Olympiad tasks in chemistry

Olympiad tasks in chemistry Grade 9

Chemistry Olympiad. Olympiad tasks in chemistry Grade 10

1. In a water-alcohol solution of zinc chloride, 1.806·10 22 chloride ions and 11.56 g of salt that did not decompose into ions were found. Determine the degree of dissociation of the salt (in%) in this solution.

An unknown alkene with a mass of 7 g adds hydrogen bromide, the volume of which is the same as the volume of methane with a mass of 2 g (N.O.) Find the molecular formula of the alkene and write the structural formulas of its isomers.

When passing 15.68 l of a mixture of methane, carbon dioxide and carbon monoxide through a solution of potassium hydroxide taken in excess, the volume of the initial mixture decreased by 8.96 l (n.o.). Complete combustion of the remaining mixture required 6.72 liters (N.O.) of oxygen.
Determine the volume fractions of the components in the initial mixture.

Salt BUT is a metal compound X which turns the colorless flame of a gas burner yellow. When heated up to 300°С BUT decomposes to form a highly water-soluble salt B. The interaction of a calcium hydroxide solution with a salt solution B or with salt solution BUT leads to precipitation AT, which decomposes when heated to 1000°C into gas G(odorless) and solid D. Gas passing G through a solution of salt B leads to the formation BUT.
Identify Substances BUT–D, write the reaction equations.

Make up reaction equations (indicate the conditions for their occurrence), with the help of which transformations can be carried out, name the substances A–D.
sodium acetate → A → B → C → D→ ethylene glycol

Chemistry Olympiad. Olympiad tasks in chemistry Grade 11

1. How can you explain such terms as "solid solutions" and "liquid crystals", which at first glance do not correspond to the traditional concepts of solutions and crystals. What are the properties of solid solutions and liquid crystals?

The chemist obtained samples of three silvery-white metals and found a way to quickly distinguish between them. To do this, he exposed the samples to the action of acids and sodium hydroxide solution. The results of his research are presented below.

Symbols: "+" - the reaction is on, "-" - the metal does not react.

Determine what metals could be obtained by the chemist and write the corresponding reaction equations.

After treating a mixture of two esters with an excess of an alkali solution, the following substances were found in the resulting aqueous solution: glycerol, sodium phenolate, sodium acetate and sodium nitrate. Determine the formulas of esters, write reaction equations and indicate the conditions for their occurrence.

BUT substitution reaction> B compound reaction> C substitution reaction> D exchange reaction> E

Make up the reaction equations according to the scheme:
Cu → CuSO 4 → Cu(OH) 2 → Cu 2 O → CuO

A gaseous hydrocarbon with a volume of 50 ml, whose hydrogen density is greater than 8, was completely oxidized to carbon dioxide and water. After condensation of water vapor, the volume of the gas mixture was 120 ml. Calculate the volume of excess oxygen if it is known that the initial and final volumes of gases were measured under the same conditions.

Answers and solutions on the site.

Task 8-1

Relative atomic mass is a dimensionless quantity.

The year was 1817. The Minister of the Weimar Duchy, the poet and philosopher Johann Goethe, his friends and relatives gathered for evening tea. Among them were Johann Döbereiner, a professor of chemistry, the wife of the duke's son, Maria Pavlovna, the sister of the Russian Tsar Alexander I, and other influential people. Döbereiner said that if all the known chemical elements were grouped according to the similarity of their properties and arranged three in a row in order of increasing atomic mass, something amazing would be found. Maria Pavlovna remarked: “God loves a trinity…”

Exercise: 1. Group these chemical elements according to their properties: lithium, chlorine, sodium, calcium, iodine, bromine, barium, potassium, strontium (3 elements in each group) and arrange them in ascending order of their atomic masses. 2. Try to guess what was so amazing about Döbereiner? (5 points)

Task 8-2

Science does not know what it owes to the imagination. Ralph Waldo Emerson

Coming up with chemical names, scientists sometimes took into account the unusual properties of a new element. What elements are named after the color and smell of simple substances or compounds? (For each example of an element given, 1 point)

Task 8-3

New ideas must be supported. Tsiolkovsky

Carry out the transformation: Oxide 1 → acid → salt 1 → base → salt 2 → salt 3.

Select the substances and write down the reaction equations. (5 points)

Task 8-4

Random discoveries are made only by trained minds. Blaise Pascal

What animals helped chemists discover chemical elements or saved their lives? (For each example given 2 points)

Task 8-5

Abysses of discoveries and wisdom await us. Tsiolkovsky

Four tubes contain colorless solutions of sodium carbonate, sodium phosphate, sodium silicate, and silver nitrate. How to identify each substance using one reagent? (7 points)

Task 8-6

You have to learn a lot to know even a little. Montesquieu

The chemical element of group IV of the Periodic system, when interacting with oxygen, forms a colorless gas, which is released when effervescent tablets are dissolved in water, when dissolved in an alkali solution, it forms an average and acid salt.

Exercise 1. Define a chemical element. 2. What compounds of this element are we talking about? 3. Make up the equations for the reactions of obtaining these compounds. (7 points)

Solving tasks of the correspondence round of the Olympiad for grade 8 with an assessment system

Task 8-1

These elements are grouped into 3 groups: Lithium, sodium, potassium, Chlorine, iodine, bromine; Calcium, barium, strontium. When arranging these elements in ascending order of atomic masses, Döbereiner found that in the “triad”, the atomic mass of the second element is approximately equal to the arithmetic mean of the atomic masses of the first and third elements

Task 8-2

For each given example of an element: Bromine - “bromos” - fetid, rubidium - colors the spectrum red, sulfur - “syrah” light yellow, etc.

Task 8-3

The chain has many solutions. For each equation that meets the condition. 1 point

Possible solution: Sulfur(VI) oxide → sulfuric acid → copper(II) sulfate → copper(II) hydroxide → copper(II) chloride → silver chloride.

Task 8-4

We are talking about Courtois, who discovered a new chemical element, iodine. He had a beloved cat, who usually sat on his master's shoulder during dinner. One day, the cat got frightened of something, jumped on the floor, but fell on the bottles that Courtois had prepared for the experiments. One of them contained a suspension of algae ash in alcohol, and the other concentrated sulfuric acid. The bottles shattered and the liquids mixed up. Clubs of blue-violet steam began to rise from the floor, which settled on objects in the form of black crystals. It was iodine.

For each episode found … 2 points

Task 8-5

This reagent is hydrochloric acid. Na 2 CO 3 + 2HCl → 2NaCl + CO 2 + H 2 O colorless gas is released Na 3 PO 4 + HCl → nothing is observed

Na 2 SiO 3 + 2HCl → H 2 SiO 3 ↓ + 2NaCl, we observe the precipitation of a colorless jelly-like precipitate

AgNO 3 + HCl → AgCl↓ + HNO 3 we observe a white precipitate

Task 8-6

1. Chemical element - carbon 2. Carbon dioxide CO 2, sodium carbonate Na 2 CO 3, sodium bicarbonate NaHCO 3 3. C + O 2 → CO 2 CO 2 + NaOH → NaHCO 3 CO 2 +2 NaOH → Na 2 CO 3 + H2O

A set of tasks for the full-time round of the school stage of the Olympiad for grade 8

Task 8-1

Who seeks will always find!

At the beginning of the 20th century, geologists encrypted on maps the places of discovery of ores of valuable metals using the coordinates of chemical elements in the Periodic system. The Arabic numeral indicated the number of the period, and the Roman numeral indicated the number of the group. In addition, the records also contained letters of the Russian alphabet - A and B. On one of the old maps of maps, the designations were found: 4VIB, 4VIIIB2, 6IB, 6IIB. Exercise: Decipher the records of geologists. (2 points)

Task 8-2

A light body that can be forged. M.V. Lomonosov

One of the chemical elements was isolated in the 16th century. German chemist and metallurgist Georgius Agricola. In Roeland's Alchemical Dictionary (1612), it is classified as a metal and called "the lightest, palest, and cheapest lead." In Russia, he was called either “nymph”, or “glaura”. The origin of the modern name is also shrouded in mystery. Some believe that it comes from the Arabic words "bi ismid" - similar to antimony. Others suggest that the name of the element is of ancient Germanic origin and means "white metal". Still others argue that the name comes from two German words - “wiese” (meadow) and “muten” (mine), since in German Saxony the element has long been mined in mines located among the meadows of the Schneeberg district. Exercise: What is this element? Describe its position in the Periodic system. Determine the composition of the highest oxide of this element.

Task 8-3

Exact logical definition of concepts is a condition of true knowledge. Socrates

Suggest different options for classifying these inorganic substances:

CO 2 , NO, NO 2 , Na 2 O, Al 2 O 3 , SiO 2 , P 2 O 5 , Cr 2 O 3 , CuO, ZnO

Task 8-4

A good idea is made up of many small ideas. Thomas Edison

Suggest several options (no more than three) for solving the chain of reactions A → B → C and make up the reaction equations for this scheme. Substances A, B, C are complex inorganic compounds belonging to different classes of compounds. (6 points)

Task 8-5

A scientific hypothesis always goes beyond the facts that served as
basis for its construction. Vernadsky

Decipher this chain of reactions, write reaction equations for it (all substances in the chain contain iron atoms): 56 → 127 → 90 → 72 → 56 (11 points)

Task 8-6

All salts are made up of some kind of acid and
any alkali ... German chemist and pharmacist O. Takheny

Give examples of salts, in the reactions of which gaseous products are released with acids and alkalis. Write the reaction equations in molecular and ionic form.

Task 8-7

Q.E.D. Euclid

In a bottle, cylinder, flask, jar are magnesium, barium hydroxide, copper sulfate (II), sulfuric acid. Magnesium and copper (II) sulfate are not in a bottle, a vessel with barium hydroxide is located between the flask and a vessel with sulfuric acid. There is no barium hydroxide in the jar or a substance that has a shine. The cylinder stands next to a vessel with a blue substance.

Exercise: Find out the contents of each vessel. Make equations for the reactions that occur when the contents of these vessels are mixed in pairs. (14 points)

Solving the tasks of the full-time round of the school stage of the Olympiad for grade 8

Task 8-1

4VIB coordinates in the Periodic system mean the 4th period and VIB - group, element chromium

4VIIIB2 - 4th period, VIIIB2 - group, element nickel 6IB - 6th period, IB - group, element - gold 6IIB - 6th period, IIB - group, element mercury. Total: 2 points.

Task 8-2

The confusion element is bismuth, low-melting reddish-white metal, the last element of the Periodic Table, which does not have natural radioactivity, serial number 83, period 6, V group, main subgroup, composition of the higher oxide Bi 2 O 5

Task 8-3

All substances given in the condition of the problem are oxides, therefore it is necessary to classify oxides according to various criteria:

A) oxides of metals and non-metals B) basic, acidic, amphoteric, others

C) salt-forming, non-salt-forming D) gaseous, solid

D) colorless, colored

Task 8-4

Options for solving the task can be different, for example: CuCl 2 →Cu(OH) 2 →CuO (no more than three options) 1) CuCl 2 + 2NaOH → Cu(OH) 2 ↓ + 2NaCl 2) Cu(OH) 2 → CuO + H2O

Task 8-5

Since all substances in the chain contain iron atoms, and the chain begins with substance 56, therefore, the cipher is based on the atomic and molecular weights of substances (Ar (Fe) = 56).

The chain of reactions looks like this: Fe → FeCl 2 → Fe(OH) 2 → FeO → Fe

1) Fe + 2HCl → FeCl 2 + H 2 2) FeCl 2 + 2NaOH → Fe(OH) 2 ↓+ 2NaCl

3) Fe(OH) 2 → FeO + H 2 O 4) 2FeO + C → 2Fe + CO 2

Task 8-6

Ammonium salts: carbonates, sulfites, sulfides and corresponding acid salts.

For example: NH 4 HCO 3 + HCl → NH 4 Cl + CO 2 + H 2 O NH 4 HCO 3 + NaOH → NH 3 + H 2 O + NaHCO 3

Task 8-7

Cylinder - Mg flask - CuSO 4 bottle - Ba(OH) 2 cans - H 2 SO 4

Mg + CuSO 4 → MgSO 4 + Cu signs of reaction - the solution becomes colorless, pink powder is released - copper Mg + H 2 SO 4 → MgSO 4 + H 2 signs of reaction - gas is released without color CuSO 4 + Ba (OH) 2 → BaSO 4 ↓ + Cu(OH) 2 ↓ signs of reaction - precipitation

Ba(OH) 2 + H 2 SO 4 → BaSO 4 ↓ + 2H 2 O reaction signs - white precipitate

A set of tasks for the full-time round of the Olympiad for grade 9

Task 9-1

Everything is good in moderation. Hippocrates

An excerpt from Haggard's story "The Pearl of the Nile": "She took one of those huge pearls out of her ear ... and ... dipped the pearl into vinegar. There was silence, the shocked guests, frozen, watched the pearl slowly dissolve in strong vinegar. There was no trace of her left, and then Cleopatra raised the goblet, twisted it, shaking the vinegar, and drank it all to the last drop.

Exercise: Explain why Cleopatra was able to drink "strong vinegar", and also make an equation for the reaction. (3 points)

Task 9-2

The universe is diversity in unity. Honore de Balzac

poor BUT quite uncomfortable in his apartment: from above they threaten to burn him B, on the right - poison poisonous D, and the one living on the left is quiet E sometimes he starts to rage and it is not at all clear what to expect from him - either he will poison or set fire to the apartment. But when E calms down, then begins to glow with a pale green light and pleases everyone. Exercise: Determine who they are A, B, D, E?

Task 9-3

Make up the reaction equations for the chain of reactions: Fe 0 → Fe +2 → Fe +3 → Fe +2 → Fe 0 → Fe +3

Task 9-4

Experience is not something that happens to you;
it is what you do with what happens to you. Sir Isaac Newton

The young chemist poured ammonia, Norwegian and calcium nitrate into the first flask; chalk was poured into the second flask, as well as Chilean and ammonium nitrate; in the third flask - sodium and Indian nitrate, as well as potash. Exercise: How many substances will be in the mixture formed by mixing the contents of: a) the first and second flasks; b) all three flasks?

Task 9-5

Science fights superstition
like light with darkness. Mendeleev D.I.

One day an alchemist told his master that he could show him a "shaitan" that takes the form of a liquid that devours gold. The alchemist showed the khan how a “shaitan” is born. He swallowed gaseous red-brown oxide others liquid oxide. At the same time, it turned out strong acid and new colorless gaseous oxide, which in the air again turned into brown gas. Then the alchemist mixed the resulting strong acid with table salt and threw a golden ring into the mixture. ring began to be covered with gas bubbles, and then ... disappeared. The Emir ordered to seal the vessel with the devil's liquid in the ground, and to imprison the alchemist in the dungeon. Mission: D prove that the alchemist is not guilty and find the chemical meaning of his actions. (7 points)

Task 9-6

There is so much truth in every natural science,
how many mathematicians are in it. Immanuel Kant

In 25 ml of a solution of 20% hydrochloric acid, 4.0 g of iron (II) sulfide was dissolved, after gas evolution, 1.68 g of iron filings were added to the solution. Calculate the mass fractions of substances in the resulting solution. (8 points)

Task 9-7

Knowledge is such a valuable thing
that it is not shameful to extract it from any source. Thomas Aquinas

An unusual blue diamond has disappeared in the house of a London jeweler. The famous detective Sherlock Holmes, urgently invited to the house, drew attention to a chlorine cylinder, the outlet tube from which was immersed in a flask with a hydrogen peroxide solution. “Where was the diamond kept?” Holmes asked. “In a small aluminum case that stood right here,” the jeweler replied and pointed to a gray pile of ash next to the burner. "Diamond can not be returned," - said Sherlock Holmes. Exercise: The jeweler sent us a diagram of the scene and asked you to explain what really happened?

Solving the tasks of the full-time round of the school stage of the Olympiad for grade 9

Task 9-1

The composition of pearls is CaCO 3, vinegar is CH 3 COOH CaCO 3 +2 CH 3 COOH → (CH 3 COO) 2 Ca + CO 2 + H 2 O

Task 9-2

Elements defined A, B, D, F: A - sulfur, B - oxygen, D - chlorine, E - phosphorus ...... 4 points

Task 9-3

Reaction equations were drawn up: 1. Fe + 2HCl → FeCl 2 + H 2 2. 2 FeCl 2 + Cl 2 → 2 FeCl 3

3. 2 FeCl 3 + Fe → 2 FeCl 2 4. 3 FeCl 2 + 2 Al → 2AlCl 3 + 3 Fe 5. 2Fe + 3Cl 2 → 2FeCl 3

Task 9-4

The formulas of these substances are determined: ammonium nitrate - NH 4 NO 3, Norwegian nitrate - Ca (NO 3) 2, calcium nitrate - Ca (NO 3) 2, Chilean nitrate - NaNO 3, Indian nitrate - KNO 3, sodium nitrate - NaNO 3 , chalk - CaCO 3, potash K 2 CO 3

Task 9-5

Substances were identified: red-brown gas is nitrogen dioxide, liquid oxide is water, when interacting with each other, they give nitric acid and nitrogen monoxide, which is easily oxidized in air to NO 2. When salt is added to nitric acid, hydrochloric acid is formed, thus , a mixture of acids (nitric and hydrochloric) is obtained - “aqua regia”, which chemically dissolves gold: Au + 4HCl + HNO 3 → H + NO + 2 H 2 O

Task 9-6

Reaction equations were drawn up: FeS + 2HCl → FeCl 2 + H 2 SFe + 2HCl → FeCl 2 + H 2

n(HCl)=0.151 mol, n(FeS)=0.0455 mol, n(Fe)=0.03 mol The substance was determined in excess: HCl - 0.06 mol - in the first reaction, in the second reaction the substances reacted completely

The amounts of substances formed were calculated: n(FeCl 2) = 0.0455 mol, n(H 2 S) = 0.0455 mol,

n (H 2) \u003d 0.03 mol The mass of the resulting solution was determined: m (r-ra) \u003d m (r-ra HCl) + m (FeS) + m (Fe) -m (H 2 S) - m (H 2 ) \u003d 27.5 + 4.0 + 1.68 - 1.55 - 0.06 \u003d 31.57 g The mass fractions of substances in solution were determined: w (FeCl 2) \u003d 17.85%.

Task 9-7

Reaction equations were drawn up: the jeweler did not turn off the tap of the chlorine cylinder, the reactions began:

Cl 2 + H 2 O 2 → 2HClO HClO + H 2 O 2 → O 2 + HCl + H 2 O 4Al + 3O 2 → 2 Al 2 O 3 C + O 2 → CO 2

A set of tasks for the full-time round of the Olympiad for grade 10

Task 10-1

What we know is limited
and what we do not know is infinite. Apuleius

Female bear butterflies signal to males with the help of an odorous substance - an attractant. It is a 2-methyl-substituted alkane. Its relative molecular weight is 254. Write the structural formula for this alkane. (2 points)

Task 10-2

Chlorophyll (from Greek chlorós - green and phýllon - leaf),
green pigment in plants
they carry out photosynthesis.

Chlorophyll is an important pigment that determines the green color of plant leaves. When burning 89.2 mg of chlorophyll in excess oxygen, the following substances are obtained: 242 mg of gas, which is usually carbonated drinks, 64.8 mg of the liquid that forms the basis of these drinks, 5.60 mg of gas, which is most in the earth's atmosphere and 4.00 mg of a white powder, which is a group II A metal oxide. Exercise: Determine the formula of chlorophyll, given that its molecule contains only one metal atom.

Task 10-3

It is not enough just to gain knowledge:
I need to find an app for them. Goethe I.V.

Decipher nominal reactions in organic chemistry:

• 13≡13 + 18 → 44
• 16 + 63 → 61 + 18
• 3 * 14 = 14 + 2 * 158 + 4 * 18 → 3 * 62 + 2 * 87 + 2 * 56
• 2 * 95 + 2 * 23 → 15 - 15 + 2 * 103. (8 points)

Task 10-4

I prefer what can be seen
hear and study. Heraclitus of Ephesus

In front of you in three boxes are: in the first - natural and fake diamonds; in the second - natural and fake pearls; in the third - a natural and fake gold product. Exercise: What chemical reactions can distinguish natural objects? (6 points)

Task 10-5

Knowledge is power, power is knowledge. Francis Bacon

Make up the reaction equations: an → in → al → ova → at → ol → diene → en.

Task 10-6

Experience is a teacher. Julius Caesar

Potassium cyanide is one of the most powerful poisons, and it must be kept in a safe under lock and key. But one morning, the laboratory assistant, having taken out a jar with the inscription KCN from the safe, found that there was no lid, the characteristic smell of almonds had disappeared, and meanwhile the volume of the contents had not decreased. The analysis showed that the substance in the jar was not potassium cyanide at all; when hydrochloric acid was added to it, a characteristic hiss was observed. Exercise: Determine what kind of substance is in the bank, and who is to blame for the loss? (5 points)

Task 10-7

Mathematics should already be loved for that,
that she puts her mind in order.
Lomonosov

When heating a sample of some solid BUT 5.6 g of solid formed B and gas AT. B dissolved in water, resulting in a solution containing 7.4 g of the substance G. AT passed through an excess of a solution of a substance D, resulting in the formation of 13.8 g of the substance E. When the latter interacts in an aqueous solution with G formed BUT and D. Define all substances. (10 points)

Solving the tasks of the full-time round of the school stage of the Olympiad for grade 10

Task 10-1

The molecular formula of alkane is determined: C 18 H 38

Task 10-2

Substances were determined: carbon dioxide, water, nitrogen, magnesium oxide The amounts of substance of elements in a molecule were determined: n(C)=5.5 mmol, n(H)=7.2 mmol, n(N)= 0.4 mmol, n (O)= 0.5 mmol, n(Mg)= 0.1 mmol The molecular formula is determined: C 55 H 72 N 4 O 5 Mg

Task 10-3

Kucherov reaction: CH≡CH + H 2 O → CH 3 - CHO Konovalov reaction: CH 4 + HNO 3 → CH 3 NO 2 + H 2 O Wagner reaction: 3CH 2 \u003d CH 2 + 2KMnO 4 + 4H 2 O → 3CH 2 OH- CH 2 OH + 2KOH + 2MnO 2

Wurtz reaction: 2 CH 3 Br + 2 Na → CH 3 - CH 3 + 2 NaBr

Task 10-4

Methods of analysis are proposed and reaction equations are drawn up:

Diamonds are properly cut diamonds and are made of pure carbon. They burn completely in an oxygen atmosphere with the formation of only carbon dioxide, fake diamonds (glass, crystal) either do not burn, or produce solid products when burned. Natural pearls are a natural variety of calcium carbonate, like all carbonates, pearls dissolve in acids: CaCO 3 +2 CH 3 COOH → (CH 3 COO) 2 Ca + CO 2 + H 2 O, fake pearls made of glass or plastic are not dissolve in acids, or do not emit CO 2. Pure gold dissolves in aqua regia, forming a yellow solution: Au + 4HCl + HNO 3 → H + NO + 2 H 2 O, fake gold products can be made from copper alloys, etc. substances, they dissolve in hydrochloric or nitric acid.

Task 10-5

Methane → acetylene → ethanal → acetic acid → ethyl acetate → ethanol → butadiene → butene

Task 10-6

The substance in the jar was identified - JISC 3 The culprits of the loss of carbon dioxide and moisture in the air. The longer the jar was open, the faster the hydrolysis of potassium cyanide proceeded, as a result of reversible hydrolysis, volatile hydrogen cyanide and hydroxide ions were obtained.

Task 10-7

All substances are defined: BUT - calcium carbonate, B - calcium oxide, AT - carbon dioxide, G - calcium hydroxide, D - potassium hydroxide, E - potassium carbonate

The reaction equations are composed: CaCO 3 → CaO + CO 2 CaO + H 2 O → Ca (OH) 2

CO 2 + 2KOH → K 2 CO 3 + H 2 OK 2 CO 3 + Ca (OH) 2 → CaCO 3 + 2KOH

The tasks were compiled using this literature:

1. Eremina E.A., Ryzhova O.N. Schoolchildren's Handbook on Chemistry Ed. NOT. Kuzmenko, V.V. Eremina M.: Exam Publishing House, 2006.- 512 p. (Series “Handbook of the student”)

2. Entertaining problems in chemistry / Ed. NOT. Deryabina. - M .: IPO "At the Nikitsky Gates", - 48 p.: ill.

Stepin B.D. Entertaining tasks and spectacular experiments in chemistry / B.D. Stepin, L.Yu. Alikberova. - 2nd ed. – M.: Bustard, 2006.

Stepin B.D., Alikberova L.Yu. Chemistry book for home reading. - M.: Chemistry, 1994

5. Tyulkov I.A., Arkhangelskaya O.V., Pavlova M.V. Preparation system for olympiads in chemistry. - M .: Pedagogical University "First of September", 2008.

6. Figurovsky N.A. The discovery of the elements and the origin of their names. - M .: publishing house "Nauka", 1970

7. Fourth Sorov Olympiad for schoolchildren 1997-1998. M.: MTSNMO, 1998. - 512 p.

Department of Education of the YaNAO Administration

Yamalo-Nenets District Institute for Advancement

qualifications of educators

City (district) Olympiad

in chemistry

Texts of tasks and answers

Salekhard

Explanatory note

Chemistry Olympiads contribute to the promotion of knowledge on the subject, reading educational and scientific and methodological literature, increase students' interest in chemistry, form professional interests and intentions. They not only develop the creative abilities of students, but also develop perseverance and perseverance in overcoming difficulties, develop skills for independent work.

The implementation of the proposed tasks requires students to have knowledge of the theoretical foundations of chemistry, the chemical properties of substances, the conditions for the occurrence and signs of chemical reactions, the transformation of substances, as well as the ability to solve problems.

The tasks are diverse both in terms of topics and in terms of complexity, they are aimed at identifying and developing the creative potential of schoolchildren.

The winners of the school round of the Olympiads take part in the Olympiad.

Allotted for tasks 4 astronomical hours.

1 place awarded to participants who score 75% ( and more) from the maximum number of points for all tasks of the Olympiad.

1. Calculate the mass of copper, iron and aluminum in the mixture, if a gas with a volume of 6.72 liters was released under the action of a mixture of 13 g of sodium hydroxide solution, and a gas with a volume of 8.96 liters (n.a.) was released under the action of hydrochloric acid without air access .

(5 points)

2. Determine the molecular formula of an alkane if it is known that it took 39 liters of oxygen to burn 6 liters of this substance. How many liters of carbon dioxide were formed?

(5 points)

3. Put the given formulas of chemicals in accordance with the areas of their application indicated in the table. (The solution can be written in the form "number - letter")

		abrasive material	BUT
		War poison
	Ca SO4 - 0.5 H2O	Baby powder
		Powder component
	K Al(SO4H2O	Bulb filler
		Oil paint pigment
	Na2 S2O3- 5 H2O	Mordant for dyeing fabrics
		Construction material
		Fixer (photo)
		refrigerant

4. An acetylene hydrocarbon containing five carbon atoms in the main chain can attach a maximum of 80 g of bromine to form reaction products weighing 104 g. Determine the structure of the acetylene hydrocarbon if it is known that it does not react with an ammonia solution of silver oxide.

(5 points)

5. Write the equations for the reactions taking place in the aquatic environment:

a) Na2SO3 + KM P O4 (in an acidic environment) → X + .....

b) X + KOH → .....

(3 points)

6. A mixture of three gases was blown up in a closed vessel. Determine the qualitative and quantitative composition of the reaction product if the first gas was obtained by the action of hydrochloric acid on 21.45 g of zinc, the second - by the decomposition of sodium nitrate weighing 25.5 g and the third - by the action of excess hydrochloric acid on manganese (IV) oxide weighing 2.61 G.

(5 points)

CHEMISTRY

GRADE 10

1. 1. Of these three metals (Cu, Fe, Al), only aluminum interacts with an alkali solution, so you can calculate the mass of aluminum in the mixture.

X mol 0.3 mol

2Al + 2NaOH + 6H2O = 2 Na [Al(OH)4] + 3H2

2 mol 3 mol

2. Determine the amount of released hydrogen substance:

ν (H2) \u003d V (H2) / Vm \u003d 6.72 / 22.4 \u003d 0.3 (mol)

1. Find the mass of aluminum.

From the equation follows:

ν (Al) : ν (H2) = 2: 3 → ν (Al) = 2 ν (H2) / 3 = 2 0.3 / 3 = 0.2 (mol)

m (Al) \u003d ν (Al) M (Al) \u003d 0.2 27 \u003d 5.4 (g).

2. From these metals, iron and aluminum react with a solution of hydrochloric acid, therefore, it is possible to calculate the volume of gas that is released during the interaction of aluminum with a quantity of substance of 0.2 mol, and then - according to the remaining volume of gas - the mass of iron contained in the mixture.

0.2 mol X mol

2Al + 6 HCl = 2 AlCl3 + 3 H2

2 mol 3 mol

We determine the amount of substance released hydrogen:

ν (H2) \u003d V (H2) / Vm \u003d 8.96 / 22.4 \u003d 0.4 (mol)

From the equation follows:

ν (Al) : ν (H2) \u003d 2: 3 → ν (H2) \u003d 3 ν (Al) / 2 \u003d 3 0.2 / 2 \u003d 0.3 (mol)

3. Find the amount of substance of the remaining hydrogen:

ν(H2) = 0.4 - 0.3 = 0.1 (mol)

4. Determine the mass of iron in the mixture:

X mol 0.1 mol

Fe + 2 HCl = FeCl2 + H2

1 mol 1 mol

From the reaction equation follows:

ν (Fe) : ν (H2) \u003d 1: 1 → ν (Fe) \u003d ν (H2) \u003d 0.1 mol.

The mass of iron is:

m (Fe) \u003d ν (Fe) M (Fe) \u003d 0.1 56 \u003d 5.6 (g).

7. Calculate the mass of copper in the mixture:

m (Cu) \u003d m mixture - [ m (Al) + m (Fe) ] \u003d 13 - (5.4 + 5.6) \u003d 2 (g).

(5 points)

2. The general equation for the combustion of alkane СnН2n+2:

СnН2n+2 + (3n +1) / 2 О2 → nСО2 + (n +1) Н2О

The volume of oxygen is 6.5 times the volume of alkane. According to Avogadro's law, this means that 6.5 moles of oxygen are required for the combustion of one mole of alkane, i.e. (3n + 1) / 2 = 6.5, whence n = 4. The alkane formula is C4H10.

It also follows from Avogadro's law that the volume of carbon dioxide is n \u003d 4 times the volume of alkane: V (CO2) \u003d 4 6 \u003d 24 (l)

Answer: C4H10., V (CO2) \u003d 24 liters.

(5 points)

		Photosensitive substance (photo)	L
		abrasive material
		Detergent component
	Ca SO4 - 0.5 H2O	Construction material
		refrigerant
		War poison
	K Al(SO4H2O	Mordant for dyeing fabrics
		Bulb filler
		Active ingredient in insulating gas masks
	Na2 S2O3- 5 H2O	Fixer (Photography)
		Oil paint pigment
		Powder component
		Baby powder
		The basis of artificial gems

(for each correct answer 0.5 points, total 7 points)

4. Two molecules of bromine can join the triple bond in acetylenic hydrocarbons:

Cn H2n-2 + 2 Br2 → Cn H2n-2 Br4

ν (Br2) = 80 / 160 = 0.5 (mol), ν (Cn H2n-2) = 0.5 / 2 = 0.25 (mol). 104 - 80 \u003d 34 (g) hydrocarbon Cn H2n-2 reacted with bromine, therefore its molar mass is M ( Cn H2n-2) = 24 / 0.25 = 96 g / mol, from which it follows,

what P = 7.

Hydrocarbon C7 H12 does not react with an ammonia solution of silver oxide, therefore the triple bond is in the middle of the chain. There is only one alkyne of composition C7 H12 with five carbon atoms in the main chain and with a triple bond in position 2 - this is 4,4 - dimethylpentine -2

CH3 - C - C ≡ C ─ CH3

(5 points)

a) 5Na2SO3 + 2KM P O4 +3 H2SO4 → 5 Na2SO4 + 2M P SO4 + K2 SO4 + 3H2O

5 SO3 2- + H2O -2e → SO4 2- + 2H +

2 M P O4 - + 8H + + 5e → M P 2+ + 4H2O

b) Of the products of the previous reaction, only manganese sulfate (substance X) reacts with alkali:

M P SO4 +2 KOH \u003d M P(OH)2 ↓ + K2 SO4

(3 points)

6. Solution:

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Thus, all the initial gases were completely consumed (0.33 mol H2, 0.15 mol O2 and 0.03 mol Cl2), while forming 0.3 mol, or 5.4 g, H2O and 0.06 mol, or 2.19 g, HCl.

5. mr-ra = ?

5.4 g+ 2.19 g = 7.59 g

So, a hydrochloric acid solution was formed with a mass fraction of 28.8% HCl.

(5 points)

Grade 11

1. Determine the formula of the limiting monohydric alcohol, if during the dehydration of a sample with a volume of 37 ml and a density of 1.4 g / ml, an alkene with a mass of 39.2 g was obtained.

(5 points)

2. Chlorine gas was passed through a hot 10% solution of formic acid weighing 75 g until the mass fractions of both acids in the solution became equal. Determine the masses of the formed acids.

(5 points)

3. Through 10 g of a mixture of benzene, phenol and aniline, a stream of dry hydrogen chloride was passed, and 2.59 g of a precipitate fell out. It was filtered and the filtrate was treated with aqueous sodium hydroxide solution. The upper organic layer was separated, its mass decreased by 4.7 g. Determine the masses of substances in the initial mixture.

(5 points)

4. Make up the reaction equations in accordance with the scheme (substances encoded with letters do not repeat; X is oxide):

A → B → sodium carbonate

X → C → dimethyl ether

acetic acid

5. When a complex substance weighing 9.7 g is burned in air, an oxide weighing 8.1 g is formed, soluble in alkalis, containing 80.2% of element (II) and a gas whose hydrogen density is 32, a decolorizing solution containing bromine weighing 16 g • Determine the starting material.

(5 points)

6. Arrange the substances in ascending order of the E–E bond energy in the molecules of ethane, hydrosine (H2N–NH2), hydrogen peroxide, and fluorine. Explain your choice.

(2 points)

CHEMISTRY

GRADE 11

Answers to tasks and problem solving

1. 1. Determine the mass of alcohol: m (Cn H2n + 1 OH) \u003d V - ρ \u003d 37- 1.4 \u003d 51.8 (g).

2. We compose the reaction equation in general form:

Cn H2n+1 OH → Cn H2n + H2O

3. The molar mass of alcohol is:

M (Cn H2n + 1 OH) \u003d (12n + 2n + 1 + 16 + 1) \u003d (14n + 18) g / mol.

4. The molar mass of an alkene is:

M (Cn H2n) \u003d 12n + 2n \u003d 14 n (g / mol).

5. Calculate the amount of substance of alcohol and alkene:

ν (Cn H2n + 1 OH) \u003d m (Cn H2n + 1 OH) / M (Cn H2n + 1 OH) \u003d 51.8 / 14n + 18 (mol),

ν(Cn H2n) = 39.2/14n (mol).

6. We find the value of n. From the equation follows:

ν (Cn H2n + 1 OH) \u003d ν (Cn H2n) or 51.8 / 14n + 18 \u003d 39.2 / 14n,

hence n = 4, hence the formula of alcohol is C4 H9 OH (butanol).

(5 points)

2. Solution: HCOOH + Cl2 = CO2 + 2 HCl

M (HCOOH) \u003d 46 g / mol, M (HCl) \u003d 36.5 g / mol

Find the mass of formic acid in solution - m (HCOOH) \u003d 75 0.1 \u003d 7.5 (g)

Let the mass of the reacted formic acid be equal to X g. Since the process takes place in the same solution, the masses of the acids also turn out to be the same (according to the condition of the problem, the reaction was carried out until the mass fraction of the formed hydrochloric acid became equal to the mass fraction of unreacted formic acid).

We express the mass of unreacted formic acid in terms of m2,

According to the reaction equation, we find the mass of hydrochloric acid formed:

m (HCl) \u003d x 2 36.5 / 46, knowing that m (HCOOH) \u003d m (HCl), we compose the equation: 7.5 - x \u003d x 2 36.5 / 46, x \u003d 2, 9y.

Find the mass of formic acid - m2 (HCOOH) \u003d 7.5 -2.9 \u003d 4.6 (g)

Answer: m (HCOOH) \u003d m (HCl) \u003d 4.6 g

(5 points)

3. Solution: When passing through a mixture of dry hydrogen chloride, a precipitate of phenylammonium chloride precipitates, which is insoluble in organic solvents:

C6H5NH2 + HCl → C6H5NH3 Cl ↓

ν (С6Н5NH3Сl) = 2.59 / 129.5 = 0.02 (mol), therefore ν (С6Н5NH2) = 0.02 mol, m (С6Н5NH2) = 0.02 ∙ 93 = 1.86 (g).

The decrease in the mass of the organic layer by 4.7 g occurred due to the reaction of phenol with sodium hydroxide: C6H5OH + NaOH → C6H5ONa + H2O.

Phenol passed into an aqueous solution in the form of sodium phenolate m (C6H5OH) = 4.7 g.

The mass of benzene in the mixture is 10 - 4.7 - 1.86 = 3.44 (g).

Answer: 1.86 g of aniline, 4.7 g of phenol, 3.44 g of benzene.

(5 points)

4. Solution: X - H2O, A - NaOH, B - NaHCO3, C - CH3OH

Reaction equations

1. 2Na + 2 H2O \u003d 2 NaOH + H2;

2. NaOH + CO2 = NaHCO3;

3. 2NaHCO3 \u003d Na2CO3 + H2O + CO2 (when heated);

4. CH3COOSH3 + H2O = CH3OH + CH3COOH;

5. 2 CH3OH → CH3 - O - CH3 + H2O;

6. (CH3CO)2 O + H2O = 2 CH3COOH.

(for each equation 0.5 points, total 3 points)

5. Solution:

1. Determine what element we are talking about.

W-in + O2 = EO + gas

100% - 80.2% = 19.8% (for oxygen in oxide).

16 wt. parts of oxygen are 19.8%

x wt. parts of the element are 80.2%

x= , i.e. BUTr (E),

therefore, the element is zinc, its oxide ZnO dissolves in alkalis.

2. Determine the formula of the resulting gas.

Mr (gas)=

Based on the relative molecular weight of the gas and its properties

decolorize bromine water, it can be assumed that this gas is SO2.

3. Determine the mass of SO2.

16 gXG

Br2 + SO2 + H2O=2HBr + H2 SO4

So, the initial substance, apparently, is ZnS, let's make sure of this.

4. Determine the mass of zinc in zinc oxide.

8.1 gx r

81 g65 g

5. Determine the mass of sulfur in sulfur oxide (IV) with a mass of 6.4 g.

6.4 gx r

64 g32 g

6. Determine the mass (Zn + S) in the substance.

6.5 g+ 3.2 g = 9.7 g

Therefore, the substance is zinc sulfide (ZnS).

(5 points)

6. In the C - N - O - F series, the number of non-bonding electron pairs increases and, consequently, the repulsion caused by them between the bound atoms of the same name increases, which facilitates the dissociation of the element-element bond. Thus, the binding energy increases in the series:

:F-F: < BUT HE < H2N - NH2 < H2C - CH2