
- April 2, 2016 (Sat) - Main exam
- April 21, 2016 (Thu) - Reserve
main stage
- June 20, 2016 (Mon) - Main exam
- June 22, 2016 (Wed) - Reserve
Changes in the USE-2016
Unlike last year, some general innovations appeared in the examination in this discipline. In particular, the number of tests that will have to be solved on basic level(from 28 to 26), and the maximum number primary points in chemistry is now 64. As for the specific features of the 2016 exam, some of the tasks have changed in the format of the answer that the student must give.
- In task number 6, you need to demonstrate whether you know the classification organic compounds, and choose 3 answers from 6 options offered in the test;
- Tests numbered 11 and 18 are designed to determine whether the student knows the genetic links between organic and inorganic compounds. The correct answer involves choosing 2 options from the 5 specified formulations;
- Tests No. 24, 25 and 26 require an answer in the form of a number that must be determined independently, while a year ago, schoolchildren had the opportunity to choose an answer from the proposed options;
- In numbers 34 and 35, students should not only choose answers, but establish a correspondence. These assignments are related to Chemical properties hydrocarbons".
In 2016, the chemistry exam includes 40 tasks General information
The exam in chemistry will last 210 minutes (3.5 hours). Examination ticket Includes 40 tasks, which are divided into three categories:
- A1–A26- relate to tasks that allow assessing the basic training of graduates. The correct answer to these tests gives you the opportunity to score 1 primary score. You should spend 1-4 minutes to complete each task;
- B1–B9- these are tests with an increased level of complexity, they will require schoolchildren to briefly formulate the correct answer and in total they will make it possible to score 18 primary points. Each task is given 5-7 minutes;
- С1–С5- belong to the category of tasks of increased complexity. In this case, the student is required to formulate a detailed answer. In total, you can get another 20 primary points for them. Each task can be given up to 10 minutes.
The minimum score in this subject must be at least 14 primary points (36 test points).
How to prepare for the exam?
To pass the statewide chemistry exam, you can download and work through the demo versions of the exam back in advance. The proposed materials give an idea of what you will have to face at the exam in 2016. Systematic work with tests will allow you to analyze gaps in knowledge. Practicing on the demo version allows students to quickly navigate through the real exam - you do not waste time calming down, focusing and understanding the wording of the questions.
Chemistry is an optional exam that graduates choose when entering higher schools for the respective specialties. Out of ten students, one passes this particular subject at will.
Students were given three hours to test. During this time, they have to deal with 40 tasks. They are traditionally divided into two parts: in the first there are 35 questions, in the second - 5 more.
All tasks are divided into three levels of difficulty:
- A - simple questions: choosing the correct answer from the options offered;
- V - elevated level difficulties. These are tasks for which the student independently forms a short answer;
- C - the most difficult tasks, involving detailed explanations on the topic.
The simplest questions give the graduate one point each if he answers them correctly. For tasks of type B, they already give 1 or 2 points, depending on their complexity, and the maximum that can be scored is 18. The most difficult tasks are estimated at 3-4 points.
Innovations in 2016
USE in chemistry is considered difficult exam. But FIPI decided that it was possible to make life a little more difficult for graduates. Therefore, in 2016, the number of the simplest questions decreased: there were 26 of them against 28.
To prepare 11th graders for the knowledge test, FIPI offers fresh demonstration materials. They clearly demonstrate the structure of the upcoming exam in chemistry. It is also worth trying your knowledge in free tests. They are based on data from previous years. For a successful passage of the exam, do not forget about textbooks, since the tests do not raise all the topics that are included in the exam.
Minimum passing score
In 2016, the minimum passing score for the Unified State Examination in Chemistry will be 64 points. With this indicator, you can enter the university.
For 2-3 months it is impossible to learn (repeat, pull up) such a complex discipline as chemistry.
There are no changes in KIM USE 2020 in chemistry.
Don't delay your preparation.
- Before starting the analysis of 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. guides in the study of the main topics and determines what knowledge and skills will be required when completing the USE tasks in chemistry. For successful delivery USE in chemistry - theory is the most important thing.
- Theory needs to be backed up practice constantly solving problems. 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 them the opportunity to decide and find out the answers. But do not rush to peek. First, decide for yourself and see how many points you have scored.
Points for each task in chemistry
- 1 point - for 1-6, 11-15, 19-21, 26-28 tasks.
- 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 require a short answer (in the form of a number or word) - tasks 1-29.
- Tasks with detailed answers - tasks 30-35.
For execution examination work Chemistry is given 3.5 hours (210 minutes).
There will be three cheat sheets on the exam. And they need to be dealt with.
This is 70% of the information that will help you successfully pass the exam in chemistry. The remaining 30% is the ability to use the provided 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 decide a lot:, training tasks even if they seem light and of the same type.
- Correctly distribute your strength and do not forget about the rest.
Dare, try and you will succeed!
Task number 1
The excited state of the atom corresponds to electronic configuration.
- 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. Therefore, such an electronic configuration corresponds to the excited state of the 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 last.
Task number 2
In which order are the elements arranged in decreasing order of 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 decreasing number electron shells(the number of electron shells corresponds to the period number Periodic system chemical elements) and during the transition to non-metals (i.e., with an increase in the number of electrons at the external level). Therefore, 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
Between atoms with the same relative electronegativity chemical bond
2) covalent polar
3) covalent non-polar
4) hydrogen
Answer: 3
Explanation:
Between atoms with the same relative electronegativity, a covalent non-polar bond is formed, since there is no shift in 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) - 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
nuclear crystal lattice It 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 modifications of sulfur (rhombic and monoclinic) have molecular crystal lattices, at the nodes of which there are cyclic crown-shaped molecules S 8 .
Lead is a metal and has a metallic crystal lattice.
Silicon has a diamond-type crystal lattice, however, due to the longer Si-Si bond length, comparison C-C inferior to diamond in hardness.
Task number 6
Among the listed substances, select three substances that are related 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, from the proposed answers, Be(OH) 2 , Zn(OH) 2 , Fe(OH) 3 belong to amphoteric hydroxides.
The Al(OH) 2 Br compound is a basic salt.
Task number 7
Are the following statements about the properties of nitrogen correct?
A. When 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 correct
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 reversible and proceeds only in the presence of catalysts.
Task number 8
Carbon monoxide (IV) reacts with each of the 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 an unstable carbonic acid, alkalis and oxides of alkali and alkaline earth metals with the formation of salts:
CO 2 + H 2 O ↔ H 2 CO 3
CO 2 + CaO → CaCO 3
Task number 9
Each of the two substances reacts with sodium hydroxide solution:
- 1.KOHCO2
- 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 (it has basic properties), therefore, interaction with an acid oxide - SO 2 and an amphoteric metal hydroxide - Al (OH) 3 is possible:
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, therefore it does not interact with salts and 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 oxide (II)
2) iron (III) hydroxide
3) iron (II) hydroxide
4) iron chloride (II)
5) iron (III) chloride
Answer: X-5; Y-2
Explanation:
Chlorine is a strong oxidizing agent (the oxidizing power of halogens increases from I 2 to F 2), oxidizes iron to Fe +3:
2Fe + 3Cl 2 → 2FeCl 3
Iron (III) chloride is a soluble salt and enters into exchange reactions with alkalis to form a precipitate - iron (III) hydroxide:
FeCl 3 + 3NaOH → Fe(OH) 3 ↓ + NaCl
Task number 12
The 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 that belong to the same class of organic compounds and differ by one or more CH 2 groups.
Glycerin and ethylene glycol are trihydric and dihydric alcohols, respectively, differ in the number of oxygen atoms, therefore they are neither isomers nor homologues.
Methanol and butanol-1 are primary alcohols with an unbranched skeleton, they differ by two CH 2 groups, therefore, they are homologues.
Propyne and ethylene belong to the classes of alkynes and alkenes, respectively, contain different amount carbon and hydrogen atoms, therefore, are neither homologues nor isomers with each other.
Propanone and propanal belong to different classes of organic compounds, but contain 3 carbon atoms, 6 hydrogen atoms and 1 oxygen atom, therefore, they are functional group isomers.
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.
The dehydration reaction is a reaction that proceeds with the elimination of a water molecule. Since butene-2 is a hydrocarbon, i.e. does not contain heteroatoms, the elimination of water 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 bromine 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:
Methyl ester of acetic acid (methyl acetate) belongs to the class of esters, undergoes acid and alkaline hydrolysis. Under the conditions of acid hydrolysis, methyl acetate is converted into acetic acid and methanol, under the 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 ways to obtain alkenes is the reaction of intramolecular dehydration of primary and secondary alcohols, which proceeds in the presence of anhydrous sulfuric acid and at temperatures above 140 o C. The splitting of a water molecule from an alcohol molecule proceeds according to the Zaitsev rule: a hydrogen atom and a hydroxyl group are split off from neighboring carbon atoms, moreover, hydrogen is split off from that carbon atom at which the smallest number of hydrogen atoms is located. Thus, intramolecular dehydration of the primary alcohol - butanol-1 leads to the formation of butene-1, 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 an unshared electron pair on the nitrogen atom, has basic properties. In addition, the basic properties of methylamine are more pronounced than those of ammonia, due to the presence of a methyl group that has a positive inductive effect. Thus, having 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 transformation scheme
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 solution of alkali 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 conditions of concentrated sulfuric acid at temperatures above 140 0 C, intramolecular dehydration occurs with the formation of ethylene and water:
All alkenes easily react with bromine:
CH 2 \u003d CH 2 + Br 2 → CH 2 Br-CH 2 Br
Task #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 temperature. The reaction proceeds by a chain mechanism with the participation of free radicals - particles with one unpaired electron:
Task number 20
For speed chemical reaction
HCOOCH 3 (l) + H 2 O (l) → HCOOH (l) + CH 3 OH (l)
does not provide influence
1) pressure increase
2) temperature increase
3) change in the concentration of HCOOCH 3
4) the use of a catalyst
Answer: 1
Explanation:
The reaction rate is affected by changes in the temperature and concentrations of the initial reagents, as well as the use of a catalyst. According to Van't Hoff's empirical rule, for every 10 degrees increase in temperature, the rate constant of a homogeneous reaction increases by 2-4 times.
The use of a catalyst also speeds up reactions, while the catalyst is not included in the composition of the products.
The starting materials and products of the reaction are in the liquid phase, therefore, a change in pressure does not affect the rate of this reaction.
Task number 21
Reduced ionic equation
Fe + 3 + 3OH - \u003d Fe (OH) 3 ↓
corresponds to the molecular reaction equation
- 1. FeCl 3 + 3NaOH \u003d Fe (OH) 3 ↓ + 3NaCl
- 2. 4Fe(OH) 2 + O 2 + 2H 2 O = 4Fe(OH) 3 ↓
- 3. FeCl 3 + 3NaHCO 3 \u003d Fe (OH) 3 ↓ + 3CO 2 + 3NaCl
- 4. 4Fe + 3O 2 + 6H 2 O \u003d 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, simple substances.
The condition for the solubility of salts and bases corresponds to the first equation, in which the salt enters into an exchange reaction with alkali to form an insoluble base and another soluble salt.
The complete ionic equation is written in the following form:
Fe +3 + 3Cl − + 3Na + + 3OH − = Fe(OH) 3 ↓ + 3Cl − + 3Na +
Task #22
Which of the following gases is toxic and has a pungent odor?
1) hydrogen
2) carbon monoxide (II)
4) carbon monoxide (IV)
Answer: 3
Explanation:
Hydrogen and carbon dioxide are non-toxic, odorless gases. Carbon monoxide and chlorine are both toxic, but unlike CO, chlorine has a strong odor.
Task #23
enters into the polymerization reaction
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 #24
To 240 g of a solution with a mass fraction of salt of 10% was added 160 ml of water. Determine the mass fraction of salt in the resulting solution. (Write down the number to the nearest integer.)
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 initial solution:
m (in-va) \u003d ω (in-va in the original solution). m (original solution) / 100% \u003d 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 1 g / ml).
The mass fraction of salt in the resulting solution will be:
Task #25
Calculate the volume of nitrogen (N.O.) produced by the complete combustion of 67.2 L (N.O.) of ammonia. (Write down the number to tenths.)
Answer: 33.6 liters
Explanation:
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 are related 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 exactly the same way:
V (N 2) \u003d 1 / 2V (NH 3)
V (N 2) \u003d 1 / 2V (NH 3) \u003d 67.2 l / 2 \u003d 33.6 l
Task #26
What volume (in NL liters) of oxygen is produced by the decomposition of 4 moles of hydrogen peroxide? (Write down the number to tenths).
Answer: 44.8 liters
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 oxygen formed is half the amount of hydrogen peroxide:
ν (O 2) \u003d 1/2 ν (H 2 O 2), therefore, ν (O 2) \u003d 4 mol / 2 \u003d 2 mol.
The volume of gases is calculated by the formula:
V = Vm ν , where V m is the molar volume of gases at n.o., equal to 22.4 l / mol
The volume of oxygen formed during the decomposition of peroxide is equal to:
V (O 2) \u003d V m ν (O 2) \u003d 22.4 l / mol 2 mol \u003d 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, contains two hydroxyl groups: CH 2 (OH)-CH 2 OH.
Carbohydrates - organic substances containing carbonyl and several hydroxyl groups, general formula 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 carbohydrate consisting of a large number monosaccharide residues, the formula of which is written as (C 6 H 10 O 5) n.
Hydrocarbons are organic substances that contain only two elements - carbon and hydrogen. The hydrocarbons of the proposed options include toluene - aromatic compound, consisting only of carbon and hydrogen atoms and not containing functional groups with heteroatoms.
Carboxylic acids are organic substances whose molecules contain a carboxyl group consisting of carbonyl and hydroxyl groups linked together. The class of carboxylic acids includes butyric (butanoic) acid - C 3 H 7 COOH.
Task #28
Establish a correspondence between the reaction equation and the change in the oxidation state of the oxidizing agent in it.
| REACTION EQUATION A) 4NH 3 + 5O 2 = 4NO + 6H 2 O B) 2Cu (NO 3) 2 \u003d 2CuO + 4NO 2 + O 2 C) 4Zn + 10HNO 3 \u003d NH 4 NO 3 + 4Zn (NO 3) 2 + 3H 2 O D) 3NO 2 + H 2 O \u003d 2HNO 3 + NO | CHANGING THE OXIDIZER DEGREE |
Answer: A-1; B-4; AT 6; G-3
Explanation:
An oxidizing agent is a substance that contains atoms that are capable of attaching electrons during a chemical reaction and thus lowering the oxidation state.
A reducing agent is a substance that contains atoms that can donate electrons during a chemical reaction and thus increase the degree of oxidation.
A) The 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 adding electrons, is reduced to an oxidation state of -2 in NO and H 2 O compounds.
B) Copper nitrate Cu (NO 3) 2 - 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. During the reaction, the nitrate anion is converted into nitrogen dioxide NO 2 (with nitrogen oxidation state +4) and oxygen O 2 (with oxidation state 0). Therefore, nitrogen is the 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, turning into ammonium nitrate, lowers the oxidation state of nitrogen from +5 (in nitric acid) to -3 (in the ammonium cation). The degree of nitrogen oxidation in the acid residues of ammonium nitrate and zinc nitrate remains unchanged; the same as that of nitrogen in HNO 3 .
D) In this reaction, nitrogen in dioxide disproportionates, i.e. simultaneously increases (from N +4 in NO 2 to N +5 in HNO 3), and lowers (from N +4 in NO 2 to N +2 in NO) its oxidation state.
Task #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 during the passage of a constant electric current through an electrolyte solution or melt. At the cathode, the reduction occurs predominantly of those cations that have the highest oxidizing activity. At the anode, those anions are oxidized first of all, which have the greatest reduction ability.
Electrolysis of aqueous solution
1) Electrolysis process 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 released at the cathode)
Zn 2+ - Pb 2+ reduction 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. 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 released at the anode)
halide ions (except F -) oxidation process 2Hal - - 2e → Hal 2 (free halogens are released)
organic acids oxidation process:
2RCOO - - 2e → R-R + 2CO 2
The overall electrolysis equation is:
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)
C) 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:
Salt hydrolysis is 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 - 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 anionic hydrolysis.
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 a weakly dissociating carboxylic acid)
The solution is alkaline (pH > 7):
C 17 H 35 COONa + H 2 O ↔ C 17 H 35 COOH + NaOH
B) Ammonium phosphate - a salt formed by weak phosphoric acid and ammonia ( weak base), therefore, undergoes hydrolysis both at the cation and at the anion.
(NH 4) 3 PO 4 → 3NH 4 + + PO 4 3-
PO 4 3- + H 2 O ↔ HPO 4 2- + OH - (formation of a weakly dissociating hydrophosphate 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 - a salt formed by a weak hydrosulfuric acid and sodium hydroxide (alkali - a strong base), therefore, undergoes anionic hydrolysis.
Na 2 S → 2Na + + S 2-
S 2- + H 2 O ↔ HS - + OH - (formation of a weakly dissociating hydrosulfide ion)
The solution is alkaline (pH > 7):
Na 2 S + H 2 O ↔ NaHS + NaOH
D) Beryllium sulfate - a salt formed by strong sulfuric acid and beryllium hydroxide (weak base), therefore, undergoes hydrolysis at the cation.
BeSO 4 → Be 2+ + SO 4 2-
Be 2+ + H 2 O ↔ Be(OH) + + H + (formation of a weakly dissociating Be(OH) + cation)
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 method of influencing an equilibrium system
MgO (solid) + CO 2 (g) ↔ MgCO 3 (solid) + Q
and a shift in chemical equilibrium as a result of this impact
Answer: A-1; B-2; IN 2; G-3Explanation:
This reaction is in chemical equilibrium, i.e. in a state where the rate of the forward reaction is equal to the rate of the reverse. The shift 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 that determine the position of equilibrium:
- pressure: an increase in pressure shifts the equilibrium towards a reaction leading to a decrease in volume (conversely, a decrease in pressure shifts the equilibrium towards a reaction leading to an increase in volume)
- temperature: an increase in temperature shifts the equilibrium towards an endothermic reaction (conversely, a decrease in temperature shifts the equilibrium towards an exothermic reaction)
- concentrations of starting substances and reaction products: an increase in the concentration of the starting substances and the removal of products from the reaction sphere shift the equilibrium towards the forward reaction (on the contrary, a decrease in the concentration of the starting substances and an increase in the reaction products shift the equilibrium towards the reverse reaction)
- Catalysts do not affect the equilibrium shift, but only accelerate its achievement.
In this way,
A) since the reaction of obtaining magnesium carbonate is exothermic, a decrease in temperature will contribute to a shift in the equilibrium towards a 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 the 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 a catalyst does not affect the equilibrium shift.
Task #32
Establish a correspondence between the formula of a substance and the reagents, with each of which this substance can interact.
| SUBSTANCE FORMULA | 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) 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) Porphorite (III) oxide is an acidic oxide, therefore, it 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 oxide (III) - 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 co-proportionation reaction with iron to form iron oxide (II):
Fe 2 O 3 + Fe → 3FeO
D) Cu (OH) 2 - a water-insoluble base, 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 with 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 (an alkali is a strong base), therefore, it undergoes anionic hydrolysis.
Na 2 SO 3 → 2Na + + SO 3 2-
SO 3 2- + H 2 O ↔ HSO 3 - + OH - (formation of a low-dissociating hydrosulfite ion)
The medium of the solution is alkaline (pH > 7), the color of the phenolphthalein indicator in an alkaline medium is raspberry.
Sodium sulfate - a salt formed by strong sulfuric acid and sodium hydroxide (alkali - a strong base), does not hydrolyze. The solution medium is neutral (pH = 7), the color of the phenolphthalein indicator in a neutral medium is pale pink.
C) Na 2 SO 4 and ZnSO 4 salts can also be distinguished using 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 with 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 reacting substances and carbon-containing products of their interaction.
| REACTING 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) Propyne attaches hydrogen, in its excess turning into propane:
CH 3 -C≡CH + 2H 2 → CH 3 -CH 2 -CH 3
B) The addition of water (hydration) of alkynes in the presence of divalent mercury salts, resulting in the formation of carbonyl compounds, is the 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 breaking of the triple bond in the alkyne, resulting in the formation 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 propinide is formed and precipitates when propyne is passed through an ammonia solution of silver oxide. This reaction serves to detect alkynes 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 #35
Match the reactants with the organic matter that is the product of the reaction.
| INTERACTION PRODUCT 5) (CH 3 COO) 2 Cu |
Answer: A-4; B-6; IN 1; G-6
Explanation:
A) When ethyl alcohol is oxidized with copper (II) oxide, acetaldehyde is formed, while the oxide is reduced to metal:
B) When alcohol is exposed to concentrated sulfuric acid at a temperature above 140 0 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 hydroxyl group alcohol:
2CH 3 CH 2 OH + 2K → 2CH 3 CH 2 OK + H 2
D) In an alcohol solution of alkali, alcohols undergo an elimination reaction (cleavage). In the case of ethanol, ethylene is formed:
CH 3 CH 2 Cl + KOH (alcohol) → CH 2 \u003d CH 2 + KCl + H 2 O
Task #36
Using method electronic balance, write the reaction equation:
P 2 O 3 + HClO 3 + ... → HCl + ...
In this reaction, chloric acid is the oxidizing agent because the chlorine it contains 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 of +5, turning into orthophosphoric acid.
We compose the oxidation and reduction half-reactions:
Cl +5 + 6e → Cl −1 |2
2P +3 – 4e → 2P +5 |3
We write the redox reaction equation in the form:
3P 2 O 3 + 2HClO 3 + 9H 2 O → 2HCl + 6H 3 PO 4
Task #37
Copper was dissolved in concentrated nitric acid. The evolved gas was passed over heated zinc powder. The resulting solid was added to the sodium hydroxide solution. An excess of carbon dioxide was passed through the resulting solution, and the formation of a precipitate was observed. Write the equations for the four described reactions.
1) When copper is dissolved in concentrated nitric acid, copper is oxidized to Cu +2, and 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 heated zinc powder, zinc is oxidized, and nitrogen dioxide is reduced to molecular nitrogen (assumed by many, with reference to Wikipedia, zinc nitrate is not formed 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 a solution of sodium tetrahydroxozincate, an acid salt is formed - sodium bicarbonate, zinc hydroxide precipitates:
Na 2 + 2CO 2 → Zn(OH) 2 ↓ + 2NaHCO 3
Task #38
Write the reaction equations that can be used to carry out the following transformations:
When writing reaction equations, use the structural formulas of organic substances.
1) The most characteristic of 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 hydrogen atom at the secondary carbon atom is predominantly replaced, resulting in the formation of 2-bromobutane. This is due to the fact that a radical with an unpaired electron at the secondary carbon atom is more stable than a 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 hydrogen halide is eliminated 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 a 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 \u003d CH-CH 3 + Br 2 → CH 3 -CHBr-CHBr-CH 3
4) When interacting with a dibromo derivative, in which the halogen atoms are at neighboring carbon atoms (or at the same atom), an alcohol solution of alkali, 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 #39
A mixture of iron and zinc powders is reacted with 153 ml of a 10% hydrochloric acid solution (ρ = 1.05 g/ml). Interaction with the same weight 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 your answer, write down the reaction equations that are indicated in the condition of the problem, and give all the necessary calculations.
Answer: 46.28%
Task #40
On combustion 2.65 g organic matter received 4.48 liters of carbon dioxide (N.O.) and 2.25 g of water.
It is known that when this substance is oxidized with a sulfuric acid solution of potassium permanganate, a monobasic acid is formed and carbon dioxide is released.
Based on these conditions of the assignment:
1) make the calculations necessary to establish the molecular formula of an organic substance;
2) write down the molecular formula of the original organic matter;
3) make a structural formula of this substance, which unambiguously reflects the order of bonding of atoms in its molecule;
4) write the reaction equation for the oxidation of this substance with a sulfuric acid solution of potassium permanganate.
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
Option No. 1357842
USE in Chemistry - 2016. Main wave (Part C).
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Using the electron balance method, write the equation for the reaction:
Determine the oxidizing agent and reducing agent.
Copper(II) oxide was heated in a hydrogen atmosphere. The resulting solid was dissolved in concentrated sulfuric acid. The resulting salt reacted with potassium iodide, and the released gas was mixed with chlorine and passed through a solution of potassium hydroxide.
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Write the reaction equations that can be used to carry out the following transformations:
When writing reaction equations, use the structural formulas of organic substances.
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Zinc nitrate was heated. Some of it decomposed, and 5.6 liters of a mixture of gases were released. The solid residue weighing 64.8 g was dissolved in a strict amount of 28% sodium hydroxide solution (that is, sufficient to dissolve and without excess). Determine the mass fraction of sodium nitrate.
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