Determine which atoms of the elements indicated in the series have four electrons on the external energy level.
Answer: 3; 5
The number of electrons in the outer energy level (electronic layer) of the elements of the main subgroups is equal to the group number.
Thus, from the presented answers, silicon and carbon are suitable, because. they are in the main subgroup of the fourth group of the table D.I. Mendeleev (IVA group), i.e. Answers 3 and 5 are correct.
Of those in the series chemical elements select the three elements that are in Periodic system chemical elements D.I. Mendeleev are in the same period. Arrange the selected elements in ascending order of their metallic properties.
Write in the answer field the numbers of the selected elements in the desired sequence.
Answer: 3; 4; 1
Of the presented elements, three are in the same period - these are sodium Na, silicon Si and magnesium Mg.
When moving within one period of the Periodic Table, D.I. Mendeleev (horizontal lines) from right to left, the return of electrons located on the outer layer is facilitated, i.e. the metallic properties of the elements are enhanced. Thus, the metallic properties of sodium, silicon and magnesium are enhanced in the series Si From among the elements listed in the row, select two elements that exhibit the lowest oxidation state, equal to -4. Write down the numbers of the selected elements in the answer field. Answer: 3; 5 According to the octet rule, the atoms of chemical elements tend to have 8 electrons in their outer electronic level, like the noble gases. This can be achieved either by donating electrons of the last level, then the previous one, containing 8 electrons, becomes external, or, conversely, by adding additional electrons up to eight. Sodium and potassium are alkali metals and are in the main subgroup of the first group (IA). This means that on the outer electron layer of their atoms there is one electron each. In this regard, the loss of a single electron is energetically more favorable than the addition of seven more. With magnesium, the situation is similar, only it is in the main subgroup of the second group, that is, it has two electrons on the outer electronic level. It should be noted that sodium, potassium and magnesium are metals, and for metals, in principle, a negative oxidation state is impossible. The minimum oxidation state of any metal is zero and is observed in simple substances. The chemical elements carbon C and silicon Si are non-metals and are in the main subgroup of the fourth group (IVA). This means that there are 4 electrons on their outer electron layer. For this reason, for these elements, both the return of these electrons and the addition of four more up to a total of 8 are possible. Silicon and carbon atoms cannot attach more than 4 electrons, therefore the minimum oxidation state for them is -4. From the proposed list, select two compounds in which there is an ionic chemical bond. Answer: 1; 3 In the vast majority of cases, the presence of an ionic type of bond in a compound can be determined by the fact that its structural units simultaneously include atoms of a typical metal and non-metal atoms. On this basis, we establish that there is an ionic bond in compound number 1 - Ca (ClO 2) 2, because in its formula, one can see the atoms of the typical calcium metal and the atoms of non-metals - oxygen and chlorine. However, there are no more compounds containing both metal and non-metal atoms in this list. In addition to the above feature, the presence of an ionic bond in a compound can be said if its structural unit contains an ammonium cation (NH 4 +) or its organic analogs - cations of alkylammonium RNH 3 +, dialkylammonium R 2 NH 2 + , trialkylammonium R 3 NH + and tetraalkylammonium R 4 N + , where R is some hydrocarbon radical. For example, the ionic type of bond takes place in the compound (CH 3) 4 NCl between the cation (CH 3) 4 + and the chloride ion Cl - . Among the compounds indicated in the assignment there is ammonium chloride, in which the ionic bond is realized between the ammonium cation NH 4 + and the chloride ion Cl − . Establish a correspondence between the formula of a substance and the class / group to which this substance belongs: for each position indicated by a letter, select the corresponding position from the second column, indicated by a number. Write down the numbers of the selected connections in the answer field. Answer: A-4; B-1; AT 3 Explanation: Acid salts are called salts resulting from the incomplete replacement of mobile hydrogen atoms by a metal cation, ammonium cation or alkyl ammonium. In inorganic acids, which take place as part of the school curriculum, all hydrogen atoms are mobile, that is, they can be replaced by a metal. Examples of acidic inorganic salts among the presented list is ammonium bicarbonate NH 4 HCO 3 - the product of replacing one of the two hydrogen atoms in carbonic acid with an ammonium cation. In fact, an acid salt is a cross between a normal (medium) salt and an acid. In the case of NH 4 HCO 3 - the average between the normal salt (NH 4) 2 CO 3 and carbonic acid H 2 CO 3. In organic substances, only hydrogen atoms that are part of carboxyl groups (-COOH) or hydroxyl groups of phenols (Ar-OH) can be replaced by metal atoms. That is, for example, sodium acetate CH 3 COONa, despite the fact that not all hydrogen atoms in its molecule are replaced by metal cations, is an average, not an acid salt (!). Hydrogen atoms in organic substances, attached directly to the carbon atom, are 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 oxides of non-metals that do not form salts with basic oxides or bases, that is, they either do not react with them at all (most often), or give a different product (not a salt) in reaction with them. It is often said that non-salt-forming oxides are oxides of non-metals that do not react with bases and basic oxides. However, for the detection of non-salt-forming oxides, this approach does not always work. So, for example, CO, being a non-salt-forming oxide, reacts with basic iron (II) oxide, but with the formation of a free metal rather than a salt: CO + FeO = CO 2 + Fe Non-salt-forming oxides from the school chemistry course include non-metal oxides in the oxidation state +1 and +2. In total, there are 4 of them in the exam - these are CO, NO, N 2 O and SiO (I personally never met the last SiO in assignments). From the proposed list of substances, select two substances, with each of which iron reacts without heating. 1) zinc chloride 2) copper(II) sulfate 3) concentrated nitric acid 4) dilute hydrochloric acid 5) aluminum oxide Answer: 2; 4 Zinc chloride is a salt, and iron is a metal. The metal reacts with the salt only if it is more reactive than the one in the salt. The relative activity of metals is determined by a series of metal activity (in other words, a series of metal stresses). Iron is located to the right of zinc in the activity series of metals, which means that it is less active and is not able to displace zinc from salt. That is, the reaction of iron with substance No. 1 does not go. Copper (II) sulfate CuSO 4 will react with iron, since iron is located to the left of copper in the activity series, that is, it is a more active metal. Concentrated nitric acid, as well as concentrated sulfuric acid, are not able to react with iron, aluminum and chromium without heating due to such a phenomenon as passivation: on the surface of these metals, under the action of these acids, an insoluble salt is formed without heating, which acts as a protective shell. However, when heated, this protective shell dissolves and the reaction becomes possible. Those. since it is indicated that there is no heating, the reaction of iron with conc. HNO 3 does not leak. Hydrochloric acid, regardless of concentration, refers to non-oxidizing acids. Metals that are in the activity series to the left of hydrogen react with non-oxidizing acids with the release of hydrogen. Iron is one of these metals. Conclusion: the reaction of iron with hydrochloric acid flows. In the case of a metal and a metal oxide, the reaction, as in the case of a salt, is possible if the free metal is more active than that which is part of the oxide. Fe, according to the activity series of metals, is less active than Al. This means that Fe does not react with Al 2 O 3. From the proposed list, select two oxides that react with a solution of hydrochloric acid, but do not react
with sodium hydroxide solution. Write down the numbers of the selected substances in the answer field. Answer: 3; 4 CO is a non-salt-forming oxide, s aqueous solution alkali does not react. (It should be remembered that, nevertheless, under harsh conditions - high pressure and temperature - it still reacts with solid alkali, forming formates - salts of formic acid.) SO 3 - sulfur oxide (VI) - acid oxide, which corresponds to sulfuric acid. Acid oxides do not react with acids and other acid oxides. That is, SO 3 does not react with hydrochloric acid and reacts with a base - sodium hydroxide. Not suitable. CuO - copper (II) oxide - is classified as an oxide with predominantly basic properties. Reacts with HCl and does not react with sodium hydroxide solution. Fits MgO - magnesium oxide - is classified as a typical basic oxide. Reacts with HCl and does not react with sodium hydroxide solution. Fits ZnO, an oxide with pronounced amphoteric properties, easily reacts with both strong bases and acids (as well as acidic and basic oxides). Not suitable. Answer: 4; 2 In the reaction between two salts of inorganic acids, gas is formed only when hot solutions of nitrites and ammonium salts are mixed due to the formation of thermally unstable ammonium nitrite. For example, NH 4 Cl + KNO 2 \u003d t o \u003d\u003e N 2 + 2H 2 O + KCl However, both nitrites and ammonium salts are not on the list. This means that one of the three salts (Cu (NO 3) 2, K 2 SO 3 and Na 2 SiO 3) reacts with either an acid (HCl) or an alkali (NaOH). Among the salts of inorganic acids, only ammonium salts emit gas when interacting with alkalis: NH 4 + + OH \u003d NH 3 + H 2 O Ammonium salts, as we have already said, are not on the list. The only option left is the interaction of the salt with the acid. Salts among these substances include Cu(NO 3) 2, K 2 SO 3 and Na 2 SiO 3. The reaction of copper nitrate with hydrochloric acid does not proceed, because no gas, no precipitate, no low-dissociating substance (water or weak acid) is formed. Sodium silicate reacts with hydrochloric acid, however, due to the release of a white gelatinous precipitate of silicic acid, and not gas: Na 2 SiO 3 + 2HCl \u003d 2NaCl + H 2 SiO 3 ↓ The last option remains - the interaction of potassium sulfite and hydrochloric acid. Indeed, as a result of the ion exchange reaction between sulfite and almost any acid, unstable sulfurous acid is formed, which instantly decomposes into colorless gaseous sulfur oxide (IV) and water. 4) HCl (excess) Write in the table the numbers of the selected substances under the corresponding letters. Answer: 2; 5 CO 2 is an acidic oxide and must be treated with either a basic oxide or a base to convert it to a salt. Those. to obtain potassium carbonate from CO 2, it must be treated with either potassium oxide or potassium hydroxide. Thus, substance X is potassium oxide: K 2 O + CO 2 \u003d K 2 CO 3 Potassium bicarbonate KHCO 3, like potassium carbonate, is a salt of carbonic acid, with the only difference being that the bicarbonate is a product of incomplete substitution of hydrogen atoms in carbonic acid. To obtain an acid salt from a normal (medium) salt, one must either act on it with the same acid that formed this salt, or else act on it with an acid oxide corresponding to this acid in the presence of water. Thus reactant Y is carbon dioxide. When it is passed through an aqueous solution of potassium carbonate, the latter turns into potassium bicarbonate: K 2 CO 3 + H 2 O + CO 2 \u003d 2KHCO 3 Establish a correspondence between the reaction equation and the property of the nitrogen element that it exhibits in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number. Write in the table the numbers of the selected substances under the corresponding letters. Answer: A-4; B-2; AT 2; G-1 Explanation: A) NH 4 HCO 3 - salt, which includes the ammonium cation NH 4 +. In the ammonium cation, nitrogen always has an oxidation state of -3. As a result of the reaction, it turns into ammonia NH 3. Hydrogen almost always (except for its compounds with metals) has an oxidation state of +1. Therefore, for the ammonia molecule to be electrically neutral, nitrogen must have an oxidation state of -3. Thus, there is no change in the degree of nitrogen oxidation; it does not exhibit redox properties. B) As already shown above, nitrogen in ammonia NH 3 has an oxidation state of -3. As a result of the reaction with CuO, ammonia is converted into a simple substance N 2. In any simple substance, the oxidation state of the element with which it is formed is equal to zero. Thus, the nitrogen atom loses its negative charge, and since electrons are responsible for the negative charge, this means that they are lost by the nitrogen atom as a result of the reaction. An element that loses some of its electrons in a reaction is called a reducing agent. C) As a result of the reaction, NH 3 with an oxidation state of nitrogen equal to -3 turns into nitric oxide NO. Oxygen almost always has an oxidation state of -2. Therefore, in order for the nitric oxide molecule to be electrically neutral, the nitrogen atom must have an oxidation state of +2. This means that the nitrogen atom changed its oxidation state from -3 to +2 as a result of the reaction. This indicates the loss of 5 electrons by the nitrogen atom. That is, nitrogen, as in the case of B, is a reducing agent. D) N 2 is a simple substance. In all simple substances, the element that forms them has an oxidation state of 0. As a result of the reaction, nitrogen is converted into lithium nitride Li3N. The only oxidation state of an alkali metal other than zero (any element has an oxidation state of 0) is +1. Thus, for the Li3N structural unit to be electrically neutral, nitrogen must have an oxidation state of -3. It turns out that as a result of the reaction, nitrogen acquired a negative charge, which means the addition of electrons. Nitrogen is the oxidizing agent in this reaction. 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) 2) BaO, H 2 O, KOH 3) H 2, Cl 2, O 2 4) HBr, LiOH, CH 3 COOH 5) H 3 PO 4, BaCl 2, CuO Write in the table the numbers of the selected substances under the corresponding letters. Answer: A-3; B-2; AT 4; G-1 Explanation: A) When hydrogen gas is passed through a sulfur melt, hydrogen sulfide H 2 S is formed: H 2 + S \u003d t o \u003d\u003e H 2 S When chlorine is passed over crushed sulfur at room temperature, sulfur dichloride is formed: S + Cl 2 \u003d SCl 2 For passing the exam it is not necessary to know exactly how sulfur reacts with chlorine and, accordingly, 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 oxidizing and reducing. That is, if a strong oxidizing agent acts on sulfur, which is molecular chlorine Cl 2, it will oxidize. Sulfur burns with a blue flame in oxygen to form a gas with a pungent odor - sulfur dioxide SO 2: B) SO 3 - sulfur oxide (VI) has pronounced acid properties. For such oxides, the most characteristic reactions are interactions with water, as well as with basic and amphoteric oxides and hydroxides. In the list at number 2, we just see water, and the basic oxide BaO, and hydroxide KOH. When an acidic oxide reacts with a basic oxide, a salt of the corresponding acid and a metal that is part of the basic oxide is formed. An acidic oxide corresponds to an acid in which the acid-forming element has the same oxidation state as in the oxide. The oxide SO 3 corresponds to sulfuric acid H 2 SO 4 (both there and there the oxidation state of sulfur is +6). Thus, when SO 3 interacts with metal oxides, sulfuric acid salts will be obtained - sulfates containing the sulfate ion SO 4 2-: SO 3 + BaO = BaSO 4 When interacting with water, the acid oxide turns into the corresponding acid: SO 3 + H 2 O \u003d H 2 SO 4 And when acid oxides interact with metal hydroxides, a salt of the corresponding acid and water are formed: SO 3 + 2KOH \u003d K 2 SO 4 + H 2 O C) Zinc hydroxide Zn (OH) 2 has typical amphoteric properties, that is, it reacts both with acidic oxides and acids, and with basic oxides and alkalis. In list 4, we see both acids - hydrobromic HBr and acetic, and alkali - LiOH. Recall that water-soluble metal hydroxides are called alkalis: Zn(OH) 2 + 2HBr = ZnBr 2 + 2H 2 O Zn (OH) 2 + 2CH 3 COOH \u003d Zn (CH 3 COO) 2 + 2H 2 O Zn(OH) 2 + 2LiOH \u003d Li 2 D) Zinc bromide ZnBr 2 is a salt, soluble in water. For soluble salts, ion exchange reactions are the most common. A salt can react with another salt provided that both starting salts are soluble and a precipitate forms. Also ZnBr 2 contains bromide ion Br-. Metal halides are characterized by the fact that they are able to react with Hal 2 halogens, which are higher in the periodic table. Thus? the described types of reactions proceed with all substances of list 1: ZnBr 2 + 2AgNO 3 \u003d 2AgBr + Zn (NO 3) 2 3ZnBr 2 + 2Na 3 PO 4 = Zn 3 (PO 4) 2 + 6NaBr ZnBr 2 + Cl 2 = ZnCl 2 + Br 2 Establish a correspondence between the name of the substance and the class / group to which this substance belongs: for each position indicated by a letter, select the corresponding position indicated by a number. Write in the table the numbers of the selected substances under the corresponding letters. Answer: A-4; B-2; IN 1 Explanation: A) Methylbenzene, aka toluene, has structural formula: As you can see, the molecules of this substance consist only of carbon and hydrogen, therefore methylbenzene (toluene) refers to hydrocarbons B) The structural formula of aniline (aminobenzene) is as follows: As can be seen from the structural formula, the aniline molecule consists of an aromatic hydrocarbon radical (C 6 H 5 -) and an amino group (-NH 2), thus, aniline belongs to aromatic amines, i.e. correct answer 2. C) 3-methylbutanal. The ending "al" indicates that the substance belongs to aldehydes. The structural formula of this substance: From the proposed list, select two substances that are structural isomers of butene-1. 2) cyclobutane 4) butadiene-1,3 5) methylpropene Write down the numbers of the selected substances in the answer field. Answer: 2; 5 Explanation: Isomers are substances that have the same molecular formula and different structural, i.e. Substances that differ in the order in which atoms are combined, but with the same composition of molecules. From the proposed list, select two substances, the interaction of which with a solution of potassium permanganate will cause a change in the color of the solution. 1) cyclohexane 5) propylene Write down the numbers of the selected substances in the answer field. Answer: 3; 5 Explanation: Alkanes, as well as cycloalkanes with a ring size of 5 or more carbon atoms, are very inert and do not react with aqueous solutions of even strong oxidizing agents, such as, for example, potassium permanganate KMnO 4 and potassium dichromate K 2 Cr 2 O 7 . Thus, options 1 and 4 disappear - when cyclohexane or propane is added to an aqueous solution of potassium permanganate, a color change will not occur. Among the hydrocarbons of the homologous series of benzene, only benzene is passive to the action of aqueous solutions of oxidizing agents, all other homologues are oxidized, depending on the medium, either to carboxylic acids or to their corresponding salts. Thus, option 2 (benzene) is eliminated. The correct answers are 3 (toluene) and 5 (propylene). Both substances decolorize the purple solution of potassium permanganate due to the reactions taking place: CH 3 -CH=CH 2 + 2KMnO 4 + 2H 2 O → CH 3 -CH(OH)–CH 2 OH + 2MnO 2 + 2KOH From the proposed list, select two substances with which formaldehyde reacts. 4) Ag 2 O (NH 3 solution) 5) CH 3 DOS 3 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 proceeding along the functional group. Among the list of responses for formaldehyde, reduction reactions are typical, where hydrogen is used as a reducing agent (cat. - Pt, Pd, Ni), and oxidation - in this case, the silver mirror reaction. When reduced with hydrogen on a nickel catalyst, formaldehyde is converted to methanol: The silver mirror reaction is the reduction of silver from an ammonia solution of silver oxide. When dissolved in an aqueous solution of ammonia, silver oxide turns into a complex compound - diammine silver (I) OH hydroxide. After the addition of formaldehyde, a redox reaction occurs in which silver is reduced: From the proposed list, select two substances with which methylamine reacts. 2) chloromethane 3) hydrogen 4) sodium hydroxide 5) hydrochloric acid 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 - The following scheme of transformations of substances is given: Determine which of the given substances are substances X and Y. 5) NaOH (alcohol) Write in the table the numbers of the selected substances under the corresponding letters. Answer: 4; 2 Explanation: One of the reactions for obtaining alcohols is the hydrolysis of haloalkanes. Thus, ethanol can be obtained from chloroethane by acting on the latter with an aqueous solution of alkali - in this case, NaOH. CH 3 CH 2 Cl + NaOH (aq.) → CH 3 CH 2 OH + NaCl The next reaction is the oxidation reaction of ethyl alcohol. The oxidation of alcohols is carried out on a copper catalyst or using CuO: Establish a correspondence between the name of the substance and the product that is mainly formed during the interaction of this substance with bromine: for each position indicated by a letter, select the corresponding position indicated by a number. Answer: 5; 2; 3; 6 Explanation: For alkanes, the most characteristic reactions are free radical substitution reactions, during which a hydrogen atom is replaced by a halogen atom. Thus, by brominating ethane, one can obtain bromoethane, and by brominating isobutane, 2-bromoisobutane can be obtained: Since the small cycles of cyclopropane and cyclobutane molecules are unstable, during bromination the cycles of these molecules are opened, thus the addition reaction proceeds: In contrast to the cycles of cyclopropane and cyclobutane, the cyclohexane cycle is large, resulting in the replacement of a hydrogen atom by a bromine atom: Establish a correspondence between the reacting substances and the carbon-containing product that is formed during the interaction of these substances: for each position indicated by a letter, select the corresponding position indicated by a number. Write in the table the selected numbers under the corresponding letters. Answer: 5; 4; 6; 2 From the proposed list of types of reactions, select two types of reactions, which include the interaction of alkali metals with water. 1) catalytic 2) homogeneous 3) irreversible 4) redox 5) neutralization reaction 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 D.I. Mendeleev and are reducing agents, easily donating an electron located at the outer level. If we denote the alkali metal with the letter M, then the reaction of the alkali metal with water will look like this: 2M + 2H 2 O → 2MOH + H 2 Alkali metals are very active towards water. The reaction proceeds violently with the release a large number heat, is irreversible and does not require the use of a catalyst (non-catalytic) - a substance that accelerates the reaction and is not part of the reaction products. It should be noted that all highly exothermic reactions do not require the use of a catalyst and proceed irreversibly. Since metal and water are substances that are in different states of aggregation, then this reaction proceeds at the phase boundary, therefore, it is heterogeneous. The type of this reaction is substitution. Reactions between inorganic substances are classified as substitution reactions if a simple substance interacts with a complex one and as a result other simple and complex substances are formed. (A neutralization reaction occurs between an acid and a base, as a result of which these substances exchange their constituent parts and a salt and a low-dissociating substance are formed). As stated above, alkali metals are reducing agents, donating an electron from the outer layer, therefore, the reaction is redox. 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. 1) lowering the temperature 2) increase in ethylene concentration 3) the use of a catalyst 4) decrease in hydrogen concentration 5) pressure increase in the system Write in the answer field the numbers of the selected external influences. Answer: 1; 4 For speed chemical reaction the following factors influence: change in temperature and concentration of reagents, as well as the use of a catalyst. According to Van't Hoff's empirical rule, for every 10 degrees increase in temperature, the rate constant of a homogeneous reaction increases by 2-4 times. Therefore, a decrease in temperature also leads to a decrease in the reaction rate. The first answer is correct. As noted above, the reaction rate is also affected by a change in the concentration of reagents: if the concentration of ethylene is increased, the reaction rate will also increase, which does not meet the requirements of the problem. And a decrease in the concentration of hydrogen - the initial component, on the contrary, reduces the reaction rate. Therefore, the second option is not suitable, and the fourth is suitable. A catalyst is a substance that speeds up the rate of a chemical reaction but is not part of the products. The use of a catalyst accelerates the ethylene hydrogenation reaction, which also does not correspond to the condition of the problem, and therefore is not the right answer. When ethylene reacts with hydrogen (on Ni, Pd, Pt catalysts), ethane is formed: CH 2 \u003d CH 2 (g) + H 2 (g) → CH 3 -CH 3 (g) All components involved in the reaction and the product are gaseous substances, therefore, the pressure in the system will also affect the reaction rate. From two volumes of ethylene and hydrogen, one volume of ethane is formed, therefore, the reaction proceeds to a decrease in pressure in the system. By increasing the pressure, we will speed up the reaction. The fifth answer does not fit. Establish a correspondence between the salt formula and the products of electrolysis of an aqueous solution of this salt, which stood out on inert electrodes: for each position indicated by a letter, select the corresponding position indicated by a number. Write in the table the selected numbers under the corresponding letters. Answer: 1; 4; 3; 2 Electrolysis is a redox process that occurs on the electrodes during the passage of a constant electric current through an electrolyte solution or melt. At the cathode, the reduction occurs predominantly of those cations that have the highest oxidizing activity. At the anode, those anions are oxidized first of all, which have the greatest reduction ability. Electrolysis of aqueous solution 1) The process of electrolysis of aqueous solutions on the cathode does not depend on the material of the cathode, but depends on the position of the metal cation in the electrochemical series of voltages. For cations in a row Li + - Al 3+ recovery process: 2H 2 O + 2e → H 2 + 2OH - (H 2 is released at the cathode) Zn 2+ - Pb 2+ recovery process: Me n + + ne → Me 0 and 2H 2 O + 2e → H 2 + 2OH - (H 2 and Me will be released at the cathode) Cu 2+ - Au 3+ reduction process Me n + + ne → Me 0 (Me is released at the cathode) 2) The process of electrolysis of aqueous solutions at the anode depends on the material of the anode and on the nature of the anion. If the anode is insoluble, i.e. inert (platinum, gold, coal, graphite), the process will depend only on the nature of the anions. For anions F -, SO 4 2-, NO 3 -, PO 4 3-, OH - the oxidation process: 4OH - - 4e → O 2 + 2H 2 O or 2H 2 O - 4e → O 2 + 4H + (oxygen is released at the anode) halide ions (except F-) oxidation process 2Hal - - 2e → Hal 2 (free halogens are released ) organic acids oxidation process: 2RCOO - - 2e → R-R + 2CO 2 The overall electrolysis equation is: A) Na 3 PO 4 solution 2H 2 O → 2H 2 (at the cathode) + O 2 (at the anode) B) KCl solution 2KCl + 2H 2 O → H 2 (at the cathode) + 2KOH + Cl 2 (at the anode) C) CuBr2 solution CuBr 2 → Cu (at the cathode) + Br 2 (at the anode) D) Cu(NO3)2 solution 2Cu(NO 3) 2 + 2H 2 O → 2Cu (at the cathode) + 4HNO 3 + O 2 (at the anode) Establish a correspondence between the name of the salt and the ratio of this salt to hydrolysis: for each position indicated by a letter, select the corresponding position indicated by a number. Write in the table the selected numbers under the corresponding letters. Answer: 1; 3; 2; 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), undergoes hydrolysis by the cation. NH 4 Cl → NH 4 + + Cl - NH 4 + + H 2 O → NH 3 H 2 O + H + (formation of ammonia dissolved in water) The solution medium is acidic (pH< 7). B) Potassium sulfate (K 2 SO 4) - a salt formed by strong sulfuric acid and potassium hydroxide (alkali, i.e. strong base), does not undergo hydrolysis. K 2 SO 4 → 2K + + SO 4 2- C) Sodium carbonate (Na 2 CO 3) - a salt formed by a weak carbonic acid and sodium hydroxide (an alkali, i.e. a strong base), undergoes anion hydrolysis. CO 3 2- + H 2 O → HCO 3 - + OH - (formation of a weakly dissociating hydrocarbonate ion) The solution is alkaline (pH > 7). D) Aluminum sulfide (Al 2 S 3) - a salt formed by a weak hydrosulfide acid and aluminum hydroxide (weak base), undergoes complete hydrolysis with the formation of aluminum hydroxide and hydrogen sulfide: Al 2 S 3 + 6H 2 O → 2Al(OH) 3 + 3H 2 S The solution medium is close to neutral (pH ~ 7). Establish a correspondence between the equation of a chemical reaction and the direction of displacement of chemical equilibrium with increasing pressure in the system: for each position indicated by a letter, select the corresponding position indicated by a number. A) N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) B) 2H 2 (g) + O 2 (g) ↔ 2H 2 O (g) C) H 2 (g) + Cl 2 (g) ↔ 2HCl (g) D) SO 2 (g) + Cl 2 (g) ↔ SO 2 Cl 2 (g) 1) shifts towards a direct reaction 2) shifts towards the back reaction 3) there is no shift in equilibrium Write in the table the selected numbers under the corresponding letters. Answer: A-1; B-1; AT 3; G-1 A reaction is in chemical equilibrium when the rate of the forward reaction is equal to the rate of the reverse. The shift of equilibrium in the desired direction is achieved by changing the reaction conditions. Factors that determine the position of equilibrium: — pressure: an increase in pressure shifts the equilibrium towards a reaction leading to a decrease in volume (conversely, a decrease in pressure shifts the equilibrium towards a reaction leading to an increase in volume) — temperature: an increase in temperature shifts the equilibrium towards an endothermic reaction (conversely, a decrease in temperature shifts the equilibrium towards an exothermic reaction) — concentrations of starting substances and reaction products: an increase in the concentration of the starting substances and the removal of products from the reaction sphere shift the equilibrium towards the 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 equilibrium shift, but only accelerate its achievement A) In the first case, the reaction proceeds with a decrease in volume, since V (N 2) + 3V (H 2) > 2V (NH 3). By increasing the pressure in the system, the equilibrium will shift to the side with a smaller volume of substances, therefore, in the forward direction (in the direction of the direct reaction). B) In the second case, the reaction also proceeds with a decrease in volume, since 2V (H 2) + V (O 2) > 2V (H 2 O). By increasing the pressure in the system, the equilibrium will also shift in the direction of the direct reaction (in the direction of the product). C) In the third case, the pressure does not change during the reaction, because V (H 2) + V (Cl 2) \u003d 2V (HCl), so there is no equilibrium shift. D) In the fourth case, the reaction also proceeds with a decrease in volume, since V (SO 2) + V (Cl 2) > V (SO 2 Cl 2). By increasing the pressure in the system, the equilibrium will shift towards the formation of the product (direct reaction). 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. A) HNO 3 and H 2 O C) NaCl and BaCl 2 D) AlCl 3 and MgCl 2 Write in the table the selected numbers under the corresponding letters. Answer: A-1; B-3; AT 3; G-2 A) Nitric acid and water can be distinguished using salt - calcium carbonate CaCO 3. Calcium carbonate does not dissolve in water, and when interacting with nitric acid forms a soluble salt - calcium nitrate Ca (NO 3) 2, while the reaction is accompanied by the release of colorless carbon dioxide: CaCO 3 + 2HNO 3 → Ca(NO 3) 2 + CO 2 + H 2 O B) Potassium chloride KCl and alkali NaOH can be distinguished by a solution of copper (II) sulfate. When copper (II) sulfate interacts with KCl, the exchange reaction does not proceed, the solution contains K +, Cl -, Cu 2+ and SO 4 2- ions, which do not form poorly dissociating substances with each other. When copper (II) sulfate interacts with NaOH, an exchange reaction occurs, as a result of which copper (II) hydroxide precipitates (blue base). C) Sodium chloride NaCl and barium BaCl 2 are soluble salts, which can also be distinguished by a solution of copper (II) sulfate. When copper (II) sulfate interacts with NaCl, the exchange reaction does not proceed, the solution contains Na +, Cl -, Cu 2+ and SO 4 2- ions, which do not form poorly dissociating substances with each other. When copper (II) sulfate interacts with BaCl 2, an exchange reaction occurs, as a result of which barium sulfate BaSO 4 precipitates. D) Aluminum chloride AlCl 3 and magnesium MgCl 2 dissolve in water and behave differently when interacting with potassium hydroxide. Magnesium chloride with alkali forms a precipitate: Answer: A-4; B-2; AT 3; G-5 A) Ammonia is the most important product of the chemical industry, its production is more than 130 million tons per year. Ammonia is mainly used in the production of nitrogen fertilizers (ammonium nitrate and sulfate, urea), medicines, explosives, nitric acid, and soda. Among the proposed answers, the area of application of ammonia is the production of fertilizers (Fourth answer option). B) Methane is the simplest hydrocarbon, the most thermally stable representative of a number of saturated compounds. It is widely used as a domestic and industrial fuel, as well as a raw material for industry (Second answer). Methane is 90-98% a component of natural gas. C) Rubbers are materials that are obtained by polymerization of compounds with conjugated double bonds. Isoprene just belongs to this type of compounds and is used to obtain one of the types of rubbers: D) Low molecular weight alkenes are used to make plastics, in particular ethylene is used to make a plastic called polyethylene: n CH 2 \u003d CH 2 → (-CH 2 -CH 2 -) n 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%. Answer: 3.4 g Explanation: Let x g be the mass of potassium nitrate, which is dissolved in 150 g of the solution. Calculate the mass of potassium nitrate dissolved in 150 g of solution: m(KNO 3) \u003d 150 g 0.1 \u003d 15 g In order for the mass fraction of salt to be 12%, x g of potassium nitrate was added. In this case, the mass of the solution was (150 + x) g. We write the equation in the form: (Write down the number to tenths.) Answer: 14.4 g Explanation: As a result of the complete combustion of hydrogen sulfide, sulfur dioxide and water are formed: 2H 2 S + 3O 2 → 2SO 2 + 2H 2 O A consequence of Avogadro's law is that the volumes of gases under the same conditions are related to each other in the same way as the number of moles of these gases. Thus, according to the reaction equation: ν(O 2) = 3/2ν(H 2 S), therefore, the volumes of hydrogen sulfide and oxygen are related to each other in exactly the same way: V (O 2) \u003d 3 / 2V (H 2 S), V (O 2) \u003d 3/2 6.72 l \u003d 10.08 l, hence V (O 2) \u003d 10.08 l / 22.4 l / mol \u003d 0.45 mol Calculate the mass of oxygen required for the complete combustion of hydrogen sulfide: m(O 2) \u003d 0.45 mol 32 g / mol \u003d 14.4 g Using the electron balance method, write the equation for the reaction: Na 2 SO 3 + ... + KOH → K 2 MnO 4 + ... + H 2 O Determine the oxidizing agent and reducing agent. 2) Iron (III) sulfate - a salt soluble in water, enters into an exchange reaction with alkali, as a result of which iron (III) hydroxide precipitates (brown compound): Fe 2 (SO 4) 3 + 3NaOH → 2Fe(OH) 3 ↓ + 3Na 2 SO 4 3) Insoluble metal hydroxides decompose upon calcination to the corresponding oxides and water: 2Fe(OH) 3 → Fe 2 O 3 + 3H 2 O 4) When iron (III) oxide is heated with metallic iron, iron (II) oxide is formed (iron in the FeO compound has an intermediate oxidation state): Fe 2 O 3 + Fe → 3FeO (on heating) Write the reaction equations that can be used to carry out the following transformations: When writing reaction equations, use the structural formulas of organic substances. 1) Intramolecular dehydration occurs at temperatures above 140 o C. This occurs as a result of the elimination of a hydrogen atom from the carbon atom of the alcohol, located one through to the alcohol hydroxyl (in the β-position). CH 3 -CH 2 -CH 2 -OH → CH 2 \u003d CH-CH 3 + H 2 O (conditions - H 2 SO 4, 180 o C) Intermolecular dehydration proceeds at a temperature below 140 o C under the action of sulfuric acid and ultimately comes down to the elimination of one water molecule from two alcohol molecules. 2) Propylene refers to unsymmetrical alkenes. When hydrogen halides and water are added, a hydrogen atom is attached to a carbon atom at a multiple bond associated with a large number of hydrogen atoms: CH 2 \u003d CH-CH 3 + HCl → CH 3 -CHCl-CH 3 3) Acting with an aqueous solution of NaOH on 2-chloropropane, the halogen atom is replaced by a hydroxyl group: CH 3 -CHCl-CH 3 + NaOH (aq.) → CH 3 -CHOH-CH 3 + NaCl 4) Propylene can be obtained not only from propanol-1, but also from propanol-2 by the reaction of intramolecular dehydration at temperatures above 140 o C: CH 3 -CH(OH)-CH 3 → CH 2 \u003d CH-CH 3 + H 2 O (conditions H 2 SO 4, 180 o C) 5) In an alkaline environment, acting with a dilute aqueous solution of potassium permanganate, hydroxylation of alkenes occurs with the formation of diols: 3CH 2 \u003d CH-CH 3 + 2KMnO 4 + 4H 2 O → 3HOCH 2 -CH (OH) -CH 3 + 2MnO 2 + 2KOH Determine the mass fractions (in%) of iron (II) sulfate and aluminum sulfide in the mixture, if during the treatment of 25 g of this mixture with water a gas was released that completely reacted with 960 g of a 5% solution of copper (II) sulfate. In response, write down the reaction equations that are indicated in the condition of the problem, and give all the necessary calculations (indicate the units of measurement of the desired physical quantities). Answer: ω(Al 2 S 3) = 40%; ω(CuSO 4) = 60% When a mixture of iron (II) sulfate and aluminum sulfide is treated with water, the sulfate is simply dissolved, and the sulfide is hydrolyzed to form aluminum (III) hydroxide and hydrogen sulfide: Al 2 S 3 + 6H 2 O → 2Al(OH) 3 ↓ + 3H 2 S (I) When hydrogen sulfide is passed through a solution of copper (II) sulfate, copper (II) sulfide precipitates: CuSO 4 + H 2 S → CuS↓ + H 2 SO 4 (II) Calculate the mass and amount of substance of dissolved copper(II) sulfate: m (CuSO 4) \u003d m (p-ra) ω (CuSO 4) \u003d 960 g 0.05 \u003d 48 g; ν (CuSO 4) \u003d m (CuSO 4) / M (CuSO 4) \u003d 48 g / 160 g \u003d 0.3 mol According to the reaction equation (II) ν (CuSO 4) = ν (H 2 S) = 0.3 mol, and according to the reaction equation (III) ν (Al 2 S 3) = 1/3ν (H 2 S) = 0, 1 mol Calculate the masses of aluminum sulfide and copper (II) sulfate: m(Al 2 S 3) \u003d 0.1 mol 150 g / mol \u003d 15 g; m(CuSO4) = 25 g - 15 g = 10 g ω (Al 2 S 3) \u003d 15 g / 25 g 100% \u003d 60%; ω (CuSO 4) \u003d 10 g / 25 g 100% \u003d 40% When burning a sample of some organic compound weighing 14.8 g, 35.2 g of carbon dioxide and 18.0 g of water were obtained. It is known that the relative hydrogen vapor density of this substance is 37. 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 these conditions of the assignment: 1) do the calculations necessary to establish the molecular formula organic matter(indicate the units of measurement of the required physical quantities); 2) write down the molecular formula of the original organic matter; 3) make a structural formula of this substance, which unambiguously reflects the order of bonding of atoms in its molecule; 4) write the equation for the reaction of this substance with copper(II) oxide using the structural formula of the substance. A1. The structure of the electron shells of atoms of the elements of the first four periods:s
- Andp
—
Andd
—
elements. The electronic configuration of the atom. To solve the first task of part A, you need to know how the number of protons, electrons in an atom and ion is determined, to know how electrons are distributed over electronic levels and sublevels, to be able to write down the electronic configuration of an atom and ion, to find the number unpaired electrons in an atom and an ion, to know how the electronic configuration and valence of a chemical element are related. Examples of tasks A1 USE in chemistry: 1) The number of electrons in an atom is: Answer: An electron is negatively charged. elementary particle. Since each atom is generally electrically neutral (has no charge), the number of electrons must match the number of protons. Correct answer: 1 2) An ion, which consists of 16 protons and 18 electrons, has a charge: Answer: Again, based on the fact that an electron is a negatively charged particle, and a proton is positively charged, it is not difficult to make a calculation: +16 -18 = -2 Correct answer: 2 3) The external energy level of an atom of an element that forms the highest oxide of the EOz composition has the formula: Answer: Oxides are oxygen compounds with an oxidation state of -2. Accordingly, the element included in the EO3 oxide is in group 6, and the group number corresponds to the number of electrons in the external energy level. Further, following the rules for the distribution of electrons in orbitals, 2 electrons go to the s orbital, the remaining 4 electrons are distributed to the orbitals. Correct answer: 4 4) The configuration of the outer electron layer of the sulfur atom in the unexcited state: Answer: First, let's look at the periodic table. Sulfur is in period 3. The number of the period corresponds to the number of electron layers, for the sulfur atom it is 3. Variants 1 and 4 are omitted. Further, sulfur is in group 6, which shows that there are only 6 electrons on the outer electron layer of the sulfur atom. Further, according to the electron distribution rule, the s sublevel is filled first, and then p. The S orbital has a maximum of 2 electrons, p - 6. Based on this, the configuration of the outer layer of the sulfur atom in the unexcited state is 3s 2 3p 4. Correct answer: 3 5) The electronic configuration 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 in the ground state has an atom: Answer: We immediately look at the external electronic level - 4s 1 . Based on this, we can determine that the element has 4 electronic levels, and is in period 4, and has 1 electron in the outer level, which corresponds to the first group. Looking at the periodic table, you can find only 1 such element, and that will be potassium. Correct answer: 3 You can also try the test. Methodology for solving problems in chemistry When solving problems, you need to be guided by a few simple rules: In order to successfully prepare in chemistry, one should carefully consider the solutions to the problems given in the text, as well as independently solve a sufficient number of them. It is in the process of solving problems that the main theoretical provisions of the chemistry course will be fixed. It is necessary to solve problems throughout the entire time of studying chemistry and preparing for the exam. You can use the tasks on this page, or you can download a good collection of tasks and exercises with the solution of typical and complicated tasks (M. I. Lebedeva, I. A. Ankudimova): download. Mole, molar mass Molar mass is the ratio of the mass of a substance to the amount of a substance, i.e. М(х) = m(x)/ν(x), (1) where M(x) is the molar mass of substance X, m(x) is the mass of substance X, ν(x) is the amount of substance X. The SI unit for molar mass is kg/mol, but g/mol is commonly used. The unit of mass is g, kg. The SI unit for the amount of a substance is the mole. Any chemistry problem solved through the amount of matter. Remember the basic formula: ν(x) = m(x)/ М(х) = V(x)/V m = N/N A , (2) where V(x) is the volume of substance Х(l), Vm is the molar volume of gas (l/mol), N is the number of particles, N A is the Avogadro constant. 1. Determine the mass sodium iodide NaI amount of substance 0.6 mol. Given: ν(NaI)= 0.6 mol. Find: m(NaI) =? Solution. The molar mass of sodium iodide is: M(NaI) = M(Na) + M(I) = 23 + 127 = 150 g/mol Determine the mass of NaI: m(NaI) = ν(NaI) M(NaI) = 0.6 150 = 90 g. 2. Determine the amount of substance atomic boron contained in sodium tetraborate Na 2 B 4 O 7 weighing 40.4 g. Given: m(Na 2 B 4 O 7) \u003d 40.4 g. Find: ν(B)=? Solution. The molar mass of sodium tetraborate is 202 g/mol. Determine the amount of substance Na 2 B 4 O 7: ν (Na 2 B 4 O 7) \u003d m (Na 2 B 4 O 7) / M (Na 2 B 4 O 7) \u003d 40.4 / 202 \u003d 0.2 mol. Recall that 1 mol of sodium tetraborate molecule contains 2 mol of sodium atoms, 4 mol of boron atoms and 7 mol of oxygen atoms (see the formula of sodium tetraborate). Then the amount of atomic boron substance is: ν (B) \u003d 4 ν (Na 2 B 4 O 7) \u003d 4 0.2 \u003d 0.8 mol. Calculations by chemical formulas. Mass share. The mass fraction of a substance is the ratio of the mass of a given substance in the system to the mass of the entire system, i.e. ω(X) =m(X)/m, where ω(X) is the mass fraction of substance X, m(X) is the mass of substance X, m is the mass of the entire system. Mass fraction is a dimensionless quantity. It is expressed as a fraction of a unit or as a percentage. For example, the mass fraction of atomic oxygen is 0.42, or 42%, i.e. ω(O)=0.42. The mass fraction of atomic chlorine in sodium chloride is 0.607, or 60.7%, i.e. ω(Cl)=0.607. 3. Determine the mass fraction water of crystallization in barium chloride dihydrate BaCl 2 2H 2 O. Solution: The molar mass of BaCl 2 2H 2 O is: M (BaCl 2 2H 2 O) \u003d 137+ 2 35.5 + 2 18 \u003d 244 g / mol From the formula BaCl 2 2H 2 O it follows that 1 mol of barium chloride dihydrate contains 2 mol of H 2 O. From this we can determine the mass of water contained in BaCl 2 2H 2 O: m(H 2 O) \u003d 2 18 \u003d 36 g. We find the mass fraction of water of crystallization in barium chloride dihydrate BaCl 2 2H 2 O. ω (H 2 O) \u003d m (H 2 O) / m (BaCl 2 2H 2 O) \u003d 36/244 \u003d 0.1475 \u003d 14.75%. 4. From a rock sample weighing 25 g containing the mineral argentite Ag 2 S, silver weighing 5.4 g was isolated. Determine the mass fraction argentite in the sample. Given: m(Ag)=5.4 g; m = 25 g. Find: ω(Ag 2 S) =? Solution: we determine the amount of silver substance in argentite: ν (Ag) \u003d m (Ag) / M (Ag) \u003d 5.4 / 108 \u003d 0.05 mol. From the formula Ag 2 S it follows that the amount of argentite substance is half the amount of silver substance. Determine the amount of argentite substance: ν (Ag 2 S) \u003d 0.5 ν (Ag) \u003d 0.5 0.05 \u003d 0.025 mol We calculate the mass of argentite: m (Ag 2 S) \u003d ν (Ag 2 S) M (Ag 2 S) \u003d 0.025 248 \u003d 6.2 g. Now we determine the mass fraction of argentite in a rock sample, weighing 25 g. ω (Ag 2 S) \u003d m (Ag 2 S) / m \u003d 6.2 / 25 \u003d 0.248 \u003d 24.8%. Derivation of compound formulas 5. Determine the simplest compound formula potassium with manganese and oxygen, if the mass fractions of elements in this substance are 24.7, 34.8 and 40.5%, respectively. Given: ω(K)=24.7%; ω(Mn)=34.8%; ω(O)=40.5%. Find: compound formula. Solution: for calculations, we select the mass of the compound, equal to 100 g, i.e. m=100 g. Masses of potassium, manganese and oxygen will be: m (K) = m ω (K); m (K) \u003d 100 0.247 \u003d 24.7 g; m (Mn) = m ω(Mn); m (Mn) = 100 0.348 = 34.8 g; m (O) = m ω(O); m (O) \u003d 100 0.405 \u003d 40.5 g. We determine the amount of substances of atomic potassium, manganese and oxygen: ν (K) \u003d m (K) / M (K) \u003d 24.7 / 39 \u003d 0.63 mol ν (Mn) \u003d m (Mn) / M (Mn) \u003d 34.8 / 55 \u003d 0.63 mol ν (O) \u003d m (O) / M (O) \u003d 40.5 / 16 \u003d 2.5 mol We find the ratio of the amounts of substances: ν(K) : ν(Mn) : ν(O) = 0.63: 0.63: 2.5. Dividing the right side of the equation by a smaller number (0.63) we get: ν(K) : ν(Mn) : ν(O) = 1: 1: 4. Therefore, the simplest formula of the KMnO 4 compound. 6. During the combustion of 1.3 g of the substance, 4.4 g of carbon monoxide (IV) and 0.9 g of water were formed. Find the molecular formula substance if its hydrogen density is 39. Given: m(in-va) \u003d 1.3 g; m(CO 2)=4.4 g; m(H 2 O)=0.9 g; D H2 \u003d 39. Find: the formula of the substance. Solution: Assume that the substance you are looking for contains carbon, hydrogen and oxygen, because during its combustion, CO 2 and H 2 O were formed. Then it is necessary to find the amounts of substances CO 2 and H 2 O in order to determine the amounts of substances of atomic carbon, hydrogen and oxygen. ν (CO 2) \u003d m (CO 2) / M (CO 2) \u003d 4.4 / 44 \u003d 0.1 mol; ν (H 2 O) \u003d m (H 2 O) / M (H 2 O) \u003d 0.9 / 18 \u003d 0.05 mol. We determine the amount of substances of atomic carbon and hydrogen: ν(C)= ν(CO 2); v(C)=0.1 mol; ν(H)= 2 ν(H 2 O); ν (H) \u003d 2 0.05 \u003d 0.1 mol. Therefore, the masses of carbon and hydrogen will be equal: m(C) = ν(C) M(C) = 0.1 12 = 1.2 g; m (H) \u003d ν (H) M (H) \u003d 0.1 1 \u003d 0.1 g. We determine the qualitative composition of the substance: m (in-va) \u003d m (C) + m (H) \u003d 1.2 + 0.1 \u003d 1.3 g. Consequently, the substance consists only of carbon and hydrogen (see the condition of the problem). Let us now determine its molecular weight, based on the given in the condition tasks density of a substance with respect to hydrogen. M (in-va) \u003d 2 D H2 \u003d 2 39 \u003d 78 g / mol. ν(C) : ν(H) = 0.1: 0.1 Dividing the right side of the equation by the number 0.1, we get: ν(C) : ν(H) = 1: 1 Let's take the number of carbon (or hydrogen) atoms as "x", then, multiplying "x" by the atomic masses of carbon and hydrogen and equating this sum molecular weight substances, we solve the equation: 12x + x \u003d 78. Hence x \u003d 6. Therefore, the formula of the substance C 6 H 6 is benzene. Molar volume of gases. Laws of ideal gases. Volume fraction. The molar volume of a gas is equal to the ratio of the volume of gas to the amount of substance of this gas, i.e. Vm = V(X)/ ν(x), where V m is the molar volume of gas - a constant value for any gas under given conditions; V(X) is the volume of gas X; ν(x) - the amount of gas substance X. The molar volume of gases under normal conditions (normal pressure p n \u003d 101 325 Pa ≈ 101.3 kPa and temperature Tn \u003d 273.15 K ≈ 273 K) is V m \u003d 22.4 l /mol. In calculations involving gases, it is often necessary to switch from these conditions to normal conditions or vice versa. In this case, it is convenient to use the formula following from the combined gas law of Boyle-Mariotte and Gay-Lussac: ──── = ─── (3) Where p is pressure; V is the volume; T is the temperature in the Kelvin scale; the index "n" indicates normal conditions. The composition of gas mixtures is often expressed using a volume fraction - the ratio of the volume of a given component to the total volume of the system, i.e. where φ(X) is the volume fraction of the X component; V(X) is the volume of the X component; V is the volume of the system. The volume fraction is a dimensionless quantity, it is expressed in fractions of a unit or as a percentage. 7. What volume takes at a temperature of 20 ° C and a pressure of 250 kPa ammonia weighing 51 g? Given: m(NH 3)=51 g; p=250 kPa; t=20°C. Find: V(NH 3) \u003d? Solution: determine the amount of ammonia substance: ν (NH 3) \u003d m (NH 3) / M (NH 3) \u003d 51/17 \u003d 3 mol. The volume of ammonia under normal conditions is: V (NH 3) \u003d V m ν (NH 3) \u003d 22.4 3 \u003d 67.2 l. Using formula (3), we bring the volume of ammonia to these conditions [temperature T \u003d (273 + 20) K \u003d 293 K]: p n TV n (NH 3) 101.3 293 67.2 V (NH 3) \u003d ──────── \u003d ────────── \u003d 29.2 l. 8. Determine volume, which will take under normal conditions a gas mixture containing hydrogen, weighing 1.4 g and nitrogen, weighing 5.6 g. Given: m(N 2)=5.6 g; m(H2)=1.4; Well. Find: V(mixture)=? Solution: find the amount of substance hydrogen and nitrogen: ν (N 2) \u003d m (N 2) / M (N 2) \u003d 5.6 / 28 \u003d 0.2 mol ν (H 2) \u003d m (H 2) / M (H 2) \u003d 1.4 / 2 \u003d 0.7 mol Since under normal conditions these gases do not interact with each other, the volume of the gas mixture will be equal to the sum of the volumes of gases, i.e. V (mixtures) \u003d V (N 2) + V (H 2) \u003d V m ν (N 2) + V m ν (H 2) \u003d 22.4 0.2 + 22.4 0.7 \u003d 20.16 l.
Calculations for chemical equations(stoichiometric calculations) are based on the law of conservation of mass of substances. However, in real chemical processes, due to an incomplete reaction and various losses of substances, the mass of the resulting products is often less than that which should be formed in accordance with the law of conservation of the mass of substances. The yield of the reaction product (or the mass fraction of the yield) is the ratio of the mass of the actually obtained product, expressed as a percentage, to its mass, which should be formed in accordance with the theoretical calculation, i.e. η = /m(X) (4) Where η is the product yield, %; m p (X) - the mass of the product X obtained in the real process; m(X) is the calculated mass of substance X. In those tasks where the product yield is not specified, it is assumed that it is quantitative (theoretical), i.e. η=100%. 9. What mass of phosphorus should be burned for getting phosphorus oxide (V) weighing 7.1 g? Given: m(P 2 O 5) \u003d 7.1 g. Find: m(P) =? Solution: we write the equation for the combustion reaction of phosphorus and arrange the stoichiometric coefficients. 4P+ 5O 2 = 2P 2 O 5 We determine the amount of substance P 2 O 5 obtained in the reaction. ν (P 2 O 5) \u003d m (P 2 O 5) / M (P 2 O 5) \u003d 7.1 / 142 \u003d 0.05 mol. It follows from the reaction equation that ν (P 2 O 5) \u003d 2 ν (P), therefore, the amount of phosphorus substance required in the reaction is: ν (P 2 O 5) \u003d 2 ν (P) \u003d 2 0.05 \u003d 0.1 mol. From here we find the mass of phosphorus: m(Р) = ν(Р) М(Р) = 0.1 31 = 3.1 g. 10. Magnesium weighing 6 g and zinc weighing 6.5 g were dissolved in an excess of hydrochloric acid. What volume hydrogen, measured under normal conditions, stand out wherein? Given: m(Mg)=6 g; m(Zn)=6.5 g; Well. Find: V(H 2) =? Solution: we write down the reaction equations for the interaction of magnesium and zinc with hydrochloric acid and arrange the stoichiometric coefficients. Zn + 2 HCl \u003d ZnCl 2 + H 2 Mg + 2 HCl \u003d MgCl 2 + H 2 We determine the amount of magnesium and zinc substances that reacted with hydrochloric acid. ν(Mg) \u003d m (Mg) / M (Mg) \u003d 6/24 \u003d 0.25 mol ν (Zn) \u003d m (Zn) / M (Zn) \u003d 6.5 / 65 \u003d 0.1 mol. It follows from the reaction equations that the amount of the substance of the metal and hydrogen are equal, i.e. ν (Mg) \u003d ν (H 2); ν (Zn) \u003d ν (H 2), we determine the amount of hydrogen resulting from two reactions: ν (Н 2) \u003d ν (Mg) + ν (Zn) \u003d 0.25 + 0.1 \u003d 0.35 mol. We calculate the volume of hydrogen released as a result of the reaction: V (H 2) \u003d V m ν (H 2) \u003d 22.4 0.35 \u003d 7.84 l. 11. When passing hydrogen sulfide with a volume of 2.8 liters (normal conditions) through an excess of copper (II) sulfate solution, a precipitate was formed weighing 11.4 g. Determine the exit reaction product. Given: V(H 2 S)=2.8 l; m(precipitate)= 11.4 g; Well. Find: η =? Solution: we write the reaction equation for the interaction of hydrogen sulfide and copper (II) sulfate. H 2 S + CuSO 4 \u003d CuS ↓ + H 2 SO 4 Determine the amount of hydrogen sulfide substance involved in the reaction. ν (H 2 S) \u003d V (H 2 S) / V m \u003d 2.8 / 22.4 \u003d 0.125 mol. It follows from the reaction equation that ν (H 2 S) \u003d ν (СuS) \u003d 0.125 mol. So you can find the theoretical mass of CuS. m(CuS) \u003d ν (CuS) M (CuS) \u003d 0.125 96 \u003d 12 g. Now we determine the product yield using formula (4): η = /m(X)= 11.4 100/ 12 = 95%. 12. What weight ammonium chloride is formed by the interaction of hydrogen chloride weighing 7.3 g with ammonia weighing 5.1 g? What gas will be left in excess? Determine the mass of the excess. Given: m(HCl)=7.3 g; m(NH 3) \u003d 5.1 g. Find: m(NH 4 Cl) =? m(excess) =? Solution: write the reaction equation. HCl + NH 3 \u003d NH 4 Cl This task is for "excess" and "deficiency". We calculate the amount of hydrogen chloride and ammonia and determine which gas is in excess. ν(HCl) \u003d m (HCl) / M (HCl) \u003d 7.3 / 36.5 \u003d 0.2 mol; ν (NH 3) \u003d m (NH 3) / M (NH 3) \u003d 5.1 / 17 \u003d 0.3 mol. Ammonia is in excess, so the calculation is based on the deficiency, i.e. by hydrogen chloride. It follows from the reaction equation that ν (HCl) \u003d ν (NH 4 Cl) \u003d 0.2 mol. Determine the mass of ammonium chloride. m (NH 4 Cl) \u003d ν (NH 4 Cl) M (NH 4 Cl) \u003d 0.2 53.5 \u003d 10.7 g. We determined that ammonia is in excess (according to the amount of substance, the excess is 0.1 mol). Calculate the mass of excess ammonia. m (NH 3) \u003d ν (NH 3) M (NH 3) \u003d 0.1 17 \u003d 1.7 g. 13. Technical calcium carbide weighing 20 g was treated with an excess of water, obtaining acetylene, which, when passed through an excess bromine water 1,1,2,2-tetrabromoethane was formed with a mass of 86.5 g. Determine mass fraction SaS 2 in technical carbide. Given: m = 20 g; m(C 2 H 2 Br 4) \u003d 86.5 g. Find: ω (CaC 2) =? Solution: we write down the equations of interaction of calcium carbide with water and acetylene with bromine water and arrange the stoichiometric coefficients. CaC 2 +2 H 2 O \u003d Ca (OH) 2 + C 2 H 2 C 2 H 2 +2 Br 2 \u003d C 2 H 2 Br 4 Find the amount of substance tetrabromoethane. ν (C 2 H 2 Br 4) \u003d m (C 2 H 2 Br 4) / M (C 2 H 2 Br 4) \u003d 86.5 / 346 \u003d 0.25 mol. It follows from the reaction equations that ν (C 2 H 2 Br 4) \u003d ν (C 2 H 2) \u003d ν (CaC 2) \u003d 0.25 mol. From here we can find the mass of pure calcium carbide (without impurities). m (CaC 2) \u003d ν (CaC 2) M (CaC 2) \u003d 0.25 64 \u003d 16 g. We determine the mass fraction of CaC 2 in technical carbide. ω (CaC 2) \u003d m (CaC 2) / m \u003d 16/20 \u003d 0.8 \u003d 80%. Solutions. Mass fraction of the solution component 14. Sulfur weighing 1.8 g was dissolved in benzene with a volume of 170 ml. The density of benzene is 0.88 g / ml. Determine mass fraction sulfur in solution. Given: V(C 6 H 6) =170 ml; m(S) = 1.8 g; ρ(C 6 C 6)=0.88 g/ml. Find: ω(S) =? Solution: to find the mass fraction of sulfur in the solution, it is necessary to calculate the mass of the solution. Determine the mass of benzene. m (C 6 C 6) \u003d ρ (C 6 C 6) V (C 6 H 6) \u003d 0.88 170 \u003d 149.6 g. Find the total mass of the solution. m (solution) \u003d m (C 6 C 6) + m (S) \u003d 149.6 + 1.8 \u003d 151.4 g. Calculate the mass fraction of sulfur. ω(S) =m(S)/m=1.8 /151.4 = 0.0119 = 1.19%. 15. Iron sulfate FeSO 4 7H 2 O weighing 3.5 g was dissolved in water weighing 40 g. Determine mass fraction of iron sulfate (II) in the resulting solution. Given: m(H 2 O)=40 g; m (FeSO 4 7H 2 O) \u003d 3.5 g. Find: ω(FeSO 4) =? Solution: find the mass of FeSO 4 contained in FeSO 4 7H 2 O. To do this, calculate the amount of substance FeSO 4 7H 2 O. ν (FeSO 4 7H 2 O) \u003d m (FeSO 4 7H 2 O) / M (FeSO 4 7H 2 O) \u003d 3.5 / 278 \u003d 0.0125 mol From the formula of ferrous sulfate it follows that ν (FeSO 4) \u003d ν (FeSO 4 7H 2 O) \u003d 0.0125 mol. Calculate the mass of FeSO 4: m (FeSO 4) \u003d ν (FeSO 4) M (FeSO 4) \u003d 0.0125 152 \u003d 1.91 g. Given that the mass of the solution consists of the mass of ferrous sulfate (3.5 g) and the mass of water (40 g), we calculate the mass fraction of ferrous sulfate in the solution. ω (FeSO 4) \u003d m (FeSO 4) / m \u003d 1.91 / 43.5 \u003d 0.044 \u003d 4.4%. Tasks for independent solution
SUBSTANCE FORMULA
REAGENTS
A) S
1) AgNO 3, Na 3 PO 4, Cl 2
SALT FORMULA
ELECTROLYSIS PRODUCTS
REACTION EQUATION
DIRECTION OF SHIFT OF CHEMICAL EQUILIBRIUM
SUBSTANCE FORMULA
REAGENT
Calculations by chemical equations
