Ege kimy in chemistry. Additional materials and equipment

For tasks 1-3, use the following row chemical elements. The answer in tasks 1-3 is a sequence of numbers, under which the chemical elements in this row are indicated.

1.S
2. Na
3 Al
4. Si
5.Mg

Task number 1

Determine the atoms of which of the elements indicated in the series in the ground state contain one unpaired electron.

Answer: 23

Explanation:

Let's write down the electronic formula for each of the indicated chemical elements and draw the electron-graphic formula of the last electronic level:

1) S: 1s 2 2s 2 2p 6 3s 2 3p 4

2) Na: 1s 2 2s 2 2p 6 3s 1

3) Al: 1s 2 2s 2 2p 6 3s 2 3p 1

4) Si: 1s 2 2s 2 2p 6 3s 2 3p 2

5) Mg: 1s 2 2s 2 2p 6 3s 2

Task number 2

From the chemical elements indicated in the row, select three metal elements. Arrange the selected elements in ascending order of restorative properties.

Write in the answer field the numbers of the selected elements in the desired sequence.

Answer: 352

Explanation:

In the main subgroups of the periodic table, metals are located under the boron-astatine diagonal, as well as in secondary subgroups. Thus, the metals from this list include Na, Al, and Mg.

metallic and therefore restorative properties elements increase when moving to the left in the period and down in the subgroup. Thus, the metallic properties of the metals listed above increase in the series Al, Mg, Na

Task number 3

From among the elements indicated in the row, select two elements that, in combination with oxygen, exhibit an oxidation state of +4.

Write down the numbers of the selected elements in the answer field.

Answer: 14

Explanation:

The main oxidation states of elements from the list presented in complex substances:

Sulfur - "-2", "+4" and "+6"

Sodium Na - "+1" (single)

Aluminum Al - "+3" (the only one)

Silicon Si - "-4", "+4"

Magnesium Mg - "+2" (single)

Task number 4

From the proposed list of substances, select two substances in which an ionic chemical bond is present.

1. KCl
2. KNO 3
3.H3BO3
4.H2SO4
5. PCl 3

Answer: 12

Explanation:

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.

Based on this criterion, ion type bonding takes place in the compounds KCl and KNO 3 .

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 - .

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 indicated by a number.

BUT	B	AT

Answer: 241

Explanation:

N 2 O 3 - non-metal oxide. All non-metal oxides except N 2 O, NO, SiO and CO are acidic.

Al 2 O 3 - metal oxide in the oxidation state +3. Metal oxides in the oxidation state +3, +4, as well as BeO, ZnO, SnO and PbO, are amphoteric.

HClO 4 is a typical representative of acids, because. during dissociation in an aqueous solution, only H + cations are formed from cations:

HClO 4 \u003d H + + ClO 4 -

Task number 6

From the proposed list of substances, select two substances, with each of which zinc interacts.

1) nitric acid (solution)

2) iron(II) hydroxide

3) magnesium sulfate (solution)

4) sodium hydroxide (solution)

5) aluminum chloride (solution)

Write down the numbers of the selected substances in the answer field.

Answer: 14

Explanation:

1) Nitric acid is a strong oxidizing agent and reacts with all metals except platinum and gold.

2) Iron hydroxide (ll) is an insoluble base. Metals do not react with insoluble hydroxides at all, and only three metals react with soluble (alkalis) - Be, Zn, Al.

3) Magnesium sulfate - more salt active metal than zinc, and therefore the reaction does not proceed.

4) Sodium hydroxide - alkali (soluble metal hydroxide). Only Be, Zn, Al work with metal alkalis.

5) AlCl 3 - a salt of a more active metal than zinc, i.e. reaction is not possible.

Task number 7

From the proposed list of substances, select two oxides that react with water.

1.BaO
2. CuO
3. NO
4 SO3
5.PbO2

Write down the numbers of the selected substances in the answer field.

Answer: 14

Explanation:

Of the oxides, only oxides of alkali and alkaline earth metals, as well as all acid oxides except SiO 2, react with water.

Thus, answer options 1 and 4 are suitable:

BaO + H 2 O \u003d Ba (OH) 2

SO 3 + H 2 O \u003d H 2 SO 4

Task number 8

1) hydrogen bromide

3) sodium nitrate

4) sulfur oxide (IV)

5) aluminum chloride

Write in the table the selected numbers under the corresponding letters.

Answer: 52

Explanation:

Salts among these substances are only sodium nitrate and aluminum chloride. All nitrates, like sodium salts, are soluble, and therefore sodium nitrate cannot precipitate in principle with any of the reagents. Therefore, salt X can only be aluminum chloride.

A common mistake among those who pass the exam in chemistry is a misunderstanding that in an aqueous solution ammonia forms a weak base - ammonium hydroxide due to the reaction:

NH 3 + H 2 O<=>NH4OH

In this regard, an aqueous solution of ammonia gives a precipitate when mixed with solutions of metal salts that form insoluble hydroxides:

3NH 3 + 3H 2 O + AlCl 3 \u003d Al (OH) 3 + 3NH 4 Cl

Task number 9

In a given transformation scheme

Cu X> CuCl2 Y>Cui

substances X and Y are:

1. AgI
2. I 2
3.Cl2
4.HCl
5.KI

Answer: 35

Explanation:

Copper is a metal located in the activity series to the right of hydrogen, i.e. does not react with acids (except H 2 SO 4 (conc.) and HNO 3). Thus, the formation of copper (ll) chloride is possible in our case only by reaction with chlorine:

Cu + Cl 2 = CuCl 2

Iodide ions (I -) cannot coexist in the same solution with divalent copper ions, because are oxidized:

Cu 2+ + 3I - \u003d CuI + I 2

Task number 10

Establish a correspondence between the reaction equation and an oxidizing agent in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.

Answer: 1433

Explanation:

An oxidizing agent in a reaction is a substance that contains an element that lowers its oxidation state.

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.

Answer: 1215

Explanation:

A) Cu(NO 3) 2 + NaOH and Cu(NO 3) 2 + Ba(OH) 2 - similar interactions. Salt with metal hydroxide reacts if the starting materials are soluble, and the products contain a precipitate, a gas, or a low-dissociating substance. Both for the first and for the second reaction, both requirements are met:

Cu(NO 3) 2 + 2NaOH = 2NaNO 3 + Cu(OH) 2 ↓

Cu(NO 3) 2 + Ba(OH) 2 = Na(NO 3) 2 + Cu(OH) 2 ↓

Cu (NO 3) 2 + Mg - the salt reacts with the metal if the free metal is more active than what is included in the salt. Magnesium in the activity series is located to the left of copper, which indicates its greater activity, therefore, the reaction proceeds:

Cu(NO 3) 2 + Mg = Mg(NO 3) 2 + Cu

B) Al (OH) 3 - metal hydroxide in the oxidation state +3. Metal hydroxides in the oxidation state +3, +4, and also, as exceptions, the hydroxides Be (OH) 2 and Zn (OH) 2, are amphoteric.

By definition, amphoteric hydroxides are those that react with alkalis and almost all soluble acids. For this reason, we can immediately conclude that answer 2 is appropriate:

Al(OH) 3 + 3HCl = AlCl 3 + 3H 2 O

Al (OH) 3 + LiOH (solution) \u003d Li or Al (OH) 3 + LiOH (solid) \u003d to \u003d\u003e LiAlO 2 + 2H 2 O

2Al(OH) 3 + 3H 2 SO 4 = Al 2 (SO 4) 3 + 6H 2 O

C) ZnCl 2 + NaOH and ZnCl 2 + Ba (OH) 2 - interaction of the "salt + metal hydroxide" type. The explanation is given in p.A.

ZnCl 2 + 2NaOH = Zn(OH) 2 + 2NaCl

ZnCl 2 + Ba(OH) 2 = Zn(OH) 2 + BaCl 2

It should be noted that with an excess of NaOH and Ba (OH) 2:

ZnCl 2 + 4NaOH \u003d Na 2 + 2NaCl

ZnCl 2 + 2Ba(OH) 2 = Ba + BaCl 2

D) Br 2, O 2 are strong oxidizing agents. Of the metals, they do not react only with silver, platinum, gold:

Cu + Br2 t° > CuBr2

2Cu + O2 t° > 2CuO

HNO 3 is an acid with strong oxidizing properties, because oxidizes not with hydrogen cations, but with an acid-forming element - nitrogen N +5. Reacts with all metals except platinum and gold:

4HNO 3 (conc.) + Cu \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O

8HNO 3 (razb.) + 3Cu \u003d 3Cu (NO 3) 2 + 2NO + 4H 2 O

Task number 12

Match between general formula homologous series and the name of the substance belonging to this series: 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.

BUT	B	AT

Answer: 231

Explanation:

Task number 13

From the proposed list of substances, select two substances that are isomers of cyclopentane.

1) 2-methylbutane

2) 1,2-dimethylcyclopropane

3) pentene-2

4) hexene-2

5) cyclopentene

Write down the numbers of the selected substances in the answer field.

Answer: 23

Explanation:

Cyclopentane has the molecular formula C 5 H 10 . Let's write the structural and molecular formulas of the substances listed in the condition

Substance name	Structural formula	Molecular formula
cyclopentane		C 5 H 10
2-methylbutane
1,2-dimethylcyclopropane		C 5 H 10
		C 5 H 10

cyclopentene

Task number 14

From the proposed list of substances, select two substances, each of which reacts with a solution of potassium permanganate.

1) methylbenzene

2) cyclohexane

3) methyl propane

Write down the numbers of the selected substances in the answer field.

Answer: 15

Explanation:

From hydrocarbons with aqueous solution potassium permanganate, those that contain C \u003d C or C \u003d C bonds in their structural formula, as well as benzene homologues (except for benzene itself) react.

Thus methylbenzene and styrene are suitable.

Task number 15

From the proposed list of substances, select two substances with which phenol interacts.

1) hydrochloric acid

2) sodium hydroxide

4) nitric acid

5) sodium sulfate

Write down the numbers of the selected substances in the answer field.

Answer: 24

Explanation:

Phenol has weak acid properties more pronounced than in alcohols. For this reason, phenols, unlike alcohols, react with alkalis:

C 6 H 5 OH + NaOH = C 6 H 5 ONa + H 2 O

Phenol contains in its molecule a hydroxyl group directly attached to the benzene ring. The hydroxy group is an orientant of the first kind, that is, it facilitates substitution reactions in the ortho and para positions:

Task number 16

From the proposed list of substances, select two substances that undergo hydrolysis.

1) glucose

2) sucrose

3) fructose

5) starch

Write down the numbers of the selected substances in the answer field.

Answer: 25

Explanation:

All of these substances are carbohydrates. Monosaccharides do not undergo hydrolysis from carbohydrates. Glucose, fructose, and ribose are monosaccharides, sucrose is a disaccharide, and starch is a polysaccharide. Consequently, sucrose and starch from the specified list are subjected to hydrolysis.

Task number 17

The following scheme of transformations of substances is given:

1,2-dibromoethane → X → bromoethane → Y → ethyl formate

Determine which of the following substances are substances X and Y.

2) ethanal

4) chloroethane

5) acetylene

Write in the table the numbers of the selected substances under the corresponding letters.

Task number 18

Establish a correspondence between the name of the starting 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.

Write in the table the selected numbers under the corresponding letters.

BUT	B	AT	G

Answer: 2134

Explanation:

Substitution at the secondary carbon atom proceeds to a greater extent than at the primary. Thus, the main product of propane bromination is 2-bromopropane and not 1-bromopropane:

Cyclohexane is a cycloalkane with a ring size of more than 4 carbon atoms. Cycloalkanes with a ring size of more than 4 carbon atoms, when interacting with halogens, enter into a substitution reaction with preservation of the cycle:

Cyclopropane and cyclobutane are cycloalkanes with minimum size cycles predominantly enter into addition reactions, accompanied by ring rupture:

The substitution of hydrogen atoms at the tertiary carbon atom occurs to a greater extent than at the secondary and primary. Thus, the bromination of isobutane proceeds mainly as follows:

Task #19

Establish a correspondence between the reaction scheme and the organic substance that is the product of this reaction: 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.

BUT	B	AT	G

Answer: 6134

Explanation:

Heating aldehydes with freshly precipitated copper hydroxide results in the oxidation of the aldehyde group to a carboxyl group:

Aldehydes and ketones are reduced by hydrogen in the presence of nickel, platinum or palladium to alcohols:

Primary and secondary alcohols are oxidized by hot CuO to aldehydes and ketones, respectively:

Under the action of concentrated sulfuric acid on ethanol during heating, the formation of two different products is possible. When heated to a temperature below 140 ° C, intermolecular dehydration predominantly occurs with the formation of diethyl ether, and when heated above 140 ° C, intramolecular dehydration occurs, as a result of which ethylene is formed:

Task number 20

From the proposed list of substances, select two substances whose thermal decomposition reaction is redox.

1) aluminum nitrate

2) potassium bicarbonate

3) aluminum hydroxide

4) ammonium carbonate

5) ammonium nitrate

Write down the numbers of the selected substances in the answer field.

Answer: 15

Explanation:

Redox reactions are such reactions as a result of which the chemical one or more chemical elements change their oxidation state.

Decomposition reactions of absolutely all nitrates are redox reactions. Metal nitrates from Mg to Cu inclusive decompose to metal oxide, nitrogen dioxide and molecular oxygen:

All metal bicarbonates decompose already with slight heating (60 ° C) to metal carbonate, carbon dioxide and water. In this case, there is no change in oxidation states:

Insoluble oxides decompose when heated. The reaction in this case is not a redox reaction, because not a single chemical element changes its oxidation state as a result of it:

Ammonium carbonate decomposes when heated into carbon dioxide, water and ammonia. The reaction is not redox:

Ammonium nitrate decomposes into nitric oxide (I) and water. The reaction refers to OVR:

Task number 21

From the proposed list, select two external influences that lead to an increase in the rate of the reaction of nitrogen with hydrogen.

1) lowering the temperature

2) pressure increase in the system

5) use of an inhibitor

Write in the answer field the numbers of the selected external influences.

Answer: 24

Explanation:

1) lowering the temperature:

The rate of any reaction decreases with decreasing temperature.

2) pressure increase in the system:

An increase in pressure increases the rate of any reaction in which at least one gaseous substance takes part.

3) decrease in hydrogen concentration

Decreasing the concentration always slows down the rate of the reaction.

4) increase in nitrogen concentration

Increasing the concentration of reactants always increases the rate of the reaction

5) use of an inhibitor

Inhibitors are substances that slow down the rate of a reaction.

Task #22

Establish a correspondence between the formula of a substance and the products of electrolysis of an aqueous solution of this substance 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.

BUT	B	AT	G

Answer: 5251

Explanation:

A) NaBr → Na + + Br -

Na + cations and water molecules compete for the cathode.

2H 2 O + 2e - → H 2 + 2OH -

2Cl - -2e → Cl 2

B) Mg (NO 3) 2 → Mg 2+ + 2NO 3 -

Mg 2+ cations and water molecules compete for the cathode.

Alkali metal cations, as well as magnesium and aluminum, are not able to recover in an aqueous solution due to their high activity. For this reason, instead of them, water molecules are restored in accordance with the equation:

2H 2 O + 2e - → H 2 + 2OH -

Anions NO 3 - and water molecules compete for the anode.

2H 2 O - 4e - → O 2 + 4H +

So the answer is 2 (hydrogen and oxygen).

C) AlCl 3 → Al 3+ + 3Cl -

2H 2 O + 2e - → H 2 + 2OH -

Anions Cl - and water molecules compete for the anode.

Anions consisting of one chemical element (except F -) win competition from water molecules for oxidation at the anode:

2Cl - -2e → Cl 2

Thus answer 5 (hydrogen and halogen) is appropriate.

D) CuSO 4 → Cu 2+ + SO 4 2-

Metal cations to the right of hydrogen in the activity series are easily reduced in an aqueous solution:

Cu 2+ + 2e → Cu 0

Acid residues containing an acid-forming element in the highest degree oxidation, lose competition to water molecules for oxidation at the anode:

2H 2 O - 4e - → O 2 + 4H +

Thus, answer 1 (oxygen and metal) is appropriate.

Task #23

Establish a correspondence between the name of the salt and the medium of the aqueous solution of this salt: 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.

BUT	B	AT	G

Answer: 3312

Explanation:

A) iron (III) sulfate - Fe 2 (SO 4) 3

formed by a weak "base" Fe(OH) 3 and strong acid H2SO4. Conclusion - acidic environment

B) chromium (III) chloride - CrCl 3

formed by the weak "base" Cr(OH) 3 and the strong acid HCl. Conclusion - acidic environment

C) sodium sulfate - Na 2 SO 4

Formed by the strong base NaOH and the strong acid H 2 SO 4 . Conclusion - neutral environment

D) sodium sulfide - Na 2 S

Formed by the strong base NaOH and the weak acid H2S. Conclusion - the environment is alkaline.

Task #24

Establish a correspondence between the method of influencing an equilibrium system

CO (g) + Cl 2 (g) COCl 2 (g) + Q

and the direction of the chemical equilibrium shift as a result of this impact: 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.

BUT	B	AT	G

Answer: 3113

Explanation:

Equilibrium shift under external impact on the system occurs in such a way as to minimize the effect of this external impact (Le Chatelier's principle).

A) An increase in the concentration of CO leads to a shift in the equilibrium towards a direct reaction, since as a result of it the amount of CO decreases.

B) An increase in temperature will shift the equilibrium towards an endothermic reaction. Since the forward reaction is exothermic (+Q), the equilibrium will shift towards the reverse reaction.

C) A decrease in pressure will shift the equilibrium in the direction of the reaction as a result of which an increase in the amount of gases occurs. As a result of the reverse reaction, more gases are formed than as a result of the forward reaction. Thus, the equilibrium will shift in the direction of the reverse reaction.

D) An increase in the concentration of chlorine leads to a shift in the equilibrium towards a direct reaction, since as a result of it the amount of chlorine decreases.

Task #25

Establish a correspondence between two substances and a reagent with which these substances can be distinguished: for each position indicated by a letter, select the corresponding position indicated by a number.

Answer: 3454

Explanation:

It is possible to distinguish two substances with the help of a third only if these two substances interact with it in different ways, and, most importantly, these differences are outwardly distinguishable.

A) Solutions of FeSO 4 and FeCl 2 can be distinguished using a solution of barium nitrate. In the case of FeSO 4, a white precipitate of barium sulfate is formed:

FeSO 4 + BaCl 2 = BaSO 4 ↓ + FeCl 2

In the case of FeCl 2, there are no visible signs of interaction, since the reaction does not proceed.

B) Solutions Na 3 PO 4 and Na 2 SO 4 can be distinguished using a solution of MgCl 2. A solution of Na 2 SO 4 does not enter into the reaction, and in the case of Na 3 PO 4 a white precipitate of magnesium phosphate precipitates:

2Na 3 PO 4 + 3MgCl 2 = Mg 3 (PO 4) 2 ↓ + 6NaCl

C) KOH and Ca(OH) 2 solutions can be distinguished using a Na 2 CO 3 solution. KOH does not react with Na 2 CO 3, but Ca (OH) 2 gives a white precipitate of calcium carbonate with Na 2 CO 3:

Ca(OH) 2 + Na 2 CO 3 = CaCO 3 ↓ + 2NaOH

D) KOH and KCl solutions can be distinguished using a MgCl 2 solution. KCl does not react with MgCl 2, and mixing solutions of KOH and MgCl 2 leads to the formation of a white precipitate of magnesium hydroxide:

MgCl 2 + 2KOH \u003d Mg (OH) 2 ↓ + 2KCl

Task #26

Establish a correspondence between the substance and its scope: for each position indicated by a letter, select the corresponding position indicated by a number.

Write in the table the selected numbers under the corresponding letters.

BUT	B	AT	G

Answer: 2331

Explanation:

Ammonia - used in the production of nitrogenous fertilizers. In particular, ammonia is a raw material for the production nitric acid, from which, in turn, fertilizers are obtained - sodium, potassium and ammonium nitrate (NaNO 3, KNO 3, NH 4 NO 3).

Carbon tetrachloride and acetone are used as solvents.

Ethylene is used to produce high-molecular compounds (polymers), namely polyethylene.

The answer to tasks 27-29 is a number. Write this number in the answer field in the text of the work, while observing the specified degree of accuracy. Then transfer this number to the ANSWER FORM No. 1 to the right of the number of the corresponding task, starting from the first cell. Write each character in a separate box in accordance with the samples given in the form. Units physical quantities no need to write.

Task number 27

What mass of potassium hydroxide must be dissolved in 150 g of water to obtain a solution with a mass fraction of alkali of 25%? (Write down the number to the nearest integer.)

Answer: 50

Explanation:

Let the mass of potassium hydroxide, which must be dissolved in 150 g of water, be x g. Then the mass of the resulting solution will be (150 + x) g, and the mass fraction of alkali in such a solution can be expressed as x / (150 + x). From the condition, we know that the mass fraction of potassium hydroxide is 0.25 (or 25%). Thus, the following equation is true:

x/(150+x) = 0.25

Thus, the mass that must be dissolved in 150 g of water to obtain a solution with a mass fraction of alkali of 25% is 50 g.

Task #28

In a reaction whose thermochemical equation

MgO (tv.) + CO 2 (g) → MgCO 3 (tv.) + 102 kJ,

entered 88 g of carbon dioxide. How much heat will be released in this case? (Write down the number to the nearest integer.)

Answer: ___________________________ kJ.

Answer: 204

Explanation:

Calculate the amount of carbon dioxide substance:

n (CO 2) \u003d n (CO 2) / M (CO 2) \u003d 88/44 \u003d 2 mol,

According to the reaction equation, the interaction of 1 mol of CO 2 with magnesium oxide releases 102 kJ. In our case, the amount of carbon dioxide is 2 mol. Denoting the amount of heat released in this case as x kJ, we can write the following proportion:

1 mol CO 2 - 102 kJ

2 mol CO 2 - x kJ

Therefore, the following equation is valid:

1 ∙ x = 2 ∙ 102

Thus, the amount of heat that will be released when 88 g of carbon dioxide participates in the reaction with magnesium oxide is 204 kJ.

Task #29

Determine the mass of zinc that reacts with hydrochloric acid to obtain 2.24 L (n.o.s.) of hydrogen. (Write down the number to tenths.)

Answer: ___________________________

Answer: 6.5

Explanation:

Let's write the reaction equation:

Zn + 2HCl \u003d ZnCl 2 + H 2

Calculate the amount of hydrogen substance:

n (H 2) \u003d V (H 2) / V m \u003d 2.24 / 22.4 \u003d 0.1 mol.

Since there are equal coefficients in front of zinc and hydrogen in the reaction equation, this means that the amounts of zinc substances that entered into the reaction and hydrogen formed as a result of it are also equal, i.e.

n (Zn) \u003d n (H 2) \u003d 0.1 mol, therefore:

m(Zn) = n(Zn) ∙ M(Zn) = 0.1 ∙ 65 = 6.5 g.

Do not forget to transfer all answers to the answer sheet No. 1 in accordance with the instructions for doing the work.

Task number 33

Sodium bicarbonate weighing 43.34 g was calcined to constant weight. The residue was dissolved in excess hydrochloric acid. The resulting gas was passed through 100 g of a 10% sodium hydroxide solution. Determine the composition and mass of the formed salt, its mass fraction in the solution. In your answer, write down the reaction equations that are indicated in the condition of the problem, and give all the necessary calculations (indicate the units of measurement of the required physical quantities).

Answer:

Explanation:

Sodium bicarbonate, when heated, decomposes in accordance with the equation:

2NaHCO 3 → Na 2 CO 3 + CO 2 + H 2 O (I)

The resulting solid residue obviously consists only of sodium carbonate. When sodium carbonate is dissolved in hydrochloric acid, the following reaction occurs:

Na 2 CO 3 + 2HCl → 2NaCl + CO 2 + H 2 O (II)

Calculate the amount of substance of sodium bicarbonate and sodium carbonate:

n (NaHCO 3) \u003d m (NaHCO 3) / M (NaHCO 3) \u003d 43.34 g / 84 g / mol ≈ 0.516 mol,

Consequently,

n (Na 2 CO 3) \u003d 0.516 mol / 2 \u003d 0.258 mol.

Calculate the amount of carbon dioxide formed by reaction (II):

n(CO 2) \u003d n (Na 2 CO 3) \u003d 0.258 mol.

Calculate the mass of pure sodium hydroxide and its amount of substance:

m(NaOH) = m solution (NaOH) ∙ ω(NaOH)/100% = 100 g ∙ 10%/100% = 10 g;

n (NaOH) \u003d m (NaOH) / M (NaOH) \u003d 10/40 \u003d 0.25 mol.

The interaction of carbon dioxide with sodium hydroxide, depending on their proportions, can proceed in accordance with two different equations:

2NaOH + CO 2 \u003d Na 2 CO 3 + H 2 O (with an excess of alkali)

NaOH + CO 2 = NaHCO 3 (with excess carbon dioxide)

From the presented equations it follows that only the average salt is obtained with a ratio of n (NaOH) / n (CO 2) ≥ 2, but only acidic, with a ratio of n (NaOH) / n (CO 2) ≤ 1.

According to calculations, ν (CO 2) > ν (NaOH), therefore:

n(NaOH)/n(CO 2) ≤ 1

Those. the interaction of carbon dioxide with sodium hydroxide occurs exclusively with the formation of an acid salt, i.e. according to the equation:

NaOH + CO 2 \u003d NaHCO 3 (III)

The calculation is carried out by the lack of alkali. According to the reaction equation (III):

n (NaHCO 3) \u003d n (NaOH) \u003d 0.25 mol, therefore:

m (NaHCO 3) \u003d 0.25 mol ∙ 84 g / mol \u003d 21 g.

The mass of the resulting solution will be the sum of the mass of the alkali solution and the mass of carbon dioxide absorbed by it.

From the reaction equation it follows that reacted, i.e. only 0.25 mol CO 2 out of 0.258 mol was absorbed. Then the mass of absorbed CO 2 is:

m(CO 2) \u003d 0.25 mol ∙ 44 g / mol \u003d 11 g.

Then, the mass of the solution is:

m (r-ra) \u003d m (r-ra NaOH) + m (CO 2) \u003d 100 g + 11 g \u003d 111 g,

and the mass fraction of sodium bicarbonate in solution will thus be equal to:

ω(NaHCO 3) \u003d 21 g / 111 g ∙ 100% ≈ 18.92%.

Task number 34

On combustion 16.2 g organic matter non-cyclic structure received 26.88 l (n.o.) of carbon dioxide and 16.2 g of water. It is known that 1 mol of this organic substance in the presence of a catalyst adds only 1 mol of water and this substance does not react with an ammonia solution of silver oxide.

Based on these conditions of the problem:

1) make the calculations necessary to establish the molecular formula of an organic substance;

2) write down the molecular formula of the organic substance;

3) make a structural formula of organic matter, which unambiguously reflects the order of bonding of atoms in its molecule;

4) write the reaction equation for the hydration of organic matter.

Answer:

Explanation:

1) To determine the elemental composition, we calculate the amounts of carbon dioxide, water and then the masses of the elements included in them:

n(CO 2) \u003d 26.88 l / 22.4 l / mol \u003d 1.2 mol;

n(CO 2) \u003d n (C) \u003d 1.2 mol; m(C) \u003d 1.2 mol ∙ 12 g / mol \u003d 14.4 g.

n(H 2 O) \u003d 16.2 g / 18 g / mol \u003d 0.9 mol; n(H) \u003d 0.9 mol ∙ 2 \u003d 1.8 mol; m(H) = 1.8 g.

m (org. in-va) \u003d m (C) + m (H) \u003d 16.2 g, therefore, there is no oxygen in organic matter.

General formula organic compound- C x H y .

x: y = ν(C) : ν(H) = 1.2: 1.8 = 1: 1.5 = 2: 3 = 4: 6

In this way the simplest formula substances C 4 H 6 . The true formula of a substance may coincide with the simplest one, or it may differ from it by an integer number of times. Those. be, for example, C 8 H 12 , C 12 H 18, etc.

The condition says that the hydrocarbon is non-cyclic and one of its molecules can attach only one molecule of water. This is possible if there is only one multiple bond (double or triple) in the structural formula of the substance. Since the desired hydrocarbon is non-cyclic, it is obvious that one multiple bond can only be for a substance with the formula C 4 H 6 . In the case of other hydrocarbons with a higher molecular weight the number of multiple bonds is greater than one everywhere. Thus, the molecular formula of the substance C 4 H 6 coincides with the simplest.

2) The molecular formula of organic matter is C 4 H 6.

3) From hydrocarbons, alkynes interact with an ammonia solution of silver oxide, in which the triple bond is located at the end of the molecule. In order for there to be no interaction with an ammonia solution of silver oxide, the alkyne of the composition C 4 H 6 must have the following structure:

CH 3 -C≡C-CH 3

4) Hydration of alkynes proceeds in the presence of divalent mercury salts.

To complete tasks 1-3, use the following row of chemical elements. The answer in tasks 1-3 is a sequence of numbers, under which the chemical elements in this row are indicated.

1) Na 2) K 3) Si 4) Mg 5) C

Task number 1

Determine which atoms of the elements indicated in the series have four electrons on the external energy level.

Answer: 3; 5

The number of electrons in the outer energy level (electronic layer) of the elements of the main subgroups is equal to the group number.

Thus, from the presented answers, silicon and carbon are suitable, because. they are in the main subgroup of the fourth group of the table D.I. Mendeleev (IVA group), i.e. Answers 3 and 5 are correct.

Task number 2

From the chemical elements indicated in the series, select three elements that are in the Periodic Table of Chemical Elements D.I. Mendeleev are in the same period. Arrange the selected elements in ascending order of their metallic properties.

Write in the answer field the numbers of the selected elements in the desired sequence.

Answer: 3; four; one

Three of the presented elements are in the same period - sodium Na, silicon Si and magnesium Mg.

When moving within one period of the Periodic Table, D.I. Mendeleev (horizontal lines) from right to left, the return of electrons located on the outer layer is facilitated, i.e. the metallic properties of the elements are enhanced. Thus, the metallic properties of sodium, silicon and magnesium are enhanced in the series Si

Task number 3

From among the elements listed in the row, select two elements that exhibit the lowest oxidation state, equal to -4.

Write down the numbers of the selected elements in the answer field.

Answer: 3; 5

According to the octet rule, the atoms of chemical elements tend to have 8 electrons in their outer electronic level, like the noble gases. This can be achieved either by donating electrons of the last level, then the previous one, containing 8 electrons, becomes external, or, conversely, by adding additional electrons up to eight. Sodium and potassium are alkali metals and are in the main subgroup of the first group (IA). This means that on the outer electron layer of their atoms there is one electron each. In this regard, the loss of a single electron is energetically more favorable than the addition of seven more. With magnesium, the situation is similar, only it is in the main subgroup of the second group, that is, it has two electrons on the outer electronic level. It should be noted that sodium, potassium and magnesium are metals, and for metals, in principle, a negative oxidation state is impossible. The minimum oxidation state of any metal is zero and is observed in simple substances.

The chemical elements carbon C and silicon Si are non-metals and are in the main subgroup of the fourth group (IVA). This means that there are 4 electrons on their outer electron layer. For this reason, for these elements, both the return of these electrons and the addition of four more up to a total of 8 are possible. Silicon and carbon atoms cannot attach more than 4 electrons, therefore the minimum oxidation state for them is -4.

Task number 4

From the proposed list, select two compounds in which there is an ionic chemical bond.

1. Ca(ClO 2) 2
2. HClO 3
3.NH4Cl
4. HClO 4
5.Cl2O7

Answer: 1; 3

On this basis, we establish that there is an ionic bond in compound number 1 - Ca(ClO 2) 2, because in its formula, one can see atoms of a typical calcium metal and atoms of non-metals - oxygen and chlorine.

However, there are no more compounds containing both metal and non-metal atoms in this list.

Among the compounds indicated in the assignment there is ammonium chloride, in which the ionic bond is realized between the ammonium cation NH 4 + and the chloride ion Cl − .

Task number 5

Write down the numbers of the selected connections in the answer field.

Answer: A-4; B-1; AT 3

Explanation:

Acid salts are salts resulting from the incomplete replacement of mobile hydrogen atoms by a metal cation, ammonium cation or alkyl ammonium.

In not organic acids ah, 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 acidic 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 H2CO3.

In organic substances, only hydrogen atoms that are part of carboxyl groups (-COOH) or hydroxyl groups of phenols (Ar-OH) can be replaced by metal atoms. That is, for example, sodium acetate CH 3 COONa, despite the fact that not all hydrogen atoms in its molecule are replaced by metal cations, is an average, not an acid salt (!). Hydrogen atoms in organic substances, attached directly to the carbon atom, are almost never able to be replaced by metal atoms, with the exception of hydrogen atoms in the triple C≡C bond.

Non-salt-forming oxides are oxides of non-metals that do not form salts with basic oxides or bases, that is, they either do not react with them at all (most often), or give a different product (not a salt) in reaction with them. It is often said that non-salt-forming oxides are oxides of non-metals that do not react with bases and basic oxides. However, for the detection of non-salt-forming oxides, this approach does not always work. So, for example, CO, being a non-salt-forming oxide, reacts with basic iron (II) oxide, but with the formation of a free metal rather than a salt:

CO + FeO = CO 2 + Fe

Non-salt-forming oxides from the school chemistry course include non-metal oxides in the oxidation state +1 and +2. In total, they are found in the USE 4 - these are CO, NO, N 2 O and SiO (I personally never met the last SiO in assignments).

Task number 6

From the proposed list of substances, select two substances, with each of which iron reacts without heating.

zinc chloride
copper(II) sulfate
concentrated nitric acid
dilute hydrochloric acid
aluminium oxide

Answer: 2; four

Zinc chloride is a salt, and iron is a metal. The metal reacts with the salt only if it is more reactive than the one in the salt. The relative activity of metals is determined by a series of metal activity (in other words, a series of metal stresses). Iron is located to the right of zinc in the activity series of metals, which means that it is less active and is not able to displace zinc from salt. That is, the reaction of iron with substance No. 1 does not go.

Copper (II) sulfate CuSO 4 will react with iron, since iron is located to the left of copper in the activity series, that is, it is a more active metal.

Concentrated nitric acid, as well as concentrated sulfuric acid, are not able to react with iron, aluminum and chromium without heating due to such a phenomenon as passivation: on the surface of these metals, under the action of these acids, an insoluble salt is formed without heating, which acts as a protective shell. However, when heated, this protective shell dissolves and the reaction becomes possible. Those. since it is indicated that there is no heating, the reaction of iron with conc. HNO 3 does not leak.

Hydrochloric acid, regardless of concentration, refers to non-oxidizing acids. Metals that are in the activity series to the left of hydrogen react with non-oxidizing acids with the release of hydrogen. Iron is one of these metals. Conclusion: the reaction of iron with hydrochloric acid proceeds.

In the case of a metal and a metal oxide, the reaction, as in the case of a salt, is possible if the free metal is more active than that which is part of the oxide. Fe, according to the activity series of metals, is less active than Al. This means that Fe does not react with Al 2 O 3.

Task number 7

From the proposed list, select two oxides that react with a solution of hydrochloric acid, but do not react with sodium hydroxide solution.

1. CO
2 SO 3
3. CuO
4. MgO
5. ZnO

Write down the numbers of the selected substances in the answer field.

Answer: 3; four

CO is a non-salt-forming oxide; it does not react with an aqueous solution of alkali.

(It should be remembered that, nevertheless, under harsh conditions - high pressure and temperature - it still reacts with solid alkali, forming formates - salts of formic acid.)

SO 3 - sulfur oxide (VI) - acid oxide, which corresponds to sulphuric 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 is an oxide with pronounced amphoteric properties- easily reacts with both strong bases and acids (as well as acidic and basic oxides). Not suitable.

Task number 8

1.KOH
2.HCl
3. Cu(NO 3) 2
4.K2SO3
5. Na 2 SiO 3

Answer: 4; 2

In the reaction between two salts of inorganic acids, gas is formed only when hot solutions of nitrites and ammonium salts are mixed due to the formation of thermally unstable ammonium nitrite. For example,

NH 4 Cl + KNO 2 \u003d t o \u003d\u003e N 2 + 2H 2 O + KCl

However, both nitrites and ammonium salts are not on the list.

This means that one of the three salts (Cu (NO 3) 2, K 2 SO 3 and Na 2 SiO 3) reacts with either an acid (HCl) or an alkali (NaOH).

Among the salts of inorganic acids, only ammonium salts emit gas when interacting with alkalis:

NH 4 + + OH \u003d NH 3 + H 2 O

Ammonium salts, as we have already said, are not on the list. The only option left is the interaction of the salt with the acid.

Salts among these substances include Cu(NO 3) 2, K 2 SO 3 and Na 2 SiO 3. The reaction of copper nitrate with hydrochloric acid does not proceed, because no gas, no precipitate, no low-dissociating substance (water or weak acid) is formed. Sodium silicate reacts with hydrochloric acid, however, due to the release of a white gelatinous precipitate of silicic acid, and not gas:

Na 2 SiO 3 + 2HCl \u003d 2NaCl + H 2 SiO 3 ↓

The last option remains - the interaction of potassium sulfite and hydrochloric acid. Indeed, as a result of the ion exchange reaction between sulfite and almost any acid, unstable sulfurous acid is formed, which instantly decomposes into colorless gaseous sulfur oxide (IV) and water.

Task number 9

1. KCl (solution)
2.K2O
3.H2
4. HCl (excess)
5. CO 2 (solution)

Write in the table the numbers of the selected substances under the corresponding letters.

Answer: 2; 5

CO 2 is an acidic oxide and must be treated with either a basic oxide or a base to convert it to a salt. Those. to obtain potassium carbonate from CO 2, it must be treated with either potassium oxide or potassium hydroxide. Thus, substance X is potassium oxide:

K 2 O + CO 2 \u003d K 2 CO 3

Potassium bicarbonate KHCO 3, like potassium carbonate, is a salt of carbonic acid, with the only difference being that the bicarbonate is a product of incomplete substitution of hydrogen atoms in carbonic acid. To obtain an acid salt from a normal (medium) salt, one must either act on it with the same acid that formed this salt, or else act on it with an acid oxide corresponding to this acid in the presence of water. Thus reactant Y is carbon dioxide. When it is passed through an aqueous solution of potassium carbonate, the latter turns into potassium bicarbonate:

K 2 CO 3 + H 2 O + CO 2 \u003d 2KHCO 3

Task number 10

Establish a correspondence between the reaction equation and the property of the nitrogen element that it exhibits in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.

Write in the table the numbers of the selected substances under the corresponding letters.

Answer: A-4; B-2; IN 2; G-1

A) NH 4 HCO 3 - salt, which includes the ammonium cation NH 4 +. In the ammonium cation, nitrogen always has an oxidation state of -3. As a result of the reaction, it turns into ammonia NH 3. Hydrogen almost always (except for its compounds with metals) has an oxidation state of +1. Therefore, for the ammonia molecule to be electrically neutral, nitrogen must have an oxidation state of -3. Thus, there is no change in the degree of nitrogen oxidation; it does not exhibit redox properties.

B) As already shown above, nitrogen in ammonia NH 3 has an oxidation state of -3. As a result of the reaction with CuO, ammonia is converted into a simple substance N 2. In any simple substance, the oxidation state of the element with which it is formed is equal to zero. Thus, the nitrogen atom loses its negative charge, and since electrons are responsible for the negative charge, this means that they are lost by the nitrogen atom as a result of the reaction. An element that loses some of its electrons in a reaction is called a reducing agent.

C) As a result of the reaction, NH 3 with an oxidation state of nitrogen equal to -3 turns into nitric oxide NO. Oxygen almost always has an oxidation state of -2. Therefore, in order for the nitric oxide molecule to be electrically neutral, the nitrogen atom must have an oxidation state of +2. This means that the nitrogen atom changed its oxidation state from -3 to +2 as a result of the reaction. This indicates the loss of 5 electrons by the nitrogen atom. That is, nitrogen, as in the case of B, is a reducing agent.

D) N 2 is a simple substance. In all simple substances, the element that forms them has an oxidation state of 0. As a result of the reaction, nitrogen is converted into lithium nitride Li3N. The only oxidation state of an alkali metal other than zero (any element has an oxidation state of 0) is +1. Thus, for the Li3N structural unit to be electrically neutral, nitrogen must have an oxidation state of -3. It turns out that as a result of the reaction, nitrogen acquired a negative charge, which means the addition of electrons. Nitrogen is the oxidizing agent in this reaction.

Task number 11

SUBSTANCE FORMULA

REAGENTS

D) ZnBr 2 (solution)

1) AgNO 3, Na 3 PO 4, Cl 2

2) BaO, H 2 O, KOH

3) H 2, Cl 2, O 2

4) HBr, LiOH, CH 3 COOH

5) H 3 PO 4, BaCl 2, CuO

Write in the table the numbers of the selected substances under the corresponding letters.

Answer: A-3; B-2; AT 4; G-1

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 fundamental level remember that sulfur reacts with chlorine. Chlorine is a strong oxidizing agent, sulfur often exhibits a dual function - both oxidizing and reducing. That is, if a strong oxidizing agent acts on sulfur, which is molecular chlorine Cl 2, it will oxidize.

Sulfur burns with a blue flame in oxygen to form a gas with a pungent odor - sulfur dioxide SO 2:

B) SO 3 - sulfur oxide (VI) has pronounced acidic properties. For such oxides, the most characteristic reactions are interactions with water, as well as with basic and amphoteric oxides and hydroxides. In the list at number 2, we just see water, and the basic oxide BaO, and hydroxide KOH.

When an acid oxide reacts with a basic oxide, a salt of the corresponding acid and a metal that is part of the basic oxide is formed. An acidic oxide corresponds to an acid in which the acid-forming element has the same oxidation state as in the oxide. The oxide SO 3 corresponds to sulfuric acid H 2 SO 4 (both there and there the oxidation state of sulfur is +6). Thus, when SO 3 interacts with metal oxides, sulfuric acid salts will be obtained - sulfates containing the sulfate ion SO 4 2-:

SO 3 + BaO = BaSO 4

When interacting with water, the acid oxide turns into the corresponding acid:

SO 3 + H 2 O \u003d H 2 SO 4

And when acid oxides interact with metal hydroxides, a salt of the corresponding acid and water are formed:

SO 3 + 2KOH \u003d K 2 SO 4 + H 2 O

C) Zinc hydroxide Zn (OH) 2 has typical amphoteric properties, that is, it reacts both with acidic oxides and acids, and with basic oxides and alkalis. In list 4, we see both acids - hydrobromic HBr and acetic, and alkali - LiOH. Recall that water-soluble metal hydroxides are called alkalis:

Zn(OH) 2 + 2HBr = ZnBr 2 + 2H 2 O

Zn (OH) 2 + 2CH 3 COOH \u003d Zn (CH 3 COO) 2 + 2H 2 O

Zn(OH) 2 + 2LiOH \u003d Li 2

D) Zinc bromide ZnBr 2 is a salt, soluble in water. For soluble salts, ion exchange reactions are the most common. A salt can react with another salt provided that both starting salts are soluble and a precipitate forms. Also ZnBr 2 contains bromide ion Br-. Metal halides are characterized by the fact that they are able to react with Hal 2 halogens, which are higher in the periodic table. In this way? the described types of reactions proceed with all substances of list 1:

ZnBr 2 + 2AgNO 3 \u003d 2AgBr + Zn (NO 3) 2

3ZnBr 2 + 2Na 3 PO 4 = Zn 3 (PO 4) 2 + 6NaBr

ZnBr 2 + Cl 2 = ZnCl 2 + Br 2

Task number 12

Establish a correspondence between the name of the substance and the class / group to which this substance belongs: for each position indicated by a letter, select the corresponding position indicated by a number.

Write in the table the numbers of the selected substances under the corresponding letters.

Answer: A-4; B-2; IN 1

Explanation:

A) Methylbenzene, also known as toluene, has the structural formula:

As you can see, the molecules of this substance consist only of carbon and hydrogen, therefore methylbenzene (toluene) refers to hydrocarbons

B) The structural formula of aniline (aminobenzene) is as follows:

As can be seen from the structural formula, the aniline molecule consists of an aromatic hydrocarbon radical (C 6 H 5 -) and an amino group (-NH 2), thus, aniline belongs to aromatic amines, i.e. correct answer 2.

C) 3-methylbutanal. The ending "al" indicates that the substance belongs to aldehydes. The structural formula of this substance:

Task number 13

From the proposed list, select two substances that are structural isomers of butene-1.

butane
cyclobutane
butin-2
butadiene-1,3
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.

Task number 14

From the proposed list, select two substances, the interaction of which with a solution of potassium permanganate will cause a change in the color of the solution.

cyclohexane
benzene
toluene
propane
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 or to carboxylic acids, or to their corresponding salts. Thus, option 2 (benzene) is eliminated.

The correct answers are 3 (toluene) and 5 (propylene). Both substances decolorize the purple solution of potassium permanganate due to the reactions taking place:

CH 3 -CH=CH 2 + 2KMnO 4 + 2H 2 O → CH 3 -CH(OH)–CH 2 OH + 2MnO 2 + 2KOH

Task number 15

From the proposed list, select two substances with which formaldehyde reacts.

1. Cu
2. N 2
3.H2
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; four

Explanation:

Formaldehyde belongs to the class of aldehydes - oxygen-containing organic compounds that have an aldehyde group at the end of the molecule:

Typical reactions of aldehydes are oxidation and reduction reactions proceeding along the functional group.

Among the list of responses for formaldehyde, reduction reactions are typical, where hydrogen is used as a reducing agent (cat. - Pt, Pd, Ni), and oxidation - in this case, the silver mirror reaction.

When reduced with hydrogen on a nickel catalyst, formaldehyde is converted to methanol:

The silver mirror reaction is the reduction of silver from an ammonia solution of silver oxide. When dissolved in an aqueous solution of ammonia, silver oxide turns into a complex compound - diammine silver (I) OH hydroxide. After the addition of formaldehyde, a redox reaction occurs in which silver is reduced:

Task number 16

From the proposed list, select two substances with which methylamine reacts.

propane
chloromethane
hydrogen
sodium hydroxide
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. A characteristic feature of amines is the presence of a lone electron pair on the nitrogen atom, as a result of which amines exhibit the properties of bases and act as nucleophiles in reactions. Thus, in this regard, from the proposed answers, methylamine as a base and nucleophile reacts with chloromethane and hydrochloric acid:

CH 3 NH 2 + CH 3 Cl → (CH 3) 2 NH 2 + Cl -

CH 3 NH 2 + HCl → CH 3 NH 3 + Cl -

Task number 17

The following scheme of transformations of substances is given:

Determine which of the given substances are substances X and Y.

1.H2
2. CuO
3. Cu(OH) 2
4. NaOH (H 2 O)
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 ethyl alcohol. The oxidation of alcohols is carried out on a copper catalyst or using CuO:

Task number 18

Establish a correspondence between the name of the substance and the product that is mainly formed during the interaction of this substance with bromine: for each position indicated by a letter, select the corresponding position indicated by a number.

Answer: 5; 2; 3; 6

Explanation:

For alkanes, the most characteristic reactions are free radical substitution reactions, during which a hydrogen atom is replaced by a halogen atom. Thus, by brominating ethane, one can obtain bromoethane, and by brominating isobutane, 2-bromoisobutane can be obtained:

Since the small cycles of cyclopropane and cyclobutane molecules are unstable, during bromination the cycles of these molecules are opened, thus the addition reaction proceeds:

Unlike the cyclopropane and cyclobutane cycles, the cyclohexane cycle large sizes, resulting in the replacement of a hydrogen atom by a bromine atom:

Task #19

Establish a correspondence between the reacting substances and the carbon-containing product that is formed during the interaction of these substances: for each position indicated by a letter, select the corresponding position indicated by a number.

Write in the table the selected numbers under the corresponding letters.

Answer: 5; four; 6; 2

Task number 20

From the proposed list of types of reactions, select two types of reactions, which include the interaction of alkali metals with water.

catalytic
homogeneous
irreversible
redox
neutralization reaction

Write down the numbers of the selected types of reactions in the answer field.

Answer: 3; four

Alkali metals (Li, Na, K, Rb, Cs, Fr) are located in the main subgroup of group I of the table D.I. Mendeleev and are reducing agents, easily donating an electron located at the outer level.

If we denote the alkali metal with the letter M, then the reaction of the alkali metal with water will look like this:

2M + 2H 2 O → 2MOH + H 2

Alkali metals are very active towards water. The reaction proceeds violently with the release of a large amount of heat, is irreversible and does not require the use of a catalyst (non-catalytic) - a substance that accelerates the reaction and is not part of the reaction products. It should be noted that all highly exothermic reactions do not require the use of a catalyst and proceed irreversibly.

Since metal and water are substances that are in different states of aggregation, 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 substance. (A neutralization reaction occurs between an acid and a base, as a result of which these substances exchange their constituent parts and a salt and a low-dissociating substance are formed).

As mentioned above, alkali metals are reducing agents, donating an electron from the outer layer, therefore, the reaction is redox.

Task number 21

From the proposed list of external influences, select two influences that lead to a decrease in the rate of the reaction of ethylene with hydrogen.

temperature drop
increase in ethylene concentration
use of a catalyst
decrease in hydrogen concentration
pressure increase in the system

Write in the answer field the numbers of the selected external influences.

Answer: 1; four

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 rule of thumb, for every 10 degrees increase in temperature, the rate constant is homogeneous reaction increases by 2-4 times. Therefore, a decrease in temperature also leads to a decrease in the reaction rate. The first answer is correct.

As noted above, the reaction rate is also affected by a change in the concentration of reagents: if the concentration of ethylene is increased, the reaction rate will also increase, which does not meet the requirements of the problem. And a decrease in the concentration of hydrogen - the initial component, on the contrary, reduces the reaction rate. Therefore, the second option is not suitable, but the fourth one is.

A catalyst is a substance that speeds up the rate of a chemical reaction but is not part of the products. The use of a catalyst accelerates the ethylene hydrogenation reaction, which also does not correspond to the condition of the problem, and therefore is not the right answer.

When ethylene reacts with hydrogen (on Ni, Pd, Pt catalysts), ethane is formed:

CH 2 \u003d CH 2 (g) + H 2 (g) → CH 3 -CH 3 (g)

All components involved in the reaction and the product are gaseous substances, therefore, the pressure in the system will also affect the reaction rate. From two volumes of ethylene and hydrogen, one volume of ethane is formed, therefore, the reaction proceeds to a decrease in pressure in the system. By increasing the pressure, we will speed up the reaction. The fifth answer does not fit.

Task #22

Establish a correspondence between the formula of salt and the products of electrolysis of an aqueous solution of this salt, which stood out on inert electrodes: for each position,

SALT FORMULA

ELECTROLYSIS PRODUCTS

Write in the table the selected numbers under the corresponding letters.

Answer: 1; four; 3; 2

Electrolysis is a redox process that occurs on the electrodes during the passage of a constant electric current through an electrolyte solution or melt. At the cathode, the reduction occurs predominantly of those cations that have the highest oxidizing activity. At the anode, those anions are oxidized first of all, which have the greatest reduction ability.

Electrolysis of aqueous solution

1) The process of electrolysis of aqueous solutions on the cathode does not depend on the material of the cathode, but depends on the position of the metal cation in the electrochemical series of voltages.

For cations in a row

Li + - Al 3+ recovery process:

2H 2 O + 2e → H 2 + 2OH - (H 2 is released at the cathode)

Zn 2+ - Pb 2+ recovery process:

Me n + + ne → Me 0 and 2H 2 O + 2e → H 2 + 2OH - (H 2 and Me will be released at the cathode)

Cu 2+ - Au 3+ reduction process Me n + + ne → Me 0 (Me is released at the cathode)

2) The process of electrolysis of aqueous solutions at the anode depends on the material of the anode and on the nature of the anion. If the anode is insoluble, i.e. inert (platinum, gold, coal, graphite), the process will depend only on the nature of the anions.

For anions F -, SO 4 2-, NO 3 -, PO 4 3-, OH - the oxidation process:

4OH - - 4e → O 2 + 2H 2 O or 2H 2 O - 4e → O 2 + 4H + (oxygen is released at the anode) halide ions (except F-) oxidation process 2Hal - - 2e → Hal 2 (free halogens are released ) organic acids oxidation process:

2RCOO - - 2e → R-R + 2CO 2

The overall electrolysis equation is:

A) Na 3 PO 4 solution

2H 2 O → 2H 2 (at the cathode) + O 2 (at the anode)

B) KCl solution

2KCl + 2H 2 O → H 2 (at the cathode) + 2KOH + Cl 2 (at the anode)

C) CuBr2 solution

CuBr 2 → Cu (at the cathode) + Br 2 (at the anode)

D) Cu(NO3)2 solution

2Cu(NO 3) 2 + 2H 2 O → 2Cu (at the cathode) + 4HNO 3 + O 2 (at the anode)

Task #23

Establish a correspondence between the name of the salt and the ratio of this salt to hydrolysis: for each position indicated by a letter, select the corresponding position indicated by a number.

Write in the table the selected numbers under the corresponding letters.

Answer: 1; 3; 2; four

Hydrolysis of salts - the interaction of salts with water, leading to the addition of the hydrogen cation H + of the water molecule to the anion of the acid residue and (or) the hydroxyl group OH - of the water molecule to the metal cation. Salts formed by cations corresponding to weak bases and anions corresponding to weak acids undergo hydrolysis.

A) Ammonium chloride (NH 4 Cl) - a salt formed by strong hydrochloric acid and ammonia ( weak base), undergoes hydrolysis at the cation.

NH 4 Cl → NH 4 + + Cl -

NH 4 + + H 2 O → NH 3 H 2 O + H + (formation of ammonia dissolved in water)

The solution medium is acidic (pH< 7).

B) Potassium sulfate (K 2 SO 4) - a salt formed by strong sulfuric acid and potassium hydroxide (alkali, i.e. strong base), does not undergo hydrolysis.

K 2 SO 4 → 2K + + SO 4 2-

C) Sodium carbonate (Na 2 CO 3) - a salt formed by a weak carbonic acid and sodium hydroxide (an alkali, i.e. a strong base), undergoes anion hydrolysis.

CO 3 2- + H 2 O → HCO 3 - + OH - (formation of a weakly dissociating hydrocarbonate ion)

The solution is alkaline (pH > 7).

D) Aluminum sulfide (Al 2 S 3) - a salt formed by a weak hydrosulfide acid and aluminum hydroxide (weak base), undergoes complete hydrolysis with the formation of aluminum hydroxide and hydrogen sulfide:

Al 2 S 3 + 6H 2 O → 2Al(OH) 3 + 3H 2 S

The solution medium is close to neutral (pH ~ 7).

Task #24

Establish a correspondence between the equation of a chemical reaction and the direction of displacement of chemical equilibrium with increasing pressure in the system: for each position indicated by a letter, select the corresponding position indicated by a number.

REACTION EQUATION

A) N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g)

B) 2H 2 (g) + O 2 (g) ↔ 2H 2 O (g)

C) H 2 (g) + Cl 2 (g) ↔ 2HCl (g)

D) SO 2 (g) + Cl 2 (g) ↔ SO 2 Cl 2 (g)

DIRECTION OF SHIFT OF CHEMICAL EQUILIBRIUM

1) shifts towards a direct reaction

2) shifts towards the back reaction

3) there is no shift in equilibrium

Write in the table the selected numbers under the corresponding letters.

Answer: A-1; B-1; AT 3; G-1

The reaction is in chemical equilibrium when the rate of the forward reaction is equal to the rate of the reverse. The shift of equilibrium in the desired direction is achieved by changing the reaction conditions.

Factors that determine the position of equilibrium:

- pressure: an increase in pressure shifts the equilibrium towards a reaction leading to a decrease in volume (conversely, a decrease in pressure shifts the equilibrium towards a reaction leading to an increase in volume)

- temperature: an increase in temperature shifts the equilibrium towards an endothermic reaction (conversely, a decrease in temperature shifts the equilibrium towards an exothermic reaction)

- concentrations of starting substances and reaction products: an increase in the concentration of the starting substances and the removal of products from the reaction sphere shift the equilibrium towards the forward reaction (on the contrary, a decrease in the concentration of the starting substances and an increase in the reaction products shift the equilibrium towards the reverse reaction)

- Catalysts do not affect the equilibrium shift, but only accelerate its achievement

A) In the first case, the reaction proceeds with a decrease in volume, since V (N 2) + 3V (H 2) > 2V (NH 3). By increasing the pressure in the system, the equilibrium will shift to the side with a smaller volume of substances, therefore, in the forward direction (in the direction of the direct reaction).

B) In the second case, the reaction also proceeds with a decrease in volume, since 2V (H 2) + V (O 2) > 2V (H 2 O). By increasing the pressure in the system, the equilibrium will also shift in the direction of the direct reaction (in the direction of the product).

C) In the third case, the pressure does not change during the reaction, because V (H 2) + V (Cl 2) \u003d 2V (HCl), so there is no equilibrium shift.

D) In the fourth case, the reaction also proceeds with a decrease in volume, since V (SO 2) + V (Cl 2) > V (SO 2 Cl 2). By increasing the pressure in the system, the equilibrium will shift towards the formation of the product (direct reaction).

Task #25

Establish a correspondence between the formulas of substances and a reagent with which you can distinguish their aqueous solutions: for each position indicated by a letter, select the corresponding position indicated by a number.

SUBSTANCE FORMULA

A) HNO 3 and H 2 O

C) NaCl and BaCl 2

D) AlCl 3 and MgCl 2

Write in the table the selected numbers under the corresponding letters.

Answer: A-1; B-3; AT 3; G-2

A) Nitric acid and water can be distinguished using salt - calcium carbonate CaCO 3. Calcium carbonate does not dissolve in water, and when interacting with nitric acid forms a soluble salt - calcium nitrate Ca (NO 3) 2, while the reaction is accompanied by the release of 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:

MgCl 2 + 2KOH → Mg(OH) 2 ↓ + 2KCl

When alkali interacts with aluminum chloride, a precipitate first forms, which then dissolves to form a complex salt - potassium tetrahydroxoaluminate:

AlCl 3 + 4KOH → K + 3KCl

Task #26

Establish a correspondence between the substance and its scope: for each position indicated by a letter, select the corresponding position indicated by a number.

Write in the table the selected numbers under the corresponding letters.

Answer: A-4; B-2; AT 3; G-5

A) Ammonia is the most important product of the chemical industry, its production is more than 130 million tons per year. Ammonia is mainly used in the production of nitrogen fertilizers (ammonium nitrate and sulfate, urea), medicines, explosives, nitric acid, and soda. Among the proposed answers, the area of application of ammonia is the production of fertilizers (Fourth answer option).

B) Methane is the simplest hydrocarbon, the most thermally stable representative of a number of saturated compounds. It is widely used as a domestic and industrial fuel, as well as a raw material for industry (Second answer). Methane is 90-98% a component of natural gas.

C) Rubbers are materials that are obtained by polymerization of compounds with conjugated double bonds. Isoprene just belongs to this type of compounds and is used to obtain one of the types of rubbers:

D) Low molecular weight alkenes are used to make plastics, in particular ethylene is used to make a plastic called polyethylene:

n CH 2 \u003d CH 2 → (-CH 2 -CH 2 -) n

Task number 27

Calculate the mass of potassium nitrate (in grams) that should be dissolved in 150 g of a solution with a mass fraction of this salt of 10% to obtain a solution with a mass fraction of 12%. (Write down the number to tenths.)

Answer: 3.4 g

Explanation:

Let x g be the mass of potassium nitrate, which is dissolved in 150 g of the solution. Calculate the mass of potassium nitrate dissolved in 150 g of solution:

m(KNO 3) \u003d 150 g 0.1 \u003d 15 g

In order for the mass fraction of salt to be 12%, x g of potassium nitrate was added. In this case, the mass of the solution was (150 + x) g. We write the equation in the form:

(Write down the number to tenths.)

Answer: 14.4 g

Explanation:

As a result of the complete combustion of hydrogen sulfide, sulfur dioxide and water are formed:

2H 2 S + 3O 2 → 2SO 2 + 2H 2 O

A consequence of Avogadro's law is that the volumes of gases under the same conditions are related to each other in the same way as the number of moles of these gases. Thus, according to the reaction equation:

ν(O 2) = 3/2ν(H 2 S),

therefore, the volumes of hydrogen sulfide and oxygen are related to each other in exactly the same way:

V (O 2) \u003d 3 / 2V (H 2 S),

V (O 2) \u003d 3/2 6.72 l \u003d 10.08 l, hence V (O 2) \u003d 10.08 l / 22.4 l / mol \u003d 0.45 mol

Calculate the mass of oxygen required for the complete combustion of hydrogen sulfide:

m(O 2) \u003d 0.45 mol 32 g / mol \u003d 14.4 g

Task number 30

Using the electron balance method, write the equation for the reaction:

Na 2 SO 3 + ... + KOH → K 2 MnO 4 + ... + H 2 O

Determine the oxidizing agent and reducing agent.

Mn +7 + 1e → Mn +6 │2 reduction reaction

S +4 − 2e → S +6 │1 oxidation reaction

Mn +7 (KMnO 4) - oxidizing agent, S +4 (Na 2 SO 3) - reducing agent

Na 2 SO 3 + 2KMnO 4 + 2KOH → 2K 2 MnO 4 + Na 2 SO 4 + H 2 O

Task number 31

Iron was dissolved in hot concentrated sulfuric acid. The resulting salt was treated with an excess of sodium hydroxide solution. The brown precipitate formed was filtered off and dried. The resulting substance was heated with iron.

Write the equations for the four described reactions.

1) Iron, like aluminum and chromium, does not react with concentrated sulfuric acid, becoming covered with a protective oxide film. The reaction occurs only when heated with the release of sulfur dioxide:

2Fe + 6H 2 SO 4 → Fe 2 (SO 4) 2 + 3SO 2 + 6H 2 O (on heating)

2) Iron (III) sulfate - a salt soluble in water, enters into an exchange reaction with alkali, as a result of which iron (III) hydroxide precipitates (brown compound):

Fe 2 (SO 4) 3 + 3NaOH → 2Fe(OH) 3 ↓ + 3Na 2 SO 4

3) Insoluble metal hydroxides decompose upon calcination to the corresponding oxides and water:

2Fe(OH) 3 → Fe 2 O 3 + 3H 2 O

4) When iron (III) oxide is heated with metallic iron, iron (II) oxide is formed (iron in the FeO compound has an intermediate oxidation state):

Fe 2 O 3 + Fe → 3FeO (on heating)

Task #32

Write the reaction equations that can be used to carry out the following transformations:

When writing reaction equations, use the structural formulas of organic substances.

1) Intramolecular dehydration occurs at temperatures above 140 o C. This occurs as a result of the elimination of a hydrogen atom from the carbon atom of the alcohol, located one through to the alcohol hydroxyl (in the β-position).

CH 3 -CH 2 -CH 2 -OH → CH 2 \u003d CH-CH 3 + H 2 O (conditions - H 2 SO 4, 180 o C)

Intermolecular dehydration proceeds at a temperature below 140 o C under the action of sulfuric acid and ultimately comes down to the elimination of one water molecule from two alcohol molecules.

2) Propylene refers to unsymmetrical alkenes. When hydrogen halides and water are added, a hydrogen atom is added to the carbon atom at the multiple bond associated with a large number 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

Task number 33

Determine the mass fractions (in%) of iron (II) sulfate and aluminum sulfide in the mixture, if during the treatment of 25 g of this mixture with water a gas was released that completely reacted with 960 g of a 5% solution of copper (II) sulfate.

In response, write down the reaction equations that are indicated in the condition of the problem, and give all the necessary calculations (indicate the units of measurement of the required physical quantities).

Answer: ω(Al 2 S 3) = 40%; ω(CuSO 4) = 60%

When a mixture of iron (II) sulfate and aluminum sulfide is treated with water, the sulfate is simply dissolved, and the sulfide is hydrolyzed to form aluminum (III) hydroxide and hydrogen sulfide:

Al 2 S 3 + 6H 2 O → 2Al(OH) 3 ↓ + 3H 2 S (I)

When hydrogen sulfide is passed through a solution of copper (II) sulfate, copper (II) sulfide precipitates:

CuSO 4 + H 2 S → CuS↓ + H 2 SO 4 (II)

Calculate the mass and amount of substance of dissolved copper(II) sulfate:

m (CuSO 4) \u003d m (p-ra) ω (CuSO 4) \u003d 960 g 0.05 \u003d 48 g; ν (CuSO 4) \u003d m (CuSO 4) / M (CuSO 4) \u003d 48 g / 160 g \u003d 0.3 mol

According to the reaction equation (II) ν (CuSO 4) = ν (H 2 S) = 0.3 mol, and according to the reaction equation (III) ν (Al 2 S 3) = 1/3ν (H 2 S) = 0, 1 mol

Calculate the masses of aluminum sulfide and copper (II) sulfate:

m(Al 2 S 3) \u003d 0.1 mol 150 g / mol \u003d 15 g; m(CuSO4) = 25 g - 15 g = 10 g

ω (Al 2 S 3) \u003d 15 g / 25 g 100% \u003d 60%; ω (CuSO 4) \u003d 10 g / 25 g 100% \u003d 40%

Task number 34

When burning a sample of some organic compound weighing 14.8 g, 35.2 g of carbon dioxide and 18.0 g of water were obtained.

It is known that the relative hydrogen vapor density of this substance is 37. 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) make the calculations necessary to establish the molecular formula of organic matter (indicate the units of measurement of the required physical quantities);

2) write down the molecular formula of the original organic matter;

3) make a structural formula of this substance, which unambiguously reflects the order of bonding of atoms in its molecule;

4) write the equation for the reaction of this substance with copper(II) oxide using the structural formula of the substance.

The result of the Unified State Examination in Chemistry not lower than the minimum established number of points gives the right to enter universities in the specialty, where in the list entrance examinations There is a chemistry subject.

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 in order to study there, first-year students must have very good knowledge.

On the official website of FIPI, versions of the Unified State Examination in Chemistry are published every year: demonstration, early period. It is these options that give an idea of the structure of the future exam and the level of complexity of tasks and are sources of reliable information in preparing for the exam.

Early version of the exam in chemistry 2017

Year	Download early version
2017	variantpo himii
2016	download

Demonstration version of the Unified State Examination in Chemistry 2017 from FIPI

Task variant + answers	Download demo
Specification	demo variant himiya ege
Codifier	codifier

AT USE options in chemistry in 2017, there are changes compared to the KIM of the last 2016, so it is advisable to train according to the current version, and for the diverse development of graduates, use the options from previous years.

Additional materials and equipment

The following materials are attached to each version of the USE examination paper in chemistry:

− periodic system chemical elements D.I. Mendeleev;

− table of solubility of salts, acids and bases in water;

− electrochemical series of voltages of metals.

It is allowed to use a non-programmable calculator during the examination work. The list of additional devices and materials, the use of which is allowed for the Unified State Examination, 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 tests in the chosen specialty
(direction of training).

The list of entrance examinations in universities for all specialties (training areas) is determined by the order of the Russian Ministry of Education and Science. Each university chooses from this list those or other subjects that are indicated in its admission rules. You need to familiarize yourself with this information on the websites of selected universities before applying for participation in the Unified State Examination with a list of selected subjects.

Specification
control measuring materials
for holding in 2017 a unified state exam
in chemistry

1. Appointment of KIM USE

The Unified State Examination (hereinafter referred to as the Unified State Examination) is a form of objective assessment of the quality of training of persons who have mastered the educational programs of the secondary general education, using tasks of a standardized form (control measuring materials).

The exam is held in accordance with federal law dated December 29, 2012 No. 273-FZ “On Education in the Russian Federation”.

Control measuring materials allow you to set the level of development of graduates 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 educational organizations middle vocational education and educational organizations of higher professional education as the results of entrance examinations in chemistry.

2. Documents defining the content of KIM USE

3. Approaches to the selection of content, the development of the structure of the KIM USE

The basis of approaches to the development of KIM USE 2017 in chemistry was those general methodological guidelines that were determined during the formation of examination models of previous years. The essence of these settings is as follows.

KIM are focused on testing the assimilation of the knowledge system, which is considered as an invariant core of the content of existing programs in chemistry for educational organizations. In the standard, this system of knowledge is presented in the form of requirements for the preparation of graduates. These requirements correspond to the level of presentation in the KIM of the content elements being checked.
In order to ensure the possibility of a differentiated assessment of the educational achievements of graduates of the KIM USE, they check the development of the main educational programs in chemistry at three levels of difficulty: basic, advanced and high. Educational material, on the basis of which tasks are built, is selected on the basis of its significance for the general education of secondary school graduates.
The fulfillment of the tasks of the examination work involves the implementation of a certain set of actions. Among them, the most indicative are, for example, such as: to identify the classification features of substances and reactions; determine the degree of oxidation of chemical elements according to the formulas of their compounds; explain the essence of a particular process, the relationship of the composition, structure and properties of substances. The ability of the examinee to carry out various actions when performing work is considered as an indicator of the assimilation of the studied material with the necessary depth of understanding.
The equivalence of all variants of the examination work is ensured by maintaining the same ratio of the number of tasks that test the assimilation of the main elements of the content of the key sections of the chemistry course.

4. The structure of KIM USE

Each version of the examination work is built according to a single plan: the work consists of two parts, including 40 tasks. Part 1 contains 35 tasks with a short answer, including 26 tasks of the basic level of complexity (the serial numbers of these tasks: 1, 2, 3, 4, ... 26) and 9 tasks advanced level complexity (the serial numbers of these tasks: 27, 28, 29, ... 35).

Part 2 contains 5 tasks high level complexity, with a detailed answer (the serial numbers of these tasks: 36, 37, 38, 39, 40).

Early version of the exam in chemistry 2017

Demonstration version of the Unified State Examination in Chemistry 2017 from FIPI

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