White powders in two numbered cups. The "behavior" of the powder, when placed in a glass of water, sinks in water, gradually dissolves. gradually dissolves. Gas condensation occurs

The solution of qualitative problems for the determination of substances in bottles without labels involves a number of operations, according to the results of which it is possible to determine which substance is in a particular bottle.

The first step in the solution is a thought experiment, which is a plan of action and their intended results. To record a thought experiment, a special matrix table is used, in it the formulas of the determined substances are indicated horizontally and vertically. At the intersection of the formulas of the interacting substances, the expected results of observations are recorded: - gas evolution, - precipitation, changes in color, odor or the absence of visible changes are indicated. If, according to the condition of the problem, it is possible to use additional reagents, then it is better to record the results of their use before drawing up a table - the number of substances to be determined in the table can be reduced in this way.
The solution to the problem will, therefore, consist of the following stages:
- preliminary discussion of individual reactions and external characteristics of substances;
- recording formulas and expected results of pairwise reactions in a table,
- conducting an experiment in accordance with the table (in the case of an experimental problem);
- analysis of the results of reactions and their correlation with specific substances;
- the formulation of the answer to the problem.

It should be emphasized that a thought experiment and reality do not always completely coincide, since real reactions are carried out at a certain concentration, temperature, illumination (for example, under electric light, AgCl and AgBr are identical). Thought experiment often leaves out many little things. For example, Br 2 / aq is perfectly discolored with solutions of Na 2 CO 3, Na 2 SiO 3, CH 3 COONa; the formation of a precipitate of Ag 3 PO 4 does not proceed in a strongly acidic environment, since the acid itself does not give this reaction; glycerol forms a complex with Cu (OH) 2, but does not form with (CuOH) 2 SO 4, if there is no excess alkali, etc. The real situation does not always agree with the theoretical prediction, and in this chapter the table-matrix of the "ideal" and "realities" will sometimes be different. And in order to understand what is really happening, look for every opportunity to work with your hands experimentally in a lesson or elective (remember the safety requirements).

Example 1. Numbered flasks contain solutions of the following substances: silver nitrate, hydrochloric acid, silver sulfate, lead nitrate, ammonia and sodium hydroxide. Without using other reagents, determine in which bottle the solution of which substance is.

Solution. To solve the problem, we will compose a matrix table, into which we will enter in the corresponding squares below the diagonal crossing it, the observation data of the results of merging substances from one test tube to another.

Observation of the results of sequential infusion of the contents of one numbered test tubes to all others:

1 + 2 - white precipitate falls out; ;
1 + 3 - no visible changes are observed;

Substances	1. AgNO 3,	2. НСl	3. Pb (NO 3) 2,	4. NH 4 OH	5. NaOH
1. AgNO 3	X	AgCl white	-	the precipitate is dissolved	Ag 2 O brown
2. НСl	White	X	PbCl 2 white,	-	_
3. Pb (NO 3) 2	-	white PbCl 2	X	Pb (OH) 2 turbidity)	Pb (OH) 2 white
4. NH 4 OH	-	-	(clouding)		-
S. NaOH	brown	-	White	-	X

1 + 4 - depending on the order of draining the solutions, a precipitate may form;
1 + 5 - a brown precipitate falls out;
2 + 3- white precipitate is formed;
2 + 4 - no visible changes are observed;
2 + 5 - no visible changes are observed;
3 + 4 - turbidity is observed;
3 + 5 - white precipitate falls out;
4 + 5 - no visible changes are observed.

Let us write further the equations of the reactions taking place in those cases when changes are observed in the reaction system (gas evolution, precipitate, color change) and we enter the formula of the observed substance and the corresponding square of the matrix table above the diagonal crossing it:

I. 1 + 2:	AgNO 3 + HCl		AgCl + HNO 3;
II. 1 + 5:	2AgNO 3 + 2NaOH		Ag 2 O + 2NaNO 3 + H 2 O;
			brown (2AgOH Ag 2 O + H 2 O)
III. 2 + 3:	2HCl + Pb (NO 3) 2		PbCl 2 + 2HNO 3;
			White
IV. 3 + 4:	Pb (NO 3) 2 + 2NH 4 OH		Pb (OH) 2 + 2NH 4 NO 3;
			turbidity
V. 3 + 5:	Pb (NO 3) 2 + 2NaOH		Pb (OH) 2 + 2NaNO 3
			White

(when lead nitrate is poured into an excess of alkali, the precipitate can immediately dissolve).
Thus, on the basis of five experiments, we distinguish between substances in numbered test tubes.

Example 2. Eight numbered test tubes (from 1 to 8) without labels contain dry substances: silver nitrate (1), aluminum chloride (2), sodium sulfide (3), barium chloride (4), potassium nitrate (5), phosphate potassium (6), as well as solutions of sulfuric (7) and hydrochloric (8) acids. How, without any additional reagents, except water, to distinguish between these substances?

Solution. First of all, let's dissolve the solids in water and note the test tubes where they ended up. Let's compose a matrix table (as in the previous example), in which we will enter the observation data of the results of merging the substances of some test tubes with others below and above the diagonal crossing it. On the right side of the table, we introduce an additional column "general observation result", which we will fill in after completing all experiments and summing up the observation results horizontally from left to right (see, for example, p. 178).

1+2:	3AgNO 3 + A1C1,		3AgCl white	+ Al (NO 3) 3;
1 + 3:	2AgNO 3 + Na 2 S		Ag 2 S black	+ 2NaNO 3;
1 + 4:	2AgNO 3 + BaCl 2		2AgCl white	+ Ba (NO 3) 2;
1 + 6:	3AgN0 3 + K 3 PO 4		Ag 3 PO 4 yellow	+ 3KNO 3;
1 + 7:	2AgNO 3 + H 2 SO 4		Ag, SO 4 white	+ 2HNO S;
1 + 8:	AgNO 3 + HCl		AgCl white	+ HNO 3;
2 + 3:	2AlCl 3 + 3Na 2 S + 6H 2 O		2Al (OH) 3,	+ 3H 2 S + 6NaCl;
		(Na 2 S + H 2 O NaOH + NaHS, hydrolysis);
2 + 6:	AlCl 3 + K 3 PO 4		A1PO 4 white	+ 3KCl;
3 + 7:	Na 2 S + H 2 SO 4		Na 2 SO 4	+ H 2 S
3 + 8:	Na 2 S + 2HCl		-2NaCl	+ H 2 S;
4 + 6:	3BaCl 2 + 2K 3 PO 4		Ba 3 (PO 4) 2 white	+ 6KC1;
4 + 7	BaCl 2 + H 2 SO 4		BaSO 4 white	+ 2HC1.

Visible changes do not occur only with potassium nitrate.

By the number of times a precipitate is precipitated and gas is released, all reagents are uniquely determined. In addition, BaCl 2 and K 3 PO 4 are distinguished by the color of the precipitate with AgNO 3: AgCl is white, and Ag 3 PO 4 is yellow. In this problem, the solution can be simpler - any of the acid solutions allows you to immediately isolate sodium sulfide, it determines silver nitrate and aluminum chloride. Among the remaining three solids, barium chloride and potassium phosphate are determined by silver nitrate; hydrochloric and sulfuric acids are distinguished by barium chloride.

Example 3. Four unlabeled tubes contain benzene, chlorhexane, hexane and hexene. Using the minimum quantities and numbers of reagents, propose a method for determining each of the specified substances.

Solution. The substances to be determined do not react with each other, there is no point in compiling a table of pairwise reactions.
There are several methods for the determination of these substances, one of them is given below.
Only hexene immediately decolours bromine water:

C 6 H 12 + Br 2 = C 6 H 12 Br 2.

Chlorhexane can be distinguished from hexane by passing the products of their combustion through a solution of silver nitrate (in the case of chlorhexane, a white precipitate of silver chloride precipitates, insoluble in nitric acid, unlike silver carbonate):

2C 6 H 14 + 19O 2 = 12CO 2 + 14H 2 O;
C 6 H 13 Cl + 9O 2 = 6CO 2 + 6H 2 O + HC1;
HCl + AgNO 3 = AgCl + HNO 3.

Benzene differs from hexane in freezing in ice water (C 6 H 6 m.p. = + 5.5 ° C, and C 6 H 14 m.p. = -95.3 ° C).

1. Equal volumes are poured into two identical beakers: water into one, and a dilute sulfuric acid solution into the other. How, without having any chemical reagents at hand, can you distinguish between these liquids (you cannot taste the solutions)?

2. Four test tubes contain powders of copper (II) oxide, iron (III) oxide, silver, iron. How can these substances be recognized using only one chemical? Recognition by appearance is excluded.

3. Four numbered test tubes contain dry copper (II) oxide, carbon black, sodium chloride, and barium chloride. How, using the minimum amount of reagents, can you determine which test tube contains which substance? Justify and confirm the answer with the equations of the corresponding chemical reactions.

4. Six test tubes without inscriptions contain anhydrous compounds: phosphorus (V) oxide, sodium chloride, copper sulfate, aluminum chloride, aluminum sulfide, ammonium chloride. How can you determine the contents of each tube if you only have a set of empty tubes, water and a burner? Suggest a plan for the analysis.

5 ... Four unmarked test tubes contain aqueous solutions of sodium hydroxide, hydrochloric acid, potash, and aluminum sulfate. Suggest a way to determine the contents of each tube without using additional reagents.

6 ... Numbered test tubes contain solutions of sodium hydroxide, sulfuric acid, sodium sulfate, and phenolphthalein. How to distinguish between these solutions without using additional reagents?

7. In cans without labels, there are the following individual substances: powders of iron, zinc, calcium carbonate, potassium carbonate, sodium sulfate, sodium chloride, sodium nitrate, as well as solutions of sodium hydroxide and barium hydroxide. There are no other chemicals at your disposal, including water. Make a plan for identifying the contents of each jar.

8 ... Four numbered cans without labels contain solid phosphorus (V) oxide (1), calcium oxide (2), lead nitrate (3), calcium chloride (4). Determine which of the jars is each from of these compounds, if it is known that substances (1) and (2) react violently with water, and substances (3) and (4) dissolve in water, and the resulting solutions (1) and (3) can react with all other solutions with the formation of precipitation.

9 ... Five test tubes without labels contain solutions of hydroxide, sulfide, chloride, sodium iodide and ammonia. How can these substances be determined using one additional reagent? Give the equations of chemical reactions.

10. How to recognize solutions of sodium chloride, ammonium chloride, barium hydroxide, sodium hydroxide, which are in vessels without labels, using only these solutions?

11. ... Eight numbered test tubes contain aqueous solutions of hydrochloric acid, sodium hydroxide, sodium sulfate, sodium carbonate, ammonium chloride, lead nitrate, barium chloride, silver nitrate. Using indicator paper and carrying out any reactions between solutions in test tubes, establish what substance is contained in each of them.

12. Two test tubes contain sodium hydroxide and aluminum sulfate solutions. How to distinguish them, if possible, without the use of additional substances, with only one empty tube or even without it?

13. Five numbered test tubes contain solutions of potassium permanganate, sodium sulfide, bromine water, toluene, and benzene. How to distinguish them using only the named reagents? Use the characteristic features of each of the five substances to detect (list them); give a plan for the analysis. Write schemes of the necessary reactions.

14. Six unnamed bottles contain glycerin, an aqueous solution of glucose, butyraldehyde (butanal), hexene-1, an aqueous solution of sodium acetate, and 1,2-dichloroethane. With only anhydrous sodium hydroxide and copper sulfate as additional chemicals, determine what is in each bottle.

1. To determine water and sulfuric acid, you can use the difference in physical properties: boiling and freezing points, density, electrical conductivity, refractive index, etc. The biggest difference will be in electrical conductivity.

2. Let's add hydrochloric acid to the powders in test tubes. Silver will not react. When iron dissolves, gas will be released: Fe + 2HCl = FeCl 2 + H 2
Iron (III) oxide and copper (II) oxide dissolve without gas evolution, forming yellow-brown and blue-green solutions: Fe 2 O 3 + 6HCl = 2FeCl 3 + 3H 2 O; CuO + 2HCl = CuCl 2 + H 2 O.

3. CuO and C are black, NaCl and BaBr 2 are white. The only reagent can be, for example, dilute sulfuric acid H 2 SO 4:

CuO + H 2 SO 4 = CuSO 4 + H 2 O (blue solution); BaCl 2 + H 2 SO 4 = BaSO 4 + 2HCl (white precipitate).
Diluted sulfuric acid does not interact with soot and NaCl.

4 ... We put a small amount of each of the substances in water:

CuSO 4 + 5H 2 O = CuSO 4 5H 2 O	(a blue solution and crystals are formed);
Al 2 S 3 + 6H 2 O = 2Al (OH) 3 + 3H 2 S	(a precipitate forms and a gas with an unpleasant odor is released);
AlCl 3 + 6H 2 O = A1C1 3 6H 2 O + Q AlCl 3 + H 2 O AlOHCl 2 + HCl AlOHC1 2 + H 2 0 = Al (OH) 2 Cl + HCl A1 (OH) 2 C1 + H 2 O = A1 (OH) 2 + HCl	(a violent reaction proceeds, precipitates of basic salts and aluminum hydroxide are formed);
P 2 O 5 + H 2 O = 2HPO 3 HPO 3 + H 2 O = H 3 PO 4	(violent reaction with the release of a large amount of heat, a clear solution is formed).

Two substances - sodium chloride and ammonium chloride - dissolve without reacting with water; they can be distinguished by heating dry salts (ammonium chloride sublimes without residue): NH 4 Cl NH 3 + HCl; or by coloring the flame with solutions of these salts (sodium compounds paint the flame yellow).

5. Let's compose a table of pairwise interactions of the indicated reagents

Substances	1. NaOH	2 HCl	3.K 2 CO 3	4. Al 2 (SO 4) 3	Overall observation result
1, NaOH		-	-	Al (OH) 3	1 sediment
2. HC1	_		CO 2	__	1 gas
3.K 2 CO 3	-	CO 2		Al (OH) 3 CO 2	1 sediment and 2 gases
4. Al 2 (S0 4) 3	A1 (OH) 3	-	A1 (OH) 3 CO 2		2 sediment and 1 gas
NaOH + HCl = NaCl + H 2 O
K 2 CO 3 + 2HC1 = 2KS1 + H 2 O + CO 2

3K 2 CO 3 + Al 2 (SO 4) 3 + 3H 2 O = 2 Al (OH) 3 + 3CO 2 + 3K 2 SO 4;

Based on the presented table, all substances can be determined by the number of precipitation and gas evolution.

6. All solutions are mixed in pairs. A pair of solutions, giving a raspberry color, - NaOH and phenolphthalein. The raspberry solution is added to the two remaining tubes. Where the color disappears, there is sulfuric acid, in the other, sodium sulfate. It remains to distinguish between NaOH and phenolphthalein (tubes 1 and 2).
A. From tube 1 add a drop of solution to a large amount of solution 2.
B. From test tube 2 - add a drop of solution to a large amount of solution 1. In both cases, raspberry color.
To solutions A and B add 2 drops of sulfuric acid solution. Where the color disappears, there was a drop of NaOH. (If the color disappears in solution A, then NaOH - in test tube 1).

Substances	Fe	Zn	CaCO 3	K 2 CO 3	Na 2 SO 4	NaCl	NaNO 3
Wa (OH) 2				sediment	sediment	solution	solution
NaOH		hydrogen evolution possible		solution	solution	solution	solution
There is no precipitate in the case of two salts in Ba (OH) 2 and in the case of four salts in NaOH	dark powders (soluble in alkalis - Zn, insoluble in alkalis - Fe)		CaCO 3 gives a precipitate with both alkalis	give one draft at a time, differ in color of the flame: K + - violet, Na + - yellow		do not give precipitation; differ in behavior when heated (NaNO 3 melts, and then decomposes with the release of O 2, then NO 2

8 ... React violently with water: P 2 O 5 and CaO with the formation of H 3 PO 4 and Ca (OH) 2, respectively:

P 2 O 5 + 3H 2 O = 2H 3 PO 4, CaO + H 2 O = Ca (OH) 2.
Substances (3) and (4) -Pb (NO 3) 2 and CaCl 2 - dissolve in water. Solutions can react with each other as follows:

Substances	1.H 3 PO 4	2. Ca (OH) 2,	3. Pb (NO 3) 2	4. CaCl 2
1.H 3 PO 4		CaHPO 4	PbHPO 4	CaHPO 4
2. Ca (OH) 2	SaNRO 4		Pb (OH) 2	-
3. Pb (NO 3) 2	РbНРО 4	Pb (OH) 2		PbCl 2
4. CaC1 2	CaHPO 4		PbCl 2

Thus, solution 1 (H 3 PO 4) forms precipitates with all other solutions upon interaction. Solution 3 - Pb (NO 3) 2 also forms precipitates with all other solutions. Substances: I -P 2 O 5, II -CaO, III -Pb (NO 3) 2, IV-CaCl 2.
In the general case, the majority of precipitation will depend on the order of draining the solutions and the excess of one of them (in a large excess of Н 3 РО 4, lead and calcium phosphates are soluble).

9. The problem has several solutions, two of which are given below.
a. Add a solution of copper sulfate to all test tubes:
2NaOH + CuSO 4 = Na 2 SO 4 + Cu (OH) 2 (blue precipitate);
Na 2 S + CuSO 4 = Na 2 SO 4 + CuS (black precipitate);
NaCl + CuSO 4 (no changes in the diluted solution);
4NaI + 2CuSO 4 = 2Na 2 SO 4 + 2CuI + I 2 (brown precipitate);
4NH 3 + CuSO 4 = Cu (NH 3) 4 SO 4 (blue solution or blue precipitate, soluble in excess of ammonia solution).

b. Add silver nitrate solution to all test tubes:
2NaOH + 2AgNO 3 = 2NaNO 3 + H 2 O + Ag 2 O (brown precipitate);
Na 2 S + 2AgNO 3 = 2NaNO 3 + Ag 2 S (black precipitate);
NaCl + AgNO 3 = NaNO 3 + AgCl (white precipitate);
NaI + AgNO 3 = NaNO 3 + AgI (yellow precipitate);
2NH 3 + 2AgNO 3 + H 2 O = 2NH 4 NO 3 + Ag 2 O (brown precipitate).
Ag 2 O dissolves in an excess of ammonia solution: Ag 2 0 + 4NH 3 + H 2 O = 2OH.

10 ... To recognize these substances, all solutions should be reacted with each other:

Substances	1. NaCl	2. NH 4 C1	3. Ba (OH),	4. NaOH	Overall observation result
1. NaCl		___	_	_	no interaction
2. NH 4 Cl	_	X	NH 3	NH 3	gas is released in two cases
3. Wa (OH) 2	-	NH 3	X	-
4. NaOH	-	NH 3	-	X	in one case, gas is released

NaOH and Ba (OH) 2 can be distinguished by different coloration of the flame (Na + is colored yellow, and Ba 2 + is green).

11. Determine the acidity of solutions using indicator paper:
1) acidic environment - HCl, NH 4 C1, Pb (NO 3) 2;
2) neutral medium - Na 2 SO 4, ВаС1 2, AgNO 3;
3) alkaline medium - Na 2 CO 3, NaOH. We draw up a table.

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1 ALL-RUSSIAN OLYMPIAD OF SCHOOLBOYS IN CHEMISTRY uch. SCHOOL STAGE. 10 CLASS Tasks, answers and assessment criteria The final grade out of 6 tasks includes 5 solutions for which the participant scored the highest points, that is, one of the tasks with the lowest score is not taken into account. 1. Ten powders In ten numbered glasses, powders of the following substances were dispensed: copper, copper (ii) oxide, charcoal, red phosphorus, sulfur, iron, sodium chloride, sugar, chalk, malachite (basic copper (ii) carbonate). The students investigated the properties of the given powdered substances, the results of their observations were presented in the table. Glass number Color of test substance 1 white 2 white 3 white 4 yellow 5 red 6 dark red 7 green 8 dark gray 9 black 10 black “Behavior” of the powder when placed in a glass of water gradually dissolves gradually dissolves particles float on the water surface, do not dissolve particles float on the surface of the water, do not dissolve Changes observed when the test powder is heated in a spoon with the help of an alcohol lamp practically does not change melts, darkens, gradually charred practically does not change melts, burns with a bluish flame gradually turns black burns with a bright white flame gradually turns black darkens, particles in the flame glowing begins to smolder practically does not change 1

2 1. Determine which glass contains each of the substances issued for research. Justify the answer. 2. Which of the dispensed substances react with hydrochloric acid to produce gas? Write the appropriate reaction equations. 3. It is known that the density of substances in glasses 4 and 9 is greater than the density of water, that is, these substances must sink in water. However, powders of these substances float on the surface of the water. Suggest a possible explanation for this fact. 4. The three dispensed substances are known to conduct an electric current. What are these substances? A solution of what substance conducts an electric current? 1. Glass 1 contains sodium chloride. White color, soluble in water, practically does not change in air when heated. 2 sugar; white, soluble in water, melts and gradually charred when heated. 3 chalk; white, in water. 4 sulfur; yellow color, characteristic burning. 5 copper; Red color; the appearance of a black color when heated in air due to the formation of copper oxide (ii). 6 red phosphorus; dark red color; characteristic burning. 7 malachite; green color; the appearance of a black color during thermal decomposition due to the formation of copper oxide (ii). 8 iron; dark gray color; darkening when heated. 9 charcoal; black color; smolders when heated in air. 10 copper (ii) oxide; black color; no change when heated. 0.5 points for each correct definition and reasonable justification. Maximum 5 points. 2. Gaseous substances are released during the interaction of hydrochloric acid with chalk, malachite and iron: CaCO 3 + 2HCl = CaCl 2 + CO 2 + H 2 O (CuOH) 2 CO 3 + 4HCl = 2CuCl 2 + CO 2 + 3H 2 O Fe + 2HCl = FeCl 2 + H 2 For each equation 3. In glasses 4 and 9, respectively, are powders of sulfur and charcoal. Charcoal particles are penetrated by capillaries filled with air, so their average density is less than 1 g / ml. In addition, the surface of coal, like the surface of sulfur, is not wetted by water, that is, it is hydrophobic. Small particles of these substances are held on the surface of the water by the force of surface tension. 4. Electric current is conducted by copper, iron and coal. Sodium chloride solution conducts electric current, since NaCl is an electrolyte. 2

3 2. Derivation of the formula for organic matter Organic compound A contains 39.73% carbon and 7.28% hydrogen by weight. Determine the molecular formula of substance A and establish its structural formula if it is known that it contains a quaternary carbon atom and the vapor density in air is 5.2. Name organic compound A according to the systematic nomenclature. Suggest a way to get A. 1) Because the sum of the mass fractions is not 100%, therefore, there is still some residue in the molecule, the content of which is:, 73 7.28 = 52.99%. Molar mass of a substance: M (A) = D air M air = 5.2 29 = 151 g / mol. The number of hydrogen atoms in the molecule A: 151 0.0728 / 1 = 11. The number of carbon atoms in the molecule A: 151 0.3973 / 12 = 5. The molar mass of the residue is 151 0.5299 = 80 g / mol, which corresponds to one atom bromine, therefore, the molecular formula of the substance A C 5 H 11 Br. 2) A contains a quaternary carbon atom, so A has the following structure: H 3 CC CH 2 Br 1-bromo-2,2-dimethylpropane 3) Method of obtaining A: H 3 CC + Br2 hv H 3 CC CH 2 Br + HBr Grading system: 1) Determination of the number of carbon atoms Determination of the number of hydrogen atoms Determination of bromine 2 points Molecular formula 2) Structure 2 points Title 3) The reaction equation for obtaining 2 points 3

4 3. Three hydrocarbons Mass fraction of carbon in three hydrocarbons is 85.7%. Establish the molecular formulas of these hydrocarbons if their density by air is 0.97; 1.43; 1.93. Give the structural formulas of the isomers of these hydrocarbons and name them in accordance with the rules of international nomenclature. a) Determination of molar masses of hydrocarbons: M = D (CH) 29; air x y M1 0, g; M g g 2 1,; М3 1, mol mol mol b) Determination of the simplest formula of the desired hydrocarbons: 100 g of the substance C x H y contain 85.7 g of carbon and 14.3 g of hydrogen. The ratio of the amount of carbon to hydrogen for the desired hydrocarbons 85.7 14.3 is: x: y C: H: 7.14: 14.30 1: 2. Therefore, the simplest formula for the desired hydrocarbons is CH 2; M (CH 2) = 14 g / mol c) Determination of the molecular formulas of the desired hydrocarbons and the reduction of the structural formulas of their isomers: 1st hydrocarbon 28: 14 = 2; n = 2 C 2 H 4 ethylene. Has no isomers. 2nd hydrocarbon 42: 14 = 3; n = 3 C 3 H 6 Isomers C 3 H 6: CH 2 = CH CH 3 propene cyclopropane 3rd hydrocarbon 56: 14 = 4; n = 4 С 4 Н 8 Isomers: СН 3 СН 2 СН = СН 2 butene-1 H H C C H 3 C cis-butene-2 ​​trans-butene-2 ​​4

5 CH 3 C (CH 3) = CH 2 methylpropene cyclobutane methylcyclopropane Determination of the simplest formula of hydrocarbons. Determination of the molecular formulas of hydrocarbons and reduction of the structural formulas of their isomers: 1st hydrocarbon 2 points (for the molecular formula and for the structure of ethylene) 2nd hydrocarbon 2 points (for the molecular formula and 0.5 points for each structure) 3rd hydrocarbon 5 points (for the molecular formula, 0.5 points for each structure and additionally if cis-trans isomerism is taken into account). (If the molecular formulas are correctly determined without using the simplest formula, then for the simplest formula is added to the result.) 4. Transformations of a non-metal Select a suitable non-metal and carry out transformations for it: simple substance X hydrogen compound higher oxide simple substance salt Write the equations of the corresponding reactions. Silicon and phosphorus are suitable. Chain for silicon. Si Mg 2 Si SiH 4 SiO 2 Si Na 2 SiO 3 Reaction equations: Si + 2Mg = Mg 2 Si Mg 2 Si + 4HCl = SiH 4 + 2MgCl 2 SiH 4 + 2O 2 = SiO 2 + 2H 2 O SiO 2 + C = Si + CO 2 Si + 2NaOH + H 2 O = Na 2 SiO 3 + 2H 2 Each reaction equation is 2 points. 5

6 5. Right sides with coefficients Reconstruct the left side of the equations: = Na 2 SO 4 + 2Ag + 2HNO 3. = Na 2 S + 3Na 2 SO = 3Na 2 SO 4 + 2MnO 2 + 2KOH. +. = POCl 3 + SOCl 2 ot H 2 SO 4 Na 2 SO 3 + H 2 O + 2AgNO 3 = Na 2 SO 4 + 2Ag + 2HNO 3 4Na 2 SO 3 = Na 2 S + 3Na 2 SO 4 3Na 2 SO 3 + H 2 O + 2KMnO 4 = 3Na 2 SO 4 + 2MnO 2 + 2KOH SO 2 + PCl 5 = POCl 3 + SOCl 2 ot 2SO 2 + 2H 2 O + O 2 2H 2 SO 4 2 points for the equation. 6. Magic powder While disassembling reagents in the laboratory, the young chemist found an unsigned jar of white odorless powder. To study its properties, the young chemist carefully weighed 10.00 grams and divided them into exactly 5 parts, with each of the parts he carried out the following experiments: Experiment number The course of the experiment Added to water, and then added a few drops of litmus solution Dissolved in water. Then he added an excess of potassium carbonate. Gently introduced a portion of the sample into the burner flame. Dissolved in water. Then an excess of barium chloride was added. Dissolved in water. Then an excess of potassium hydroxide was added. Observations Well soluble in water. The solution turned red. Violent gas evolution. The burner flame turned purple. 3.43 g of white precipitate, insoluble in acids and alkalis, dropped out. The test tube was heated. No visible signs of reaction were observed 1. Determine the composition of the white powder. Confirm the answer by calculation. 2. For experiments 2, 4, 5, give the corresponding reaction equation. 3. What happens when white powder is heated? Give the reaction equation. 6

7 1. Violet coloring of the burner flame indicates that the required powder is potassium salt. The precipitation of a white precipitate with an excess of barium chloride is a qualitative reaction to the sulfate ion. But potassium sulfate (K 2 SO 4) has a neutral environment (the salt is formed by a strong base and a strong acid), and according to experiment 1 litmus colors the salt solution red, which indicates an acidic reaction. Therefore, the desired salt is potassium hydrogen sulfate, KHSO 4. Let's check this by calculating: KHSO 4 + BaCl 2 BaSO 4 + HCl + KCl, since the young chemist divided the initial sample of 10.00 g into five equal parts, so 2 entered the reaction, 00 g salt: 2 g (KHSO 4) (BaSO 4) 0.0147 mol; 136 g / mol m (baso 4) 0.0147 mol 233 g / mol 3.43 g The resulting mass of barium sulfate coincides with the experimental results, therefore the white powder is really KHSO Reaction equations: 2KHSO 4 + K 2 CO 3 2K 2 SO 4 + CO 2 + H 2 O KHSO 4 + BaCl 2 BaSO 4 + HCl + KCl KHSO 4 + KOH K 2 SO 4 + H 2 O 3. The equation of the decomposition reaction: t 2KHSO 4 K 2 S 2 O 7 + H 2 O System evaluation: 1) Conclusion on the presence of potassium ions Conclusion on the presence of sulfate ions Calculation 2 points Formula of salt 2) 3 equations for y 3 points 3) Equation of the decomposition reaction 2 points 7

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Undivided powders are written out with a total weight of 5 to 100 g. The amount of powder per dose is indicated in the signature. Medicinal substances are prescribed in undivided powders that are not potent and do not require an exact dosage. They are used more often externally, less often internally. For external use, the finest powders are preferable, since they do not have a local irritating effect and have a larger adsorbent surface compared to conventional powders.

A. Simple undivided powders Simple undivided powders are composed of a single drug substance.

Discharging rules

When prescribing such powders, after the designation Rp .: indicate the name of the medicinal substance in the genitive case with a capital letter and its total amount in grams. The second line begins with D. S., followed by the signature. The name of the dosage form is not indicated in the prescription.

Rp .: Kalii permanganatis 5.0

D. S. For the preparation of solutions.

WRITE OUT:

1.30.0 magnesium sulfate (Magnesii sulfas). Take 1 tablespoon at a time, dissolved in 2/3 glass of water.

20.0 powder of anesthesin (Anesthesinum). Prescribe for application to the wound.

25.0 streptocide powder (Streptocidum). Assign to be applied to affected areas.

4.50.0 magnesium oxide (Magnesii oxidum). Assign 1/4 teaspoon 2 times a day.

5. 5.0 boric acid (Acidum boricum). Take for washing, after dissolving in 250 ml of water.

B. Complex undivided powders Complex undivided powders consist of two or more medicinal substances.

Discharging rules

When prescribing such powders, after the designation Rp .: indicate the name of one medicinal substance in the genitive case with a capital letter and its total amount in grams or units of action. On the second line - the name of the next medicinal substance in the genitive case with a capital letter and its total amount in grams or units of action, etc. Then M. f is indicated. pulvis (Mix to make a powder). This is followed by the D. S. designation and signature.

Rp .: Benzylpenicillinum-natrii 125,000 ED Aethazoli 5.0 M. f. pulvis

D. S. At 1/4 of the powder every 4 hours for blowing in.

WRITE OUT:

Separated powders

Separated powders are divided into individual doses at pharmacies or pharmaceutical factory. The average mass of the separated powder usually ranges from 0.3 to 0.5, but should not be less than 0.1.

A. Simple separated powders

Simple separated powders are composed of a single drug substance.

Discharging rules

When prescribing such powders, after the designation Rp .: indicate the name of the medicinal substance in the genitive case with a capital letter and its amount in grams. The second line gives an indication of the amount of powders: D. t. d N .... (Give such doses in number ...). The third line is the signature (S.).

Rp .: Pancreatini 0.6 D. t. d N. 24 S. 1 powder 3 times a day before meals.

WRITE OUT:

1.10 powders bromized (Bromisovalum) 0.5 each. Prescribe 1 powder half an hour before bedtime.

2.12 powders of quinine hydrochloride (Chinini hydrochlo-ridum) 100 mg each. Assign 1 powder 3 times a day.

3.6 powders of pancreatin (Pancreatinum) 600 mg each. Assign 1 powder 3 times a day after meals.

4.12 powders of bromcamphora (Bromcamphora) 250 mg each. Assign 1 powder 3 times a day.

5.12 powders of sulginum (Sulginum), 500 mg each. Assign 1 powder 4 times a day.

B. Complex separated powders

Complex separated powders are composed of several medicinal substances.

Discharging rules

When prescribing such powders, after the designation Rp.i, indicate the name of one medicinal substance in the genitive case with a capital letter and its amount in grams. On the second line - the name of the next medicinal substance in the genitive case with a capital letter and its amount in grams, etc. Further, M. f is indicated. pulvis (Mix to make a powder). Then an indication of the amount of powders is given: D. t. d. N .... (Give such doses in number ...). The last line is the signature (S.).

Rp .: Codeini phosphatis 0.015 Natrii hydrocarbonatis 0.3 M. f. pulvis D.tdN. 10 S. 1 powder 3 times a day

WRITE OUT:

1.30 powders containing 0.2 ascorbic acid (Acidum ascorbinicum) and 0.01 thiamine bromide (Tiamini bromidum). Assign 1 powder 3 times a day.

2.12 powders containing 20 mg of ethylmorphine hydrochloride (Aethylmorphini hydrochloridum) and 400 mg of sodium bicarbonate (Natrii hydrocarbonas). Assign 1 powder 2 times a day.

3.20 powders, each containing 300 mg of tanalbin (Tannal-binum) and bismuth subnitrate (Bismuthi subnitras). Assign 1 powder 4 times a day.

4.15 powders containing 0.1 each Acrichinum and Bigumalum. Assign 1 powder 2 times a day.

5.14 powders containing 0.015 ifodeini phosphas codeine phosphate each) and 0.25 terpine hydrate (Terpini hydratum). Assign 1 powder 2 times a day.

B. When prescribing powders for children or when prescribing potent medicinal substances, the dose of which is less than 0.1, in order to increase the mass of the powder, indifferent substances (for example, sugar - Saccharum) are added in an amount of 0.2-0.3 to obtain an average mass of the powder.

Rp .: Dibazoli 0.02 Sacchari 0.3 M. f. pulvis D.tdN. 10 S. 1 powder 3 times a day.

WRITE OUT:

1.6 powders of quinine hydrochloride (Chinini hydrochlo-ridum) no 30 mg. Assign 1 powder 2 times a day.

30 powders containing 0.01 riboflavin (Riboflavinum). Assign 1 powder 3 times a day.

20 powders containing 30 mg of rutin (Rutinum) and 50 mg of ascorbic acid (Acidum ascorbinicum). Assign 1 powder 3 times a day.

4.10 powders containing 20 mg of papaverine hydrochloride (Papaverini hydrochloridum) and 3 mg of Platyphyllini hydrotartras. Assign 1 powder 2 times a day.

5.15 powders containing 5 mg of diphenhydramine (Dimedrolum). Assign 1 powder 3 times a day.

D. Vegetable powders

Discharging rules

The recipe for powders of plant origin begins with the name of the dosage form in the genitive singular with a capital letter (Pulveris), then the part of the plant is indicated in the genitive case with a lowercase letter and its name is also in the genitive case with a capital letter.

To powders of plant origin (from leaves, roots, etc.), indifferent substances are added if the mass of the powder is less than 0.05.

Rp.-. Pulveris radicis Rhei 0.6 D. t. d. N. 24 S. 1 powder per night.

WRITE OUT:

10 powders from foxglove leaves (folia Digitalis), 40 mg each. Assign 1 powder 3 times a day.

20 herba Thermopsidis powders, 100 mg each. Assign 1 powder 5 times a day.

25 sea onion powders (bulbum Scillae) no.50 mg. Assign 1 powder 4 times a day.

4.6 powders from herba Gnaphalii uliginosi togashoi (herba Gnaphalii) 0.2 each. Take 1 powder 3 times daily before meals, dissolving in 1/4 cup of warm water.

Lessons 1-2. Safety rules when working in a chemical room.

1. Why is it strictly forbidden to taste substances, smell substances from the neck of a bottle, pinch the hole with your finger while mixing substances in a test tube?
Because there may be poisonous substances or acids.

2. Why is it possible to pour and pour substances only over a table or a special tray, and remove spilled or spilled substances only with a special cloth (tampon)?
Because these can be substances that interact with each other, or poisonous substances.

3. Why should the experiments be carried out only with the amount of substances indicated in the guidelines?
Large quantities of substances can lead the reaction in a different direction.

4. Why light the burner only with a match or a splinter, and not with a lighter or burning paper?
So that there is no fire.

5. Why can't you bend low over the flame?
You can get burned.

6. Why, when heating a test tube with a solution, does it need to be warmed up first?
To prevent the test tube from cracking.

7. Why should the opening of the test tube be directed away from itself and the neighbor during heating?
So that if the liquid accidentally boils, it does not splash on people.

8. While doing the work, the student violated the safety rules and left a bottle with a reagent (for example, an acid solution) open. What can happen in this situation?
All acids are dangerous, acid can evaporate - poisoning with acid vapors is possible.

9. While fixing the test tube or flask in the tripod leg, the student violated the installation rules and the test tube (flask) burst. What should the student do in this situation?
Carefully, with gloves, remove the fragments, collect the spilled liquid with special tampons.

10. During the heating process, the test tube with the reaction mixture burst. Why could this happen? What should the student do?
The tube may have been heated unevenly. Carefully, use gloves to collect the fragments.

Chemistry as part of natural science. The concept of a substance.

Complete the diagram:

1. Remember and write down the chemical products you know (at least five). Where are they used?

2. What substances, known to you, are used in agriculture. For what?
Fertilizers - to increase soil fertility.
In medicine - preservatives for preserving drugs.
In construction - limestone (CaCO3).

3. List the substances known to you that are part of a living organism. What is their biological role?

4. Replace the blanks with the terms "substance" or "body":
1) Under normal conditions body has a shape and volume.
2) Substance can be solid, liquid or gaseous.
3) Substance has thermal conductivity.

5. Underline the names of substances with one line, and physical bodies with two.
Substances: water, iron, aluminum, sugar, ice, granite block, starch, protein.
Physical bodies: drop, nail, spoon, snowflake, pill, aspirin, grain.

6. Properties of a substance are: signs by which one substance differs from another.

7. Insert words - transparent, colorless, white, colored, cloudy - in sentences with the meaning:
1) The sugar solution is colorless.
2) The glass for sunglasses is colored and transparent.
3) The iodine solution is colored and transparent.
4) If you grind the chalk and stir it in water, you get a cloudy and white suspension.

8. Using reference material and personal experience, complete Tables 1 and 2.

9. In two numbered cups, white powders - icing sugar and chalk. How can these substances be distinguished? Describe the experiment.
If you add water to both glasses, then the sugar substance will dissolve, but the chalk will not. A glass with sugar will contain a colorless transparent liquid.

in the 2017-2018 academic year

School stage

8th grade

Dear participant!

‒

‒

We wish you success!

Task 1. (8 points)

TEST. Choose one correct answer

1... A carbon compound that plays a major role in its natural cycle:

A) carbon monoxide; B) soot; B) oil; D) methane; D) carbon dioxide.

2. The cleanest water listed in the list is:

A) plumbing; B) spring; C) rain;

D) well; D) mineral.

3. From the listed chemical and physicochemical processes, select one that does not require a high temperature:

A) firing; B) calcination; C) fermentation;

D) sintering; D) fusion.

4. Among the listed metal materials used for the manufacture of prize medals, tokens and coins, the alloy is

A) gold; B) silver; B) bronze; D) nickel; D) aluminum.

5. Which of the following operations is not used in a chemical laboratory for the separation and purification of substances?

A) recrystallization; B) hypothermia; B) distillation; D) sublimation; E) re-precipitation.

6. Starting from the top left cell and moving horizontally (left or right) or vertically (up or down), go through all the cells so that the letters given in the cells form a rule on precautions when handling chemicals. Each cell can only be used once.

7. Solve the crossword puzzle by filling it in with the Russian names of chemical elements. The key word is the surname of the great Russian scientist, one of the founders of the atomic-molecular doctrine.

1) C, 2) O, 3) Al, 4) N, 5) Zn, 6) I, 7) P, 8) H, 9) Pb

Task 2. (8 points)

1) Complete the phrases:(a) The composition of an individual substance, in contrast to the composition of a mixture __________ and can be expressed chemically __________; (b) __________, unlike __________, boils with a constant __________.

2) Which of the two liquids - acetone and milk - is an individual substance, and which is a mixture?

3) You need to prove that the substance you have chosen (one of the two in item 2) is a mixture. Briefly describe your actions.

Task 3. (8 points)

Common substance “This complex substance is widespread in nature. Found across the globe. Odorless. At atmospheric pressure, a substance can only be in a gaseous and solid state. Many scientists believe that this substance has an effect on the rise in the temperature of our planet. It is used in various industries, including the food industry. Used to extinguish fires. However, in a chemical laboratory, they cannot extinguish burning metals such as magnesium. Children are very fond of drinks made with this substance. But the constant consumption of such drinks can irritate the walls of the stomach. "

1) Identify the substance based on its description.

2) What names of this substance do you know?

3) Give examples of application known to you and name the sources of formation of this substance.

Task 4. (8 points)

In the process of breathing, a person consumes oxygen and exhales carbon dioxide. The contents of these gases in the inhaled and exhaled air are shown in the table.

	O 2 (% by volume)	CO 2 (% by volume)
Inhaled
Exhaled

The volume of inhalation-exhalation is 0.5 liters, the frequency of normal breathing is 15 breaths per minute.

1) How many liters of oxygen does a person consume per hour and how much does carbon dioxide emit?

2) There are 20 people in a class with a volume of 100 m 3. The windows and doors are closed. What is the CO 2 volumetric content of the air after a 45 minute lesson? (Absolutely safe content - up to 0.1%).

Task 5 (10 points)

Powders of the following substances were dispensed in five numbered glasses: copper, copper (II) oxide, charcoal, red phosphorus and sulfur.

The students investigated the properties of the given powdered substances, the results of their observations were presented in the table.

Glass number	Substance color		Changes observed when the test powder is heated in air
		floats on the surface of the water	starts to smolder
		drowning in water	does not change
		floats on the surface of the water	melts, burns with a bluish flame, when burning, a colorless gas with a pungent odor is formed
	Dark red	4 sinks in water	burns with a bright white flame, when burning, a thick white smoke is formed
		drowning in water	gradually turns black

1) Determine which glass contains each of the substances issued for research. Justify the answer.

2) Write the equations of the reactions that occur with the participation of the substances given out when they are heated in air.

3) It is known that the density of substances in glasses No. 1 and No. 3 is greater than the density of water, that is, these substances must sink in water. However, powders of these substances float on the surface of the water. Suggest a possible explanation for this.

All-Russian chemistry Olympiad for schoolchildren

in the 2017-2018 academic year

School stage

Grade 9

Dear participant!

When completing tasks, you have to perform certain work, which is best organized as follows:

‒ read the assignment carefully;

‒ if you are answering a theoretical question or solving a situational problem, consider and formulate a specific answer (the answer should be short, enter its content in the provided field, write clearly and legibly).

For each correct answer, you can receive a number of points determined by the jury members, but not higher than the specified maximum mark.

When performing tasks, you can use a calculator, periodic table and solubility table. Tasks are considered completed if you handed them over to the person in charge of the audience on time.

We wish you success!

Task 1. (6 points)

Which particle contains 11 protons, 10 electrons and 7 neutrons? Determine its composition, charge, relative molecular weight. Write the formulas for the two compounds that contain this particle.

Task 2. (10 points)

The following substances are given: copper (II) sulfate, barium chloride, iron (III) oxide, carbon (IV) oxide, sodium oxide, silver, iron, sodium carbonate, water. Which of these substances will react with each other directly or in aqueous solution at room temperature? Give the equations for five possible reactions. For each reaction, indicate what type it belongs to.

Task 3. (10 points)

Calcium chips weighing 4.0 g were calcined in air and then thrown into water. When the shavings were dissolved in water, 560 ml of gas (n.a.) was released, which practically does not dissolve in water.

1) Write down the reaction equations.

2) Determine by how many grams the mass of chips increased during ignition.

3) Calculate the composition of the calcined shavings in mass percent.

Write the reaction equations, with the help of which, using simple substances calcium, phosphorus and oxygen, you can get calcium phosphate.

Task 4.(8 points)

To dissolve 7.8 g of metal, 40 ml of 20% hydrochloric acid (density 1.095 g / ml) is required, and a divalent metal salt is formed. The evolved hydrogen completely reacts with 6.4 g of trivalent metal oxide. Determine which metals were used in these reactions.

Task 5. (8 points)

Four numbered test tubes contain solutions of barium chloride, sodium carbonate, potassium sulfate, and hydrochloric acid. Suggest a way to identify substances without the use of additional reagents. Write down the reaction equations.

School stage

Grade 10

Dear participant!

When completing tasks, you have to perform certain work, which is best organized as follows:

‒ read the assignment carefully;

‒ if you are answering a theoretical question or solving a situational problem, consider and formulate a specific answer (the answer should be short, enter its content in the provided field, write clearly and legibly).

For each correct answer, you can receive a number of points determined by the jury members, but not higher than the specified maximum mark.

When performing tasks, you can use a calculator, periodic table and solubility table. Tasks are considered completed if you handed them over to the person in charge of the audience on time.

We wish you success!

Task 1. (10 points)

In ten numbered glasses, powders of the following substances were dispensed: copper, copper (II) oxide, charcoal, red phosphorus, sulfur, iron, sodium chloride, sugar, chalk, malachite (basic copper (II) carbonate). The students investigated the properties of the given powdered substances, the results of their observations were presented in the table.

Glass number	Test substance color	"Behavior" of the powder when placed in a glass of water	Changes observed when the test powder is heated in a spoon with an alcohol lamp
			practically does not change
		sinks in water, gradually dissolves	melts, darkens, gradually charred
		sinks in water, does not dissolve	practically does not change
			melts, burns with a bluish flame
		sinks in water, does not dissolve	gradually turns black
	Dark red	sinks in water, does not dissolve	burns with a bright white flame
		sinks in water, does not dissolve	gradually turns black
	dark grey	sinks in water, does not dissolve	darkens, particles in the flame glow
		particles float on the surface of the water, do not dissolve	starts to smolder
		sinks in water, does not dissolve	practically does not change

1. Determine which glass contains each of the substances issued for research. Justify the answer.

2. Which of the dispensed substances react with hydrochloric acid to produce gas? Write the appropriate reaction equations.

3. It is known that the density of substances in glasses No. 4 and No. 9 is greater than the density of water, that is, these substances must sink in water. However, powders of these substances float on the surface of the water. Suggest a possible explanation for this fact.

4. The three dispensed substances are known to conduct an electric current. What are these substances? A solution of what substance conducts an electric current?

Task 2. (7 points)

Make up all isomers of dichloroalkene composition C 3 H 4 Cl 2

Task 3. (10 points)

Organic compound A contains 39.73% carbon and 7.28% hydrogen by weight. Determine the molecular formula of substance A and establish its structural formula if it is known that it contains a quaternary carbon atom and the vapor density in air is 5.2. Name organic compound A according to the systematic nomenclature. Suggest a way to get A.

Assignment 4 (10 points)

Rebuild the left side of the equations:

…. +…. +…. = Na 2 SO 4 + 2Ag ↓ + 2HNO 3

…. = Na 2 S + 3Na 2 SO 4

…. +…. +…. = 3Na 2 SO 4 + 2MnO 2 ↓ + 2KOH

…. +…. = POCl 3 + SOCl 2

…. +…. +…. → 2H 2 SO 4

Task 5. (10 points)

While disassembling reagents in the laboratory, the young chemist found an unsigned can of odorless white powder. To study its properties, the young chemist carefully weighed 10.00 grams and divided them into exactly 5 parts, with each of the parts he carried out the following experiments:

Experience number	Experiment progress	Observations
		Let's well dissolve in water. The solution turns red
		Vigorous gas evolution
	Carefully put part of the sample into the burner flame	The burner flame turns purple
		Dropped 3.43 g of white precipitate, insoluble in acids and alkalis
		The tube is hot. No visible signs of reaction were observed

1. Determine the composition of the white powder. Confirm the answer by calculation.

2. For experiments 2, 4, 5, give the corresponding reaction equation.

3. What happens when white powder is heated? Give the equation for the reaction of the blow.

All-Russian chemistry Olympiad for schoolchildren in 2017-2018 academic year year

School stage

Grade 11

Dear participant!

When completing tasks, you have to perform certain work, which is best organized as follows:

‒ read the assignment carefully;

‒ if you are answering a theoretical question or solving a situational problem, consider and formulate a specific answer (the answer should be short, enter its content in the provided field, write clearly and legibly).

For each correct answer, you can receive a number of points determined by the jury members, but not higher than the specified maximum mark.

When performing tasks, you can use a calculator, periodic table and solubility table. Tasks are considered completed if you handed them over to the person in charge of the audience on time.

We wish you success!

Task 1. (10 points)

Two elements that are in the same period and in the same group of the periodic table (in its short version) form with each other a single stable binary compound with a mass fraction of one of the elements of 25.6%. This compound is readily soluble in water, and when gaseous ammonia is passed into its solution, a white precipitate forms, gradually darkening in air. Name the elements, determine the formula of the substance and write the reaction equations.

Task 2. (10 points)

How to obtain benzoic ethyl ester C 6 H 5 COOC 2 H 5 from natural limestone according to the following scheme:

CaCO 3 → CaC 2 → C 2 H 2 →… → C 6 H 5 C 2 H 5 → C 6 H 5 COOH → C 6 H 5 COOC 2 H 5

Write the reaction equations, indicate the reaction conditions.

Task 3. (10 points)

A white solid, self-decomposing at room temperature or under mechanical action, has the following elemental composition: ω (N) = 45.16%, ω (O) = 51.61%, ω (H) = 3.23%. The substance is highly soluble in water and is a weak diacid.

A. Establish the formula of the substance, name it, write the equation for the dissociation of the acid.

B. Draw the structural formula of the acid.

C. Write down the reaction equations: a) thermal decomposition of a given acid, b) its interaction with atmospheric oxygen, c) its interaction with alkali

Task 4. (8 points)

The young chemist Vasya decided to investigate a certain alloy that he inherited from his grandmother. To begin with, Vasya tried to dissolve the alloy in hydrochloric acid, but found that no dissolution occurred during this process. Then he tried to dissolve it in hot concentrated nitric acid. In this case, the alloy collapsed, the solution turned blue, but a colored precipitate remained at the bottom, which did not dissolve even after prolonged heating in nitric acid. Vasya filtered the precipitate and dried it. Placing the powder in a crucible and heating it until melting, and then cooling it down, Vasya immediately realized which substance was an insoluble precipitate.

1. What two metals does the alloy that Vasya studied consist of?

2. How to dissolve the precipitate formed when the alloy is heated in nitric acid? Give the reaction equation.

3. How to isolate the second component of the alloy from the blue solution obtained after the reaction with nitric acid? Give the required reaction equations

Task 5. (10 points)

A student of the 8th grade during the practical work "Obtaining oxygen and studying its properties" assembled a device for obtaining oxygen by displacing water. At the same time, he violated one of the requirements of the instructions - he did not place a piece of cotton wool in a test tube near the gas outlet tube. When potassium permanganate was heated, the water in the crystallizer turned red-violet. While collecting oxygen, part of the colored solution got into the bottle with gas. The disciple burned sulfur in it. In this case, the red-violet color of the solution disappeared, and a colorless solution was formed. Having decided to investigate the resulting solution, the student poured into it a part of the colored solution from the crystallizer. And again the color changed - a dark brown precipitate of an unknown substance fell out.

1. Write down the reaction equation for the decomposition of potassium permanganate.

2. What substance got into the crystallizer with water?

3. Why was the solution discolored when burning sulfur? Write down the reaction equation.

4. Name the substance that precipitated. Write down the reaction equation.

Keys

All-Russian chemistry Olympiad for schoolchildren in 2017-2018 academic year year

School stage

Grade 8 (max. 42 points)

Task 1. (8 points)

Test 2.5score (0.5 point for each task)

6. Rule -Chemicals cannot be tasted. - 1 point

7. Crossword 4.5 points(0.5 point for each element)

1 - Carbon, 2- oxygen, 3-aluminum, 4-nitrogen, 5-zinc, 6 - iodine, 7- phosphorus, 8- hydrogen, 9 - lead.

Task 2. (8 points)

1) (a) The composition of an individual substance, in contrast to the composition of a mixture, is constant and can be expressed by a chemical formula; (b) an individual substance, unlike a mixture of substances, boils at a constant temperature. (4 points)

2) Acetone is an individual substance, milk is a mixture. (2 points)

3) Place the drops of both liquids in the microscope. The milk will be uneven under the microscope. This is a mixture. The acetone will be homogeneous under the microscope.

Another possible solution: acetone boils at a constant temperature. Water evaporates from milk during boiling, a film forms on the milk surface - foam. Other reasonable evidence is also accepted. (2 points)

Task 3. (8 points)

1. Named the substance - carbon dioxide (carbon monoxide (IV)) (2 points ). A possible answer - water - is wrong. Water does not irritate the stomach.

2. Dry ice, carbon dioxide, carbonic anhydride (3 points: 1 point for each answer).

3. Carbon dioxide is used in the production of carbonated drinks, sugar production, when extinguishing fires as a refrigerant, etc. It is formed during the respiration of animal organisms, fermentation, decay of organic residues, in the production of quicklime, combustion of organic substances (peat, wood, natural gas, kerosene , gasoline, etc.) . (One point per example, but no more than 3 points).

Task 4. (8 points)

1) In an hour, a person takes 900 breaths and 450 liters of air pass through the lungs. (1 point) Not all of the inhaled oxygen is consumed, but only 21% - 16.5% = 4.5% of the air volume, that is, approximately 20 liters. (1 point )

The same amount of carbon dioxide is emitted as oxygen consumed, 20 liters. (1 point)

2) In 45 minutes (3/4 hours), 1 person emits 15 liters of CO2. (1 point)

20 people emit 300 liters of CO2 . (1 point)

Initially, the air contained 0.03% of 100 m3, 30 liters of CO2, (1 point)

after the lesson it became 330 liters. CO2 content: 330 l / (100,000 l) 100% = 0.33% (2 points ) This content exceeds the safe threshold, therefore the classroom must be ventilated.

Note. The calculation in the second question uses the answer to the first question. If in the first question the wrong number was received, but then the correct actions were performed with it in the second paragraph, the maximum score is given for this point, despite the incorrect answer.

Task 5. (10 points)

1) Glass No. 1 contains coal powder. Black in color, smolders in air when heated.

No. 2 - copper (II) oxide; has a black color, does not change when heated.

No. 3 - sulfur; yellow color, characteristic combustion with the formation of sulfur dioxide.

No. 4 - red phosphorus; dark red color, characteristic combustion with the formation of phosphorus (V) oxide.

No. 5 - copper; Red color; the appearance of a black color on heating due to the formation of copper (II) oxide.

(0.5 points for each correct definition and another 0.5 points for reasonable justification.Total - 5 points)

2) C + O2 = CO 2 S + O2 = SO 2 4P + 5O 2 = 2P 2 O 5 2Cu + O 2 = 2CuO (1 point for each equation Total - 4 points)

3) The glasses No. 1 and No. 3 contain, respectively, powders of charcoal and sulfur. Charcoal particles are penetrated by capillaries filled with air, so their average density is less than 1 g / ml. In addition, the surface of coal, like the surface of sulfur, is not wetted by water, that is, it is hydrophobic. Small particles of these substances are held on the surface of the water by the force of surface tension. (1 point)

Keys

All-Russian chemistry Olympiad for schoolchildren in 2017-2018 academic year year

School stage

Grade 9 (max. 42 points)

Task 1. (6 points)

1. There are 1 more protons than electrons. Therefore, the particle has a charge of +1. There are fewer neutrons than protons, therefore, the particle contains hydrogen atoms, in which there are no neutrons at all. 11 - 7 = 4 - this is the minimum number of H atoms. Without hydrogen there will be 7 protons and 7 neutrons - this is a nitrogen atom-14: 14N. Particle composition: 14NH4 + - ammonium ion ( 2 points )

Charge: 11 - 10 = +1 (1 point)

Relative molecular weight: 11 + 7 = 18 or 14 + 4 = 18 (1 point)

Formulas: NH4Cl, (NH4) 2CO3 or other ammonium salts (2 points)

Task 2 (10 points)

Possible reactions:

Na 2 O + H 2 O = 2NaOH compounds

Na 2 O + CO 2 = Na 2 CO 3 compounds

BaCl 2 + CuSO 4 = BaSO 4 + CuCl 2 exchange

2CuSO 4 + 2Na 2 CO 3 + H 2 O = Cu 2 (OH) 2 CO 3 + CO 2 + 2Na 2 SO 4 exchange

Fe + CuSO 4 = Cu + FeSO 4 substitution

Na 2 CO 3 + CO 2 + H 2 O = 2NaHCO 3 compounds

Na 2 O + H 2 O + CuSO 4 = Cu (OH) 2 + Na 2 SO 4 compounds and exchange

2NaOH + CO 2 = Na 2 CO 3 + H 2 O exchange

BaCl 2 + Na 2 CO 3 = BaCO 3 + 2NaCl exchange

For each of the five equations - 2 points (1 point for substances, 0.5 points for the coefficients, 0.5 points for the type of reaction).

(other formulations of the answer are allowed that do not distort its meaning)

Task 3 (10 points)

When calcined shavings are calcined, the reaction occurs: 2Ca + O 2 = 2CaO (The condition that the gas is practically insoluble in water excludes the reaction of calcium with nitrogen, which can lead to calcium nitride hydrolyzing to form NH 3.) Since calcium melts at a high temperature, and the reaction product is also refractory, metal oxidation initially occurs only from the surface. Calcined shavings are metal coated with an oxide layer on the outside. When placed in water, both metal and oxide react with it: CaO + H 2 O = Ca (OH) 2; Ca + 2H 2 O = Ca (OH) 2 + H 2.

2) The amount of the metal substance that has not reacted with oxygen is equal to the amount of the substance of the released gas (hydrogen): n (Ca) = n (H 2) = 0.56 / 22.4 = 0.025 mol. In total, n (Ca) = 4/40 = 0.1 mol in the original shavings. Thus, 0.1 - 0.025 = 0.075 mol of calcium entered into a reaction with oxygen, which is m (Ca) = 0.075 * 40 = 3 g. The increase in the mass of the shavings is associated with the addition of oxygen. The mass of oxygen reacted with calcium is equal to m (O 2) = 32 * 0.0375 = 1.2 g. So, the mass of the shavings after calcination increased by 1.2 g.

3. Calcined shavings consist of calcium (0.025 mol) weighing 1 g and calcium oxide (0.075 mol) weighing 4.2 g. Composition in weight percent: Ca - 19.2%; CaO - 80.8%. Grading system:

1. For each reaction equation, 1 point - 3 points

2. Due to the calculation of the amount of substance hydrogen - 1 point

For the correct answer - 3 points

3. For the correct answer - 3 points

Task 4 (8 points)

1) Determine the amount of hydrogen substance

m (HCl) = w ρ v = 0.2 1.095 40 = 8.76 g

ν (HCl) = m in-va / M in-va = 8.76 / 36.5 = 0.24 mol (2 points)

2) Me + 2HCl = MeCl 2 + H2

a) ν (Me) = ν (H 2) = 0.5ν (HCl) = 0.5 0.24 = 0.12 mol

b) M (Me) = m in-va / ν = 7.8 / 0.12 = 65g / mol (2 points)

Metal - zinc (1 point)

3) Me 2 O 3 + 3 H 2 = 2Me + H 2 O

a) ν (Ме 2 О 3) = 1 / 3ν (H 2) = 0.12 / 3 = 0.04 mol

b) M (Me 2 O 3) = m in-va / ν = 6.4 / 0.04 = 160 g / mol

160 = 2Ame + 3 16 Ame = 56 (2 points)

Metal - iron (1 point)

Assignment 5 (8 points)

Thought experiment table compiled

		a white precipitate forms	a white precipitate forms	without changes
	a white precipitate forms		Without changes	colorless and odorless gas is emitted
	a white precipitate forms	Without changes		Without changes
	Without changes	Colorless and odorless gas is released	Without changes

The reaction equations are given in molecular and ionic form:

BaCl 2 + Na 2 CO 3 → BaCO 3 ↓ + 2NaCl;

Na 2 CO 3 + 2HCl → 2NaCl + CO 2 + H 2 O

BaCl 2 + K 2 SO 4 = BaSO 4 ↓ + 2KCl;

Grading guidelines

For compiling a table - 1 point

For the thought experiment table - 4 points

For each correctly composed molecular equation, 1 point (3 equations) - 3 score

Keys

All-Russian chemistry Olympiad for schoolchildren in 2017-2018 academic year year

School stage

Grade 10 (max. 47 points)

Task 1. (10 points)

1. Glass No. 1 contains sodium chloride. White color, soluble in water, practically does not change in air when heated.

No. 2 - sugar; white, soluble in water, melts and gradually charred when heated.

No. 3 - chalk; white, insoluble in water.

No. 4 - sulfur; yellow color, characteristic burning.

No. 5 - copper; Red color; the appearance of a black color when heated in air due to the formation of copper (II) oxide.

No. 6 - red phosphorus; dark red color; characteristic burning.

No. 7 - malachite; green color; the appearance of a black color during thermal decomposition due to the formation of copper (II) oxide.

No. 8 - iron; dark gray color; darkening when heated.

No. 9 - charcoal; black color; smolders when heated in air.

No. 10 - copper (II) oxide; black color; no change when heated.

0.5 points for each correct definition and reasonable justification. Maximum - 5 points.

2. Gaseous substances are released during the interaction of hydrochloric acid with chalk, malachite and iron:

CaCO 3 + 2HCl = CaCl 2 + CO 2 + H 2 O

(CuOH) 2 CO 3 + 4HCl = 2CuCl 2 + CO 2 + 3H 2 O

Fe + 2HCl = FeCl 2 + H 2

3 points - 1 point for each equation

3. The glasses No. 4 and No. 9 contain, respectively, powders of sulfur and charcoal. Charcoal particles are penetrated by capillaries filled with air, so their average density is less than 1 g / ml. In addition, the surface of coal, like the surface of sulfur, is not wetted by water, that is, it is hydrophobic. Small particles of these substances are held on the surface of the water by the force of surface tension. 1 point

4. Electric current is conducted by copper, iron and coal. Sodium chloride solution conducts electric current, since NaCl is an electrolyte. 1 point

Task 2. (7 points)

Trans-1,2-dichloropropene

Cis-1,2-dichloropropene

1,1-dichloropropene

2,3 - dichloropropene

Trans-1,3-dichloropropene

Cis-1,3-dichloropropene

3,3-dichloropropene

7 points: 0.5 points for each structure, 0.5 points for each name.

Task 3. (10 points)

1) Because the sum of the mass fractions is not 100%, therefore, there is still some residue in the molecule, the content of which is equal to:

100 – 39,73 – 7,28 = 52,99 %.

Molar mass of a substance: M (A) = D air * M air = 5.2 * 29 = 151 g / mol.

The number of hydrogen atoms in a molecule A: 151 * 0.0728 / 1 = 11.

The number of carbon atoms in a molecule A: 151 * 0.3973 / 12 = 5.

The molar mass of the residue is 151 × 0.5299 = 80 g / mol, which corresponds to one bromine atom, therefore, the molecular formula of the substance A- C 5 H 11 Br.

2) A contains a quaternary carbon atom, so A has the following structure:

CH 3 - C - CH 2 Br 1-bromo-2,2-dimethylpropane

3) Method of obtaining A:

CH 3 -C (CH 3) 2 -CH 3 + Br 2 = CH 3 -C (CH 3) 2 –CH 2 Br + HBr

Grading system:

1) Determination of the number of carbon atoms 1 point

Determination of the number of hydrogen atoms 1 point

Determination of bromine 2 points

Molecular Formula 1 point

2) Structure 2 points

Name 1 point

3) The equation of the reaction of obtaining 2 points.

Task 4. (10 points)

Na 2 SO 3 + H 2 O + 2AgNO 3 = Na 2 SO 4 + 2Ag ↓ + 2HNO 3

4Na 2 SO 3 = Na 2 S + 3Na 2 SO 4

3Na 2 SO 3 + H 2 O + 2KMnO 4 = 3Na 2 SO 4 + 2MnO 2 ↓ + 2KOH

SO 2 + PCl 5 = POCl 3 + SOCl 2

2SO 2 + 2H 2 O + O2 = 2H 2 SO 4

For each equation - 2 points

Task 5. (10 points)

1. Violet coloring of the burner flame indicates that the required powder is a potassium salt. The precipitation of a white precipitate with an excess of barium chloride is a qualitative reaction to the sulfate ion. But potassium sulfate (K 2 SO 4) has a neutral environment (the salt is formed by a strong base and a strong acid), and according to experiment No. 1 litmus colors the salt solution red, which indicates an acidic reaction. Therefore, the desired salt is potassium hydrogen sulfate, KHSO 4. Let's check it by calculation: KHSO 4 + BaCl 2 → BaSO 4 ↓ + HCl + KCl

since the young chemist divided the initial sample of 10.00 g into five equal parts, it means that 2.00 g of salt entered into the reaction:

n (KHSO 4) = n (BaSO 4) = 2g / 136g / mol = 0.0147mol;

m (BaSO 4) = 0.0147 mol * 233 g / mol = 3.43 g.

The resulting mass of barium sulfate coincides with the experimental results, therefore the white powder is really KHSO 4.

2. Equations of reactions:

2KHSO 4 + K 2 CO 3 → 2K 2 SO 4 + CO 2 + H 2 O

KHSO 4 + BaCl 2 → BaSO 4 ↓ + HCl + KCl

KHSO 4 + KOH → K 2 SO 4 + H 2 O

3. The equation of the decomposition reaction: 2KHSO 4 = K 2 S 2 O 7 + H 2 O

Grading system:

1) Conclusion about the presence of potassium ions - 1 point

Conclusion on the presence of sulfate ions - 1 point

Payment - 2 points

Salt Formula - 1 point

2) 3 equations by 1 point - 3 points

3) The equation of the decomposition reaction - 2 points

Keys

All-Russian chemistry Olympiad for schoolchildren in 2017-2018 academic year year

School stage

Grade 11 (max. 48 points)

Task 1. (10 points)

Since the elements are in the same period and in the same group of the periodic system, one of them is in the main subgroup, and the other is in the side, that is, it is a d-metal. Judging by the solubility in water, the substance is a halide, which means that the metal is in a side subgroup of the seventh group of the periodic system. Judging by the properties, it is manganese, and the substance is MnBr 2.

Indeed, the mass fraction of manganese in it is 55: 215 ≈ 0.256 = 25.6%. Elements - Mn and Br, substance - MnBr 2 (6 points: 2 points for each element, 2 points for the substance).

Reaction equations:

MnBr 2 + 2NH 3 + 2H 2 O = Mn (OH) 2 ↓ + 2NH 4 Br;

2Mn (OH) 2 + O 2 = 2MnO 2 ↓ + 2H 2 O (4 points: 2 points per equation).

Task 2. (10 points)

CaCO 3 + 4C = CaC 2 + 3CO (calcination);

CaC 2 + 2H 2 O = Ca (OH) 2 + C 2 H 2;

3C 2 H 2 = C 6 H 6 (heating, catalyst - coal);

C 2 H 2 + 2H 2 = C 2 H 6 (when heated, the catalyst is platinum);

C 2 H 6 + Cl 2 = C 2 H 5 Cl (under illumination);

C 6 H 6 + C 2 H 5 Cl = C 6 H 5 C 2 H 5 + HCl (catalyst - aluminum chloride);

C 6 H 5 C 2 H 5 + 2K 2 Cr 2 O 7 + 8H 2 SO 4 = C 6 H 5 COOH + CO 2 + 2K 2 SO 4 + 2Cr 2 (SO 4) 3 + 10H 2 O; C 2 H 5 Cl + NaOH = C 2 H 5 OH + NaCl;

C 6 H 5 COOH + C 2 H 5 OH = C 6 H 5 COOC 2 H 5 + H 2 O (when heated, the catalyst is H 2 SO 4). Grading Scheme:

for the correct transition from limestone to acetylene- 3 points;

for obtaining benzene from acetylene- 1 point;

for the production of benzoic acid from benzene- 2 points;

for obtaining ester in one way or another from benzoic acid- 4 points .

Task 3. (10 points)

A. Establishing the formula of a substance.

Let's designate the formula H x N y O z - x: y: z = 3.23 / 1: 45.16 / 14: 51.61 / 16 = 1: 1: 1;

The simplest formula is HNO, but according to the condition it is a dibasic acid, so it is logical to assume that its formula H 2 N 2 O 2 is a nitrous acid. Dissociation equation H 2 N 2 O 2 ↔ H + + HN 2 O 2 - ↔ 2H + + N 2 O 2 -2 (5 points)

B. Structural formula H-O- N = N - O-H (2 points)

B. Decomposition: H 2 N 2 O 2 → H 2 O + N 2 O

Oxidation with air oxygen: 2H 2 N 2 O 2 + 3O 2 (air) = 2HNO 2 + 2HNO 3 Neutralization with alkali: H 2 N 2 O 2 + 2NaOH = Na 2 N 2 O 2 + 2 H 2 O (3 points)

(other formulations of the answer are allowed that do not distort its meaning)

Task 4. (8 points)

1. Copper (by the color of the solution) and gold (insolubility in nitric acid and a characteristic form of a compact metal) (4 points: 2 points per element)

2. Dissolving in aqua regia (1 point)

Reaction equation:

Au + HNO 3 (conc.) + 4HCl (conc.) = H + NO + 2H 2 O (2 points) (Variants with hydrochloric acid and chlorine, selenic acid, a mixture of nitric and hydrofluoric acids, etc. are also suitable - evaluate with a full score.)

3. Any reasonable method, for example: Fe + Cu (NO 3) 2 = Cu + Fe (NO 3) 2 (1 point).

Assignment 5. (10 points)

1.2KMnO 4 = K 2 MnO 4 + MnO 2 + O 2 (2 points)

2.Potassium permanganate particles got into the crystallizer with oxygen flow (1 point) 3. S + O 2 = SO 2 (1 score )

2KMnO 4 + 5 SO 2 + 2H 2 O = K 2 SO 4 + 2MnSO 4 + 2H 2 SO 4 (2 score )

4. Sediment - manganese dioxide MnO 2 (2 points)

2KMnO 4 + 3MnSO 4 + 2H 2 O = 5MnO 2 ↓ + K 2 SO 4 + 2H 2 SO 4 (2 score )