Unified State Exam Result in chemistry not lower than the minimum established number of points gives the right to enter universities for specialties, where in the list entrance examinations there is a subject of chemistry.
Universities do not have the right to set a minimum threshold for chemistry below 36 points. Prestigious universities tend to set their minimum threshold much higher. Because first-year students must have very good knowledge to study there.
On the official website of the FIPI, each year, variants of the exam in chemistry are published: demo, early period... It is these options that give an idea of the structure of the future exam and the level of complexity of tasks and are sources of reliable information in preparation for the exam.
Early version of the exam in chemistry 2017
| Year | Download early version |
| 2017 | variant po himii |
| 2016 | download |
Demonstration version of the exam in chemistry 2017 from FIPI
| Option of tasks + answers | Download demo version |
| Specification | demo variant himiya ege |
| Codifier | kodifikator |
IN variants of the exam in chemistry in 2017, there are changes compared to the CMM of the last 2016, so it is advisable to conduct training according to the current version, and use the options of previous years for the diverse development of graduates.
Additional materials and equipment
For each option examination work The exam in chemistry is accompanied by the following materials:
- periodic system of chemical elements D.I. Mendeleev;
- table of solubility of salts, acids and bases in water;
- electrochemical series of metal voltages.
During the examination work, it is allowed to use a non-programmable calculator. The list of additional devices and materials, the use of which is permitted for the Unified State Exam, is approved by the order of the Ministry of Education and Science of Russia.
For those who want to continue their education at a university, the choice of subjects should depend on the list of entrance examinations in the chosen specialty
(direction of training).
The list of entrance examinations in universities for all specialties (areas of training) is determined by the order of the Ministry of Education and Science of Russia. Each university chooses 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.
For tasks 1-3, use the following row of chemical elements. The answer in tasks 1–3 is a sequence of numbers, under which the chemical elements in this row are indicated.
1) 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; five
The number of electrons on the external energy level (electronic layer) of the elements of the main subgroups is equal to the group number.
Thus, silicon and carbon are suitable from the options presented. 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 listed in the row, 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; one
Three of the elements presented in one period are sodium Na, silicon Si and magnesium Mg.
When driving within one period Periodic table DI. Mendeleev (horizontal lines) from right to left facilitates the return of electrons located on the outer layer, i.e. the metallic properties of the elements are enhanced. Thus, the metallic properties of sodium, silicon, and magnesium are enhanced in the Si Task number 3 From among the elements listed in the row, select two elements that exhibit the lowest oxidation state of –4. Write down the numbers of the selected elements in the answer field. Answer: 3; five According to the octet rule, atoms of chemical elements tend to have 8 electrons at their outer electronic level, like noble gases. This can be achieved either by giving up electrons of the last level, then the previous one, containing 8 electrons, becomes external, or, conversely, by attaching 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 there is one electron each on the outer electron layer of their atoms. In this regard, the loss of a single electron is energetically more favorable than the addition of seven more. The situation is similar with magnesium, only it is in the main subgroup of the second group, that is, it has two electrons at the external electronic level. It should be noted that sodium, potassium and magnesium belong to 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 release of these electrons and the addition of four more 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 an ionic chemical bond is present. Answer: 1; 3 In the overwhelming majority of cases, it is possible to determine the presence of an ionic type of bond in a compound by the fact that atoms of a typical metal and atoms of a non-metal are simultaneously included in its structural units. On this basis, we establish that there is an ionic bond in the compound under number 1 - Ca (ClO 2) 2, since in its formula you can see the atoms of a 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 sign, the presence of an ionic bond in a compound can be said if its structural unit contains an ammonium cation (NH 4 +) or its organic analogs - alkylammonium cations RNH 3 +, dialkylammonium R 2 NH 2 +, trialkylammonium R 3 NH + and tetraalkylammonium R 4 N +, where R is some hydrocarbon radical. For example, the ionic type of bond occurs in the compound (CH 3) 4 NCl between the cation (CH 3) 4 + and the chloride ion Cl -. Among the compounds specified in the task 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 incomplete replacement of mobile hydrogen atoms with a metal cation, ammonium or alkylammonium cation. In inorganic acids, which take place in 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 the replacement of one of two hydrogen atoms in carbonic acid with an ammonium cation. Basically, acidic salt is a cross between normal (medium) salt and 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 in its molecule not all hydrogen atoms are replaced by metal cations, is a medium, not an acidic salt (!). Hydrogen atoms in organic substances, attached directly to a carbon atom, are practically 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 non-metal oxides 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 salt) in reaction with them. It is often said that non-salt-forming oxides are nonmetal oxides that do not react with bases and basic oxides. However, this approach does not always work for the detection of non-salt-forming oxides. So, for example, CO, being a non-salt-forming oxide, reacts with the basic iron (II) oxide, but with the formation of not 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. All of them are found in the exam 4 - these are CO, NO, N 2 O and SiO (the last SiO I personally never met in tasks). Task number 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. The metal reacts with the salt only if it is more active than that which is part of the salt. The relative activity of metals is determined by a number of metal activity (in another way, a number of metal stresses). Iron in the row of metal activity is located to the right of zinc, 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 to the left of copper in the range of activity, that is, it is a more active metal. Concentrated nitric and concentrated sulfuric acids are not capable of reacting without heating with iron, aluminum and chromium in view of such a phenomenon as passivation: on the surface of these metals under the action of these acids, a salt insoluble without heating is formed, 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, belongs to non-oxidizing acids. Metals in the order of activity to the left of hydrogen react with non-oxidizing acids with the evolution of hydrogen. Iron belongs to such 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 series of metal activities, 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 hydrochloric acid solution, but do not react
with sodium hydroxide solution. Write down the numbers of the selected substances in the answer field. Answer: 3; 4 CO - non-salt-forming oxide, with aqueous solution alkali does not react. (It should be remembered that, nevertheless, under harsh conditions - high pressure and temperature - it nevertheless 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 - belongs to oxides with predominantly basic properties. Reacts with HCl and does not react with sodium hydroxide solution. Fits MgO - magnesium oxide - belongs to the typical basic oxides. 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 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 = t o => N 2 + 2H 2 O + KCl However, the list does not include either nitrites or ammonium salts. This means that one of the three salts (Cu (NO 3) 2, K 2 SO 3 and Na 2 SiO 3) reacts either with an acid (HCl) or with an alkali (NaOH). Among the salts of inorganic acids, only ammonium salts release gas when interacting with alkalis: NH 4 + + OH = NH 3 + H 2 O Ammonium salts, as we said, are not on the list. There is only a variant of the interaction of the salt with the acid. The 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 neither gas, nor sediment, nor 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 = 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 practically any acid, unstable sulfurous acid is formed, which instantly decomposes into colorless gaseous sulfur (IV) oxide and water. Task number 9 Write down the numbers of the selected substances in the table under the appropriate letters. Answer: 2; five 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 acted upon 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 that bicarbonate is a product of incomplete replacement of hydrogen atoms in carbonic acid. To obtain an acidic salt from a normal (average) salt, one must either act on it with the same acid that formed this salt, or else act with an acidic 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 transforms into potassium bicarbonate: K 2 CO 3 + H 2 O + CO 2 = 2KHCO 3 Task number 10 Establish a correspondence between the reaction equation and the property of the nitrogen element that it manifests 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 in the table under the appropriate letters. Answer: A-4; B-2; AT 2; G-1 A) NH 4 HCO 3 - salt, which 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 is converted to 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, no change in the oxidation state of nitrogen occurs; 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 is converted 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 their loss by the nitrogen atom as a result of the reaction. An element that, as a result of the reaction, loses some of its electrons, is called a reducing agent. B) As a result of the reaction of NH 3 with a nitrogen oxidation state of -3, it turns into nitrogen oxide NO. Oxygen almost always has an oxidation state of -2. Therefore, in order for the nitrogen 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 the loss of 5 electrons by the nitrogen atom. That is, nitrogen, as in case 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 structural unit of Li3N 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 in this reaction is an oxidizing agent. 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 down the numbers of the selected substances in the table under the appropriate letters. Answer: A-3; B-2; AT 4; G-1 A) When gaseous hydrogen is passed through the sulfur melt, 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 exam to know exactly how sulfur reacts with chlorine and, accordingly, you do not need to 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 oxidative and reductive. That is, if sulfur is acted upon by a strong oxidizing agent, which is molecular chlorine Cl 2, it will oxidize. Sulfur burns with a blue flame in oxygen with the formation of a gas with a pungent odor - sulfur dioxide SO 2: B) SO 3 - sulfur (VI) oxide has pronounced acidic properties. For such oxides, the most typical reactions are reactions 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 interacts with a basic oxide, a salt of the corresponding acid and the metal that is part of the basic oxide is formed. An acidic oxide corresponds to that acid in which the acid-forming element has the same oxidation state as in the oxide. Sulfuric acid H 2 SO 4 corresponds to SO 3 oxide (both there and there the oxidation state of sulfur is +6). Thus, when SO 3 interacts with metal oxides, sulfuric acid salts - sulfates containing sulfate ion SO 4 2- will be obtained: SO 3 + BaO = BaSO 4 When interacting with water, the acidic oxide turns into the corresponding acid: SO 3 + H 2 O = H 2 SO 4 And when acidic oxides react 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 with both 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 alkalis are water-soluble metal hydroxides: 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 most common. The salt can be reacted with another salt, provided that both starting salts are soluble and a precipitate forms. Also ZnBr 2 contains bromide ion Br-. For metal halides, it is characteristic that they are capable of reacting with Hal 2 halogens, which are higher in the periodic table. Thus? the described types of reactions proceed with all substances of 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 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 down the numbers of the selected substances in the table under the appropriate letters. Answer: A-4; B-2; IN 1 Explanation: A) Methylbenzene, aka 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 you can see from the structural formula, aniline molecule consists of an aromatic hydrocarbon radical (C 6 H 5 -) and an amino group (-NH 2), thus, aniline refers 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 butene-1. Write down the numbers of the selected substances in the answer field. Answer: 2; five Explanation: Isomers are substances that have the same molecular formula and different structural, i.e. substances that differ in the order of connection of atoms, but with the same composition of molecules. Task number 14 From the proposed list, select two substances, when interacting with a solution of potassium permanganate, a change in the color of the solution will be observed. Write down the numbers of the selected substances in the answer field. Answer: 3; five 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, the 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 discolor the purple potassium permanganate solution 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 number 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 with 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 to methanol: The silver mirror reaction is the reaction of reducing silver from an ammoniacal solution of silver oxide. When dissolved in an aqueous solution of ammonia, silver oxide turns into complex compound- hydroxide of diamminesilver (I) OH. After the addition of formaldehyde, a redox reaction takes place, in which silver is reduced: Task number 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; five Explanation: Methylamine is the simplest to represent organic compounds 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, of the proposed answer options, 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 specified substances are substances X and Y. Write down the numbers of the selected substances in the table under the appropriate letters. Answer: 4; 2 Explanation: One of the reactions for obtaining alcohols is the reaction of 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. 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, which is predominantly formed by 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 typical reactions are free radical substitution reactions, during which a hydrogen atom is replaced by a halogen atom. Thus, brominating ethane, one can obtain bromoethane, and by brominating isobutane, 2-bromisobutane: Since the small cycles of the molecules of cyclopropane and cyclobutane are unstable, the cycles of these molecules open upon bromination, thus the addition reaction proceeds: In contrast to the cyclopropane and cyclobutane cycles, the cyclohexane cycle is large, as a result of which the hydrogen atom is replaced by a bromine atom: Task number 19 Establish a correspondence between the reactants 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 number 20 From the proposed list of reaction types, 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; 4 Alkali metals (Li, Na, K, Rb, Cs, Fr) are located in the main subgroup of group I of the table of D.I. Mendeleev and are reducing agents, easily donating an electron located at the external level. If you designate an alkali metal with the letter M, then the reaction of an alkali metal with water will look like this: 2M + 2H 2 O → 2MOH + H 2 Alkali metals are highly reactive 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 aggregate states, then this reaction occurs at the interface, therefore, is heterogeneous. The type of this reaction is substitution. Reactions between inorganic substances are referred to as substitution reactions if a simple substance interacts with a complex one and, as a result, other simple and complex substances are formed. (The neutralization reaction takes place between the acid and the base, as a result of which these substances exchange their constituents and form a salt and a low-dissociating substance). 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 down the numbers of the selected external influences in the answer field. Answer: 1; 4 The following factors affect 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 rule of thumb, for every 10 degrees, the rate constant for 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 fine. As noted above, the reaction rate is also influenced by a change in the concentration of reagents: if the concentration of ethylene is increased, the reaction rate will also increase, which does not correspond to the requirement of the problem. A decrease in the concentration of hydrogen - the initial component, on the contrary, reduces the reaction rate. Therefore, the second option is not suitable, and the fourth is suitable. A catalyst is a substance that accelerates the rate of a chemical reaction, but is not part of the products. The use of a catalyst accelerates the progress of the ethylene hydrogenation reaction, which also does not correspond to the condition of the problem, and therefore is not the correct answer. When ethylene interacts with hydrogen (on Ni, Pd, Pt catalysts), ethane is formed: CH 2 = CH 2 (g) + H 2 (g) → CH 3 -CH 3 (g) All the components participating 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 decrease the pressure in the system. By increasing the pressure, we will speed up the reaction. The fifth answer doesn't fit. Task number 22 Establish a correspondence between the formula of the salt and the electrolysis products of an aqueous solution of this salt, which precipitated 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 when a direct electric current passes through a solution or an electrolyte melt. At the cathode, the reduction of those cations that have the highest oxidative activity occurs predominantly. At the anode, first of all, those anions that have the highest reducing ability are oxidized. Electrolysis of aqueous solution 1) The process of electrolysis of aqueous solutions at the cathode does not depend on the material of the cathode, but depends on the position of the metal cation in the electrochemical series of voltages. For cations in a row Li + - Al 3+ reduction process: 2H 2 O + 2e → H 2 + 2OH - (H 2 is evolved at the cathode) Zn 2+ - Pb 2+ recovery process: Me n + + ne → Me 0 and 2H 2 O + 2e → H 2 + 2OH - (H 2 and Me are released at the cathode) Cu 2+ - Au 3+ reduction process Me n + + ne → Me 0 (Me is released at the cathode) 2) The process of electrolysis of aqueous solutions at the anode depends on the material of the anode and on the nature of the anion. If the anode is insoluble, i.e. is inert (platinum, gold, coal, graphite), the process will depend only on the nature of the anions. For anions F -, SO 4 2-, NO 3 -, PO 4 3-, OH - the oxidation process: 4OH - - 4e → O 2 + 2H 2 O or 2H 2 O - 4e → O 2 + 4H + (oxygen is released at the anode) halide ions (except for F-) oxidation process 2Hal - - 2e → Hal 2 (free halogens are released ) organic acids oxidation process: 2RCOO - - 2e → R-R + 2CO 2 Total electrolysis equation: A) Na 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 number 23 Establish a correspondence between the name of the salt and the ratio of this salt to hydrolysis: for each position marked with a letter, select the corresponding position marked with a number. Write down the selected numbers in the table under the corresponding letters. Answer: 1; 3; 2; 4 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), is hydrolyzed 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. 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 hydrocarbonate ion) The solution medium is alkaline (pH> 7). D) Aluminum sulfide (Al 2 S 3) - a salt formed by weak hydrosulfuric 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 medium is close to neutral (pH ~ 7). Task number 24 Establish a correspondence between the chemical reaction equation 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. EQUATION OF REACTION A) N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) B) 2H 2 (d) + O 2 (d) ↔ 2H 2 O (d) C) H 2 (g) + Cl 2 (g) ↔ 2HCl (g) D) SO 2 (g) + Cl 2 (g) ↔ SO 2 Cl 2 (g) DIRECTION OF DISPLACEMENT OF CHEMICAL EQUILIBRIUM 1) shifts towards a direct reaction 2) shifts in the direction of 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. Displacement of equilibrium in the desired direction is achieved by changing the reaction conditions. Factors determining the position of equilibrium: - pressure: an increase in pressure shifts the equilibrium towards a reaction leading to a decrease in volume (conversely, a decrease in pressure shifts the equilibrium towards a reaction leading to an increase in volume) - temperature: an increase in temperature shifts the equilibrium towards an endothermic reaction (conversely, a decrease in temperature shifts the equilibrium towards an exothermic reaction) - concentration of starting 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 direct 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 displacement of equilibrium, 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 towards 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 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 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 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 number 25 Establish a correspondence between the formulas of substances and a reagent with which you can distinguish between 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 C) 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, and upon interaction with nitric acid forms a soluble salt - calcium nitrate Ca (NO 3) 2, while 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 copper (II) sulfate solution. When copper (II) sulfate interacts with KCl, the exchange reaction does not take place, 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 (base blue). C) Sodium chlorides NaCl and barium 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 take place, the solution contains ions Na +, Cl -, Cu 2+ and SO 4 2-, 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) Chlorides of aluminum 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 is first formed, which then dissolves to form a complex salt - potassium tetrahydroxoaluminate: AlCl 3 + 4KOH → K + 3KCl Task number 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 chemical industry, its production is more than 130 million tons per year. Basically, ammonia is used in the production of nitrogen fertilizers (ammonium nitrate and sulfate, urea), drugs, explosives, nitric acid, soda. Among the proposed answers, the field of application of ammonia is the production of fertilizers (Fourth answer). B) Methane is the simplest hydrocarbon, the most thermally stable representative of a number of limiting compounds. It is widely used as a household and industrial fuel, as well as a raw material for industry (second answer). Methane is 90-98% a constituent part of natural gas. C) Rubbers are materials that are obtained by polymerization of compounds with conjugated double bonds. Isoprene belongs to this type of compound 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 plastics called polyethylene: n CH 2 = CH 2 → (-CH 2 -CH 2 -) n Task number 27 Calculate the mass of potassium nitrate (in grams), which 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 down to tenths.) Answer: 3.4 g Explanation: Let x g be the mass of potassium nitrate, which is dissolved in 150 g of solution. We 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. In this case, the mass of the solution was (150 + x) g. The equation will be written in the form: (Write the number down 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 relate 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 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 us calculate the mass of oxygen required for the complete combustion of hydrogen sulfide: m (O 2) = 0.45 mol 32 g / mol = 14.4 g Task number 30 Using the electronic balance method, write the reaction equation: Na 2 SO 3 +… + KOH → K 2 MnO 4 +… + H 2 O Determine the oxidizing 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 The iron was dissolved in hot concentrated sulfuric acid. The resulting salt was treated with an excess of sodium hydroxide solution. The resulting brown precipitate was filtered off and calcined. The resulting substance was heated with iron. Write the 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 with the release of 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 alkali, as a result of which iron (III) hydroxide (brown compound) precipitates: Fe 2 (SO 4) 3 + 3NaOH → 2Fe (OH) 3 ↓ + 3Na 2 SO 4 3) Insoluble metal hydroxides decompose on 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 number 32 Write down the reaction equations with which you can carry out the following transformations: When writing reaction equations, use the structural formulas of organic substances. 1) 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 through one to the alcoholic 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 elimination of one water molecule from two alcohol molecules. 2) Propylene belongs to unsymmetrical alkenes. When hydrogen halides and water are added, a hydrogen atom is attached to a carbon atom at a multiple bond bonded to a large number hydrogen atoms: CH 2 = 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 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) In an alkaline medium, 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 number 33 Determine the mass fractions (in%) of iron (II) sulfate and aluminum sulfide in the mixture if, when treating 25 g of this mixture with water, a gas was released, which 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 provide all the necessary calculations (indicate the units of measurement of the desired physical quantities). Answer: ω (Al 2 S 3) = 40%; ω (CuSO 4) = 60% When treating a mixture of iron (II) sulfate and aluminum sulfide with water, sulfate simply dissolves, and 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) We calculate the mass and amount of the substance 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 We 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 / 25g · 100% = 60%; ω (CuSO 4) = 10 g / 25g 100% = 40% Task number 34 Burning a sample of some organic compound weighing 14.8 g yielded 35.2 g of carbon dioxide and 18.0 g of water. It is known that the relative density of the vapors of this substance in terms of hydrogen is 37. In the course of the study chemical properties of this substance, it was found that when this substance interacts with copper (II) oxide, a ketone is formed. Based on the given conditions of the assignment: 1) make the calculations necessary to establish the molecular formula of organic matter (indicate the units of measurement of the desired physical quantities); 2) write down the molecular formula of the original organic matter; 3) make up the structural formula of this substance, which unambiguously reflects the order of bonds of atoms in its molecule; 4) write the equation for the reaction of this substance with copper (II) oxide using the structural formula of the substance. 11/14/2016 on the FIPI website the approved demo options, codifiers and specifications of control measuring materials of a single state examination and the main state exam of 2017, including in chemistry. Demo versions of the exam in chemistry 2016-2015 There were significant changes in KIM in chemistry in 2017, therefore, demos of previous years are provided for review. Chemistry - significant changes: The structure of the examination work has been optimized: 1. The structure of part 1 of the CMM has been fundamentally changed: tasks with a choice of one answer are excluded; tasks are grouped into separate thematic blocks, each of which has tasks of both basic and increased difficulty levels. 2. The total number of tasks has been reduced from 40 (in 2016) to 34. 3. Changed the grading scale (from 1 to 2 points) for completing tasks basic level difficulties that test the assimilation of knowledge about the genetic relationship of inorganic and organic substances (9 and 17). 4. Maximum primary score for the performance of the work in total will be 60 points (instead of 64 points in 2016). Duration of the exam in chemistry The total duration of the examination work is 3.5 hours (210 minutes). The approximate time allotted for the completion of individual tasks is: 1) for each task of the basic level of complexity of part 1 - 2-3 minutes; 2) for each task increased level complexity of part 1 - 5–7 minutes; 3) for each task high level complexity of part 2 - 10-15 minutes. Typical test tasks in chemistry contain 10 options for sets of tasks, compiled taking into account all the features and requirements of the Unified State Exam in 2017. The purpose of the manual is to provide readers with information about the structure and content of 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 Specification 1. Appointment of KIM USE The Unified State Exam (hereinafter - the Unified State Exam) is a form of objective assessment of the quality of training of persons who have mastered educational programs of secondary general education, using tasks of a standardized form (control measuring materials). The exam is conducted in accordance with Federal law dated December 29, 2012 No. 273-FZ "On Education in the Russian Federation". Control measuring materials allow you to establish the level of development of graduates of the Federal component state standard secondary (complete) general education in chemistry, basic and specialized levels. The results of the unified state exam in chemistry are recognized by educational organizations of secondary vocational education and educational organizations of higher professional education as the results of entrance examinations in chemistry. 2. Documents defining the content of the KIM USE 3. Approaches to the selection of content, the development of the structure of the CIM USE The basis of approaches to the development of the CIM USE 2017 in chemistry was formed by those general methodological guidelines that were determined during the formation of examination models previous years... The essence of these settings is as follows. 4. The structure of the KIM USE Each version of the examination work is built according to a single plan: the work consists of two parts, including 40 tasks. Part 1 contains 35 tasks with a short answer, including 26 tasks of a basic level of difficulty (ordinal numbers of these tasks: 1, 2, 3, 4, ... 26) and 9 tasks of an increased level of difficulty (ordinal numbers of these tasks: 27, 28, 29, ... 35). Part 2 contains 5 tasks of a high level of complexity, with a detailed answer (ordinal numbers of these tasks: 36, 37, 38, 39, 40).FORMULA OF SUBSTANCE
REAGENTS
Demo version of the exam in chemistry 2017 with answers
Option of tasks + answers
Download demo
Specification
demo variant himiya ege
Codifier
kodifikator
Chemistry
Download demo + answers
2016
ege 2016
2015
ege 2015
The collection provides answers to all test options and provides solutions to all tasks of one of the options. In addition, there are samples of forms used on the exam to record answers and decisions.
The author of the assignments is a leading scientist, teacher and methodologist who is directly involved in the development of control measuring materials for the exam.
The manual is intended for teachers to prepare students for the chemistry exam, as well as high school students and graduates - for self-preparation and self-control.
There are chemical bonds in ammonium chloride:
1) ionic
2) covalent polar
3) covalent non-polar
4) hydrogen
5) metal
1) zinc chloride (solution)
2) sodium sulfate (solution)
3) diluted nitric acid
4) concentrated sulfuric acid
5) aluminum oxide
Foreword
Work instructions
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 the problems of option 10
Part 1
Part 2.
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