Specification
control measuring materials
for holding in 2017 a unified state exam
in chemistry
1. Appointment of KIM USE
The Unified State Examination (hereinafter referred to as the Unified State Examination) 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 USE is conducted in accordance with Federal Law No. 273-FZ of December 29, 2012 “On Education in the Russian Federation”.
Control measuring materials allow to establish the level of development by graduates of the Federal component of the state standard of secondary (complete) general education in chemistry, basic and profile levels.
The results of the unified state exam in chemistry are recognized by educational institutions of secondary vocational education and educational organizations of higher professional education as the results of entrance examinations in chemistry.
2. Documents defining the content of KIM USE
3. Approaches to the selection of content, the development of the structure of the KIM USE
The basis of approaches to the development of KIM USE 2017 in chemistry was those general methodological guidelines that were identified during the formation exam models 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 an invariant core of the content of existing programs in chemistry for general education organizations. In the standard, this system of knowledge is presented in the form of requirements for the preparation of graduates. These requirements correspond to the level of presentation in the KIM of the content elements being checked.
- In order to ensure the possibility of a differentiated assessment of the educational achievements of graduates of the KIM USE, they check the development of basic educational programs in chemistry at three levels of complexity: basic, advanced and high. Educational material, on the basis of which tasks are built, is selected on the basis of its significance 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 features of substances and reactions; determine the degree of oxidation of chemical elements according to the formulas of their compounds; explain the essence of a particular process, the relationship of the composition, structure and properties of substances. The ability of the examinee to carry out various actions when performing work is considered as an indicator of the assimilation of the studied material with the necessary depth of understanding.
- The equivalence of all variants of the examination work is ensured by maintaining the same ratio of the number of tasks that test the assimilation of the main elements of the content of the key sections of the chemistry course.
4. The structure of 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 short answer items, including 26 items basic level complexity (the serial numbers of these tasks: 1, 2, 3, 4, ... 26) and 9 tasks advanced level complexity (the serial numbers of these tasks: 27, 28, 29, ... 35).
Part 2 contains 5 tasks high level complexity, with a detailed answer (the serial numbers of these tasks: 36, 37, 38, 39, 40).
To complete tasks 1-3, use the following row of chemical elements. The answer in tasks 1-3 is a sequence of numbers, under which the chemical elements in this row are indicated.
1) Na 2) K 3) Si 4) Mg 5) C
Task number 1
Determine which atoms of the elements indicated in the series have four electrons on the external energy level.
Answer: 3; 5
The number of electrons in the outer energy level (electronic layer) of the elements of the main subgroups is equal to the group number.
Thus, from the presented answers, silicon and carbon are suitable, because. they are in the main subgroup of the fourth group of the table D.I. Mendeleev (IVA group), i.e. Answers 3 and 5 are correct.
Task number 2
From the chemical elements indicated in the series, select three elements that are in the Periodic Table of Chemical Elements of D.I. Mendeleev are in the same period. Arrange the selected elements in ascending order of their metallic properties.
Write in the answer field the numbers of the selected elements in the desired sequence.
Answer: 3; four; one
Three of the presented elements are in the same period - sodium Na, silicon Si and magnesium Mg.
When moving within one period of the Periodic Table, D.I. Mendeleev (horizontal lines) from right to left, the return of electrons located on the outer layer is facilitated, i.e. the metallic properties of the elements are enhanced. Thus, the metallic properties of sodium, silicon and magnesium are enhanced in the series Si Task number 3 From among the elements listed in the row, select two elements that exhibit the lowest oxidation state, equal to -4. Write down the numbers of the selected elements in the answer field. Answer: 3; 5 According to the octet rule, the atoms of chemical elements tend to have 8 electrons in their outer electronic level, like the noble gases. This can be achieved either by donating electrons of the last level, then the previous one, containing 8 electrons, becomes external, or, conversely, by adding additional electrons up to eight. Sodium and potassium are alkali metals and are in the main subgroup of the first group (IA). This means that on the outer electron layer of their atoms there is one electron each. In this regard, the loss of a single electron is energetically more favorable than the addition of seven more. With magnesium, the situation is similar, only it is in the main subgroup of the second group, that is, it has two electrons on the outer electronic level. It should be noted that sodium, potassium and magnesium are metals, and for metals, in principle, a negative oxidation state is impossible. The minimum oxidation state of any metal is zero and is observed in simple substances. The chemical elements carbon C and silicon Si are non-metals and are in the main subgroup of the fourth group (IVA). This means that there are 4 electrons on their outer electron layer. For this reason, for these elements, both the return of these electrons and the addition of four more up to a total of 8 are possible. Silicon and carbon atoms cannot attach more than 4 electrons, therefore the minimum oxidation state for them is -4. Task number 4 From the proposed list, select two compounds in which there is an ionic chemical bond. Answer: 1; 3 In the vast majority of cases, the presence of an ionic type of bond in a compound can be determined by the fact that its structural units simultaneously include atoms of a typical metal and non-metal atoms. On this basis, we establish that there is an ionic bond in compound number 1 - Ca(ClO 2) 2, because in its formula, one can see atoms of a typical calcium metal and atoms of non-metals - oxygen and chlorine. However, there are no more compounds containing both metal and non-metal atoms in this list. In addition to the above feature, the presence of an ionic bond in a compound can be said if its structural unit contains an ammonium cation (NH 4 +) or its organic analogs - alkyl ammonium RNH 3 +, dialkylammonium R 2 NH 2 +, trialkylammonium R 3 NH cations + and tetraalkylammonium R 4 N + , where R is some hydrocarbon radical. For example, the ionic type of bond takes place in the compound (CH 3) 4 NCl between the cation (CH 3) 4 + and the chloride ion Cl - . Among the compounds indicated in the assignment there is ammonium chloride, in which the ionic bond is realized between the ammonium cation NH 4 + and the chloride ion Cl − . Task number 5 Establish a correspondence between the formula of a substance and the class / group to which this substance belongs: for each position indicated by a letter, select the corresponding position from the second column, indicated by a number. Write down the numbers of the selected connections in the answer field. Answer: A-4; B-1; AT 3 Explanation: Acid salts are called salts resulting from the incomplete replacement of mobile hydrogen atoms by a metal cation, ammonium cation or alkyl ammonium. In inorganic acids, which take place as part of the school curriculum, all hydrogen atoms are mobile, that is, they can be replaced by a metal. Examples of acidic inorganic salts among the presented list is ammonium bicarbonate NH 4 HCO 3 - the product of replacing one of the two hydrogen atoms in carbonic acid with an ammonium cation. In fact, an acid salt is a cross between a normal (medium) salt and an acid. In the case of NH 4 HCO 3 - the average between the normal salt (NH 4) 2 CO 3 and carbonic acid H 2 CO 3. In organic substances, only hydrogen atoms that are part of carboxyl groups (-COOH) or hydroxyl groups of phenols (Ar-OH) can be replaced by metal atoms. That is, for example, sodium acetate CH 3 COONa, despite the fact that not all hydrogen atoms in its molecule are replaced by metal cations, is an average, not an acid salt (!). Hydrogen atoms in organic substances, attached directly to the carbon atom, are almost never able to be replaced by metal atoms, with the exception of hydrogen atoms in the triple C≡C bond. Non-salt-forming oxides are oxides of non-metals that do not form salts with basic oxides or bases, that is, they either do not react with them at all (most often), or give a different product (not a salt) in reaction with them. It is often said that non-salt-forming oxides are oxides of non-metals that do not react with bases and basic oxides. However, for the detection of non-salt-forming oxides, this approach does not always work. So, for example, CO, being a non-salt-forming oxide, reacts with basic iron (II) oxide, but with the formation of a free metal rather than a salt: CO + FeO = CO 2 + Fe Non-salt-forming oxides from the school chemistry course include non-metal oxides in the oxidation state +1 and +2. In total, they are found in the USE 4 - these are CO, NO, N 2 O and SiO (I personally never met the last SiO in assignments). Task number 6 From the proposed list of substances, select two substances, with each of which iron reacts without heating. Answer: 2; four Zinc chloride is a salt, and iron is a metal. The metal reacts with the salt only if it is more reactive than the one in the salt. The relative activity of metals is determined by a series of metal activity (in other words, a series of metal stresses). Iron is located to the right of zinc in the activity series of metals, which means that it is less active and is not able to displace zinc from salt. That is, the reaction of iron with substance No. 1 does not go. Copper (II) sulfate CuSO 4 will react with iron, since iron is located to the left of copper in the activity series, that is, it is a more active metal. Concentrated nitric acid, as well as concentrated sulfuric acid, are not able to react with iron, aluminum and chromium without heating due to such a phenomenon as passivation: on the surface of these metals, under the action of these acids, an insoluble salt is formed without heating, which acts as a protective shell. However, when heated, this protective shell dissolves and the reaction becomes possible. Those. since it is indicated that there is no heating, the reaction of iron with conc. HNO 3 does not leak. Hydrochloric acid, regardless of concentration, refers to non-oxidizing acids. Metals that are in the activity series to the left of hydrogen react with non-oxidizing acids with the release of hydrogen. Iron is one of these metals. Conclusion: the reaction of iron with hydrochloric acid proceeds. In the case of a metal and a metal oxide, the reaction, as in the case of a salt, is possible if the free metal is more active than that which is part of the oxide. Fe, according to the activity series of metals, is less active than Al. This means that Fe does not react with Al 2 O 3. Task number 7 From the proposed list, select two oxides that react with a solution of hydrochloric acid, but do not react
with sodium hydroxide solution. Write down the numbers of the selected substances in the answer field. Answer: 3; four CO is a non-salt-forming oxide; it does not react with an aqueous solution of alkali. (It should be remembered that, nevertheless, under harsh conditions - high pressure and temperature - it still reacts with solid alkali, forming formates - salts of formic acid.) SO 3 - sulfur oxide (VI) - acid oxide, which corresponds to sulfuric acid. Acid oxides do not react with acids and other acid oxides. That is, SO 3 does not react with hydrochloric acid and reacts with a base - sodium hydroxide. Not suitable. CuO - copper (II) oxide - is classified as an oxide with predominantly basic properties. Reacts with HCl and does not react with sodium hydroxide solution. Fits MgO - magnesium oxide - is classified as a typical basic oxide. Reacts with HCl and does not react with sodium hydroxide solution. Fits ZnO - an oxide with pronounced amphoteric properties - easily reacts with both strong bases and acids (as well as acidic and basic oxides). Not suitable. Task number 8 Answer: 4; 2 In the reaction between two salts of inorganic acids, gas is formed only when hot solutions of nitrites and ammonium salts are mixed due to the formation of thermally unstable ammonium nitrite. For example, NH 4 Cl + KNO 2 \u003d t o \u003d\u003e N 2 + 2H 2 O + KCl However, both nitrites and ammonium salts are not on the list. This means that one of the three salts (Cu (NO 3) 2, K 2 SO 3 and Na 2 SiO 3) reacts with either an acid (HCl) or an alkali (NaOH). Among the salts of inorganic acids, only ammonium salts emit gas when interacting with alkalis: NH 4 + + OH \u003d NH 3 + H 2 O Ammonium salts, as we have already said, are not on the list. The only option left is the interaction of the salt with the acid. Salts among these substances include Cu(NO 3) 2, K 2 SO 3 and Na 2 SiO 3. The reaction of copper nitrate with hydrochloric acid does not proceed, because no gas, no precipitate, no low-dissociating substance (water or weak acid) is formed. Sodium silicate reacts with hydrochloric acid, however, due to the release of a white gelatinous precipitate of silicic acid, and not gas: Na 2 SiO 3 + 2HCl \u003d 2NaCl + H 2 SiO 3 ↓ The last option remains - the interaction of potassium sulfite and hydrochloric acid. Indeed, as a result of the ion exchange reaction between sulfite and almost any acid, unstable sulfurous acid is formed, which instantly decomposes into colorless gaseous sulfur oxide (IV) and water. Task number 9 Write in the table the numbers of the selected substances under the corresponding letters. Answer: 2; 5 CO 2 is an acidic oxide and must be treated with either a basic oxide or a base to convert it to a salt. Those. to obtain potassium carbonate from CO 2, it must be treated with either potassium oxide or potassium hydroxide. Thus, substance X is potassium oxide: K 2 O + CO 2 \u003d K 2 CO 3 Potassium bicarbonate KHCO 3, like potassium carbonate, is a salt of carbonic acid, with the only difference being that the bicarbonate is a product of incomplete substitution of hydrogen atoms in carbonic acid. To obtain an acid salt from a normal (medium) salt, one must either act on it with the same acid that formed this salt, or else act on it with an acid oxide corresponding to this acid in the presence of water. Thus reactant Y is carbon dioxide. When it is passed through an aqueous solution of potassium carbonate, the latter turns into potassium bicarbonate: K 2 CO 3 + H 2 O + CO 2 \u003d 2KHCO 3 Task number 10 Establish a correspondence between the reaction equation and the property of the nitrogen element that it exhibits in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number. Write in the table the numbers of the selected substances under the corresponding letters. Answer: A-4; B-2; IN 2; G-1 A) NH 4 HCO 3 - salt, which includes the ammonium cation NH 4 +. In the ammonium cation, nitrogen always has an oxidation state of -3. As a result of the reaction, it turns into ammonia NH 3. Hydrogen almost always (except for its compounds with metals) has an oxidation state of +1. Therefore, for the ammonia molecule to be electrically neutral, nitrogen must have an oxidation state of -3. Thus, there is no change in the degree of nitrogen oxidation; it does not exhibit redox properties. B) As already shown above, nitrogen in ammonia NH 3 has an oxidation state of -3. As a result of the reaction with CuO, ammonia is converted into a simple substance N 2. In any simple substance, the oxidation state of the element with which it is formed is equal to zero. Thus, the nitrogen atom loses its negative charge, and since electrons are responsible for the negative charge, this means that they are lost by the nitrogen atom as a result of the reaction. An element that loses some of its electrons in a reaction is called a reducing agent. C) As a result of the reaction, NH 3 with an oxidation state of nitrogen equal to -3 turns into nitric oxide NO. Oxygen almost always has an oxidation state of -2. Therefore, in order for the nitric oxide molecule to be electrically neutral, the nitrogen atom must have an oxidation state of +2. This means that the nitrogen atom changed its oxidation state from -3 to +2 as a result of the reaction. This indicates the loss of 5 electrons by the nitrogen atom. That is, nitrogen, as in the case of B, is a reducing agent. D) N 2 is a simple substance. In all simple substances, the element that forms them has an oxidation state of 0. As a result of the reaction, nitrogen is converted into lithium nitride Li3N. The only oxidation state of an alkali metal other than zero (any element has an oxidation state of 0) is +1. Thus, for the Li3N structural unit to be electrically neutral, nitrogen must have an oxidation state of -3. It turns out that as a result of the reaction, nitrogen acquired a negative charge, which means the addition of electrons. Nitrogen is the oxidizing agent in this reaction. 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. D) ZnBr 2 (solution) 1) AgNO 3, Na 3 PO 4, Cl 2 2) BaO, H 2 O, KOH 3) H 2, Cl 2, O 2 4) HBr, LiOH, CH 3 COOH 5) H 3 PO 4, BaCl 2, CuO Write in the table the numbers of the selected substances under the corresponding letters. Answer: A-3; B-2; AT 4; G-1 A) When hydrogen gas is passed through a sulfur melt, hydrogen sulfide H 2 S is formed: H 2 + S \u003d t o \u003d\u003e H 2 S When chlorine is passed over crushed sulfur at room temperature, sulfur dichloride is formed: S + Cl 2 \u003d SCl 2 To pass the exam, you do not need to know exactly how sulfur reacts with chlorine and, accordingly, be able to write this equation. The main thing is to remember at a fundamental level that sulfur reacts with chlorine. Chlorine is a strong oxidizing agent, sulfur often exhibits a dual function - both oxidizing and reducing. That is, if a strong oxidizing agent acts on sulfur, which is molecular chlorine Cl 2, it will oxidize. Sulfur burns with a blue flame in oxygen to form a gas with a pungent odor - sulfur dioxide SO 2: B) SO 3 - sulfur oxide (VI) has pronounced acidic properties. For such oxides, the most characteristic reactions are interactions with water, as well as with basic and amphoteric oxides and hydroxides. In the list at number 2, we just see water, and the basic oxide BaO, and hydroxide KOH. When an acidic oxide reacts with a basic oxide, a salt of the corresponding acid and a metal that is part of the basic oxide is formed. An acidic oxide corresponds to an acid in which the acid-forming element has the same oxidation state as in the oxide. The oxide SO 3 corresponds to sulfuric acid H 2 SO 4 (both there and there the oxidation state of sulfur is +6). Thus, when SO 3 interacts with metal oxides, sulfuric acid salts will be obtained - sulfates containing the sulfate ion SO 4 2-: SO 3 + BaO = BaSO 4 When interacting with water, the acid oxide turns into the corresponding acid: SO 3 + H 2 O \u003d H 2 SO 4 And when acid oxides interact with metal hydroxides, a salt of the corresponding acid and water are formed: SO 3 + 2KOH \u003d K 2 SO 4 + H 2 O C) Zinc hydroxide Zn (OH) 2 has typical amphoteric properties, that is, it reacts both with acidic oxides and acids, and with basic oxides and alkalis. In list 4, we see both acids - hydrobromic HBr and acetic, and alkali - LiOH. Recall that water-soluble metal hydroxides are called alkalis: Zn(OH) 2 + 2HBr = ZnBr 2 + 2H 2 O Zn (OH) 2 + 2CH 3 COOH \u003d Zn (CH 3 COO) 2 + 2H 2 O Zn(OH) 2 + 2LiOH \u003d Li 2 D) Zinc bromide ZnBr 2 is a salt, soluble in water. For soluble salts, ion exchange reactions are the most common. A salt can react with another salt provided that both starting salts are soluble and a precipitate forms. Also ZnBr 2 contains bromide ion Br-. Metal halides are characterized by the fact that they are able to react with Hal 2 halogens, which are higher in the periodic table. In this way? the described types of reactions proceed with all substances of list 1: ZnBr 2 + 2AgNO 3 \u003d 2AgBr + Zn (NO 3) 2 3ZnBr 2 + 2Na 3 PO 4 = Zn 3 (PO 4) 2 + 6NaBr ZnBr 2 + Cl 2 = ZnCl 2 + Br 2 Task number 12 Establish a correspondence between the name of the substance and the class / group to which this substance belongs: for each position indicated by a letter, select the corresponding position indicated by a number. Write in the table the numbers of the selected substances under the corresponding letters. Answer: A-4; B-2; IN 1 Explanation: A) Methylbenzene, also known as toluene, has the structural formula: As you can see, the molecules of this substance consist only of carbon and hydrogen, therefore methylbenzene (toluene) refers to hydrocarbons B) The structural formula of aniline (aminobenzene) is as follows: As can be seen from the structural formula, the aniline molecule consists of an aromatic hydrocarbon radical (C 6 H 5 -) and an amino group (-NH 2), thus, aniline belongs to aromatic amines, i.e. correct answer 2. C) 3-methylbutanal. The ending "al" indicates that the substance belongs to aldehydes. The structural formula of this substance: Task number 13 From the proposed list, select two substances that are structural isomers of butene-1. Write down the numbers of the selected substances in the answer field. Answer: 2; 5 Explanation: Isomers are substances that have the same molecular formula and different structural, i.e. Substances that differ in the order in which atoms are combined, but with the same composition of molecules. Task number 14 From the proposed list, select two substances, the interaction of which with a solution of potassium permanganate will cause a change in the color of the solution. Write down the numbers of the selected substances in the answer field. Answer: 3; 5 Explanation: Alkanes, as well as cycloalkanes with a ring size of 5 or more carbon atoms, are very inert and do not react with aqueous solutions of even strong oxidizing agents, such as, for example, potassium permanganate KMnO 4 and potassium dichromate K 2 Cr 2 O 7 . Thus, options 1 and 4 disappear - when cyclohexane or propane is added to an aqueous solution of potassium permanganate, a color change will not occur. Among the hydrocarbons of the homologous series of benzene, only benzene is passive to the action of aqueous solutions of oxidizing agents, all other homologues are oxidized, depending on the medium, either to carboxylic acids or to their corresponding salts. Thus, option 2 (benzene) is eliminated. The correct answers are 3 (toluene) and 5 (propylene). Both substances decolorize the purple solution of potassium permanganate due to the reactions taking place: CH 3 -CH=CH 2 + 2KMnO 4 + 2H 2 O → CH 3 -CH(OH)–CH 2 OH + 2MnO 2 + 2KOH Task number 15 From the proposed list, select two substances with which formaldehyde reacts. Write down the numbers of the selected substances in the answer field. Answer: 3; four Explanation: Formaldehyde belongs to the class of aldehydes - oxygen-containing organic compounds that have an aldehyde group at the end of the molecule: Typical reactions of aldehydes are oxidation and reduction reactions proceeding along the functional group. Among the list of responses for formaldehyde, reduction reactions are typical, where hydrogen is used as a reducing agent (cat. - Pt, Pd, Ni), and oxidation - in this case, the silver mirror reaction. When reduced with hydrogen on a nickel catalyst, formaldehyde is converted to methanol: The silver mirror reaction is the reduction of silver from an ammonia solution of silver oxide. When dissolved in an aqueous solution of ammonia, silver oxide turns into a complex compound - diammine silver (I) OH hydroxide. After the addition of formaldehyde, a redox reaction occurs in which silver is reduced: Task number 16 From the proposed list, select two substances with which methylamine reacts. Write down the numbers of the selected substances in the answer field. Answer: 2; 5 Explanation: Methylamine is the simplest organic compound of the amine class. A characteristic feature of amines is the presence of a lone electron pair on the nitrogen atom, as a result of which amines exhibit the properties of bases and act as nucleophiles in reactions. Thus, in this regard, from the proposed answers, methylamine as a base and nucleophile reacts with chloromethane and hydrochloric acid: CH 3 NH 2 + CH 3 Cl → (CH 3) 2 NH 2 + Cl - CH 3 NH 2 + HCl → CH 3 NH 3 + Cl - Task number 17 The following scheme of transformations of substances is given: Determine which of the given substances are substances X and Y. Write in the table the numbers of the selected substances under the corresponding letters. Answer: 4; 2 Explanation: One of the reactions for obtaining alcohols is the hydrolysis of haloalkanes. Thus, ethanol can be obtained from chloroethane by acting on the latter with an aqueous solution of alkali - in this case, NaOH. CH 3 CH 2 Cl + NaOH (aq.) → CH 3 CH 2 OH + NaCl The next reaction is the oxidation reaction of ethyl alcohol. The oxidation of alcohols is carried out on a copper catalyst or using CuO: Task number 18 Establish a correspondence between the name of the substance and the product that is mainly formed during the interaction of this substance with bromine: for each position indicated by a letter, select the corresponding position indicated by a number. Answer: 5; 2; 3; 6 Explanation: For alkanes, the most characteristic reactions are free radical substitution reactions, during which a hydrogen atom is replaced by a halogen atom. Thus, by brominating ethane, one can obtain bromoethane, and by brominating isobutane, 2-bromoisobutane can be obtained: Since the small cycles of cyclopropane and cyclobutane molecules are unstable, during bromination the cycles of these molecules are opened, thus the addition reaction proceeds: Unlike the cyclopropane and cyclobutane cycles, the cyclohexane cycle large sizes, resulting in the replacement of a hydrogen atom by a bromine atom: Task #19 Establish a correspondence between the reacting substances and the carbon-containing product that is formed during the interaction of these substances: for each position indicated by a letter, select the corresponding position indicated by a number. Write in the table the selected numbers under the corresponding letters. Answer: 5; four; 6; 2 Task number 20 From the proposed list of types of reactions, select two types of reactions, which include the interaction of alkali metals with water. Write down the numbers of the selected types of reactions in the answer field. Answer: 3; four Alkali metals (Li, Na, K, Rb, Cs, Fr) are located in the main subgroup of group I of the table D.I. Mendeleev and are reducing agents, easily donating an electron located at the outer level. If we denote the alkali metal with the letter M, then the reaction of the alkali metal with water will look like this: 2M + 2H 2 O → 2MOH + H 2 Alkali metals are very active towards water. The reaction proceeds violently with the release of a large amount of heat, is irreversible and does not require the use of a catalyst (non-catalytic) - a substance that accelerates the reaction and is not part of the reaction products. It should be noted that all highly exothermic reactions do not require the use of a catalyst and proceed irreversibly. Since metal and water are substances that are in different states of aggregation, this reaction proceeds at the interface, therefore, it is heterogeneous. The type of this reaction is substitution. Reactions between inorganic substances are classified as substitution reactions if a simple substance interacts with a complex one and as a result other simple and complex substances are formed. (A neutralization reaction occurs between an acid and a base, as a result of which these substances exchange their constituent parts and a salt and a low-dissociating substance are formed). As mentioned above, alkali metals are reducing agents, donating an electron from the outer layer, therefore, the reaction is redox. Task number 21 From the proposed list of external influences, select two influences that lead to a decrease in the rate of the reaction of ethylene with hydrogen. Write in the answer field the numbers of the selected external influences. Answer: 1; four The following factors influence the rate of a chemical reaction: changes in temperature and concentration of reagents, as well as the use of a catalyst. According to Van't Hoff's empirical rule, for every 10 degrees increase in temperature, the rate constant of a homogeneous reaction increases by 2-4 times. Therefore, a decrease in temperature also leads to a decrease in the reaction rate. The first answer is correct. As noted above, the reaction rate is also affected by a change in the concentration of reagents: if the concentration of ethylene is increased, the reaction rate will also increase, which does not meet the requirements of the problem. And a decrease in the concentration of hydrogen - the initial component, on the contrary, reduces the reaction rate. Therefore, the second option is not suitable, but the fourth one is. A catalyst is a substance that speeds up the rate of a chemical reaction but is not part of the products. The use of a catalyst accelerates the ethylene hydrogenation reaction, which also does not correspond to the condition of the problem, and therefore is not the right answer. When ethylene reacts with hydrogen (on Ni, Pd, Pt catalysts), ethane is formed: CH 2 \u003d CH 2 (g) + H 2 (g) → CH 3 -CH 3 (g) All components involved in the reaction and the product are gaseous substances, therefore, the pressure in the system will also affect the reaction rate. From two volumes of ethylene and hydrogen, one volume of ethane is formed, therefore, the reaction proceeds to a decrease in pressure in the system. By increasing the pressure, we will speed up the reaction. The fifth answer does not fit. Task #22 Establish a correspondence between the formula of salt and the products of electrolysis of an aqueous solution of this salt, which stood out on inert electrodes: for each position, SALT FORMULA ELECTROLYSIS PRODUCTS Write in the table the selected numbers under the corresponding letters. Answer: 1; four; 3; 2 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) The process of electrolysis of aqueous solutions on 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+ recovery process: 2H 2 O + 2e → H 2 + 2OH - (H 2 is released 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 will be 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 3 PO 4 solution 2H 2 O → 2H 2 (at the cathode) + O 2 (at the anode) B) KCl solution 2KCl + 2H 2 O → H 2 (at the cathode) + 2KOH + Cl 2 (at the anode) C) CuBr2 solution CuBr 2 → Cu (at the cathode) + Br 2 (at the anode) D) Cu(NO3)2 solution 2Cu(NO 3) 2 + 2H 2 O → 2Cu (at the cathode) + 4HNO 3 + O 2 (at the anode) Task #23 Establish a correspondence between the name of the salt and the ratio of this salt to hydrolysis: for each position indicated by a letter, select the corresponding position indicated by a number. Write in the table the selected numbers under the corresponding letters. Answer: 1; 3; 2; four Hydrolysis of salts - the interaction of salts with water, leading to the addition of the hydrogen cation H + of the water molecule to the anion of the acid residue and (or) the hydroxyl group OH - of the water molecule to the metal cation. Salts formed by cations corresponding to weak bases and anions corresponding to weak acids undergo hydrolysis. A) Ammonium chloride (NH 4 Cl) - a salt formed by strong hydrochloric acid and ammonia (weak base), undergoes hydrolysis by the cation. NH 4 Cl → NH 4 + + Cl - NH 4 + + H 2 O → NH 3 H 2 O + H + (formation of ammonia dissolved in water) The solution medium is acidic (pH< 7). B) Potassium sulfate (K 2 SO 4) - a salt formed by strong sulfuric acid and potassium hydroxide (alkali, i.e. strong base), does not undergo hydrolysis. K 2 SO 4 → 2K + + SO 4 2- C) Sodium carbonate (Na 2 CO 3) - a salt formed by a weak carbonic acid and sodium hydroxide (an alkali, i.e. a strong base), undergoes anion hydrolysis. CO 3 2- + H 2 O → HCO 3 - + OH - (formation of a weakly dissociating hydrocarbonate ion) The solution is alkaline (pH > 7). D) Aluminum sulfide (Al 2 S 3) - a salt formed by a weak hydrosulfide acid and aluminum hydroxide (weak base), undergoes complete hydrolysis with the formation of aluminum hydroxide and hydrogen sulfide: Al 2 S 3 + 6H 2 O → 2Al(OH) 3 + 3H 2 S The solution medium is close to neutral (pH ~ 7). Task #24 Establish a correspondence between the equation of a chemical reaction and the direction of displacement of chemical equilibrium with increasing pressure in the system: for each position indicated by a letter, select the corresponding position indicated by a number. REACTION EQUATION A) N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) B) 2H 2 (g) + O 2 (g) ↔ 2H 2 O (g) C) H 2 (g) + Cl 2 (g) ↔ 2HCl (g) D) SO 2 (g) + Cl 2 (g) ↔ SO 2 Cl 2 (g) DIRECTION OF SHIFT OF CHEMICAL EQUILIBRIUM 1) shifts towards a direct reaction 2) shifts towards the back reaction 3) there is no shift in equilibrium Write in the table the selected numbers under the corresponding letters. Answer: A-1; B-1; AT 3; G-1 A reaction is in chemical equilibrium when 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. 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 A) In the first case, the reaction proceeds with a decrease in volume, since V (N 2) + 3V (H 2) > 2V (NH 3). By increasing the pressure in the system, the equilibrium will shift to the side with a smaller volume of substances, therefore, in the forward direction (in the direction of the direct reaction). B) In the second case, the reaction also proceeds with a decrease in volume, since 2V (H 2) + V (O 2) > 2V (H 2 O). By increasing the pressure in the system, the equilibrium will also shift in the direction of the direct reaction (in the direction of the product). C) In the third case, the pressure does not change during the reaction, because V (H 2) + V (Cl 2) \u003d 2V (HCl), so there is no equilibrium shift. D) In the fourth case, the reaction also proceeds with a decrease in volume, since V (SO 2) + V (Cl 2) > V (SO 2 Cl 2). By increasing the pressure in the system, the equilibrium will shift towards the formation of the product (direct reaction). Task #25 Establish a correspondence between the formulas of substances and a reagent with which you can distinguish their aqueous solutions: for each position indicated by a letter, select the corresponding position indicated by a number. SUBSTANCE FORMULA A) HNO 3 and H 2 O C) NaCl and BaCl 2 D) AlCl 3 and MgCl 2 Write in the table the selected numbers under the corresponding letters. Answer: A-1; B-3; AT 3; G-2 A) Nitric acid and water can be distinguished using salt - calcium carbonate CaCO 3. Calcium carbonate does not dissolve in water, and when interacting with nitric acid forms a soluble salt - calcium nitrate Ca (NO 3) 2, while the reaction is accompanied by the release of a colorless carbon dioxide: CaCO 3 + 2HNO 3 → Ca(NO 3) 2 + CO 2 + H 2 O B) Potassium chloride KCl and alkali NaOH can be distinguished by a solution of copper (II) sulfate. When copper (II) sulfate interacts with KCl, the exchange reaction does not proceed, the solution contains K +, Cl -, Cu 2+ and SO 4 2- ions, which do not form poorly dissociating substances with each other. When copper (II) sulfate interacts with NaOH, an exchange reaction occurs, as a result of which copper (II) hydroxide precipitates (base blue color). C) Sodium chloride NaCl and barium BaCl 2 are soluble salts, which can also be distinguished by a solution of copper (II) sulfate. When copper (II) sulfate interacts with NaCl, the exchange reaction does not proceed, the solution contains Na +, Cl -, Cu 2+ and SO 4 2- ions, which do not form poorly dissociating substances with each other. When copper (II) sulfate interacts with BaCl 2, an exchange reaction occurs, as a result of which barium sulfate BaSO 4 precipitates. D) Aluminum chloride AlCl 3 and magnesium MgCl 2 dissolve in water and behave differently when interacting with potassium hydroxide. Magnesium chloride with alkali forms a precipitate: MgCl 2 + 2KOH → Mg(OH) 2 ↓ + 2KCl When alkali interacts with aluminum chloride, a precipitate first forms, which then dissolves to form a complex salt - potassium tetrahydroxoaluminate: AlCl 3 + 4KOH → K + 3KCl Task #26 Establish a correspondence between the substance and its scope: for each position indicated by a letter, select the corresponding position indicated by a number. Write in the table the selected numbers under the corresponding letters. Answer: A-4; B-2; AT 3; G-5 A) Ammonia is the most important product of the chemical industry, its production is more than 130 million tons per year. Ammonia is mainly used in the production of nitrogen fertilizers (ammonium nitrate and sulfate, urea), medicines, explosives, nitric acid, and soda. Among the proposed answers, the area of application of ammonia is the production of fertilizers (Fourth answer option). B) Methane is the simplest hydrocarbon, the most thermally stable representative of a number of saturated compounds. It is widely used as a domestic and industrial fuel, as well as a raw material for industry (Second answer). Methane is 90-98% a component of natural gas. C) Rubbers are materials that are obtained by polymerization of compounds with conjugated double bonds. Isoprene just belongs to this type of compounds and is used to obtain one of the types of rubbers: D) Low molecular weight alkenes are used to make plastics, in particular ethylene is used to make a plastic called polyethylene: n CH 2 \u003d CH 2 → (-CH 2 -CH 2 -) n Task number 27 Calculate the mass of potassium nitrate (in grams) that should be dissolved in 150 g of a solution with a mass fraction of this salt of 10% to obtain a solution with a mass fraction of 12%. (Write down the number to tenths.) Answer: 3.4 g Explanation: Let x g be the mass of potassium nitrate, which is dissolved in 150 g of the solution. Calculate the mass of potassium nitrate dissolved in 150 g of solution: m(KNO 3) \u003d 150 g 0.1 \u003d 15 g In order for the mass fraction of salt to be 12%, x g of potassium nitrate was added. In this case, the mass of the solution was (150 + x) g. We write the equation in the form: (Write down the number to tenths.) Answer: 14.4 g Explanation: As a result of the complete combustion of hydrogen sulfide, sulfur dioxide and water are formed: 2H 2 S + 3O 2 → 2SO 2 + 2H 2 O A consequence of Avogadro's law is that the volumes of gases under the same conditions are related to each other in the same way as the number of moles of these gases. Thus, according to the reaction equation: ν(O 2) = 3/2ν(H 2 S), therefore, the volumes of hydrogen sulfide and oxygen are related to each other in exactly the same way: V (O 2) \u003d 3 / 2V (H 2 S), V (O 2) \u003d 3/2 6.72 l \u003d 10.08 l, hence V (O 2) \u003d 10.08 l / 22.4 l / mol \u003d 0.45 mol Calculate the mass of oxygen required for the complete combustion of hydrogen sulfide: m(O 2) \u003d 0.45 mol 32 g / mol \u003d 14.4 g Task number 30 Using the electron balance method, write the equation for the reaction: Na 2 SO 3 + ... + KOH → K 2 MnO 4 + ... + H 2 O Determine the oxidizing agent and reducing agent. Mn +7 + 1e → Mn +6 │2 reduction reaction S +4 − 2e → S +6 │1 oxidation reaction Mn +7 (KMnO 4) - oxidizing agent, S +4 (Na 2 SO 3) - reducing agent Na 2 SO 3 + 2KMnO 4 + 2KOH → 2K 2 MnO 4 + Na 2 SO 4 + H 2 O Task number 31 Iron was dissolved in hot concentrated sulfuric acid. The resulting salt was treated with an excess of sodium hydroxide solution. The brown precipitate formed was filtered off and dried. The resulting substance was heated with iron. Write the equations for the four described reactions. 1) Iron, like aluminum and chromium, does not react with concentrated sulfuric acid, becoming covered with a protective oxide film. The reaction occurs only when heated with the release of sulfur dioxide: 2Fe + 6H 2 SO 4 → Fe 2 (SO 4) 2 + 3SO 2 + 6H 2 O (on heating) 2) Iron (III) sulfate - a salt soluble in water, enters into an exchange reaction with alkali, as a result of which iron (III) hydroxide precipitates (brown compound): Fe 2 (SO 4) 3 + 3NaOH → 2Fe(OH) 3 ↓ + 3Na 2 SO 4 3) Insoluble metal hydroxides decompose upon calcination to the corresponding oxides and water: 2Fe(OH) 3 → Fe 2 O 3 + 3H 2 O 4) When iron (III) oxide is heated with metallic iron, iron (II) oxide is formed (iron in the FeO compound has an intermediate oxidation state): Fe 2 O 3 + Fe → 3FeO (on heating) Task #32 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) Intramolecular dehydration occurs at temperatures above 140 o C. This occurs as a result of the elimination of a hydrogen atom from the carbon atom of the alcohol, located one through to the alcohol hydroxyl (in the β-position). CH 3 -CH 2 -CH 2 -OH → CH 2 \u003d CH-CH 3 + H 2 O (conditions - H 2 SO 4, 180 o C) Intermolecular dehydration proceeds at a temperature below 140 o C under the action of sulfuric acid and ultimately comes down to the elimination of one water molecule from two alcohol molecules. 2) Propylene refers to unsymmetrical alkenes. When hydrogen halides and water are added, a hydrogen atom is attached to a carbon atom at a multiple bond associated with a large number of hydrogen atoms: CH 2 \u003d CH-CH 3 + HCl → CH 3 -CHCl-CH 3 3) Acting with an aqueous solution of NaOH on 2-chloropropane, the halogen atom is replaced by a hydroxyl group: CH 3 -CHCl-CH 3 + NaOH (aq.) → CH 3 -CHOH-CH 3 + NaCl 4) Propylene can be obtained not only from propanol-1, but also from propanol-2 by the reaction of intramolecular dehydration at a temperature above 140 o C: CH 3 -CH(OH)-CH 3 → CH 2 \u003d CH-CH 3 + H 2 O (conditions H 2 SO 4, 180 o C) 5) In an alkaline environment, acting with a dilute aqueous solution of potassium permanganate, hydroxylation of alkenes occurs with the formation of diols: 3CH 2 \u003d CH-CH 3 + 2KMnO 4 + 4H 2 O → 3HOCH 2 -CH (OH) -CH 3 + 2MnO 2 + 2KOH Task number 33 Determine the mass fractions (in%) of iron (II) sulfate and aluminum sulfide in the mixture, if during the treatment of 25 g of this mixture with water a gas was released that completely reacted with 960 g of a 5% solution of copper (II) sulfate. In response, write down the reaction equations that are indicated in the condition of the problem, and give all the necessary calculations (indicate the units of measurement of the required physical quantities). Answer: ω(Al 2 S 3) = 40%; ω(CuSO 4) = 60% When a mixture of iron (II) sulfate and aluminum sulfide is treated with water, the sulfate is simply dissolved, and the sulfide is hydrolyzed to form aluminum (III) hydroxide and hydrogen sulfide: Al 2 S 3 + 6H 2 O → 2Al(OH) 3 ↓ + 3H 2 S (I) When hydrogen sulfide is passed through a solution of copper (II) sulfate, copper (II) sulfide precipitates: CuSO 4 + H 2 S → CuS↓ + H 2 SO 4 (II) Calculate the mass and amount of substance of dissolved copper(II) sulfate: m (CuSO 4) \u003d m (p-ra) ω (CuSO 4) \u003d 960 g 0.05 \u003d 48 g; ν (CuSO 4) \u003d m (CuSO 4) / M (CuSO 4) \u003d 48 g / 160 g \u003d 0.3 mol According to the reaction equation (II) ν (CuSO 4) = ν (H 2 S) = 0.3 mol, and according to the reaction equation (III) ν (Al 2 S 3) = 1/3ν (H 2 S) = 0, 1 mol Calculate the masses of aluminum sulfide and copper (II) sulfate: m(Al 2 S 3) \u003d 0.1 mol 150 g / mol \u003d 15 g; m(CuSO4) = 25 g - 15 g = 10 g ω (Al 2 S 3) \u003d 15 g / 25 g 100% \u003d 60%; ω (CuSO 4) \u003d 10 g / 25 g 100% \u003d 40% Task number 34 When burning a sample of some organic compound weighing 14.8 g, 35.2 g of carbon dioxide and 18.0 g of water were obtained. It is known that the relative hydrogen vapor density of this substance is 37. During the study of the chemical properties of this substance, it was found that the interaction of this substance with copper(II) oxide forms a ketone. Based on these conditions of the assignment: 1) make the calculations necessary to establish the molecular formula of organic matter (indicate the units of measurement of the required physical quantities); 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 equation for the reaction of this substance with copper(II) oxide using the structural formula of the substance. On November 14, 2016, the approved demo options, codifiers and specifications of control measuring materials for the unified state exam and the main state exam in 2017, including in chemistry. Demo versions of the exam in chemistry 2016-2015 There were significant changes in KIM in chemistry in 2017, so demo versions of past years are given for review. Chemistry - significant changes: The structure of the examination paper has been optimized: 1. The structure of part 1 of the KIM has been fundamentally changed: tasks with a choice of one answer have been excluded; tasks are grouped into separate thematic blocks, each of which contains tasks of both basic and advanced levels of complexity. 2. Reduced the total number of tasks from 40 (in 2016) to 34. 3. The assessment scale (from 1 to 2 points) for completing tasks of the basic level of complexity, which test the assimilation of knowledge about the genetic relationship of inorganic and organic substances, has been changed (9 and 17). 4. The maximum primary score for the performance of the work as a whole will be 60 points (instead of 64 points in 2016). The duration of the exam in chemistry The total duration of the examination work is 3.5 hours (210 minutes). The approximate time allotted to complete individual tasks is: 1) for each task of the basic level of complexity of part 1 - 2-3 minutes; 2) for each task of an increased level of complexity of part 1 - 5–7 minutes; 3) for each task of a high level of complexity of part 2 - 10–15 minutes.SUBSTANCE FORMULA
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