School Olympiads in Chemistry. All-Russian Chemistry Olympiad for Schoolchildren

School stage All-Russian Chemistry Olympiad 2016.

9TH GRADE

PART 1 (test)

For each task, several answers are given, of which only one is correct. Choose the correct answer. Write down the number of the task and enter the number of the selected answer.

1. Has the highest molecular weight

1) BaCl 2 2) BaS0 4 3) Ba 3 (P0 4) 2; 4) Ba 3 R 2. (1 point)

2. A three-element substance is...

1) sulfuric acid; 2) quicklime (calcium oxide);

3) iron (III) chloride; 4) copper sulfate. (1 point)

Z. Sum of coefficients in a molecular reaction equation

(CuOH) 2 C0 3 + HC1 = CuC1 2 + C0 2 + ... (2 points)

1)10: 2)11; 3)12; 4)9.

4. Amount of substance (mol) contained in 6.255 g of phosphorus (V) chloride (2 points)

1)0,5; 2)0,3; 3)0,03; 4)0,15.

5. Mass (in grams) of a sample of aluminum nitrate, which contains 3.612∙10 23 nitrogen atoms

1)127,8; 2)42,6; 3)213; 4)14,2. (2 points)

6. Number of protons and neutrons in the nucleus of the 40 K isotope

1) p = 20, n=19; 2)p = 40, n = 19; 3)p= 19,n=21: 4)p=21,n=19. (2 points)

7. The reaction that results in the formation of a precipitate

1) KOH + HC1; 2) K 2 C0 3 + H 2 S0 4; 3) Cu(OH) 2 + HNO 3; 4) Na 2 S + Pb(N0 3) 2. (2 points)

8. When a mixture of zinc (5.2 g) and zinc carbonate (5.0 g) reacts with hydrochloric acid, gases are released in volume (n.s.) (2 points)

1)0.896 l; 2) 1.792 l; 3) 2.688 l: 4) 22.4 l.

9. 150 g of calcium chloride was dissolved in 250 ml of water. The mass fraction of salt in the solution (in percent) is equal to:

1) 60; 2) 37,5; 3) 75; 4) 62,5 (2 points)

10. The molar mass of a gas weighing 0.84 g occupying a volume of 672 ml (n.s.) is equal to

1)44; 2)28; 3)32; 4)16. (2 points)

Total 18 points

PART 2 (transformation chains)

Read the text.

A soldier was walking across the field, he decided to stop and write a letter to his family, but he had nothing to do with it. Then he found a black pebble, tried it and drew it. I wrote a letter with this stone and decided to sleep. He threw a pebble into the fire - it flared up, only smoke came out. The smoke stopped the rain and nailed it to the ground; The water flowed heavy and chalky. The water began to dry, in some places it remained, and where it dried up, there were already white pebbles lying there. The soldier woke up and was surprised at how the rain turned a black pebble into white. He took a white pebble, ran it over a white boulder, and lo and behold, he was drawing. The soldier was surprised and moved on.

Make a chain of transformations and solve it. (8 points )

(10 points)

PART 3 (tasks)

Designers of the first spaceships and submarines were faced with a problem: how to maintain a constant air composition on a ship or space station, i.e. how to get rid of excess carbon dioxide and restore oxygen supply? An elegant solution was found - we need to turn CO 2 into O 2! For this, it was proposed to use potassium superoxide (KO 2), which is formed during the combustion of potassium in oxygen. When potassium superoxide reacts with carbon dioxide, free oxygen is released (oxygen atoms are both an oxidizing agent and a reducing agent at the same time). Write down equations for the reactions discussed in the text. Knowing that a person will emit an average of 0.51 m 3 of carbon dioxide per day, calculate how much potassium superoxide should be on board the space station to ensure the livelihoods of a crew of three people for a month (30 days). (24 points)

The young chemist received from his teacher four bottles without labels containing solutions of potassium hydroxide, zinc chloride, barium chloride and potassium sulfide. The concentration of substances in each of the dispensed solutions was 0.1 mol/l. How, without using additional reagents and having only phenolphthalein paper (paper soaked in an alcohol solution of phenolphthalein), determine the contents of the bottles? Suggest a sequence of actions for the young chemist. Write down the equations for the reactions that occur. ( 40 points)

School stage of the All-Russian Chemistry Olympiad ANSWERS

9TH GRADE

PART 1Test.

Job number	Possible answer	Number of points










Total 18 points

PART 2 Chains of transformations.

Task 11

		Number of points
A chain has been created	C-CO 2 -H 2 CO 3 -CaCO 3
1 equation
2 equation	CO 2 +H 2 O=H 2 CO 3
3 equation	H 2 CO 3 + CaO = CaCO 3 + H 2 O
Total 8 points

Task 12

reaction equation 1 is written	CuSO 4 + 2KOH = Cu(OH) 2 ↓ + K 2 SO 4	2 points
reaction equation 2 is written	Cu(OH) 2 = CuO + H 2 O (heating)	2 points
reaction equation 3 is written	CuO + H 2 = Cu + H 2 O	2 points
reaction equation 4 is written	Cu + HgSO 4 = CuSO 4 + Hg	2 points
reaction equation 5 is written	CuSO 4 + Ba(NO 3) 2 = BaSO 4 ↓ +Cu(NO 3) 2	2 points
Total 10 points

PART 3 Tasks.

Task 13.

The equation for the reaction of combustion of potassium in oxygen is written		5 points
The reaction equation for the interaction of potassium superoxide with carbon dioxide has been written	4KO 2 + 2CO 2 = 2K 2 CO 3 + 3O 2	5 points
The amount of potassium superoxide that should be on board is calculated.	A person emits 0.51 m 3 = 510 l of carbon dioxide per day, which is 510 l: 22.4 l/mol = 22.77 mol; to utilize it, 2 times more potassium superoxide is required, i.e. 45.54 mol or 3.23 kg. To work for 30 days for three people, 3.23 kg x 30 days x 3 people = 290.27 kg of potassium superoxide will be required.	14 points
Total 24 points

Task 14

Phenolphthalein takes on a crimson color in an alkaline environment. The color appears in solutions of potassium hydroxide and potassium sulfide: K 2 S + NON KNS + KOH We determine which bottle contains potassium hydroxide and potassium sulfide. We add the contents of one of the bottles to the samples of solutions in which the color of phenolphthalein has not changed. If a white precipitate forms that does not dissolve in an excess of reagent solution, this is a solution of potassium sulfide, and the analyzed solution contains zinc chloride. K 2 S + ZnCl 2 = ZnS↓ + 2КCl
If a gelatinous precipitate forms, soluble in excess of the reagent, then the added solution contains potassium hydroxide, and the analyzed solution contains zinc chloride. 2KOH + ZnCl 2 = Zn(OH) 2 ↓ + 2КCl Zn(OH) 2 ↓ + 2KOH = K 2 or Zn(OH) 2 ↓ + 2KOH = K 2 ZnO 2 + 2H 2 O
The flask, to the samples of which no visible changes occur when solutions of K 2 S and KOH are added, contains barium chloride.
Total 40 points

This Olympiad is available to everyone!

For everyone who is interested in studying chemistry, we are launching a new international Olympiad. We invite students of grades 8-11 and students of the first and second years of secondary educational institutions to test their knowledge and receive well-deserved rewards for their efforts and theoretical basis.

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History of chemistry and chemical discoveries
Chemical properties of metals and non-metals
Hydrocarbons
Dispersed media and other topics

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Master the conceptual apparatus and symbolic language of chemistry.
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We hold all-Russian and international Olympiads in which schoolchildren and students participate. The Mega-Talent Olympiads in chemistry are held for students in grades 8-11 and students of the first and second years of secondary educational institutions. We reward all participants by presenting them with diplomas and certificates. Teachers receive certificates and gratitude from us for organizing the Olympiad.

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Transcript

1 1 ALL-RUSSIAN OLYMPIAD FOR SCHOOLCHILDREN IN CHEMISTRY MUNICIPAL STAGE. 9 CLASS Solutions and criteria for evaluating Olympiad tasks The five solved ones, for which the participant scored the most points, are counted in the final grade of the six proposed problems. One of the problems with the lowest score is not taken into account. 1. Properties of acidic oxides Write reaction equations in which an acidic oxide reacts with another substance in a molar ratio: a) 1: 1, b) 1: 2, c) 2: 1, d) 1: 3, e) 1: 6 (For each reaction you can choose your own acid oxide.) a) SO 3 + H 2 O = H 2 SO 4 b) CO 2 + 2KOH = K 2 CO 3 + H 2 O c) 2SO 2 + O 2 = 2SO 3 d) P 2 O 5 + 3H 2 O = 2H 3 PO 4 e) P 2 O 5 + 6KOH = 2K 3 PO 4 + 3H 2 O 2 points for each equation. In each case, several options are possible, any reasonable answers that meet the conditions are accepted. 2. Reactions with volume reduction There is a mixture of hydrogen and oxygen. The mixture is ignited with an electric spark. At what volumetric ratio of gases will the volume of the mixture after passing a spark decrease by exactly one quarter (at constant temperature and pressure)? The volume of water formed can be neglected. Give two answer options and support them with calculations. 2H 2 (g) + O 2 (g) = 2H 2 O (l) According to the reaction equation, 2 moles of water are formed from 3 moles of gases. Consequently, the amount of substance in gases decreases by 3 mol, which

2 is a quarter of the original amount, i.e. before the reaction there were 12 moles of gases. Two options are possible: 1) in excess hydrogen: 11 mol H 2 and 1 mol O 2, volume ratio H 2: O 2 = 11: 1 2) in excess oxygen: 2 mol H 2 and 10 mol O 2, volume ratio H 2: O 2 = 1: 5 Grading system: reaction equation 1 point, mention of the idea of “excess-deficiency” 1 point, each answer option 4 points, of which 3 points for calculating the composition of the mixture in moles, 1 point for the volume ratio. 3. Oxygen from Carbon Dioxide Compound alkali metal With oxygen, substance X reacts with carbon dioxide. The only reaction products are metal carbonate and oxygen in a molar ratio of 2: 3 and a mass ratio of 23: 8. Establish the formula of X and write the equations for its reactions with carbon dioxide and water. The metal can be determined by comparing the mass ratio with the molar ratio: m(m2co 3) 2 M (M2CO 3) 2 M (M2CO 3) 23 = = =, m(o) 3 M (O) from where M(M 2 CO 3) = 138 g/mol, this is K 2 CO 3, metal K. Formula X is determined by the reaction equation. Let's denote the formula X K a O b and first draw up a diagram and then an equation for the reaction. Scheme: K a O b + CO 2 2K 2 CO 3 + 3O 2. Equation: 4KO 2 + 2CO 2 = 2K 2 CO 3 + 3O 2. Substance X KO 2. Reaction equation with water: 4KO 2 + 2H 2 O = 4KOH + 3O 2. (The reaction equation with the formation of H 2 O 2 is also accepted: 2KO 2 + 2H 2 O = 2KOH + H 2 O 2 + O 2.) 2

3 3 Grading system: determination of metal 2 points, establishment of formula X 4 points, reaction equations, 2 points for each 4 points. 4. Electrons also have mass 2 Which substance, consisting of two elements, has the largest mass fraction of electrons? Briefly justify your answer using calculations or reasoned arguments. An atom consists of protons, neutrons and electrons. The number of electrons in an atom is serial number element Z. Let us assume that the masses of the proton and neutron are approximately equal, and write the expression for the mass fraction of the electron: N(e) m(e) N(e) m(e) ω(e) = = = N(e) m( e) N(p) m(p) N(n) m(n) N(e) m(e) N(p) N(n) m(p) Zm(e) 1 = = Zm(e) Am (p) A m(p) 1 Z m(e) Here N denotes the number of particles of a given type in an atom, A is the mass number. In order for the electron mass fraction to be greatest, the A/Z ratio must be smallest. It is equal to 1 for the 1 H atom; for all other atoms it is equal to 2 or more. Thus, the compound must contain H atoms and atoms of another element with the minimum (if possible) A / Z ratio. In this case, there should be as much hydrogen as possible. Of all hydrogen compounds the largest mass fraction of hydrogen in CH 4. The carbon atom 12 C is suitable as the second element since it has A / Z = 2. Answer: CH 4. Grading system: considerations that the compound must include an H atom, 4 points, considerations about CH 4 6 points (compounds with a lower mass fraction of hydrogen, for example NH 3, H 2 O, 3 points instead of 6), if only the correct answer is given without any reasoning 0 points.

4 4 5. Study of black powder The young chemist examined black powder using the means available to him. When the powder was calcined for a long time in air and in a stream of oxygen, the color of the substance became darker, and a gas with a pungent odor was released, similar to the smell of a burning match. The mass of the substance did not change during calcination. Further studies showed that the original test substance does not dissolve in dilute sulfuric acid even when heated, but the calcination product dissolves well in the acid, forming a solution blue color. Based on these observations, the young chemist came to the correct conclusion that the substance consists of two elements. a) What elements are included in the composition of the substance under study? b) What is the formula of the substance under study? c) Write down the equation for the reaction that occurs when a substance is calcined in the presence of oxygen. d) Explain using calculation why the mass solid does not change upon heating. a) The composition of the substance under study includes copper and sulfur (1.5 points for each element for a total of 3 points). b) Formula of the substance Cu 2 S (3 points; for CuS 0 points). c) Cu 2 S + 2O 2 = 2СuO + SO 2 (3 points; 3 points are also given if the correct equation for the combustion of CuS is given). d) The mass of 1 mole of Cu 2 S (160 g) is equal to the mass of 2 moles of CuO, the product of the reaction SO 2 gaseous substance(1 point).

5 5 6. Chemical fountain The following experiment was carried out in a chemistry lesson. A white crystalline substance was placed in Wurtz flask 1 (see Fig. 1), which colors the flame yellow. Using dropping funnel 2, concentrated sulfuric acid was poured into flask 1, and a colorless gas 1.26 times heavier than air immediately began to evolve. The resulting gas was collected in round-bottomed flask 3. 1) What gas was obtained in the device shown in Fig. 1? Justify your answer, confirm it with calculations and the reaction equation. 2) Why was this gas collected by displacing air rather than water? Figure 1 3) What needs to be done with flask 3 filled with gas so that a fountain begins to “glow” inside it (see Figure 2)? Explain your answer. 4) What will be observed (see Fig. 2) if you pour into crystallizer 4: a) an aqueous solution of methyl orange; b) silver nitrate solution? 5) Why should you use a round-bottomed flask rather than a flat-bottomed flask for a “fountain”? Figure 2 1) Hydrogen chloride was obtained in the device. M = 1.26 29 g/mol = 36.5 g/mol (2 points) Sulfuric acid poured into sodium chloride. It is known that sodium ions color the flame yellow, which proves the presence of sodium in the composition of this salt. (1 point) The reaction occurs in Wurtz flask 1: NaCl + H 2 SO 4 = NaHSO 4 + HCl. (1 point) (The equation is allowed: 2NaCl + H 2 SO 4 = Na 2 SO 4 + 2HCl) 2) Hydrogen chloride cannot be collected by displacing water, since it dissolves well in it. (1 point) 3) In order for the flask to “give a fountain”, it is necessary, without releasing gas, to close it with a stopper with a gas outlet tube. Then add a small amount of water into the flask, in which hydrogen chloride will dissolve,

6 and a vacuum will be created in the flask. If the end of the gas outlet tube is lowered into water, it will begin to rise into the flask under the influence of atmospheric pressure. (2 points) 4) When hydrogen chloride is dissolved in a flask, hydrochloric acid is formed. a) In hydrochloric acid, the orange color of the indicator will change to red. (1 point) b) When interacting of hydrochloric acid with a solution of silver nitrate, a white precipitate is formed, the mixture becomes cloudy: HCl + AgNO 3 = HNO 3 + AgCl. (1 point) 5) When demonstrating a “fountain”, a vacuum is created inside the flask, so its walls experience significant pressure from the outside. In the case of a round-bottomed flask, the pressure on its walls is distributed evenly. Thus, when using a round-bottomed flask, there is less chance that it will burst due to the pressure difference. It is important to note that the walls of the flask must be completely smooth and intact; there should be no cracks, scratches, or chips on them. (1 point) 6

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Participants about 250

Now the Olympiad is held in 4 stages. The holding of the All-Russian Olympiad is regulated by the Procedure approved by the Ministry of Education and Science of Russia.

Geography of the All-Russian and All-Union Olympiads in Chemistry (1965-2015)

Olympiad number	Year	City	Olympics
I			All-Russian
2.	1966	Kazan	All-Russian
3.	1967	Dnepropetrovsk	All-Union
4.	1968	Vilnius	All-Union
5.	1969	Rostov-on-Don	All-Union
6.	1970	Voronezh	All-Union
7.	1971	Minsk	All-Union
8.	1972	Ufa	All-Union
9.	1973	Baku	All-Union
10.	1974	Donetsk	All-Union
11.	1975	Vilnius	All-Union
12.	1976	Kyiv	All-Union
13.	1977	Almaty	All-Union
14.	1978	Kazan	All-Union
15.	1979	Kishinev	All-Union
16.	1980	Yerevan	All-Union
17.	1981	Frunze	All-Union
18.	1982	Tallinn	All-Union
19.	1983	Tbilisi	All-Union
20.	1984	Dushanbe	All-Union
21.	1985	Riga	All-Union
22.	1986	Donetsk	All-Union
23.	1987	Tashkent	All-Union
24.	1988	Vilnius	All-Union
25.	1989	Permian	All-Union
26.	1990	Minsk	All-Union
27.	1991	Ivanovo	All-Union
28.	1992	Samara	All-Union
29.	1993	Lipetsk	All-Russian
30.	1994	Kazan	All-Russian
31.	1995	N.-Novgorod	All-Russian
32.	1996	Samara	All-Russian
33.	1997	Kazan	All-Russian
34.	1998	Belgorod	All-Russian
35.	1999	Ryazan	All-Russian
36.	2000	Vladimir	All-Russian
37.	2001	Ulyanovsk	All-Russian
38.	2002	Velikiy Novgorod	All-Russian
39.	2003	Cheboksary	All-Russian
40.	2004	Chelyabinsk	All-Russian
41.	2005	Belgorod	All-Russian
42.	2006	Ufa	All-Russian
43.	2007	Ufa	All-Russian
44.	2008	Ufa	All-Russian
45.	2009	Arkhangelsk	All-Russian
46.	2010	Kazan	All-Russian
47.	2011	Arkhangelsk	All-Russian
48.	2012	Magnitogorsk	All-Russian
49.	2013	Kursk	All-Russian
50.	2014	Kazan	All-Russian
51.	2015	Novosibirsk	All-Russian
52.	2016	Belgorod	All-Russian
53	March 30 - April 5		All-Russian

In order to coordinate the holding of Olympiads and other intellectual competitions aimed at finding, supporting and accompanying talented children and youth during the formation period, the Russian Council for Schoolchildren Olympiads (RSOSH) was created in 2006. The work of RSOS is regulated by the Procedure for holding Olympiads for schoolchildren. This Procedure does not apply to the All-Russian Olympiad for schoolchildren.

Stages of the Olympiad

According to the Procedure for holding the All-Russian Olympiad for schoolchildren, the All-Russian Olympiad for Schoolchildren in Chemistry is held in 4 stages:

The school stage of the Olympiad is carried out according to tasks developed by the municipal subject-methodological commissions of the Olympiad, based on the content educational programs basic general and secondary general education in-depth level and corresponding focus (profile), for grades 4-11. The deadline for completing the school stage of the Olympiad is no later than November 1;
The municipal stage of the Olympiad is carried out according to tasks developed by regional subject-methodological commissions, based on the content of educational programs of basic general and secondary general education at an advanced level and the corresponding focus (profile), for grades 7-11. The deadline for the municipal stage of the Olympiad is no later than December 25;
The regional stage of the Olympiad is carried out according to tasks developed by the central subject-methodological commissions of the Olympiad, based on the content of educational programs of basic general and secondary general education at an advanced level and the corresponding focus (profile), for grades 9-11. Deadline regional stage Olympiads - no later than February 25;
The final stage of the Olympiad is carried out according to tasks developed by the central subject-methodological commissions, based on the content of educational programs of basic general and secondary general education at an advanced level and the corresponding focus (profile), for grades 9-11. The deadline for the final stage of the Olympiad is no later than April 30.

Correspondence (distance) forms of conducting stages of secondary school

The school, municipal and regional stages of the All-Russian Olympiad for schoolchildren are sometimes held in several rounds, one of which is correspondence. This tour allows you to attract as many schoolchildren as possible to participate in the Olympiad and conduct an effective selection of participants for the full-time tour. Most often, a correspondence tour is conducted via the Internet.

Organizational and scientific-methodological work

The organization and conduct of the Olympiad is coordinated by the Central Organizing Committee of the Olympiad under the leadership of the chairman. Organizational, technical and information support for the activities of the Central Organizing Committee of the Olympiad is provided by the Ministry of Education and Science of Russia. For scientific and methodological support of the Olympiad, central subject-based methodological commissions (TSMC) of the Olympiad are created. The composition of the TsPMK in chemistry is formed at the proposal of the Central Organizing Committee of the Olympiad from among teaching, scientific, scientific and pedagogical workers and is approved by the Ministry of Education and Science of Russia. The main principles of the activities of the Central Organizing Committee of the Olympiad, the central subject-methodological commissions of the Olympiad, the jury of all stages of the Olympiad are competence, objectivity, transparency, as well as compliance with professional ethics.

The All-Russian Chemistry Olympiad for schoolchildren is held in four stages. School and municipal are most often held for students in grades 7-11, in one theoretical round...

At the regional stage, only grades 9-11 can participate, and another round is added - a practical one. The final stage is held for grades 9-11 in three rounds: two theoretical and practical.

Olympiad tasks in chemistry often require participants to have additional knowledge from other areas, for example, physics, biology, geology, geography. Winners and prize-winners of the final stage of the All-Russian Olympiad receive benefits when entering universities in specialized areas.

The first Chemistry Olympiads for schoolchildren took place in Moscow and Leningrad in 1938. The first All-Union Olympiad was held in 1967.

What's new

How to participate

Inform your subject teacher about your intention to participate in the Olympiad, to the class teacher, head teacher or person responsible for competitions at school.
Find out from your school what days and where the school phase will take place.
Come to the Olympics and participate.
Find out your results and passing marks for the next stage at your school.
Participate in the municipal stage. Find out from your teacher where the Olympiad will be held.
Find out the results, carefully review your work and compare it with the criteria, ask the jury questions.
Wait for the passing scores for the regional stage.
Find out where and when the competition will take place on the website of the Olympiad organizer or from the person responsible for holding the competition in your region.
Study the requirements for holding the regional stage. Check to see if you need to bring a lab coat, rubber gloves, or goggles on the experiment tour.
Participate in the regional stage.
Wait for your results. Look at the work and compare it with the assessment criteria. Ask questions to the verification jury; if you do not agree with their answers, go to appeal.
Wait for the passing scores for the final stage to be announced.
The organizer of the Olympiad in the region must contact you and provide the necessary information about the finals.
Participate in final stage and follow safety precautions in the laboratory!