« MUNICIPAL STAGE. Grade 9 Tasks, answers, assessment criteria General instructions: if in ... "

ALL-RUSSIAN OLYMPIAD FOR SCHOOLCHILDREN

IN CHEMISTRY. 2016–2017 academic year G.

MUNICIPAL STAGE. 9 CLASS

Tasks, answers, evaluation criteria

General instructions: if the task requires calculations, they must

be given in the decision. Answer given without calculations or otherwise

justification does not count.

In the final assessment of 6 tasks, 5 solutions are counted, for which the participant scored the highest points, that is, one of the tasks with the lowest score is not taken into account.

1. (10 points) Ionic Reactions Complete the abbreviated ionic reaction equations below with coefficients. All unknown particles are indicated by dots.

a) … + 2… Cu(OH)2 b) … + 2OH– + … BaSO3 + …

c) Pb2+ + … … + 2H+

d) H+ + … CO2 + … e) 3H+ + … Al3+ + 3… For each abbreviated ionic equation, give one equation in molecular form.

Cu2+ + 2OH– Cu(OH)2 a) CuCl2 + 2KOH Cu(OH)2 + 2KCl Ba2+ + 2OH– + SO2 BaSO3 + H2O b) Ba(OH)2 + SO2 BaSO3 + H2O Pb2+ + H2S PbS + 2H+ c) Pb(NO3)2 + H2S PbS + 2HNO3 H+ + HCO3– CO2 + H2O d) HCl + NaHCO3 NaCl + CO2 + H2O 3H+ + Al(OH)3 Al3+ + 3H2O e) 3HCl + Al(OH)3 AlCl3 + 3H2O Criterion evaluation.

Each equation (abbreviated ionic or molecular) - 1 point.

Total for the task - 10 points.

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All-Russian Olympiad for Schoolchildren in Chemistry. 2016–2017 academic year G.

municipal stage. Grade 9 2. (10 points) Interaction of solutions When mixing equal masses of solutions of barium chloride and sodium carbonate, 13.79 g of precipitate X and a solution of substance Y were formed. Identify unknown substances and write down the reaction equation. Find the mass of substance Y and its mass fraction in the final solution, if it is known that no changes are observed when sulfuric acid is added to it, and the mass fraction of sodium carbonate in the initial solution is 1.7 times the mass fraction of substance Y in the final solution.

When interacting solutions of barium chloride and sodium carbonate, the reaction proceeds:

BaCl2 + Na2CO3 = BaCO3 + 2NaCl.

X – BaCO3, Y – NaCl.

Since the filtrate does not react with sulfuric acid, it contains neither excess carbonate ions nor excess barium ions. This means that both substances have reacted completely.

Let's calculate according to the reaction equation:

n(BaCO3) = 13.79 / 197 = 0.07 mol, n(Na2CO3) = 0.07 mol, m(Na2CO3) = 0.07106 = 7.42 g, n(NaCl) = 0.14 mol, m(NaCl) = 0.1458.5 = 8.19 g.

Let the mass of each of the two mixed solutions be x g, then the mass of the final solution is (2x – 13.79) g (BaCO3 precipitate is not included in the solution).

Mass fraction of sodium carbonate in the initial solution:

–  –  –

3. (10 points) Electrons of chemical bonds Give one example of molecules in which the following are involved in the formation of covalent chemical bonds:

a) all the electrons of the molecule;

b) more than half of the electrons of the molecule;

c) exactly one third of the total number of electrons in the molecule.

Justify answers. For each molecule, describe the electronic configuration of the atom with the highest atomic number.

a) Only one element capable of forming chemical bonds does not have internal electrons - this is hydrogen H. In the H2 molecule, both electrons (i.e., all the electrons of the molecule) participate in the formation of the H–H bond.

There are no other such molecules.

Electronic configuration of the hydrogen atom: 1s1.

b) The more hydrogen atoms in the molecule, the greater the proportion of electrons from their total number in the molecule participates in covalent bonds. In addition to hydrogen, it makes sense to consider only elements of the 2nd period, since they have relatively few internal electrons. We list the volatile hydrogen compounds of non-metals of the 2nd period (with the exception of boron) indicating the number of electrons:

CH4 - 10 electrons in total, 4 chemical bonds formed by 8 electrons.

8/10 1/2 fit.

NH3 - 10 electrons in total, 3 chemical bonds formed by 6 electrons.

6/10 1/2 fit.

H2O - 10 electrons in total, 2 chemical bonds formed by 4 electrons.

4/10 1/2 does not fit.

HF - 10 electrons in total, 1 chemical bond formed by 2 electrons.

2/10 1/2 does not fit.

Thus, the possible correct answers are CH4 or NH3.

Electronic configurations of central atoms:

C - 1s2 2s2 2p2, N - 1s2 2s2 2p3 (the distribution by energy levels is also accepted: 2 - 4 for carbon and 2 - 5 for nitrogen).

c) The number of electrons participating in the formation of covalent bonds is even. Then the total number of electrons in the molecule must be a multiple of 6:

6, 12, 18, etc. There are no molecules with 6 electrons (BeH2 has a non-molecular structure). The unstable C2 molecule has 12 electrons, the bond in which is © GAOU DPO TsPM. Publication on the Internet or print media without the written consent of GAOU DPO CPM is prohibited.

All-Russian Olympiad for Schoolchildren in Chemistry. 2016–2017 academic year G.

municipal stage. 9 class double. Formally, C2 is the correct answer. 18 electrons and 3 covalent bonds are in the PH3 molecule, which is the main correct answer.

The electronic configuration of the P atom is 1s2 2s2 2p6 3s2 3p3 (or 2 8 5).

Answer. a) H2; b) CH4 or NH3; c) PH3.

Evaluation criteria:

a) 3 points - of which 2 points for the formula with electron counting and 1 point for the electronic structure (only the formula without electron counting - 0 points).

b) 3 points - of which 2 points for the formula with electron counting and 1 point for the electronic structure (only the formula without electron counting - 0 points).

c) 4 points - of which 3 points for the formula with electron counting and 1 point for the electronic structure (only the formula without electron counting - 0 points).

Total for the task - 10 points.

4. (10 points) Color reactions A turquoise precipitate X weighing 7.74 g, which was released when a small amount of sodium hydroxide solution was added to an aqueous solution of copper (II) sulfate, forms 4.80 g of black powder Y on calcination, which, when heated in a current hydrogen changes color, turning into a pink-red powder Z with a mass of 3.84 g.

1. Identify substances X, Y, Z and name them. Support your answer with calculations.

2. Give the equations of all the reactions described above.

3. Write down the reactions of X with sulfuric acid and with potassium hydroxide.

1. Pink-red powder Z, formed by reduction with hydrogen, is a simple substance copper.

n(Cu) = 3.84/64 = 0.06 mol.

Substance Y is copper oxide. If we assume that the formula unit of the oxide contains one copper atom, then n(Y) = n(Cu) = 0.06 mol, then M(Y) = 4.8/0.06 = 80 g/mol, which corresponds to copper oxide CuO.

Using the same assumption, we calculate the molar mass of X. It is equal to 7.74 / 0.06 = 129 g / mol, which does not correspond to the molar mass of copper hydroxide.

It is then reasonable to assume that X is a basic salt. If the formula unit X contains two copper atoms, then M(X) = 7.74/0.03 = 258 g/mol, which corresponds to the formula Cu2(OH)2SO4.

So, X is Cu2(OH)2SO4, dihydroxydicopper(II) sulfate, or basic copper(II) sulfate, or hydroxomedi sulfate. Any of these names is accepted.

Y – CuO, copper(II) oxide.

Z - Cu, copper.

–  –  –

2CuSO4 + 2NaOH = Cu2(OH)2SO4 + Na2SO4 2.

2Cu2(OH)2SO4 = 4CuO + 2SO2 + O2 + 2H2O CuO + H2 = Cu + H2O

3. Cu2(OH)2SO4 + H2SO4 = 2CuSO4 + 2H2O Cu2(OH)2SO4 + 2KOH = 2Cu(OH)2 + K2SO4.

–  –  –

5. (10 points) Fluid analysis Two flasks with warm water were filled with 13.5 g of a colorless oxygen-containing liquid consisting of three elements. After some time, when the reaction was over, the solutions were analyzed. The resulting solutions were acidic. An excess of barium chloride solution was poured into the first flask, and 23.3 g of a white crystalline precipitate separated out. An excess of silver nitrate solution was added to the second flask. The mass of the precipitated curdled sediment was 28.7 g.

Determine the formula for the unknown liquid. Write down the equations of ongoing chemical reactions.

–  –  –

6. (10 points) School experiment In the school laboratory, the device was assembled, as shown in the figure.

Small pieces of iron(II) sulfide were placed in flask 1 and hydrochloric acid was added. The escaping gas was passed through flask 2 filled with anhydrous calcium chloride. In flask 6, concentrated sulfuric acid was added to sodium sulfite powder. The escaping gas was passed through bottle 7 with concentrated sulfuric acid. Both gases entered the flask-reactor 3, the direction of movement of the gases is shown in the figure by arrows.

Excess gases flowed through the gas outlet tube 8 into the absorption bottle 9.

When the reactor flask 3 was filled with a mixture of gases, no change was observed. However, after the clamp 5 was opened and a small amount of water was poured from the funnel 4, a reaction began in the flask 3. The space in this flask was filled with smoke, and after a while a dense yellow coating formed on its walls.

1. What gases were received in flasks 1 and 6? Support your answer with reaction equations.

2. For what purpose were the released gases passed through flasks 2 and 7?

Can flask 2 also be filled with concentrated sulfuric acid?

Explain the answer.

3. What reaction took place in flask 3 after water was added there?

What substance settled on the walls of the flask? Write the corresponding equation.

4. What substances can be used to fill the absorption bottle 9?

Give two examples of such substances and justify your answer.

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8-1.

2) slaked lime 6) water

3) gypsum 7) carbon monoxide

4) limestone (chalk, marble)

Write reaction equations confirming the amphoteric nature of oxides (including in solution and solid phase) (12 points).

8-2.

Calculate how much fat is left in the pot? How many grams of fat did the cat eat? Take the density of milk as 1 g/cm3 (6 points).

8-3. How many atoms are contained in a cube of pure gold with an edge equal to 1 mm (gold density 19.3 g / cm3) ( 10 points).

8-4. When a mixture of sulfur with 44.8 g of metal is calcined in an oxygen-free atmosphere, a divalent metal sulfide is formed. When the reaction product is dissolved in an excess of hydrochloric acid, gas X is released and 12.8 grams of an insoluble substance remains, which, when burned in excess oxygen, produces gas Y. The quantitative interaction of gas X with gas Y leads to the formation of 38.4 g of a simple substance (moreover, per mol of gas Y is formed 3 mol of a simple substance). What metal interacted with sulfur? What is the mass of sulfur contained in the initial mixture? ( 12 points).

8-5. To a mixture of nitric oxide NO and nitrogen with a volume of 100 ml was added 100 ml of air (volume fraction of oxygen 20% and volume fraction of nitrogen 80%). The final volume of the reaction mixture was 185 ml. Calculate the volume fraction (%) of nitric oxide in the initial mixture. (All gas volumes measured under normal conditions) ( 5 points).

SOLUTIONS

Municipal stage of the All-Russian Olympiad for Schoolchildren in Chemistry 2016-2017 academic year

8th grade

8-1. Write the formulas of the given substances and name them. Select those substances that are oxides. Give formulas of these oxides. Select oxides with amphoteric properties.

1) carbon dioxide 5) bauxite (alumina)

2) slaked lime 6) water

3) gypsum 7) carbon monoxide

4) limestone (chalk, marble)

Make up reaction equations confirming the amphoteric nature of oxides (including in solution and solid phase).

Solution of problem 8-1

The given substances have the following formulas: 1) carbon dioxide - CO2, slaked lime - Ca (OH) 2, gypsum - CaSO4 ∙ 2H2O, limestone (chalk, marble) - CaCO3, bauxite (alumina) - Al2O3, water - H2O, carbon monoxide – CO. (for each formula - 0.5 points).

Of these substances, oxides include: carbon dioxide (CO2), bauxite (alumina) - Al2O3, water - H2O, carbon monoxide - CO. Al2O3 and H2O have amphoteric properties. (determination of amphoteric oxides - 1.5 points).

Al2O3 can interact with both acids and alkalis:

Al2O3+2NaOH(melt)→2NaAlO2+H2O(melt) (2 points)

Al2O3+2NaOH+3Н2О→2Na (2 points)

Al2O3+2NaOH(conc) +3Н2О→2Na3

Al2O3+6HCl→2AlCl3+3H2O (1 score)

3H2O+P2O5=2H3PO4 (1 score)

H2O+Na2O=2NaOH (1 score)

Total: 12 points

8-2. After milking the cow, the hostess poured 2 liters of milk with a fat content of 4.6% into the pot. A fat fluffy cat, having slept all day, jumped on the table and licked 200 g of settled cream with a fat content of 15%.

Calculate how much fat is left in the pot? How many grams of fat did the cat eat? Take the density of milk as 1 g/cm3.

Solution of problem 8-2

1. Determine the mass of milk and the mass of fat in it:

(1 point)

(1 point)

2. We calculate the mass of fat that the cat ate:

(1 point)

The cat ate 30 g of fat out of 92 g. 62 g remained. (1 point)

3. Determine the mass of the remaining milk and its fat content:

2000-200=1800 (1 point)

(1 point)

Total: 6 points

8-3. How many atoms are contained in a cube of pure gold with an edge equal to 1 mm (density of gold is 19.3 g/cm3).

Solution of problem 8-3

The volume of a gold cube (Au) with an edge of 1 mm (0.1 cm) is

V(Au)=(0.1)3=0.001=1 10-3 cm3. (2 points)

With the density of gold ρ=19.3 g/cm3, the mass of this volume of the gold cube is

m(Au)=ρV(Au)=19.3∙10-3 g. (2 points)

The molar mass of gold M(Au)=197 g/mol. That's why:

n(Au)=m(Au):M(Au)=(19.3∙10-3):197≈1∙10-4 mol. (3 points)

In accordance with Avogadro's law, this amount of matter contains atoms

N(Au)=n(Au)NA=(1∙10-4)×(6.02∙1023)=6.02 1019 atoms. (3 points)

Total: 10 points

8-4. When a mixture of sulfur with 44.8 g of metal is calcined in an oxygen-free atmosphere, a divalent metal sulfide is formed. When the reaction product is dissolved in an excess of hydrochloric acid, gas X is released and 12.8 grams of an insoluble substance remains, which, when burned in excess oxygen, produces gas Y. The quantitative interaction of gas X with gas Y leads to the formation of 38.4 g of a simple substance (moreover, per mol of gas Y is formed 3 mol of a simple substance). What metal interacted with sulfur? What is the mass of sulfur contained in the initial mixture?

Solution of problem 8-4

1. Analyzing the condition of the task, it can be shown that sulfur in the initial mixture is in excess, and the task itself corresponds to the following set of equations of chemical reactions:

, (1) (1 point)

, (2) (1 point)

, (3) (1 point)

, (4) (1 point)

2. M(S)=32 g/mol. 38.4 g of a simple substance (sulfur) corresponds to its amount equal to

. (1 point)

Since reaction (4) proceeds quantitatively, the amount of H2S and SO2 in this reaction, respectively, is equal to

(1 point)

. (1 point)

3. Then the amount of excess sulfur present in the initial mixture and not reacted according to reaction (1) is equal to n(Sex.)=0.4 mol, and the amount of MeS, Me and S (reacted in reaction (1)) is equal to the amount of H2S :

. (2 points)

4. Since m(Me)=n(Me)M(Me), then

. (1 point)

The desired metal is iron М(Fe)=56 g/mol.

5. The total amount of sulfur in the initial mixture n(S)=n(Sex.)+n(S(1))=0.4+0.8=1.2 mol. (1 point)

The mass of sulfur contained in the initial mixture is m(S)=n(S)M(S)=1.2×32=38.4 g. (1 point)

Total: 12 points.

8-5. To a mixture of nitric oxide NO and nitrogen with a volume of 100 ml was added 100 ml of air (volume fraction of oxygen 20% and volume fraction of nitrogen 80%). The final volume of the reaction mixture was 185 ml. Calculate the volume fraction (%) of nitric oxide in the initial mixture. (All gas volumes are measured under normal conditions).

Solution of problem 8-5

1. The volume of the mixture of gases is less than the sum of the volumes of the initial gases, since the reaction

goes with the absorption of oxygen. (1 point)

2. The volume difference 200-185=15 ml is the volume of oxygen that is consumed for the oxidation of NO to NO2. Since 20 ml of oxygen was introduced into the gas mixture, and the change in volume is 15 ml, it can be argued that nitric oxide has completely reacted with oxygen. (1 point).

3. According to the reaction equation, the molar ratio of nitric oxide and oxygen is 2:1. This will determine the volume of nitric oxide in the initial mixture of gases:

where V(NO) and V(NO2) are the volumes of NO and O2, ml (n.a.)

V(NO)=2∙15=30 ml. (1 point)

4. Find the volume fraction of nitric oxide (%):

(2 points)

Total: 5 points

The school stage takes place in the city districts (on the basis of schools) from the beginning of September to the end of October.

The regional stage of the All-Russian Olympiad in Chemistry will be held in the regions of St. Petersburg on November 15, 2018 - 1 round, December 6, 2018. -2 tour, 14:00

The appeal for 1 round takes place in the districts on 12/06/2018. after round 2. Disputable issues of appeal are resolved remotely with the Jury of the regional stage by mail
If the appeal is not satisfied by you or the issue requires a decision with the Jury of the regional stage (the application is submitted from 06 to 10 December by mail This email address is being protected from spambots. You must have JavaScript enabled to view. until 12:00). The original work of the participant must be 13.12.18. in the Central Organ of St. Petersburg

Absentee appeal (for those who filed an appeal) on December 13, 2018 from 18:00 at the Central Organ of St. Petersburg (Ostrovsky Square 2a, 5th entrance, 2nd floor, room 211)

Regional stage

January 15, 9.00, theoretical tour of the State Budgetary Educational Institution Secondary School No. 167 of the Central District (Khersonskaya St., 9/11)

January 16 9.00, experimental tour, 9th grade FGBOU HE SPHFU "Saint Petersburg State Chemical and Pharmaceutical University" of the Ministry of Health of the Russian Federation (Professor Popov St., 14, lit. A);

January 16, 9.00, experimental tour, grades 10-11, St. Petersburg State University of Industrial Technologies and Design (Bolshaya Morskaya st., 18);

January 19 15:00-16:30, appeal for grades 9-11, GBOU secondary school No. 167 of the Central District (Khersonskaya st., 9/11)

Registration of participants: 20 min. before the start of the tour. The duration of each tour is 5 astronomical hours.

Passing points for participation in the regional stage in the 2018-2019 academic year: Grade 9 - 15.5 points Grade 10 - 17.25 points Grade 11 - 16.65 points

If you find yourself on the list, then register

The grounds for not admitting to participation in the regional stage of the Olympiad may be:

More than one hour late;

Absence in the registration lists.

What you need to have with you: a list of documents 2019

1) passport or birth certificate and student ID; 2) filled questionnaire and signed by parents agreement for the processing of personal data (download from the website in the documents section) ; 3) stationery (pen with black, blue or purple ink, ruler); 4) non-programmable calculator (phones and other devices are prohibited). Participants are also allowed to bring soft drinks in transparent packaging, chocolate to the audience.

Additionally for a practical tour: robe, surgical gloves, shoe covers

Change of shoes!!!

Boot coversBB

The final stage of the Chemistry Olympiad: March 17-March 23, 2019, Republic of Bashkortostan (Ufa)

Classes to prepare for the final stage of the Higher School of Chemistry (TCS) are held according to individual schedules. For questions, contact by E-mail: This email address is being protected from spambots. You must have JavaScript enabled to view.

Chemistry training camp: State Budgetary Educational Institution "Presidential Physics and Mathematics Lyceum No. 239", st. Kirochnaya d. 8B, from 03/04/2019 to 03/15/2019

March 05, 2019 at 19-00 there will be a PARENT MEETING of the members of the St. Petersburg team for the final stage of the Chemistry Olympiad! The meeting will be held at the State Budgetary Educational Institution "Presidential Physics and Mathematics Lyceum No. 239", at the address: st. Kirochnaya d. 8A, room No. 56 You must have an application filled in by the parent for sending by air (sample application) to the final stage and a photocopy of your passport. Attendance at the meeting is strictly required.

Archive

final stage!

You can bring your documents and fill out the application form at the St. Petersburg Olympiad Center at pl. Ostrovsky, d.2A, entrance 5, room. 211

June 4th and 5th from 10 am to 5 pmJune 6 and 7 from 9:30 to 15:00

If you haven't submitted yet , you must have the following documents in 2 copies:

Photocopy of TIN;

Photocopy of SNILS.

Dear winners and prize-winners of the regional stage (in astronomy, biology, mathematics, physics, chemistry and ecology) who were not present at the award ceremony.

You can get your diplomas at the St. Petersburg Olympiad Center at pl. Ostrovsky, d.2A, entrance 5, room. 211

June 4th and 5th from 10 am to 5 pmJune 6 and 7 from 9:30 to 15:00

Dear winners and prize-winners REGIONAL stage

The awarding of the winners and prize-winners of the regional stage of the All-Russian Olympiad for schoolchildren in astronomy, physics, mathematics, chemistry, biology and ecology will take place on May 21, 2018 at 17:00.

Registration starts at 16:30 at the CARNAVAL Concert Complex at 39, Nevsky Ave.

Dear winners and prize-winners FINAL stage All-Russian Olympiad for schoolchildren!

After the award, you will have to write an application for the award of the Government of St. Petersburg to the winners and prize-winners of international Olympiads and the All-Russian Olympiad for schoolchildren in 2018.

You must have the following documents with you in 2 copies:

A photocopy of the passport of a citizen of the Russian Federation (clear prints of 2,3,5 pages);

A photocopy of the diploma of the winner or prize-winner of the final stage of the All-Russian Olympiad for schoolchildren;

Certificate from an educational institution;

A photocopy of the bank card of the MIR payment system, preferably "SBERBANK" and a certificate of details for transferring to a bank card account or a photocopy of the agreement with the bank for opening an account (clear prints of pages with the candidate's full name, personal account number and branch number jar);

Photocopy of TIN;

Photocopy of SNILS;

Dear winners and prize-winners of the 84th St. Petersburg Chemistry Olympiad for schoolchildren!

In the building of the St. Petersburg State Chemical and Pharmaceutical University (SPCFU) at the address: st. Professor Popov, 14, lit. BUT.

We invite you to visit the Production Center for the development of innovative medicines and technologies.

The GMP training center includes: areas for the production of solid and soft drugs and a quality control laboratory.

The school stage takes place in the city districts (on the basis of schools) from the beginning of September to the end of October.

District Stage All-Russian Olympiad in Chemistry will be held in the districts of St. PetersburgNovember 16, 2017 - round 1 and at 14:00 and December 07, 2017 - round 2 at 14:00

Dear participants!

The meeting of the City Appeal Commission will be held on December 15 at 5:00 pm (grades 8, 9, 11) and at 6:00 pm (grade 10) at the St. Petersburg Olympiad Center (Ostrovskogo square, 2A, entrance 5, 2nd floor). In order to reduce the time to search for a job, we kindly ask participants who plan to come to the appeal, send their data (name, class, school, district) by e-mail This email address is being protected from spambots. You must have JavaScript enabled to view.(indicate in the subject of the letter - appeal in chemistry).

Passing points for the regional stage of 2018:

Grade 9 - 19 pointsGrade 10 - 19 pointsGrade 11 - 19 points

The theoretical tour took place on January 25 at GBOU No. 167 of the Central District (Khersonskaya St., 9/11) registration starts at 08:45 The duration of the tour is 5 astronomical hours.

The appeal of the theoretical round took place on January 27 at 13:00 GBOU No. 167 of the Central District (Khersonskaya St., 9/11)

The practical tour will take place on January 26: 10th and 11th grades - St. Petersburg State University of Industrial Technologies and Design (Bolshaya Morskaya st., 18);

Grade 9 - Institute of Earth Sciences of St. Petersburg State University (V.O., Sredny pr., 41/43) registration starts at 08:45 The duration of the tour is 5 astronomical hours.

The Natural Sciences Sector of the Center for Olympiads of St. Petersburg invites on September 19 at 16:30 to a meeting on the organization and holding of Olympiads in the 2016-2017 academic year methodologists or those responsible for organizing Olympiads in the region.

The meeting will take place at the address: pl. Ostrovsky, 2A(entrance under the arch of house 2A), entrance 5, floor 5, office 509

2015-2016

The awarding of the winners and prize-winners of the regional stage will take place on May 18, 2016 at 17:30 in the Concert Hall of the main building

Anichkov Palace"SPB GDTYU", Nevsky Ave., 39

Dear winners and prize-winners FINAL stage of the All-Russian Olympiad for schoolchildren!

After the award, you will have to write an application for the PNPO 2016 award. You must have the following documents with you:

1. A photocopy of the passport of a citizen of the Russian Federation (clear prints of 2,3,5 pages in 3 copies;

2. A photocopy of the diploma of the winner or prize-winner of the final stage of the All-Russian Olympiad for schoolchildren in 3 copies;

3. Certificate from an educational institution in 3 copies;

4. Copy of the first two pages of the charter of the educational institution in 3 copies;

5. A photocopy of the SBERBANK bank card and a certificate of details for transferring to the Sberbank bank card account or a photocopy of the agreement with the bank for opening an account (clear prints of pages with the candidate's full name, personal account number and bank branch number) in 3 copies; download list of documents
The results of the regional stage of the All-Russian Olympiad for schoolchildren in chemistry

Passing points for the practical tour: 8th grade - 24 points, 9th, 10th, 11th grades - 17.5 points.

Preliminary results of the theoretical round
Chemistry Theor Tour of February 27, 2016

SCHOOL STAGE

School stage of the Chemistry Olympiad in the city districts in 2015. In each of the districts of the city, it takes place on a day determined by the organizers.

REGIONAL STAGE

Preliminary passing scores for the regional stage of the All-Russian Chemistry Olympiad for schoolchildren: Grade 9 - 15 points, Grade 10 - 11 points, Grade 11 - 12.5 points.
Preliminary passing scores for the final stage of the St. Petersburg Chemistry Olympiad for schoolchildren: Grade 8 - 11 points, Grade 9 - 10 points, Grade 10 - 9 points, Grade 11 - 10.5 points.
Final passing scores will be announced after the appeal.

Appeal of disputed works:

Grade 8 December 11, 2015 from 16:30 to 18:30 GBOU secondary school No. 167 of the Central district, st. Khersonskaya, 9/11

9-11 grades December 10, 2015 from 17:00 Olympiad Center St. Petersburg pl. Ostrovskogo, d.2-a, entrance 5, 2nd floor room. 211

REGIONAL STAGE

GBOU secondary school No. 167 of the Central district, st. Khersonskaya, 9/11

RGPU them. A.I. Herzen (Embankment of the river Moika, 48 entrance from Kazanskaya street)

St. Petersburg University PT and D st. Bolshaya Morskaya 18

Dear participants of the Olympiad!

Display of works and appeal of Olympiad works in chemistry takes place in DISTRICTS cities.

If you do not agree with the results of the appeal, you can contact the City Appeals Commission 12/15/2016 at 17.00. to the Center for the Olympiads of St. Petersburg at the address: pl. Ostrovsky, d. 2A, entrance 5, 2nd floor.

To do this, the district must submit your work to the city appeals commission.

District stage of the Olympiad for schoolchildren in chemistry 2016/17 will be held in the districts of the city in two stages:

November 17, 2016 - theoretical tour (grades 8-11) and December 08, 2016 - practical tour (for grades 9-11)

WINNERS AND PRIZE WINNERS OF THE REGIONAL STAGE 2017

RESULTS OF THE PARTICIPANTS OF THE REGIONAL STAGE OF THE HIGH SCHOOL IN CHEMISTRY 2017

Passing points:

1) for the regional stage of the All-Russian Chemistry Olympiad: Grade 9 - 14.5 points; 10th grade-13.5; Grade 11 - 14.5 points
2) for the final stage of the St. Petersburg Chemistry Olympiad: Grade 8 - 10 points; Grade 9 - 11 points; Grade 10 - 10 points; Grade 11 - 11 points

List of participants of RegE in Chemistry 2017

Required documents for registration

The regional stage of the All-Russian Chemistry Olympiad will take place:

February 1, 9:00 am

February 2, 9:00, St. Petersburg State University of Industrial Technologies and Design (Bolshaya Morskaya st. 18) -10th and 11th grades;

Institute of Earth Sciences of St. Petersburg State University (V.O., Sredny pr., 41/43) - Grade 9

Congratulations!

List of winners and runners up

the final stage of the All-Russian Olympiad for schoolchildren 2016-2017 ac. years in chemistry

2018-2019 academic year

District Stage

Regional stage