New exam guidelines in chemistry. The structure of the examination paper consists of two blocks
Unified State Exam result in chemistry not lower than the minimum established number of points gives the right to enter universities in specialties where in the list entrance examinations there is a chemistry subject.
Universities do not have the right to set the minimum threshold for chemistry below 36 points. Prestigious universities, as a rule, set their minimum threshold much higher. Because to study there, first-year students must have very good knowledge.
On the official website of FIPI, versions of the Unified State Examination in chemistry are published every year: demonstration, early period. It is these options that give an idea of the structure of the future exam and the level of difficulty of the tasks and are sources of reliable information when preparing for the Unified State Exam.
Early version of the Unified State Exam in Chemistry 2017
| Year | Download early version |
| 2017 | variant po himii |
| 2016 | download |
Demo version of the Unified State Exam in Chemistry 2017 from FIPI
| Variant of tasks + answers | Download demo version |
| Specification | demo variant himiya ege |
| Codifier | codifier |
IN Unified State Exam options in chemistry in 2017 there are changes compared to the CMM of the previous 2016, so it is advisable to conduct training according to the current version, and for the diversified development of graduates to use the versions of previous years.
Additional materials and equipment
For each option exam paper The Unified State Exam in Chemistry is accompanied by the following materials:
− periodic system chemical elements DI. Mendeleev;
− table of solubility of salts, acids and bases in water;
− electrochemical series of metal voltages.
You are permitted to use a non-programmable calculator during the examination. The list of additional devices and materials, the use of which is permitted for the Unified State Examination, is approved by order of the Russian Ministry of Education and Science.
For those who want to continue their education at a university, the choice of subjects should depend on the list of entrance tests for the chosen specialty
(direction of training).
The list of entrance examinations at universities for all specialties (areas of training) is determined by order of the Russian Ministry of Education and Science. Each university selects from this list certain subjects that it indicates in its admission rules. You need to familiarize yourself with this information on the websites of the selected universities before applying for participation in the Unified State Exam with a list of selected subjects.
To complete tasks 1–3, use the following series of chemical elements. The answer in tasks 1–3 is the sequence of numbers under which the chemical elements in this series.
1) Na 2) K 3) Si 4) Mg 5) C
Task No. 1
Determine which atoms of the elements indicated in the series have four electrons at the outer 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 answer options, silicon and carbon are suitable, because they are in the main subgroup of the fourth group of the D.I. table. Mendeleev (IVA group), i.e. Answers 3 and 5 are correct.
Task No. 2
From the chemical elements indicated in the series, select three elements that are in Periodic table chemical elements D.I. Mendeleev are in the same period. Arrange the selected elements in ascending order of their metallic properties.
Write down the numbers of the selected elements in the required sequence in the answer field.
Answer: 3; 4; 1
Of the presented elements, three are found in one 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 transfer 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 increase in the Si series Task No. 3 From among the elements indicated in the series, 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, 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 from the last level, then the previous one, containing 8 electrons, becomes external, or, conversely, by adding additional electrons up to eight. Sodium and potassium belong to the alkali metals and are in the main subgroup of the first group (IA). This means that there is one electron each in the outer electron layer of their atoms. In this regard, it is energetically more favorable to lose a single electron than to gain seven more. The situation with magnesium is similar, only it is in the main subgroup of the second group, that is, it has two electrons at the outer electronic level. It should be noted that sodium, potassium and magnesium are metals, and a negative oxidation state is in principle impossible for metals. 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 their outer electron layer contains 4 electrons. For this reason, for these elements it is possible to both give up these electrons and add four more to a total of 8. Silicon and carbon atoms cannot add more than 4 electrons, so the minimum oxidation state for them is -4. Task No. 4 From the list provided, select two compounds that contain 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 atoms of a non-metal. Based on this feature, we establish that there is an ionic bond in compound number 1 - Ca(ClO 2) 2, because in its formula you can see atoms of the typical metal calcium 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 characteristic, the presence of an ionic bond in a compound can be said if its structural unit contains an ammonium cation (NH 4 +) or its organic analogues - alkylammonium cations RNH 3 +, dialkylammonium R 2 NH 2 +, trialkylammonium cations R 3 NH + and tetraalkylammonium R 4 N +, where R is some hydrocarbon radical. For example, the ionic type of bond occurs in the compound (CH 3) 4 NCl between the cation (CH 3) 4 + and the chloride ion Cl −. Among the compounds indicated in the task is ammonium chloride, in which the ionic bond is realized between the ammonium cation NH 4 + and the chloride ion Cl − . Task No. 5 Establish a correspondence between the formula of a substance and the class/group to which this substance belongs: for each position indicated by a letter, select the corresponding position 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 salts obtained as a result of incomplete replacement of mobile hydrogen atoms with a metal cation, ammonium or alkylammonium cation. In inorganic acids, which are taught 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 are ammonium bicarbonate NH 4 HCO 3 - the product of replacing one of the two hydrogen atoms in carbonic acid with an ammonium cation. Essentially, an acidic salt is a cross between a normal (average) salt and an acid. In the case of NH 4 HCO 3 - the average between normal salt (NH 4) 2 CO 3 and carbonic acid H2CO3. 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 in its molecule not all hydrogen atoms are replaced by metal cations, is an average and not an acidic salt (!). Hydrogen atoms in organic substances attached directly to a carbon atom are almost never able to be replaced by metal atoms, with the exception of hydrogen atoms at a 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, 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, this approach does not always work for identifying non-salt-forming oxides. For example, CO, being a non-salt-forming oxide, reacts with basic iron (II) oxide, but not to form a salt, but a free metal: CO + FeO = CO 2 + Fe Non-salt-forming oxides from the school chemistry course include oxides of non-metals in the oxidation state +1 and +2. In total, they are found in the Unified State Exam 4 - these are CO, NO, N 2 O and SiO (I personally have never encountered the latter SiO in tasks). Task No. 6 From the proposed list of substances, select two substances with each of which iron reacts without heating. Answer: 2; 4 Zinc chloride is a salt, and iron is a metal. A metal reacts with salt only if it is more reactive than the one in the salt. The relative activity of metals is determined by the series of metal activities (in other words, the series of metal voltages). Iron is located to the right of zinc in the activity series of metals, which means 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 occur. Copper (II) sulfate CuSO 4 will react with iron, since iron is to the left of copper in the activity series, that is, it is a more active metal. Concentrated nitric and concentrated sulfuric acids are not able to react with iron, aluminum and chromium without heating due to a phenomenon called passivation: on the surface of these metals, under the influence of these acids, a salt insoluble without heating is formed, which acts as a protective shell. However, when heated, this protective coating 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, is a non-oxidizing acid. Metals that are to the left of hydrogen in the activity series react with non-oxidizing acids and release hydrogen. Iron is one of these metals. Conclusion: the reaction of iron with hydrochloric acid occurs. In the case of a metal and a metal oxide, a 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 No. 7 From the proposed list, select two oxides that react with hydrochloric acid solution, but don't react
with sodium hydroxide solution. Write down the numbers of the selected substances in the answer field. Answer: 3; 4 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) - acidic oxide, which corresponds to sulfuric acid. Acidic oxides do not react with acids and other acidic oxides. That is, SO 3 does not react with hydrochloric acid and reacts with a base - sodium hydroxide. Doesn't fit. 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, readily reacts with both strong bases and acids (as well as acidic and basic oxides). Doesn't fit. Task No. 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 =t o => N 2 + 2H 2 O + KCl However, the list does not include both nitrites and ammonium salts. 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 = 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 salt with 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 occur, because no gas, no precipitate, no slightly dissociating substance (water or weak acid) is formed. Sodium silicate reacts with hydrochloric acid, but due to the release of a white gelatinous precipitate of silicic acid, rather than gas: Na 2 SiO 3 + 2HCl = 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 No. 9 Write down the numbers of the selected substances under the corresponding letters in the table. Answer: 2; 5 CO 2 is an acidic oxide and must be treated with either a basic oxide or a base to convert it into 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 = K 2 CO 3 Potassium bicarbonate KHCO 3, like potassium carbonate, is a salt of carbonic acid, with the only difference being that bicarbonate is a product of incomplete replacement of hydrogen atoms in carbonic acid. To obtain an acid salt from a normal (average) salt, you must either treat it with the same acid that formed this salt, or treat it with an acidic oxide corresponding to this acid in the presence of water. Thus, reactant Y is carbon dioxide. When passing it through an aqueous solution of potassium carbonate, the latter transforms into potassium bicarbonate: K 2 CO 3 + H 2 O + CO 2 = 2KHCO 3 Task No. 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 down the numbers of the selected substances under the corresponding letters in the table. Answer: A-4; B-2; AT 2; G-1 A) NH 4 HCO 3 is a salt that contains 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 an 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, i.e. it does not exhibit redox properties. B) As shown above, nitrogen in ammonia NH 3 has an oxidation state of -3. As a result of the reaction with CuO, ammonia turns into a simple substance N 2. In any simple substance, the oxidation state of the element by which it is formed is zero. Thus, the nitrogen atom loses its negative charge, and since electrons are responsible for the negative charge, this means that the nitrogen atom loses them as a result of the reaction. An element that loses some of its electrons as a result of a reaction is called a reducing agent. C) As a result of the reaction of NH 3 with the oxidation state of nitrogen equal to -3, it turns into nitric oxide NO. Oxygen almost always has an oxidation state of -2. Therefore, in order for a nitric oxide molecule to be electrically neutral, the nitrogen atom must have an oxidation state of +2. This means that the nitrogen atom as a result of the reaction changed its oxidation state from -3 to +2. This indicates that the nitrogen atom has lost 5 electrons. That is, nitrogen, as is the case with 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 (oxidation state 0 occurs for any element) 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 an oxidizing agent in this reaction. Task No. 11 Establish a correspondence between the formula of a substance and the reagents with each of which this substance can interact: for each position indicated by a letter, select the corresponding position indicated by a number. 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 down the numbers of the selected substances under the corresponding letters in the table. Answer: A-3; B-2; AT 4; G-1 A) When hydrogen gas is passed through molten sulfur, hydrogen sulfide H 2 S is formed: H 2 + S =t o => H 2 S When chlorine is passed over crushed sulfur at room temperature, sulfur dichloride is formed: S + Cl 2 = SCl 2 For passing the Unified State Exam you don’t 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 sulfur is exposed to a strong oxidizing agent, which is molecular chlorine Cl2, it will oxidize. Sulfur burns with a blue flame in oxygen to form a gas with a pungent odor - sulfur dioxide SO2: B) SO 3 - sulfur oxide (VI) has pronounced acidic properties. For such oxides, the most characteristic reactions are reactions with water, as well as with basic and amphoteric oxides and hydroxides. In the list at number 2 we see water, the main oxide BaO, and the hydroxide KOH. When an acidic oxide interacts with a basic oxide, a salt of the corresponding acid and the metal that is part of the basic oxide is formed. Any acidic oxide corresponds to the acid in which acid-forming element has the same oxidation state as in the oxide. The oxide SO 3 corresponds to sulfuric acid H 2 SO 4 (in both cases, 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 reacting with water, an acidic oxide is converted into the corresponding acid: SO 3 + H 2 O = H 2 SO 4 And when acidic oxides interact with metal hydroxides, a salt of the corresponding acid and water are formed: SO 3 + 2KOH = 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 acid, and alkali - LiOH. Let us recall that alkalis are metal hydroxides soluble in water: Zn(OH) 2 + 2HBr = ZnBr 2 + 2H 2 O Zn(OH) 2 + 2CH 3 COOH = Zn(CH 3 COO) 2 + 2H 2 O Zn(OH) 2 + 2LiOH = 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 salts are soluble and a precipitate is formed. ZnBr 2 also contains bromide ion Br-. It is characteristic of metal halides that they are capable of reacting with Hal 2 halogens, which are higher in the periodic table. Thus? the described types of reactions occur with all substances in list 1: ZnBr 2 + 2AgNO 3 = 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 No. 12 Establish a correspondence between the name of a substance and the class/group to which this substance belongs: for each position indicated by a letter, select the corresponding position indicated by a number. Write down the numbers of the selected substances under the corresponding letters in the table. 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) is a hydrocarbon 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 the aromatic amines, i.e. correct answer 2. B) 3-methylbutanal. The ending “al” indicates that the substance is an aldehyde. Structural formula of this substance: Task No. 13 From the proposed list, select two substances that are structural isomers of 1-butene. 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 a different structure, i.e. substances that differ in the order of connection of atoms, but with the same composition of molecules. Task No. 14 From the proposed list, select two substances that, when interacting 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 are eliminated - when adding cyclohexane or propane to an aqueous solution of potassium permanganate, no color change will occur. Among hydrocarbons homologous series benzene, only benzene is passive to the action of aqueous solutions of oxidizing agents; all other homologues are oxidized, depending on the environment, 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 discolor the purple solution of potassium permanganate due to the following reactions: CH 3 -CH=CH 2 + 2KMnO 4 + 2H 2 O → CH 3 -CH(OH)–CH 2 OH + 2MnO 2 + 2KOH Task No. 15 From the list provided, select two substances with which formaldehyde reacts. Write down the numbers of the selected substances in the answer field. Answer: 3; 4 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 occurring along the functional group. Among the list of answers for formaldehyde, reduction reactions are characteristic, where hydrogen is used as a reducing agent (cat. – Pt, Pd, Ni), and oxidation – in this case, the reaction of a silver mirror. When reduced with hydrogen on a nickel catalyst, formaldehyde is converted into methanol: The silver mirror reaction is the reduction reaction of silver from an ammonia solution of silver oxide. When dissolved in an aqueous solution of ammonia, silver oxide turns into complex compound– diammine silver hydroxide (I) OH. After adding formaldehyde, a redox reaction occurs in which silver is reduced: Task No. 16 From the list provided, 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. Characteristic feature 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 No. 17 The following scheme of substance transformations is specified: Determine which of the indicated substances are substances X and Y. Write down the numbers of the selected substances under the corresponding letters in the table. Answer: 4; 2 Explanation: One of the reactions for producing alcohols is the hydrolysis reaction of haloalkanes. Thus, ethanol can be obtained from chloroethane by treating 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 No. 18 Establish a correspondence between the name of the substance and the product, which is mainly formed when this substance reacts with bromine: for each position indicated by a letter, select the corresponding position indicated by a number. 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 you can get bromoethane, and by brominating isobutane you can get 2-bromoisobutane: Since the small rings of cyclopropane and cyclobutane molecules are unstable, during bromination the rings of these molecules open, thus an addition reaction occurs: In contrast to the cycles of cyclopropane and cyclobutane, the cyclohexane cycle is large, resulting in the replacement of a hydrogen atom with a bromine atom: Task No. 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 down the selected numbers in the table under the corresponding letters. Answer: 5; 4; 6; 2 Task No. 20 From the proposed list of reaction types, select two reaction types, which include the interaction of alkali metals with water. Write down the numbers of the selected reaction types in the answer field. Answer: 3; 4 Alkali metals (Li, Na, K, Rb, Cs, Fr) are located in the main subgroup of group I of the D.I. table. Mendeleev and are reducing agents, easily donating an electron located on the outer level. If we denote the alkali metal by 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 reactive towards water. The reaction proceeds rapidly 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 in different states of aggregation, this reaction occurs at the phase boundary and, therefore, 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 components and a salt and a slightly 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 No. 21 From the proposed list of external influences, select two influences that lead to a decrease in the rate of reaction of ethylene with hydrogen. Write down the numbers of the selected external influences in the answer field. Answer: 1; 4 For speed chemical reaction the following factors influence: changes in temperature and concentration of reagents, as well as the use of a catalyst. According to Van't Hoff's rule of thumb, with every 10 degree increase in temperature, the rate constant of a homogeneous reaction increases by 2-4 times. Consequently, 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 changes 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 task. A decrease in the concentration of hydrogen, the starting component, on the contrary, reduces the reaction rate. Therefore, the second option is not suitable, but the fourth is suitable. A catalyst is a substance that accelerates the rate of a chemical reaction, but is not part of the product. The use of a catalyst accelerates the reaction of ethylene hydrogenation, which also does not correspond to the conditions of the problem, and therefore is not the correct answer. When ethylene reacts with hydrogen (on Ni, Pd, Pt catalysts), ethane is formed: CH 2 =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 is to reduce the pressure in the system. By increasing the pressure, we will speed up the reaction. The fifth answer is not correct. Task No. 22 Establish a correspondence between the formula of the salt and the electrolysis products of an aqueous solution of this salt, which were released on the inert electrodes: to each position, SALT FORMULA ELECTROLYSIS PRODUCTS Write down the selected numbers in the table under the corresponding letters. Answer: 1; 4; 3; 2 Electrolysis is a redox process that occurs on the electrodes during the passage of a constant electric current through a solution or molten electrolyte. At the cathode, the reduction of those cations that have the greatest oxidative activity occurs predominantly. At the anode, those anions that have the greatest reducing ability are oxidized first. Electrolysis of aqueous solution 1) The process of electrolysis of aqueous solutions at the cathode does not depend on the cathode material, but depends on the position of the metal cation in the electrochemical voltage series. For cations in a series 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 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 anode material and the nature of the anion. If the anode is insoluble, i.e. inert (platinum, gold, coal, graphite), then the process will depend only on the nature of the anions. For anions F − , SO 4 2- , NO 3 − , PO 4 3- , OH − 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 acid oxidation process: 2RCOO − - 2e → R-R + 2CO 2 Summary equation electrolysis: 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) B) 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 No. 23 Establish a correspondence between the name of the salt and the relationship of this salt to hydrolysis: for each position indicated by a letter, select the corresponding position indicated by a number. Write down the selected numbers in the table under the corresponding letters. Answer: 1; 3; 2; 4 Hydrolysis of salts is the interaction of salts with water, leading to the addition of the hydrogen cation H + of a water molecule to the anion of the acid residue and (or) the hydroxyl group OH - of a 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) is a salt formed by strong hydrochloric acid and ammonia ( weak foundation), undergoes hydrolysis at 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 environment is acidic (pH< 7). B) Potassium sulfate (K 2 SO 4) - a salt formed by strong sulfuric acid and potassium hydroxide (alkali, i.e. a 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 weak carbonic acid and sodium hydroxide (alkali, i.e. a strong base), undergoes hydrolysis at the anion. CO 3 2- + H 2 O → HCO 3 - + OH - (formation of weakly dissociating bicarbonate ion) The solution medium 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 to form aluminum hydroxide and hydrogen sulfide: Al 2 S 3 + 6H 2 O → 2Al(OH) 3 + 3H 2 S The solution environment is close to neutral (pH ~ 7). Task No. 24 Establish a correspondence between the equation of a chemical reaction and the direction of displacement of the 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) B) H 2 (g) + Cl 2 (g) ↔ 2HCl (g) D) SO 2 (g) + Cl 2 (g) ↔ SO 2 Cl 2 (g) DIRECTION OF CHEMICAL EQUILIBRIUM SHIFT 1) shifts towards the direct reaction 2) shifts towards the reverse reaction 3) there is no shift in equilibrium Write down the selected numbers in the table under the corresponding letters. Answer: A-1; B-1; AT 3; G-1 The reaction is in chemical equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction. Shifting the equilibrium in the desired direction is achieved by changing the reaction conditions. Factors determining the equilibrium position: - 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 shifts the equilibrium towards the forward reaction (conversely, a decrease in the concentration of the starting substances and an increase in the reaction products shifts the equilibrium towards the reverse reaction) - catalysts do not affect the shift in equilibrium, but only accelerate its achievement A) In the first case, the reaction occurs 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 (towards the direct reaction). B) In the second case, the reaction also occurs 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 towards the direct reaction (towards the product). C) In the third case, the pressure does not change during the reaction, because V(H 2) + V(Cl 2) = 2V(HCl), so the equilibrium does not shift. D) In the fourth case, the reaction also occurs 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 No. 25 Establish a correspondence between the formulas of substances and the reagent with which you can distinguish their aqueous solutions: for each position indicated by a letter, select the corresponding position indicated by a number. FORMULAS OF SUBSTANCES A) HNO 3 and H 2 O B) NaCl and BaCl 2 D) AlCl 3 and MgCl 2 Write down the selected numbers in the table under the corresponding letters. Answer: A-1; B-3; AT 3; G-2 A) Nitric acid and water can be distinguished using a salt - calcium carbonate CaCO 3. Calcium carbonate does not dissolve in water, but when interacting with nitric acid forms a soluble salt - calcium nitrate Ca(NO 3) 2, and the reaction is accompanied by the release of 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 occur; the solution contains ions K +, Cl -, Cu 2+ and SO 4 2-, which do not form low-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 (the base blue color). C) Sodium chloride NaCl and barium chloride BaCl 2 are soluble salts that can also be distinguished by a solution of copper (II) sulfate. When copper (II) sulfate interacts with NaCl, the exchange reaction does not occur; the solution contains Na +, Cl -, Cu 2+ and SO 4 2- ions, which do not form low-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 chlorides AlCl 3 and magnesium chlorides 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 reacts with aluminum chloride, a precipitate is first formed, which then dissolves to form a complex salt - potassium tetrahydroxoaluminate: AlCl 3 + 4KOH → K + 3KCl Task No. 26 Establish a correspondence between the substance and its area of application: for each position indicated by a letter, select the corresponding position indicated by a number. Write down the selected numbers in the table under the corresponding letters. 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, soda. Among the proposed answer options, 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 domestic and industrial fuel, as well as raw material for industry (Second answer). Methane is 90-98% a component of natural gas. C) Rubbers are materials obtained by polymerization of compounds with conjugated double bonds. Isoprene is one of these types of compounds and is used to produce one of the types of rubbers: D) Low molecular weight alkenes are used to produce plastics, in particular ethylene is used to produce a plastic called polyethylene: n CH 2 =CH 2 → (-CH 2 -CH 2 -) n Task No. 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 the number to the nearest tenth.) Answer: 3.4 g Explanation: Let x g be the mass of potassium nitrate that is dissolved in 150 g of solution. Let's calculate the mass of potassium nitrate dissolved in 150 g of solution: m(KNO 3) = 150 g 0.1 = 15 g In order for the mass fraction of salt to be 12%, x g of potassium nitrate was added. The mass of the solution was (150 + x) g. We write the equation in the form: (Write the number to the nearest tenth.) Answer: 14.4 g Explanation: As a result of 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 relate to each other in exactly the same way: V(O 2) = 3/2V(H 2 S), V(O 2) = 3/2 · 6.72 l = 10.08 l, hence V(O 2) = 10.08 l/22.4 l/mol = 0.45 mol Let's calculate the mass of oxygen required for complete combustion of hydrogen sulfide: m(O 2) = 0.45 mol 32 g/mol = 14.4 g Task No. 30 Using method electronic balance, write the reaction equation: Na 2 SO 3 + … + KOH → K 2 MnO 4 + … + H 2 O Identify the oxidizing agent and the 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 No. 31 Iron was dissolved in hot concentrated sulfuric acid. The resulting salt was treated with an excess of sodium hydroxide solution. The brown precipitate that formed was filtered and calcined. The resulting substance was heated with iron. Write equations for the four reactions described. 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, releasing sulfur dioxide: 2Fe + 6H 2 SO 4 → Fe 2 (SO 4) 2 + 3SO 2 + 6H 2 O (when heated) 2) Iron (III) sulfate is a water-soluble salt that enters into an exchange reaction with an alkali, as a result of which iron (III) hydroxide precipitates (a 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 (when heated) Task No. 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 abstraction of a hydrogen atom from the carbon atom of the alcohol, located one after the other to the alcohol hydroxyl (in the β-position). CH 3 -CH 2 -CH 2 -OH → CH 2 =CH-CH 3 + H 2 O (conditions - H 2 SO 4, 180 o C) Intermolecular dehydration occurs at temperatures below 140 o C under the action of sulfuric acid and ultimately comes down to the splitting of one water molecule from two alcohol molecules. 2) Propylene is an unsymmetrical alkene. When adding hydrogen halides and water, a hydrogen atom is added to the carbon atom at the multiple bond associated with a large number hydrogen atoms: CH 2 =CH-CH 3 + HCl → CH 3 -CHCl-CH 3 3) By treating 2-chloropropane with an aqueous solution of NaOH, 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 temperatures above 140 o C: CH 3 -CH(OH)-CH 3 → CH 2 =CH-CH 3 + H 2 O (conditions H 2 SO 4, 180 o C) 5) B alkaline environment acting with a dilute aqueous solution of potassium permanganate, hydroxylation of alkenes occurs with the formation of diols: 3CH 2 =CH-CH 3 + 2KMnO 4 + 4H 2 O → 3HOCH 2 -CH(OH)-CH 3 + 2MnO 2 + 2KOH Task No. 33 Define mass fractions(in%) of iron (II) sulfide and aluminum sulfide in a mixture, if when treating 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 problem statement and provide 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 sulfide simply dissolves and the sulfide hydrolyzes 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) Let's calculate the mass and amount of dissolved copper(II) sulfate: m(CuSO 4) = m(solution) ω(CuSO 4) = 960 g 0.05 = 48 g; ν(CuSO 4) = m(CuSO 4)/M(CuSO 4) = 48 g/160 g = 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 mole Let's calculate the masses of aluminum sulfide and copper (II) sulfate: m(Al 2 S 3) = 0.1 mol · 150 g/mol = 15 g; m(CuSO4) = 25 g – 15 g = 10 g ω(Al 2 S 3) = 15 g/25 g 100% = 60%; ω(CuSO 4) = 10 g/25 g 100% = 40% Task No. 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 vapor density of this substance with respect to hydrogen is 37. During the study chemical properties of this substance, it has been established that when this substance interacts with copper(II) oxide, a ketone is formed. Based on the data of the task conditions: 1) make the calculations necessary to establish the molecular formula of an organic substance (indicate the units of measurement of the required physical quantities); 2) write down the molecular formula of the original organic substance; 3) draw up a structural formula of this substance, which unambiguously reflects the order of bonds of atoms in its molecule; 4) write the equation for the reaction of this substance with copper(II) oxide using the structural formula of the substance. In 2-3 months it is impossible to learn (repeat, improve) such a complex discipline as chemistry. There are no changes to the 2020 Unified State Exam KIM in chemistry.
Don't put off preparing for later. Total: 60 points. 3.5 hours (210 minutes) are allotted to complete the examination paper in chemistry. There will be three cheat sheets on the exam. And you need to understand them This is 70% of the information that will help you pass the chemistry exam successfully. The remaining 30% is the ability to use the provided cheat sheets. Dare, try and you will succeed! Typical test tasks in chemistry contain 10 versions of sets of tasks, compiled taking into account all the features and requirements of the Unified State Exam in 2017. The purpose of the manual is to provide readers with information about the structure and content of the 2017 KIM in chemistry, the degree of difficulty of the tasks. Examples. From the proposed list of substances, select two substances with each of which copper reacts. CONTENT 1.
The approach to structuring part 1 of the examination paper will be fundamentally changed. It is assumed that, unlike examination model past years, the structure of Part 1 of the work will include several thematic blocks, each of which will present tasks of both basic and advanced levels of complexity. Within each thematic block, tasks will be arranged in increasing order of the number of actions required to complete them. Thus, the structure of Part 1 of the examination paper will be more consistent with the structure of the chemistry course itself. This structuring of Part 1 of KIM will help examinees during work to more effectively concentrate their attention on the use of what knowledge, concepts and laws of chemistry and in what relationship the completion of tasks that test mastery will require educational material a certain section of a chemistry course. 2.
There will be noticeable changes in approaches to task design basic level difficulties. These can be tasks with a single context, with the choice of two correct answers out of five, three out of six, tasks “to establish correspondence between the positions of two sets,” as well as calculation tasks. 3.
Increasing the differentiating ability of tasks makes it objective to raise the question of reducing the total number of tasks in the examination paper. It is expected that the total number of exam tasks will be reduced from 40 to 34. This will be done mainly by streamlining the optimal number of those tasks, the implementation of which involved the use of similar types of activities. An example of such tasks, in particular, are tasks aimed at testing the chemical properties of salts, acids, bases, and the conditions for ion exchange reactions. 4.
A change in the format of tasks and their number will inevitably be associated with an adjustment in the grading scale for some tasks, which, in turn, will cause a change in the primary total score for completing the work as a whole, presumably in the range from 58 to 60 (instead of the previous 64 points). The consequence of the planned changes in the examination model as a whole should be an increase in the objectivity of testing the formation of a number of subject and meta-subject skills, which are an important indicator of the success of mastering the subject. We are talking, in particular, about such skills as: applying knowledge in the system, combining knowledge about chemical processes with an understanding of the mathematical relationship between various physical quantities, independently assess the correctness of the implementation of educational and educational-practical tasks, etc.FORMULA OF THE SUBSTANCE
REAGENTS
Points for each chemistry task
Structure of the examination paper consists of two blocks:
The collection contains answers to all test options and provides solutions to all tasks of one of the options. In addition, samples of forms used in the Unified State Exam for recording answers and solutions are provided.
The author of the assignments is a leading scientist, teacher and methodologist who is directly involved in the development of tests measuring materials Unified State Exam.
The manual is intended for teachers to prepare students for the chemistry exam, as well as for high school students and graduates - for self-preparation and self-control.
Ammonium chloride contains chemical bonds:
1) ionic
2) covalent polar
3) covalent non-polar
4) hydrogen
5) metal
1) zinc chloride (solution)
2) sodium sulfate (solution)
3) dilute nitric acid
4) concentrated sulfuric acid
5) aluminum oxide
Preface
Instructions for performing the work
OPTION 1
Part 1
Part 2
OPTION 2
Part 1
Part 2
OPTION 3
Part 1
Part 2
OPTION 4
Part 1
Part 2
OPTION 5
Part 1
Part 2
OPTION 6
Part 1
Part 2
OPTION 7
Part 1
Part 2
OPTION 8
Part 1
Part 2
OPTION 9
Part 1
Part 2
OPTION 10
Part 1
Part 2
ANSWERS AND SOLUTIONS
Answers to the tasks of part 1
Solutions and answers to the tasks of part 2
Solving problems of option 10
Part 1
Part 2.
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The following changes will be made in the 2017 Unified State Exam KIMs: