Assignments for the school stage Olympiads for schoolchildren in chemistry, grade 11.
Exercise 1.
There are ten elements in the Periodic Table, in the designation of which the second letter is "e". For each of these elements, typical reactions are given below. The capital letter is indicated by the sign “ ? ". Arrange the elements in their places.
1) D + T → ? e + n;
2) ? eO 4 → ? e + 2O 2;
3) ? e + O 3, F 2, etc. → does not go;
4) 2Au + 6H 2 ? eO 4 (load) → 3 ? eO 2 + Au 2 ( ? eO 4) 3 + 6H 2 0;
5) ? eCl 2 + 4NaOH → Na 2 [ ? e (OH) 4] + 2NaCl;
6) ? eO 3 + 3H 2 O → H 6 ? eO 6;
7) ? e (OH) 4 + 4HCl → ? eCl 3 + 1 / 2Cl 2 + 4H 2 0;
8) ? eCl 4 + Li → ? eH 4 + Li;
9) ? eCl 2 + K 3 [ ? e (CN) 6] → K ? e [ ? e (CN) 6] ↓ (blue precipitate);
10) 2H ? eO 4 + 7H 2 S → ? e 2 S 7 ↓ + 8H 2 0.
The task 2.
The chemist took samples of three silver-white metals and found a way to quickly distinguish 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.
Legend: "+" - the reaction is in progress, « - "- the metal does not react. Determine which metals could be obtained by a chemist and write the corresponding reaction equations.
Task 3.
A small piece of a plastic disposable cup was heated without air access to 400 ° C. As a result of heating, a hydrocarbon was obtained X (carbon content 92.26% by weight, its vapor density for oxygen 3.25). It is known that during the oxidation of a hydrocarbon X With a solution of potassium permanganate in an acidic medium, benzoic acid is formed as the only organic product.
1. Calculate the molecular formula H.
2. Give the structural formula and the name of the hydrocarbon X... What is the name of the starting polymer?
3. Write the reaction equation (with all products and stoichiometric coefficients) for hydrocarbon oxidation X potassium permanganate solution acidified with sulfuric acid.
Task 4.
Young chemist Petya received an assignment from his mother to buy 1 liter of food vinegar (mass fraction of acetic acid CH3COOH 9%) for home canning in the store. Arriving at the store, he found that only acetic essence was on sale (the mass fraction of acetic acid was 70%). Petya decided that he could make edible vinegar out of it on his own. At home, in the reference book, he managed to find the density values \u200b\u200bfor a 9% solution of acetic acid - 1.012 g / ml and for a 70% solution - 1.069 g / ml. Of the equipment Petya has only measuring cylinders of various sizes.
What safety rules should be followed when preparing dilute solutions from concentrated acids?
Which of the substances available at home should Pete have on hand if acid gets on the skin? Name this substance and reflect its effect on the acid in the reaction equation.
What volume of vinegar essence should Petya measure in order to prepare 1 liter of 9% acetic acid solution?
Task 5.
The laboratory contains iron, hydrochloric acid, caustic soda, calcium carbonate, copper (II) oxide. Using these substances, as well as the products of their interaction, give at least 10 equations for the reactions of obtaining new inorganic substances.
Task 6.
When 2.8 g of an alloy of copper and silver were dissolved in concentrated nitric acid, 5.28 g of a mixture of nitrates was formed. Determine the mass fraction of metals in the alloy.
Grade 11 Answers
Exercise 1.
Answer:
Points |
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Even if the decisive cannot determine all the elements by reactions, this can be done using the Periodic Table, given that the maximum oxidation state of an element cannot be greater than the group number. | |
1) D + T → He + n; | |
2) XeO 4 → Xe + 2O 2; | |
3)Ne + O 3, F 2, etc. → does not go; | |
4) 2Au + 6H 2 SeO 4 (load) → 3 SeO 2 + Au 2 ( SeO 4) 3 + 6H 2 0; | |
5) BeCl 2 + 4NaOH → Na 2 [ Be(OH) 4] + 2NaCl; | |
6) TeO 3 + 3H 2 O → H 6 TeO 6; | |
7) Ce(OH) 4 + 4HCl → CeCl 3 + 1 / 2Cl 2 + 4H 2 0; | |
8) GeCl 4 + Li → GeH 4 + Li; | |
9) FeCl 2 + K 3 → K Fe[Fe(CN) 6] ↓ (blue precipitate); | |
10) 2H ReO 4 + 7H 2 S → Re 2 S 7 ↓ + 8H 2 0. | |
Maximum score |
Task 2.
By the combination of properties, i.e. according to the behavior in reactions with acids and sodium hydroxide, we conclude: I metal - silver, (copper does not match in color), metal II - aluminum, metal III - zinc. Reaction equations: 2Al + 6HCl \u003d 2AlCl 3 + 3H 2 Zn + 2HCl \u003d ZnCl 2 + 2H 2 Ag + 2HNO 3 (conc.) \u003d AgNO 3 + NO 2 + H 2 OAl + HNO 3 (conc.) No reaction Zn + 4HNO 3 (conc.) + Zn (NO 3) 2 + 2NO 2 + 2H 2 OZn + 2NaOH + 2H 2 O \u003d Na 2 + H 2 2Al + 6NaOH + 6H 2 O \u003d 2Na 3 + 3H 2 (7 points)
Task 3.
Answer:
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(other formulations of the answer are allowed that do not distort its meaning) | Points |
1.Mr (X) \u003d 3.2532 \u003d 104 amu. Let's find the molecular formula of the hydrocarbon X: C: H \u003d 0.9226 / 12.01: 0.0774 / 1.008 \u003d 1: 1, taking into account the molecular weight, we get C 8 H 8. | |
2. Since the oxidation of hydrocarbon X with a solution of potassium permanganate in an acidic medium forms benzoic acid (C 6 H 5 COOH) as the only organic product, its molecule contains a benzene ring with one substituent. Subtracting the C 6 H 5 fragment from the gross formula C 8 H 8, we obtain the C 2 H 3 substituent. The only possible substituent is vinyl, and hydrocarbon X is styrene (vinyl benzene). Consequently, the polymer from which the disposable cup was made is polystyrene. | |
3. The equation of the reaction of styrene oxidation with a solution of KMnO 4 acidified with H 2 SO 4: 2KMnO 4 + 3H 2 SO 4 → + CO 2 + 2 MnSO 4 + K 2 SO 4 + 4H 2 O | |
All response elements are written incorrectly | |
Maximum score |
Task 4.
Pour water into acid (1 point).
Baking soda or sodium bicarbonate (1 point).
NaHCO3 + CH3COOH \u003d CH3COONa + H2O neutralization reaction (2 points).
The calculated mass of acetic acid in a 9% solution is 91.08 g (1 point).
The calculated mass of the vinegar essence solution is 130.1 g (1 point).
The volume of vinegar essence is calculated - 121.7 ml or ≈ 122 ml (1 point).
Total: 7 points.
Task 5.
Possible answers:
Fe + 2HCl \u003d FeCl 2 + H 2
HCl + NaOH \u003d NaCl + H 2 O
2HCl + CaCO 3 \u003d CaCl 2 + H 2 O + CO 2
CuO + 2HCl \u003d CuCl 2 + H 2 O
2NaOH + CO 2 \u003d Na 2 CO 3 + H 2 O; NaOH + CO 2 \u003d NaHCO 3 + H 2 O
CuO + H 2 \u003d Cu + H 2 O
FeCl 2 + 2NaOH \u003d Fe (OH) 2 + 2NaCl
3Fe + 4H 2 O \u003d Fe 3 O 4 + 4H 2
CaCO 3 \u003d CaO + CO 2
CuO + CO 2 \u003d CuCO 3
Fe 3 O 4 + 8HCl \u003d FeCl 2 + 2FeCl 3 + 4H 2 O, etc.
Task 6.
Reaction equations:
Cu + 4 HNO 3 \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O
Ag + 2HNO 3 \u003d AgNO 3 + NO 2 + H 2 O (1 point)
We introduce the designations: n (Cu) \u003d x mol n (Ag) \u003d y mol; then
a) m (Cu) \u003d 64x, m (Ag) \u003d 108y
m (mixture) \u003d 64x + 108y \u003d 2.8
b) according to equation (1) n (Cu (NO 3) 2 \u003d x, m (Cu (NO 3) 2 \u003d 188x;
c) according to equation (2) n (AgNO 3) \u003d y, m (AgNO 3) \u003d 170y
d) m (mixtures) \u003d 188x + 170y \u003d 5.28 (2 points)
3) compose and solve the system of equations:
64x + 108y \u003d 2.8 x \u003d 0.01 mol Cu
188x + 170y \u003d 5.28 y \u003d 0.02 mol Ag (2 points)
4) calculate the mass fractions of the mixture components:
a) m (Cu) \u003d 0.01 * 64 \u003d 0.64g. w (Cu) \u003d 0.64 / 2.8 \u003d 0.2286 or 22.86%
m (Ag) \u003d 0.02 * 108 \u003d 2.16g. w (Ag) \u003d 2.16 / 2.8 \u003d 0.7714 or 77.14% (2 points)
maximum score - 7 points
Tasks for the school stage of the chemistry Olympiad, grade 10
Exercise 1.
What gases can be obtained, having at its disposal the following substances:
sodium chloride, sulfuric acid, ammonium nitrate, water, ammonium nitrite, hydrochloric acid, potassium permanganate, sodium hydroxide, aluminum carbide, calcium carbide and sodium sulfite?
Write down all the equations of possible reactions, indicate the conditions for their occurrence.
Task 2.
For three chemical elements A, B and C, it is known that they take part in the following transformations:
С 2 + 3В 2 2СВ 3
4СВ 3 + 5А 2 4СА + 6В 2 А
4СВ 3 + 3А 2 \u003d 2С 2 + 6В 2 А
C 2 + A 2 \u003d 2 CA
4CA 2 + 2B 2 A + A 2 \u003d 4BCA 3
What elements are discussed in the assignment? Write down the reaction equations.
The task3.
When 11.5 g of a mixture of aluminum, magnesium and copper were dissolved in hydrochloric acid, 5.6 liters (NU) of gas were released. The insoluble residue was transferred to a solution with concentrated nitric acid. At the same time, 4.48 l (standard) of gas was released. Determine the mass fractions (in%) of the components of the initial mixture.
Task 4.
The reaction scheme is given:
KI + KMnO 4 + H 2 SO 4 → I 2 + K 2 SO 4 + MnSO 4 + H 2 O
1. Determine the oxidation states of the elements.
2. Make the equation of electronic balance
3. Determine the oxidizing and reducing agent
4. Place the coefficients in this equation.
5. List the areas of application of the substance, the formula of which is KMnO 4
Task 5.
There are nine elements in the periodic table, the names of which in Russian are nouns "not masculine". For each of these elements, typical reactions are given below, in which they are encrypted with the letter " E". Define the elements for each reaction:
1) 2H 2 E+ EO 2 → 3 E+ 2H 2 O;
2) ECl 3 + KI → ECl 2 + KCl +1/2 I 2;
3) ENO 3 + KCl → ECl ↓ + KNO 3;
4) E + 3HNO 3 conc. + 4HCl conc. → H [ ECl 4] + 3NO 2 + 3H 2 O;
5) ECl 2 + 4NH 3 → [ E(NH 3) 4] Cl 2 (blue);
E(catalyst), 800 ° С
6) 4NH 3 + 5O 2 4NO + 6H 2 O;
7) 2 E + O 2 2EO
8) ECl 2 + Cl 2 → ECl 4;
9) E 2 O 3 + 4 HNO 3 conc. + ( x-2) H 2 O → E 2 O 5 x H 2 O + 4NO 2
Task 6.
As a result of the reaction of calcium phosphate with magnesium at
heating, two substances are formed, one of which
interacts with water, while a colorless poisonous
gas with a garlic smell. The latter is oxidized
oxygen in the air.
Write the equations of all the specified chemical processes, name
their products.
Calculate the volume of air (n.o.) required to oxidize the gas,
if 2.4 g of magnesium was used in the first of these reactions.
Grade 10 Answers
Exercise 1.
1) 2NaCl (TV) + H 2 SO 4 (conc.) Na 2 SO 4 + 2HCl
or NaCl (solid) + H 2 SO 4 (conc.) NaHSO 4 + HCl
NH 4 NO 3 \u003d N 2 O + 2H 2 O
8th grade
What chemical elements are named after countries? List at least four titles.
Which element was first exposed to the sun?
Indicate the number of protons and neutrons contained in the nuclei of atoms, the elements you named.
Number of points –10
A raindrop has a mass of about 10 -4 g. Calculate the number of water molecules and the total number of atoms of all the elements contained in this drop.
Number of points - 10
What is the percentage of 35 Сl and 37 Сl isotopes in natural chlorine, which has a relative molecular weight of 70.90?
Number of points - 10
You are given a mixture of the following substances: iron, soot, copper, chalk, table salt.
Suggest a plan for separating these substances using water and hydrochloric acid.
What laboratory equipment will be required to separate this mixture?
Write the reaction equations to be used in the separation.
Calculate the mass of chalk in the mixture by the volume of gas evolved in 5.6 liters.
Number of points - 20
Indicate the chemical formulas of gases: nitrogen, hydrogen chloride, hydrogen, ammonia, chlorine, carbon monoxide, hydrogen sulfide, carbon dioxide. Which of these gases are simple substances, oxides, have a color, a characteristic odor, are poisonous? Fill out the answer in the form of a table using the signs "+" and "-".
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carbon dioxide |
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Number of points - 10
Olympiad tasks for the school stage of the All-Russian Olympiad
schoolchildren in chemistry 2011-2012
Grade 9
9- 1. Write the reaction equations for the following transformations:
Zn → ZnS → H 2 S → S → SO 2 → SO 3 → H 2 SO 4 → BaSO 4
Specify the reaction conditions; consider one of the reactions as redox.
Number of points - 10
9-2. A sample of technical iron (II) sulfide weighing 5 g, containing 5% metallic iron, interacts with hydrochloric acid. Calculate the volume of evolved gaseous products (at standard conditions) and the volumetric composition of the gas mixture.
9-3 ... With partial dissolution of 4 g of a mixture of copper and magnesium sawdust in an excess of hydrochloric acid, 1.12 liters of hydrogen (n.o.) were released. Set the composition of the initial mixture in percent by weight
Number of points - 10
9- 4. Liquid laboratory waste containing acids must be neutralized with alkali or soda.
1. Determine the mass of sodium hydroxide and sodium carbonate required to neutralize waste containing 0.60 mol of hydrochloric acid.
2. What volume of gas (n.u.) will be released when the specified amount of waste is neutralized?
3. How much silver nitrate would be required to precipitate the chloride ions contained in 0.6 mol of hydrochloric acid?
Number of points -20
9-5. It is known that four test tubes contain solutions of nitric acid, potassium carbonate, silver nitrate, and barium chloride. How to determine the contents of each tube without using other reagents? Make an experiment plan and write the reaction equations.
Number of points - 10
Olympiad tasks for the school stage of the All-Russian Olympiad
schoolchildren in chemistry 2011-2012
Grade 10
When 11.5 g of a mixture of aluminum, magnesium and copper were dissolved in hydrochloric acid, 5.6 liters (NU) of gas were released. The insoluble residue was transferred to a solution with concentrated nitric acid. At the same time, 4.48 l (standard) of gas was released. Determine the mass fractions (in%) of the components of the initial mixture.
numberpoints – 10.
The reaction scheme is given:
KI + KMnO 4 + H 2 SO 4 → I 2 + K 2 SO 4 + MnSO 4 + H 2 O
1. Determine the oxidation states of the elements.
2. Make the equation of electronic balance
3. Determine the oxidizing and reducing agent
4. Place the coefficients in this equation.
5. List the areas of application of the substance, the formula of which is KMnO 4
numberpoints-7.
Write the reaction equations corresponding to the transformation scheme:
t °, Pt KMnO 4, H 2 O g. HBr 2KOH (alcohol), t °
C 2 H 5 Cl → C 3 H 8 ─── → X 1 ───── → X 2 ───── → X 3 ───────── → X 4
numberpoints – 10.
A mixture of alkene with hydrogen with a total volume of 26.88 L (NU) was passed over a platinum catalyst at 200 ° C. At the same time, 75% of the alkene reacted, and the volume of the mixture decreased to 20.16 L (NU). When passing the initial mixture through a bottle with bromine water, all the hydrocarbon reacted, and the weight of the bottle increased by 16.8 g. Determine the composition of the initial mixture (in% by volume) and the structure of the initial alkene.
Number of points - 10
10-5. In four test tubes without inscriptions there are 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.
Number of points - 10
Olympiad tasks for the school stage of the All-Russian Olympiad
schoolchildren in chemistry 2011-2012
Grade 11
11-1. The laboratory contains iron, hydrochloric acid, caustic soda, calcium carbonate, copper (II) oxide. Using these substances, as well as the products of their interaction, give at least 10 equations for the reactions of obtaining new inorganic substances
Number of points - 10
11-2. How many isomers does C 5 H 12. Write down their structural formulas and give each substance a name according to the substituent nomenclature. Which of these isomers has the highest boiling point. Calculate the relative vapor density of this compound over air.
Number of points - 7
In 264 g of a solution with a mass fraction of Hg 2 (NO 3) 2 equal to 20% placed iron filings. After some time, the mass fraction of mercury (I) nitrate in the solution became 6% .
1. What is the mass of mercury obtained as a result of the reaction?
2. What is the mass fraction of iron nitrate in the resulting solution?
Number of points - 20
11-4. Propionic acid is contaminated with formic acid and propyl alcohol. When an excess of potassium bicarbonate was added to 150 g of this acid, 44.8 liters of gas (NU) were evolved. When an excess of ammonia solution of silver oxide was added to the same amount of acid, a precipitate weighing 2.16 g was formed.
1. Give the equations of all reactions.
2. Determine the mass fraction of impurities in the acid.
Number of points - 20
11-5. In test tubes there are concentrated solutions of the following acids: oxalic, sulfuric, hydroiodic, phosphorous, nitric, formic.
1. How to establish the contents of each tube without using any other reagents?
2. Write the reaction equations.
3. Indicate signs of reactions
Number of points - 20
Answers to the tasks of the school stage of the All-Russian Olympiad
schoolchildren in chemistry
2011
8th grade
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Points |
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Ruthenium (Ru) - named after Russia; 44 protons, 57 neutrons. |
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Polonius (Po) - in honor of Poland; 84 protons, 37 neutrons. |
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France (Fr) - in honor of France; 87 protons, 35 neutrons. |
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Germanium (Ge) - in honor of Germany; 32 protons, 40 neutrons. |
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Helium (He) - open to the sun; protons 2, neutrons 2. |
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Maximum score |
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(other formulations of the answer are allowed that do not distort its meaning) |
Points |
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Let's calculate the number of moles of water - n (H 2 O) \u003d m (H 2 O): M (H 2 O) \u003d 10 -4 g: 18 g / mol \u003d 5.56 · 10 -6 mol. |
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One mole of water contains N A \u003d 6.02 · 10 23 water molecules. |
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Then 5.56 10 -6 mol contains the number of water molecules equal to N (H 2 O) \u003d N A n (H 2 O) \u003d 6.02 10 23 5.56 10 -6 \u003d 33.5 10 17 (molecules) |
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Water is made up of three atoms: two hydrogen atoms and one oxygen atom. |
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The total number of all atoms contained in a raindrop is N ∑ \u003d 3N (H 2 O) \u003d 3 33.5 10 17 \u003d 100.4 10 17 \u003d 10 19 (atoms). |
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All response elements are written incorrectly |
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Maximum score |
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(other formulations of the answer are allowed that do not distort its meaning) |
Points |
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Let's take for X the percentage of 35 Cl isotope. |
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Then the percentage of the isotope 37 Cl will be (100 - X) |
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The mass of 35 Cl isotope atoms is 35X. |
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The mass of atoms of the 37 Сl isotope is 37 (100 - X) |
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Let's make the equation: 35X + 37 (100 - X) \u003d 35.45 |
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ω (35 Cl) \u003d 77.5%, ω (37 Cl) \u003d 22.5%. |
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All response elements are written incorrectly |
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Maximum score |
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(other formulations of the answer are allowed that do not distort its meaning) |
Points |
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Separate the iron with a magnet. Put the remaining mixture in water - table salt will dissolve, soot will be on the surface. We filter the solution. The soot will remain on the filter. Evaporate the filtrate, it will be NaCl. We treat copper and chalk with hydrochloric acid. Chalk (CaCO 3) dissolve, and the copper remains. |
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The following equipment was required for the separation: a magnet, a filtering device (a rack with a ring, a funnel, a filter, a glass rod, a collection of filtrate (a glass)), porcelain steaming cup, electric stove, beaker for dissolving. |
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3) CaCO 3 + 2HCl \u003d CaCl 2 + H 2 O + CO 2 |
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Let us find the amount of substance in the CO 2 gas: n (CO 2) \u003d 5.6 L: 22.4 mol / L \u003d 0.25 mol; n (CaCO 3) \u003d n (CO 2) \u003d 0.25 mol; m (CaCO 3) \u003d 0.25 mol 100 g / mol \u003d 25 g. |
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Maximum score |
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Index |
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greyish |
carbon dioxide |
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Chemical |
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substance |
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Evaluation guidelines |
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Chemical formula (8 1 point) |
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Simple substance (3 0.1 point) |
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Oxide (2 0.1 point) |
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Characteristic odor (4 0.125 points) |
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Poisonous (5 0.1 point) |
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Grade 9
9-2. 1. Under the action of acid on technical iron (II) sulfate with an admixture of iron, reactions occur:
FeS + 2HCl \u003d FeCl 2 + H 2 S (1 point)
Fe + 2HCl \u003d FeCl 2 + H 2 (1 point)
2. The amount of FeS and Fe in the weighed portion of the original sample (95% and 5%, respectively) is equal to:
n (FeS) \u003d 5 ∙ 0.95 / 88 \u003d 5.4 ∙ 10 -2 mol (1 point)
n (Fe) \u003d 5 ∙ 0.05 / 56 \u003d 4.48 ∙ 10 -3 mol (1 point)
3. The amount of gaseous products of the reactions H 2 S and H 2 obtained from the initial sample, in accordance with the given chemical equations, will be:
n (H 2 S) \u003d 5.4 ∙ 10 -2 mol
n (H 2) \u003d 4.48 ∙ 10 -3 mol (2 points)
4. Find the volume of evolved gaseous products:
V (H 2 S) \u003d 5.4 ∙ 10 -2 mol ∙ 22.4 \u003d 1.21 (l)
V (H 2) \u003d 4.48 ∙ 10 -3 mol ∙ 22.4 \u003d 0.1 (L) (2 points)
5. We calculate the volumetric composition of the gas mixture:
V total \u003d 1.21 + 0.1 \u003d 1.31 (l)
φ (H 2 S) \u003d 1.21 / 1.31 \u003d 0.9237 or 92.37%
φ (H 2) \u003d 0.1 / 1.31 \u003d 0.0763 or 7.63% (2 points)
Total: 10 points
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(other formulations of the answer are allowed that do not distort its meaning) |
Points |
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Let's compose the reaction equations: 1) HСl + NaOH \u003d NaCl + H 2 O (1) 2HCl + Na 2 CO 3 \u003d 2NaCl + H 2 O + CO 2 (2) |
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To neutralize 0.60 mol HCl in accordance with the equation reaction (1) requires 0.60 mol of NaOH, since n (HCl) \u003d n (NaOH); n (Na 2 CO 3) \u003d 1 / 2n (HCl) \u003d 0.60 mol: 2 \u003d 0.30 mol - equation (2). m (NaOH) \u003d 0.60 mol * 40 g / mol \u003d 24 g; m (Na 2 CO 3) \u003d 0.30 mol 106 g / mol \u003d 31.8 g. |
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3) Let's calculate the volume of carbon dioxide released during neutralization according to reaction (2): n (CO 2) \u003d 1 / 2n (HCl) \u003d 0.60 mol: 2 \u003d 0.30 mol; V (CO 2) \u003d n (CO 2) V M \u003d 0.30 mol 22.4 L / mol \u003d 6.72 L. |
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4) Cl - + AgNO 3 \u003d AgCl ↓ + NO 3 - n (AgNO 3) \u003d n (Cl -) \u003d 0.60 mol; m (AgNO 3) \u003d 0.60 mol · 170 g / mol \u003d 102 g |
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Maximum score |
Grade 10
1) Copper does not interact with hydrochloric acid.
Cu + 4 HNO 3 \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O
n (NO 2) \u003d 4.48 / 22.4 \u003d 0.2 mol;
n (Cu) \u003d 0.1 mol; m (Cu) \u003d 64 x 0.1 \u003d 6.4 g 1 point
x mol 1.5 x mol
2) 2Al + 6HCl \u003d 2AlCl 3 + 3H 2
y mol y mol
Mg + 2HCl \u003d MgCl 2 + H 2 2 points
3) Determination of the amount of hydrogen substance: n (H 2) \u003d 5.6 / 22.4 \u003d 0.25 mol; 1 point
4) Compiled a system with two unknowns:
24x + 27y \u003d 5.1 2 points
5) The system with two unknowns is solved (the values \u200b\u200bof "x" and "y" are determined) 2 points
x \u003d 0.1; y \u003d 0.1
6) The masses and mass fractions of the mixture components were determined.
m (Al) \u003d 2.7g; m (Mg) \u003d 2.4g 1 point
ώ (Al) \u003d 2.7 / 11.5 \u003d 0.2348 or 23.48%
ώ (Mg) \u003d 2.4 / 11.5 \u003d 0.2087 or 20.87%. 1 point
ώ (Cu) \u003d 6.4 / 11.5 \u003d 0.5565 or 55.65%
Total: 10 points
1) The oxidation states of the elements have been determined
1-1 +1 +7 -2 +1 +6 -2 0 +1 +6 -2 +2 +6 -2 +1 -2
KI + KMnO 4 + H 2 SO 4 → I 2 + K 2 SO 4 + MnSO 4 + H 2 O 1 point
2) The equation of the electronic balance was compiled and the oxidant and
reducing agent:
3) Coefficients in the reaction equation are arranged:
2KMnO 4 + 10KI + 8H 2 SO 4 \u003d 5I 2 + 6K 2 SO 4 + 2MnSO 4 + 8H 2 O
4) The areas of application of potassium permanganate are listed:
(laboratory method for producing oxygen and chlorine, chemical analysis (permanganatometry), oxidation of organic substances, etc.) only 2 points
Total: 7 points
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(other formulations of the answer are allowed that do not distort its meaning) |
Points |
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1.C 2 H 5 Cl + CH 3 Cl + Mg → MgCl 2 + CH 3 -CH 2 -CH 3 (+ C 4 H 10 + C 2 H 6) |
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2.CH 3 -CH 2 -CH 3 ─── → H 3 C-CH \u003d CH 2 + H 2 |
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3.H 3 C-CH \u003d CH 2 + 2KMnO 4 + 4H 2 O → 3H 3 C-CH (OH) -CH 2 OH + 2MnO 2 + 2KOH |
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4.H 3 C-CH (OH) -CH 2 OH + 2HBr → 2H 2 O + H 3 C-CH (Br) -CH 2 Br |
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5.H 3 C-CH (Br) -CH 2 Br + 2 KOH (alcohol) → H 3 C-C≡CH + 2KBr + 2H 2 O |
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All response elements are written incorrectly |
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Maximum score |
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(other formulations of the answer are allowed that do not distort its meaning) |
Points |
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C n H 2n + H 2 \u003d C n H 2n + 2 |
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2 moles of gases (hydrocarbon and hydrogen) enter into the reaction, and one mole (alkane) is formed. Thus, the decrease in the volume of the mixture is equal to the volume of hydrogen that entered into the reaction, or the volume of the reacted alkene. |
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This volume is 26.88 - 20.16 \u003d 6.72 (l), that is, 0.3 mol. |
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Since 75% of the alkene reacted, there was a total of 0.4 mol. |
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When passing through bromine water, the mass of the flask increased by the mass of the hydrocarbon, i.e. 0.4 mol of alkene have a mass of 16.8 g. Molar mass 16.8 / 0.4 \u003d 42 (g / mol). |
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Alken with such a molar mass - C 3 H 6, propene: H 3 C-CH \u003d CH 2 |
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The composition of the mixture: 0.4 mol of propene occupy a volume of 8.96 liters. This is 33% (1/3) of the volume of the mixture. The rest - 67% (2/3) - hydrogen. |
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All response elements are written incorrectly |
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Maximum score |
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(other formulations of the answer are allowed that do not distort its meaning) |
Points |
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Let's compile a table of possible pairwise interactions of substances, as a result of which we will (or will not) watch certain signs of reactions. |
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Substances |
4. Аl 2 (SO 4) 3 |
Overall observation result |
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Al (OH) 3 |
1 sediment and |
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4. Al 2 (S0 4) 3 |
A1 (OH) 3 |
2 draft and |
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NaOH + HCl \u003d NaCl + H 2 O |
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K 2 CO 3 + 2HC1 \u003d 2KS1 + H 2 O + CO 2 |
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3K 2 CO 3 + Al 2 (SO 4) 3 + 3H 2 O \u003d 2Al (OH) 3 + 3CO 2 + 3K 2 SO 4; |
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Al 2 (SO 4) 3 + 6NaOH \u003d 2Al (OH) 3 + 3Na 2 SO 4 Al (OH) 3 + NaOH + 2H 2 O \u003d Na (the presence of sediment depends on the order of draining and excess and alkali) |
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Based on the presented table, all substances can be determined by the number of precipitation and gas evolution. |
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All response elements are written incorrectly |
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Maximum score |
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Grade 11
Possible answers:
Fe + 2HCl \u003d FeCl 2 + H 2
HCl + NaOH \u003d NaCl + H 2 O
2HCl + CaCO 3 \u003d CaCl 2 + H 2 O + CO 2
CuO + 2HCl \u003d CuCl 2 + H 2 O
2NaOH + CO 2 \u003d Na 2 CO 3 + H 2 O; NaOH + CO 2 \u003d NaHCO 3 + H 2 O
CuO + H 2 \u003d Cu + H 2 O
FeCl 2 + 2NaOH \u003d Fe (OH) 2 + 2NaCl
3Fe + 4H 2 O \u003d Fe 3 O 4 + 4H 2
CaCO 3 \u003d CaO + CO 2
CuO + CO 2 \u003d CuCO 3
Fe 3 O 4 + 8HCl \u003d FeCl 2 + 2FeCl 3 + 4H 2 O, etc.
CH 3 -CH 2 -CH 2 -CH 2 -CH 3 - Pentane (0.5 + 0.5) \u003d 1 point
CH 3 -CH-CH 2 -CH 3 - 2-methylbutane (0.5 + 0.5) \u003d 1 point
CH 3 -C-CH 3 - 2,2-dimethylpropane (0.5 + 0.5) \u003d 1 point
Pentane has a higher boiling point. the length of the molecule is greater and therefore the intermolecular forces are also greater (1 + 1 \u003d 2 points)
D air \u003d M C 5 H 12 / M air M C 5 H 12 \u003d 72 g / mol M air \u003d 29 g / mol (1 point)
D air \u003d 72/29 \u003d 2.48 (1 point) 2 points total for the task
Total: 7 points
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(other formulations of the answer are allowed that do not distort its meaning) |
Points |
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Hg 2 (NO 3) 2 + Fe \u003d Fe (NO 3) 2 + 2Hg ↓ (1) |
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M (Fe (NO 3) 2 \u003d 180 g / mol; M (Hg) \u003d 201 g / mol; M (Fe) \u003d 56 g / mol; M (Hg 2 (NO 3) 2 \u003d 526 g / mol. |
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Find the mass of Hg 2 (NO 3) 2 in the original solution: m (Hg 2 (NO 3) 2) \u003d 0.2 264 \u003d 52.8 (g) |
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The mass of the solution changes during the reaction. The mass of the solution increases by the mass of the reacted iron and decreases by the mass of mercury precipitated. |
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Let be X g of iron reacted. Let us find the mass of the resulting mercury according to the reaction equation (1): 56 g Fe - 2 201 g Hg m 1 (Hg) \u003d 7.18 X X - m 1 |
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The mass of the resulting solution: m (p-pa) \u003d 264 + X - 7.18X \u003d 264 - 6.18 X (d) |
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Find the mass of Hg 2 (NO 3) 2 in the resulting solution: m (Hg 2 (NO 3) 2) \u003d 0.06 (264 - 6.18 X) = 15,84 – 0,37X |
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Find the mass of Hg 2 (NO 3) 2, which entered into the reaction: m (Hg 2 (NO 3) 2) \u003d 52.8 - (15.84 - 0.37 X) = 36,96 + 0,37X |
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Find the value Xaccording to the reaction equation (1), solving the proportion: 56 g Fe react with 526 g Hg 2 (NO 3) 2 X - (36.96 + 0.37 X) X \u003d 4.1; m (Fe) \u003d 4.1 g. |
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The mass of the resulting mercury is 29.4 g (7.18 4.1) |
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The mass of the resulting solution is 238.7 g (264 - 6.18 4.1) |
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Find the mass of the resulting iron (II) nitrate: 56 g Fe - 180 g Fe (NO 3) 2 4.1 g - X X \u003d 13.18; m (Fe (NO 3) 2) \u003d 13.18 g. |
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Find the mass fraction of iron (II) nitrate in the resulting solution: ω (Fe (NO 3) 2) \u003d m (Fe (NO 3) 2): m (p-pa) \u003d 13.18: 238.7 \u003d 0,055 (5,5 %) |
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All response elements are written incorrectly |
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Maximum score |
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(other formulations of the answer are allowed that do not distort its meaning) |
Points |
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C 2 H 5 COOH + KHSO 3 \u003d C 2 H 5 COOK + H 2 O + CO 2 (1) |
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НСООН + КНСО 3 \u003d НСООК + Н 2 О + СО 2 (2) |
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НСООН + Ag 2 O \u003d 2Ag ↓ + Н 2 О + СО 2 (3) |
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M (C 2 H 5 COOH) \u003d 74 g / mol; M (HCOOH) \u003d 46 g / mol; M (Ag) \u003d 108 g / mol |
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Based on the mass of isolated silver (equation 3), we find the mass of formic acid: 45 g HCOOH - 2 × 108 g Ag X \u003d 0.46 g; m (HCOOH) \u003d 0.46 g. |
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Let us find the volume of CO 2 released when potassium bicarbonate interacts with formic acid (equation 2): 46 g HCOOH - 22.4 l CO 2 X \u003d 0.224 l; V (CO 2) \u003d 0.224 l. |
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Therefore, when interacting with propionic acid (equation 1), 44.576 l of CO 2 (44.8 - 0.224) |
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Find the mass of propionic acid: 74 g C 2 H 5 COOH - 22.4 l CO 2 X \u003d 147.26 g; m (C 2 H 5 COOH) \u003d 147.26 g. X - 44.576 l |
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Let's find the mass of propyl alcohol: m (C 3 H 7 OH) \u003d 150 - m (C 2 H 5 COOH) - m (HCOOH) \u003d 150 –147.26 - 0.46 \u003d 2.28 (g) |
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The mass fraction of formic acid in propionic acid solution is equal to: ω (HCOOH) \u003d m (HCOOH): m (mixtures) \u003d 0.46: 150 \u003d 0.0031 (0.31%) |
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The mass fraction of propyl alcohol in a propionic acid solution is equal to: ω (C 3 H 7 OH) \u003d m (C 3 H 7 OH): m (mixtures) \u003d 2.28: 150 \u003d 0.0152 (1.52%) |
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All response elements are written incorrectly |
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Maximum score |
11-5.
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(other formulations of the answer are allowed that do not distort its meaning) |
Points |
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Concentrated sulfuric acid as a strong oxidant will interact with hydroiodic acid, which exhibits reducing properties: 8HI + H 2 SO 4 \u003d 4I 2 ↓ + H 2 S + 4H 2 O A brown iodine precipitate is precipitated and hydrogen sulfide is released with an unpleasant smell of rotten eggs. |
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A specific property of formic acid is its ability to decompose into carbon monoxide (II) and water under the action of concentrated sulfuric acid, which exhibits water-removing properties: НСООН CO + H 2 O a colorless gas with an unpleasant odor is emitted |
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When heated, concentrated sulfuric acid removes water from oxalic acid: t 0, H 2 SO 4 (k) H 2 C 2 O 4 CO 2 + CO + H 2 O colorless gases are emitted |
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Phosphorous acid as a reducing agent can react with nitric acid, which is a strong oxidizing agent: H 3 PO 3 + 2HNO 3 \u003d H 3 PO 4 + 2NO 2 + H 2 O brown gas is released |
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Nitric and sulfuric acids, which are strong oxidizing agents, do not react with each other |
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All response elements are written incorrectly |
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Maximum score |
... – this is a single, interconnected process. Consider reaction dissolving zinc in saline acid: Zn0 ... which the oxidizes to water and nitrogen. Write the equation reactions, determine by calculation for standard states substances direction reactions at ...
Factory # 8 was assigned its own factory indexes to other people's guns (Bolshevik, Hotchkiss, Maxim, Reinmetall, etc.), thus the Lender system
DocumentOf substancesdissolved in water... In sea mines for of this traditionally used sugar. ... by plan Work status Possible execution at cash equipped. at replenished, ... which will be stand on bearings, not babbit, as Now. it ...
Thematic planning of chemistry lessons in grades 8-11 6
Thematic planning8. What kind inorganic substances available, using water, air, sulfur and calcium. Write the equation reactions and indicate the conditions for their flow. nine. When ...
Tasks for the school stage of the Olympiad for schoolchildren in chemistry, grade 11.
Exercise 1.
There are ten elements in the Periodic Table, in the designation of which the second letter is "e". For each of these elements, typical reactions are given below. The capital letter is indicated by the sign “? ". Arrange the elements in their places.
1) D + T →? e + n;
2)? eO 4 →? e + 2O 2;
3)? e + O 3, F 2, etc. → does not go;
4) 2Au + 6H 2? eO 4 (load) → 3? eO 2 + Au 2 (? eO 4) 3 + 6H 2 0;
five) ? eCl 2 + 4NaOH → Na 2 [? e (OH) 4] + 2NaCl;
6)? eO 3 + 3H 2 O → H 6? eO 6;
7)? e (OH) 4 + 4HCl →? eCl 3 + 1 / 2Cl 2 + 4H 2 0;
8) ? eCl 4 + Li →? eH 4 + Li;
nine) ? eCl 2 + K 3 [? e (CN) 6] → K? e [? e (CN) 6 ] ↓ (blue precipitate);
10) 2H? eO 4 + 7H 2 S →? e 2 S 7 ↓ + 8H 2 0.
Task 2.
The chemist took samples of three silver-white metals and found a way to quickly distinguish 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.
Legend: "+" - the reaction is in progress, "-" - the metal does not react.
Determine which metals could be obtained by a chemist and write the corresponding reaction equations.
Task 3.
A small piece of a plastic disposable cup was heated without air access to 400 ° C. As a result of heating, a hydrocarbon was obtainedX (carbon content 92.26% by weight, its vapor density by oxygen 3.25). It is known that during the oxidation of a hydrocarbonX With a solution of potassium permanganate in an acidic medium, benzoic acid is formed as the only organic product.
1. Calculate the molecular formulaH.
2. Give the structural formula and the name of the hydrocarbonX ... What is the name of the starting polymer?
3. Write the reaction equation (with all products and stoichiometric coefficients) for hydrocarbon oxidationX potassium permanganate solution acidified with sulfuric acid.
Task 4.
Young chemist Petya got an assignment from his mother to buy 1 liter of food vinegar (mass fraction of acetic acid CH3COOH 9%) for home canning in the store. Arriving at the store, he found that only acetic essence was on sale (the mass fraction of acetic acid was 70%). Petya decided that he could make edible vinegar out of it on his own. At home, in the reference book, he managed to find the density values \u200b\u200bfor a 9% solution of acetic acid - 1.012 g / ml and for a 70% solution - 1.069 g / ml. Of the equipment Petya has only measuring cylinders of various sizes.
- What safety rules should be followed when preparing dilute solutions from concentrated acids?
- Which of the substances available at home should Pete have on hand if acid gets on the skin? Name this substance and reflect its effect on the acid in the reaction equation.
- What volume of vinegar essence should Petya measure in order to prepare 1 liter of 9% acetic acid solution?
Task 5.
The laboratory contains iron, hydrochloric acid, caustic soda, calcium carbonate, copper (II) oxide. Using these substances, as well as the products of their interaction, give at least 10 equations for the reactions of obtaining new inorganic substances.
Task 6.
When 2.8 g of an alloy of copper and silver were dissolved in concentrated nitric acid, 5.28 g of a mixture of nitrates was formed. Determine the mass fraction of metals in the alloy.
Grade 11 Answers
Exercise 1.
Answer:
Points |
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Even if the decisive cannot determine all the elements by reactions, this can be done using the Periodic Table, given that the maximum oxidation state of an element cannot be greater than the group number. | |
1) D + T → He + n; | |
2) Xe O 4 → Xe + 2O 2; | |
3) Ne + O 3, F 2, etc. → does not go; | |
4) 2Au + 6H 2 Se O 4 (heating) → 3 Se O 2 + Au 2 (Se O 4) 3 + 6H 2 0; | |
5) Be Cl 2 + 4NaOH → Na 2 [Be (OH) 4] + 2NaCl; | |
6) Te O 3 + 3H 2 O → H 6 Te O 6; | |
7) Ce (OH) 4 + 4HCl → Ce Cl 3 + 1 / 2Cl 2 + 4H 2 0; | |
8) Ge Cl 4 + Li → Ge H 4 + Li; | |
9) Fe Cl 2 + K 3 → K Fe [Fe (CN) 6 ] ↓ (blue precipitate); | |
10) 2H Re O 4 + 7H 2 S → Re 2 S 7 ↓ + 8H 2 0. | |
Maximum score |
Task 2.
By the combination of properties, i.e. according to the behavior in reactions with acids and sodium hydroxide, we conclude: I metal - silver, (copper does not match in color), metal II - aluminum, metal III - zinc.
Reaction equations:
2Al + 6HCl \u003d 2AlCl3 + 3H 2
Zn + 2HCl \u003d ZnCl 2 + 2H 2
Ag + 2HNO 3 (conc.) \u003d AgNO 3 + NO 2 + H 2 O
Al + HNO 3 (conc.) No reaction
Zn + 4HNO3 (conc.) + Zn (NO 3) 2 + 2NO 2 + 2H 2 O
Zn + 2NaOH + 2H 2 O \u003d Na 2 + H 2
2Al + 6NaOH + 6H 2 O \u003d 2Na 3 + 3H 2 (7 points)
Task 3.
Answer:
(other formulations of the answer are allowed that do not distort its meaning) | Points |
1.Mr (X) \u003d 3.2532 \u003d 104 amu. Let's find the molecular formula of the hydrocarbon X: C: H \u003d 0.9226 / 12.01: 0.0774 / 1.008 \u003d 1: 1, taking into account the molecular weight, we get C8 H 8. | |
2. Since the oxidation of hydrocarbon X with a solution of potassium permanganate in an acidic medium forms benzoic acid as the only organic product (C6 H 5 COOH), then its molecule contains a benzene ring with one substituent. Subtracting from the gross formula C8 H 8 fragment C 6 H 5 , we get substituent C2 H 3 ... The only possible substituent is vinyl, and hydrocarbon X is styrene (vinylbenzene). Consequently, the polymer from which the disposable cup was made is polystyrene. | |
3. The equation of the reaction of styrene oxidation with a KMnO solution4 acidified with H2 SO 4: 2KMnO 4 + 3H 2 SO 4 → + CO 2 + 2 MnSO 4 + K 2 SO 4 + 4H 2 O | |
All response elements are written incorrectly | |
Maximum score |
Task 4.
- Pour water into acid (1 point).
- Baking soda or sodium bicarbonate (1 point).
NaHCO3 + CH3COOH \u003d CH3COONa + H2O neutralization reaction (2 points).
- The calculated mass of acetic acid in a 9% solution is 91.08 g (1 point).
The calculated mass of the vinegar essence solution is 130.1 g (1 point).
The volume of vinegar essence is calculated - 121.7 ml or ≈ 122 ml (1 point).
Total: 7 points.
Task 5.
Possible answers:
Fe + 2HCl \u003d FeCl 2 + H 2
HCl + NaOH \u003d NaCl + H 2 O
2HCl + CaCO 3 \u003d CaCl 2 + H 2 O + CO 2
CuO + 2HCl \u003d CuCl 2 + H 2 O
2NaOH + CO 2 \u003d Na 2 CO 3 + H 2 O; NaOH + CO 2 \u003d NaHCO 3 + H 2 O
CuO + H 2 \u003d Cu + H 2 O
FeCl 2 + 2NaOH \u003d Fe (OH) 2 + 2NaCl
3Fe + 4H 2 O \u003d Fe 3 O 4 + 4H 2
CaCO 3 \u003d CaO + CO 2
CuO + CO 2 \u003d CuCO 3
Fe 3 O 4 + 8HCl \u003d FeCl 2 + 2FeCl 3 + 4H 2 O, etc.
Task 6.
- Reaction equations:
Cu + 4 HNO 3 \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O
Ag + 2HNO 3 \u003d AgNO 3 + NO 2 + H 2 O (1 point)
- We enter the designations: n (Cu) \u003d x mol n (Ag) \u003d y mol; then
a) m (Cu) \u003d 64x, m (Ag) \u003d 108y
m (mixture) \u003d 64x + 108y \u003d 2.8
b) according to equation (1) n (Cu (NO3) 2 \u003d x, m (Cu (NO 3) 2 \u003d 188x;
c) according to equation (2) n (AgNO3) \u003d y, m (AgNO 3) \u003d 170y
d) m (mixtures) \u003d 188x + 170y \u003d 5.28 (2 points)
3) compose and solve the system of equations:
64x + 108y \u003d 2.8 x \u003d 0.01 mol Cu
188x + 170y \u003d 5.28 y \u003d 0.02 mol Ag (2 points)
4) calculate the mass fractions of the mixture components:
a) m (Cu) \u003d 0.01 * 64 \u003d 0.64g. w (Cu) \u003d 0.64 / 2.8 \u003d 0.2286 or 22.86%
M (Ag) \u003d 0.02 * 108 \u003d 2.16g. w (Ag) \u003d 2.16 / 2.8 \u003d 0.7714 or 77.14% (2 points)
Maximum score - 7 points
Tasks for the school stage of the chemistry Olympiad, grade 10
Exercise 1.
What gases can be obtained with the following substances at your disposal:
sodium chloride, sulfuric acid, ammonium nitrate, water, ammonium nitrite, hydrochloric acid, potassium permanganate, sodium hydroxide, aluminum carbide, calcium carbide and sodium sulfite?
Write down all the equations of possible reactions, indicate the conditions for their occurrence.
Task 2.
For three chemical elements A, B and C, it is known that they take part in the following transformations:
- С 2 + 3В 2 2СВ 3
- 4SV 3 + 5A 2 4CA + 6V 2 A
- 4SV 3 + 3A 2 \u003d 2C 2 + 6V 2 A
- C 2 + A 2 \u003d 2 CA
- 4CA 2 + 2B 2 A + A 2 \u003d 4BCA 3
What elements are discussed in the assignment? Write down the reaction equations.
Task 3.
When 11.5 g of a mixture of aluminum, magnesium and copper were dissolved in hydrochloric acid, 5.6 liters (NU) of gas were released. The insoluble residue was transferred to a solution with concentrated nitric acid. At the same time, 4.48 liters (standard) of gas were released. Determine the mass fractions (in%) of the components of the initial mixture.
Task 4.
The reaction scheme is given:
KI + KMnO 4 + H 2 SO 4 → I 2 + K 2 SO 4 + MnSO 4 + H 2 O
1. Determine the oxidation states of the elements.
2. Make the equation of electronic balance
3. Determine the oxidizing and reducing agent
4. Place the coefficients in this equation.
5. List the areas of application of the substance, the formula of which is KMnO4
Task 5.
There are nine elements in the periodic table, the names of which in Russian are nouns "not masculine". For each of these elements, typical reactions are given below, in which they are encrypted with the letter "E ". Define the elements for each reaction:
1) 2H 2 E + E O 2 → 3 E + 2H 2 O;
2) E Cl 3 + KI → E Cl 2 + KCl +1/2 I 2;
3) E NO 3 + KCl → E Cl ↓ + KNO 3;
4) E + 3HNO 3conc. + 4HCl conc. → H [E Cl 4] + 3NO 2 + 3H 2 O;
5) E Cl 2 + 4NH 3 → [E (NH 3) 4] Cl 2 (blue);
E (catalyst), 800 ° С
300 ° C
7) 2 E + O 2 2EO
8) E Cl 2 + Cl 2 → E Cl 4;
9) E 2 O 3 + 4 HNO 3conc. + (x -2) H 2 O → E 2 O 5 x H 2 O + 4NO 2
Task 6.
As a result of the reaction of calcium phosphate with magnesium at
heating, two substances are formed, one of which
interacts with water, while a colorless poisonous
gas with a garlic smell. The latter is oxidized
oxygen in the air.
Write the equations of all the specified chemical processes, name
their products.
Calculate the volume of air (n.o.) required to oxidize the gas,
if 2.4 g of magnesium was used in the first of these reactions.
Grade 10 Answers
Exercise 1.
1) 2NaCl (TV) + H 2 SO 4 (conc.) Na 2 SO 4 + 2HCl
Or NaCl (TV) + H 2 SO 4 (conc.) NaHSO 4 + HCl
- NH 4 NO 3 \u003d N 2 O + 2H 2 O
- NH 4 NO 2 \u003d N 2 + 2H 2 O
- 2KMnO 4 \u003d K 2 MnO 4 + MnO 2 + O 2
- 2KMnO 4 + 16HCl \u003d 2KCl + 2MnCl 2 + 5Cl 2 + 8H 2 O
or 2NaCl + 2H 2 O 2NaOH + H 2 + Cl 2
- Al 4 C 3 + 12H 2 O \u003d 4Al (OH) 3 + 3CH 4
- CaC 2 + 2H 2 O \u003d Ca (OH) 2 + C 2 H 2
- 2H 2 O 2H 2 + O 2
- NH 4 NO 3 + NaOH \u003d NaNO 3 + NH 3 + H 2 O
- Na 2 SO 3 + H 2 SO 4 \u003d Na 2 SO 4 + H 2 O + SO 2
Total: 10 points
Task 2.
- N 2 + 3H 2 2NH 3
- 4NH 3 + 5O 2 4NO + 6H 2 O
- 4NH 3 + 3O 2 \u003d 2N 2 + 6H 2 O
- N 2 + O 2 \u003d 2NO
- 4NO 2 + 2H 2 O + O 2 \u003d 4HNO 3
A - oxygen, B - hydrogen, C - nitrogen.
For each reaction equation 1 point
Total: 5 points
Assignment 3.
1) Copper does not interact with hydrochloric acid.
4.48 l
Cu + 4 HNO 3 \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O
n (NO 2) \u003d 4.48 / 22.4 \u003d 0.2 mol;
N (Cu) \u003d 0.1 mol; m (Cu) \u003d 64 x 0.1 \u003d 6.4 g 1 point
X mol 1.5 x mol
2) 2Al + 6HCl \u003d 2AlCl 3 + 3H 2
Y mol y mol
Mg + 2HCl \u003d MgCl 2 + H 2 2 points
3) Determination of the amount of hydrogen substance: n (H2 ) \u003d 5.6 / 22.4 \u003d 0.25 mol; 1 point
4) Compiled a system with two unknowns:
x + 1.5y \u003d 0.25
24x + 27y \u003d 5.1 2 points
5) The system with two unknowns is solved (the values \u200b\u200bof "x" and "y" are determined) 2 points
X \u003d 0.1; y \u003d 0.1
6) The masses and mass fractions of the mixture components were determined.
M (Al) \u003d 2.7g; m (Mg) \u003d 2.4g 1 point
ώ (Al) \u003d 2.7 / 11.5 \u003d 0.2348 or 23.48%
ώ (Mg) \u003d 2.4 / 11.5 \u003d 0.2087 or 20.87%. 1 point
ώ (Cu) \u003d 6.4 / 11.5 \u003d 0.5565 or 55.65%
Total: 10 points
Task 4.
1) The oxidation states of the elements have been determined
1-1 +1 +7 -2 +1 +6 -2 0 +1 +6 -2 +2 +6 -2 +1 -2
KI + KMnO 4 + H 2 SO 4 → I 2 + K 2 SO 4 + MnSO 4 + H 2 O 1 point
2) The equation of the electronic balance was compiled and the oxidant and
Reducing agent:
2 points
3) Coefficients in the reaction equation are arranged:
2KMnO 4 + 10KI + 8H 2 SO 4 \u003d 5I 2 + 6K 2 SO 4 + 2MnSO 4 + 8H 2 O 2 points
4) The areas of application of potassium permanganate are listed:
(laboratory method for producing oxygen and chlorine, chemical analysis (permanganatometry), oxidation of organic substances, etc.) only 2 points
Total: 7 points
Task 5.
Answer:
(other formulations of the answer are allowed that do not distort its meaning) | Points |
These elements are sulfur, iron, silver, gold, copper, platinum, mercury, tin, and antimony. All of them, except for sulfur, are metals, and all are easily recognizable by their characteristic transformations, or by the oxidation states characteristic only of their position in the Periodic Table. | |
1) 2H 2 S + SO 2 → 3S + 2H 2 O; | |
2) FeCl 3 + KI → FeCl 2 + KCl +1/2 I 2; | |
3) AgNO 3 + KCl → AgCl ↓ + KNO 3; | |
4) Au + 3HNO 3conc. + 4HCl conc. → H + 3NO 2 + 3H 2 O; | |
5) CuCl 2 + 4NH 3 → Cl 2 (blue); | |
Pt (catalyst), 800 ° С 6) 4NH 3 + 5O 2 4NO + 6H 2 O; | |
300 ° C 400 ° C 7) 2Hg + O 2 2Hg O 2Hg + O 2; | |
8) SnCl 2 + Cl 2 → SnCl 4; | |
9) Sb 2 O 3 + 4 HNO 3conc. + (x-2) H 2 O → Sb 2 O 5 x H 2 O + 4NO 2 | |
All response elements are written incorrectly | |
Maximum score |
Task 6.
Ca 3 (PO 4) 2 + 8 Mg \u003d Ca 3 P 2 + 8 MgO (calcium phosphide) (1 point)
Ca 3 P 2 + 6 H 2 O \u003d 3 Ca (OH) 2 + 2 PH 3 (d) (phosphine) (1 point)
2 PH 3 + 4 O 2 \u003d 2 H 3 PO 4 (ortho-phosphoric acid) (1 point)
n (Mg) \u003d 0.1 mol. n (О 2) \u003d 0.05 mol. V (О 2) \u003d 1.12 l. (2 points)
V (air containing 21% oxygen) \u003d 5.3 liters.
Maximum score - 5 points
Tasks for the school stage of the Olympiad in chemistry, grade 9
Exercise 1.
Task 2.
Task 3.
When solving problems, we are often faced with the expression of the molar mass of a substance in terms of the molar mass of another substance, most often air and hydrogen.
Imagine that you have discovered a new planet. An analysis of the composition of the atmosphere of this planet is given below: 48% N2, 19% O 2, 15% Cl 2 , 8% Ne, 7% Ar, 2.9% Kr and 0.1% CO2 .
Calculate the molar mass of your planet's air and the density of each constituent of the atmosphere over that planet's air.
Task 4.
Task 6.
To obtain a sodium chloride solution, the previously calculated mass of sodium carbonate was dissolved in 5.0% hydrochloric acid.
Determine the mass fraction of sodium chloride in the resulting solution.
Answers to assignments grade 9
Exercise 1.
The bottles without labels contain the following substances: dry silver nitrate, sodium sulfide, calcium chloride; powders of silver and aluminum, as well as hydrochloric and concentrated nitric acids. You have at your disposal water, a burner and any number of test tubes.
Make up the reaction equations and indicate the signs by which you can determine each of the indicated substances.
Number of points –10
Answer:
(other formulations of the answer are allowed that do not distort its meaning) | Points |
1) AgNO 3 + HCl \u003d AgCl ↓ + HNO 3 White | |
2) CaCl2 + 2AgNO3 \u003d 2AgCl ↓ + Ca (NO3 ) 2 White | |
3) Na2 S + 2HCl \u003d 2NaCl + H2 S Rotten egg smell | |
4) 2Al + 6HCl \u003d 2AlCl3 + 3H2 Colorless, odorless | |
5) Ag + 2HNO3 \u003d AgNO3 + NO2 + H2 O Brown, with a pungent odor | |
6) Na2 S + 4HNO3 \u003d 2NaNO3 + S ↓ + 2NO2 + 2H2 O Yellow | |
All response elements are written incorrectly | |
Maximum score |
Task 2.
When passing sulfur (IV) oxide through a solution of potassium permanganate, a solution was formed in which the mass fraction of sulfuric acid is 5%. Calculate the mass fractions of the remaining reaction products in the resulting solution.
Number of points –10
Answer:
(other formulations of the answer are allowed that do not distort its meaning) | Points |
1) Let's compose the reaction equation: 5SO2 + 2KMnO4 + 2H2 O \u003d K2 SO4 + 2MnSO4 + 2H2 SO4 | |
2) Find the mass of 2 mol H2 SO4 - m (H2 SO4 ) \u003d 2 mol ∙ 98 g / mol \u003d 196 g. | |
3) Find the mass of 2 mol MnSO4 - m (MnSO4 ) \u003d 2 mol ∙ 151 g / mol \u003d 302 g. | |
4) Find the mass of 1 mol K2 SO4 - m (K2 SO4 ) \u003d 1 mol ∙ 174 g / mol \u003d 174 g. | |
5) The mass fraction of each substance is equal to: ω (in-va) \u003d m (in-va): m (solution). Since all these substances are in one solution (i.e., the mass of the solution is the same for them), the ratio of their mass fractions is equal to the ratio of the masses: ω (K2 SO4 ): ω (H2 SO4 ) \u003d m (K2 SO4 ): m (H2 SO4 ) = 174: 196; whence ω (K2 SO4 ) \u003d 0.05 ∙ (174: 196) \u003d 0.04 or 4.4%. | |
6) ω (MnSO4 ): ω (H2 SO4 ) \u003d m (MnSO4 ): m (H2 SO4 ) \u003d 302: 196, whence ω (MnSO4 ) \u003d 0.05 ∙ (302: 196) \u003d 0.077 or 7.7%. | |
All response elements are written incorrectly | |
Maximum score |
Task 3.
M (planetary air) \u003d 0.48 * 28 + 0.19 * 32 + 0.15 * 71 + 0.08 * 20 + 0.07 * 40 + 0.029 * 83 + 0.001 * 44 \u003d
37.021g / mol... (2 points)
D (N2 )=28/37.021=0.7563
D (O2 ) \u003d 0.8644; D (Cl2 ) \u003d 1.9178; D (Ne) \u003d 0.5402; D (Ar) \u003d 1.0805; D (Kr) \u003d 2.2420; D (CO2 )=1.1858. (8 points)
Maximum score - 10
Task 4.
Consider a 100 g sample of ethyl mercaptan.
In such a sample, m (C) \u003d 38.7 g, m (H) \u003d 9.6 g, m (S) \u003d 51.7 g. (1 point)
n (C) \u003d 38.7 g: 12 g / mol \u003d 3.2 mol(1 point)
n (H) \u003d 9.6 g: 1 g / mol \u003d 9.6 mol (1 point)
n (S) \u003d 51.7 g: 32 g / mol \u003d 1.6 mol(1 point)
n (C): n (H): n (S) \u003d 3.2: 9.6: 1.6 \u003d 2: 6: 1 \u003d\u003e C2H6S(2 points)
Maximum score - 6
Task 5.
3Fe + 2O2
\u003d Fe3
O4
Fe3
O4
+ 8HCl \u003d FeCl2
+ 2FeCl3
+ 4H2
O
FeCl2
+ 2NaOH \u003d Fe (OH)2
↓ + 2NaCl
FeCl3
+ 3NaOH \u003d Fe (OH)3
↓ + 3NaCl
2Fe (OH)3
+ Fe (OH)2
→ Fe3
O4
+ 4H2
O (t °, inert atmosphere)
Substances: A - Fe3
O4
; В and С - FeCl2
or FeCl3
; D and E - Fe (OH)2
or Fe (OH)3
.
Maximum score -5
Task 6.
106 73 117 44
Na2 CO3 + 2 HCl \u003d 2 NaCl + CO2 + H2 O
X 5 Z Y
Take 100 g of 5% hydrochloric acid.
With 5 g of HCl, X \u003d (106 * 5) / 73 \u003d 7.26 g of Na2 CO3 , while the solution is released
Y \u003d (44 * 5) / 73 \u003d 3.01 g CO2 It turns out sodium chloride Z \u003d (117 * 5) / 73 \u003d 8.01 g
Mass fraction of NaCl 100% (8.01 / (100 + 7.26 - 3.01)) \u003d7,7 %
Maximum score -6 points
According to Le Chatelier's principle: if an external influence is exerted on a system in equilibrium, then as a result of the processes occurring in it, the equilibrium position shifts to the side, weakening this effect; with a decrease in pressure, the balance in a reversible process shifts towards an increase in pressure.
a) N 2 O 4 (g) ⇄ 2NO 2 (g),
the process proceeds with an increase in volume (1
b) 2NO (g) + О 2 (g) ⇄ 2NO 2 (g),
the process proceeds with a decrease in volume (3\u003e 2), i.e. an increase in pressure, therefore, the equilibrium shifts in the initial substances.
c) 3Fe 2 O 3 (c) + CO (g) ⇄ 2Fe 3 O 4 (c) + CO 2 (g),
the process proceeds without changing the volume, a decrease in pressure does not affect the state of equilibrium.
Example 8.
Explain why flaming calcium metal cannot be extinguished with water. Write the reaction equations.
Decision:
Metallic calcium reacts with water, so adding water to burning calcium will only intensify the process. Let's write the equation of the reaction of interaction of calcium with water:
Ca + 2H 2 O \u003d Ca (OH) 2 + H 2.
Example 9.
Write the reaction equations in molecular and ionic form, with the help of which you can carry out the following transformations: Cl - Сl 2 Cl - AgCl.
Decision:
1. Interaction of manganese (IV) oxide with concentrated hydrochloric acid:
molecular form:
4HCl + MnO 2 \u003d MnCl 2 + Cl 2 + 2H 2 O,
ionic form
4H + + 2Cl - + MnO 2 \u003d Mn 2+ + Cl 2 + 2H 2 O.
Sodium Chloride Venture Electrolysis:
molecular form:
molecular form:
2NaCl 2Na + Cl 2,
ionic form
2Na + + 2Cl - 2Na + Cl 2.
2. Interaction of chlorine with sodium bromide solution:
molecular form:
Cl 2 + 2NaBr \u003d 2NaCl + Br 2,
ionic form:
Cl 2 + 2Br - \u003d 2Cl - + Br 2.
Reaction between metallic sodium and chlorine gas:
molecular form:
2Na + Cl 2 \u003d 2NaCl.
3. Interaction of sodium chloride solution with silver nitrate solution:
NaCl + AgNO 3 \u003d AgCl + NaNO 3,
ionic form:
Cl - + Ag + \u003d AgCl.
Example 10.
Decision:
1. Interaction of copper with dilute nitric acid:
Cu + 6HNO 3 \u003d Cu (NO 3) 2 + 4NO + H 2 O.
2. Oxidation of copper with oxygen:
2Cu + O 2 \u003d 2CuO.
3. Interaction of copper with concentrated hydrochloric acid in the presence of oxygen:
2Cu + O 2 + 4HCl \u003d 2CuCl 2 + 2H 2 O.
4. Interaction of copper (II) chloride solution with sodium hydroxide solution:
CuCl 2 + 2NaOH \u003d Cu (OH) 2 + 2NaCl.
5. Thermal decomposition of copper (II) hydroxide:
Cu (OH) 2 \u003d CuO + H 2 O.
6. Interaction of copper (II) nitrate with sodium hydroxide solution:
Cu (NO 3) 2 + 2NaOH \u003d Cu (OH) 2 + 2NaNO 3.
7. Thermal decomposition of copper (II) nitrate:
2Cu (NO 3) 2 \u003d 2CuO + 4NO 2 + O 2.
8. Interaction of copper (II) oxide with hydrochloric acid:
CuO + 2HCl \u003d CuCl 2 + H 2 O.
Tasks for independent solution
Option 1
Write the reaction equations showing the properties of aluminum hydroxide.
Show how, with the help of one reagent, to determine which of the flasks contains dry salts: sodium chloride, sodium carbonate, sodium sulfide. Write down the equations of the corresponding reactions.
Get basic copper carbonate from metallic copper using as few reagents as possible. Write down the equations of the corresponding reactions.
To neutralize 10.0 g of a solution containing a mixture of hydrochloric and hydrobromic acids, 2.5 g of a 3.2% NaOH solution was required, and when the solution of the same mass was exposed to a solution of silver nitrate, 0.3315 g of precipitate fell out. Determine the mass fraction (%) of acids in the original solution.
Write the equations for the interaction of iron (III) hydroxide with concentrated hydrochloric and dilute sulfuric and nitric acids.
Calculate the amount of technical zinc containing 96% zinc and 27.5% HCl solution should be consumed to obtain 1 ton of 45% zinc chloride solution.
Indicate which element is oxidized and which is reduced in the following reactions: a) NH 3 + O 2 N 2 + H 2 O; b) KI + Cu (NO 3) 2 CuI + I 2 + KNO 3. Place the coefficients, indicate the transition of electrons.
In one collective farm, for each hectare, phosphorus fertilizers - 60 kg (in terms of P 2 O 5), potash fertilizers - 150 kg (in terms of K 2 O) and copper sulfate - 10 kg were applied for hemp. Assuming, for simplicity, that the latter does not contain impurities, indicate how many moles of each of the remaining oxides are in 1 mole of copper (II) oxide.
Indicate how to change the temperature and pressure (increase or decrease) so that the equilibrium in the decomposition reaction of calcium carbonate: CaCO 3 (k) ⇄ CaO (k) + CO 2 (g) - 178 kJ to shift towards the decomposition products.
Explain why sodium and potassium hydroxide solutions destroy glassware, especially when boiled for a long time. Write the reaction equations.
Write the reaction equations in molecular and ionic forms, with which you can carry out the following transformations: CO 3 2- CaCO 3 Ca 2+ CaSO 4.
Write the reaction equations with which you can carry out the following transformations:
Option 2
Write the equations for the interaction of dilute and concentrated sulfuric acid: a) with copper; b) with zinc; c) with lead.
Show, by means of which reactions it is possible to distinguish solutions of sulfuric, nitric and hydrochloric acids. Write down the equations of the corresponding reactions.
You have been given substances: calcium nitrate, sulfuric acid, caustic soda, potassium carbonate. How, using only these reagents, to obtain sodium nitrate in two ways? Write down the equations of the corresponding reactions.
Sulfur and aluminum contained in the mixture interacted with each other. The reaction product was treated with hot water. Part of the evolved gas was passed through chlorine water, the precipitate formed was separated, and an excess of silver nitrate was added to the solution. 8.61 g of a white curdled precipitate was formed. The other part of the gas was passed through 145 ml of a 10% copper sulfate solution (density 1.1 g / ml), as a result of which the concentration of copper sulfate in the solution became 6.09%. Calculate the mass of sulfur that has reacted. Write down the equations for all reactions.
Indicate which "limes" are known. Write down their chemical composition and obtaining equations.
Calculate how many grams of chromium can be obtained by reacting chromium (III) oxide with silicon, the mass of which is 10 g. The product yield is 90%.
Determine the oxidation state of each element and arrange the coefficients in the following schemes: a) Fe + FeI 3 FeI 2; b) H 2 S + I 2 + H 2 O H 2 SO 4 + HI.
In one collective farm, fertilizers were applied for spring wheat at the rate of: ammonium nitrate - 150 kg, superphosphate (containing 30% of digestible Р 2 О 5) - 300 kg and potassium chloride - 100 kg per hectare. Calculate how much this is in terms of nitrogen, calcium dihydrogen phosphate and potassium oxide.
Indicate how the increase in pressure will affect the equilibrium in the systems: a) SO 2 (g) + Cl 2 (g) ⇄ SO 2 Cl 2 (g); b) H 2 (d) + Br 2 (d) ⇄ 2HBr (d).
Show how to chemically remove corrosion products (aluminum oxide and hydroxide) from an aluminum product without causing damage to the metal. Write the reaction equations.
Show which of the mixtures: a) metal oxide and hydroxide; b) metal and metal oxide - when interacting with water, they give a solution of only one substance. Give examples, write reaction equations.
Write the reaction equations with which you can carry out the following transformations:
Ca Ca (OH) 2 CaCO 3 CaO Ca (OH) 2 CaCl 2 Ca.
Option 3
Write down the ionic exchange equations if:
one of the resulting substances dissociates little into ions, the second is insoluble;
one of the resulting substances is soluble, the second is not;
the reaction is reversible;
one of the resulting substances is soluble, the second is released in the form of a volatile substance.
Three test tubes contain dry substances: calcium oxide, aluminum oxide, phosphorus oxide. Show with what reagents you can distinguish these substances. Write the reaction equations.
Concentrated hydrochloric acid, water, manganese dioxide, copper and zinc in the form of thin wires were dispensed. How, using these substances, to obtain zinc chloride and copper (II) chloride in the form of crystalline hydrates? Describe the progress of the work, make up the equations of chemical reactions, indicate the conditions for their course.
When burning 0.896 L (standard) of a mixture of CO and CO2 in an excess of oxygen, 0.112 L of oxygen was consumed, the resulting gas mixture was passed through a solution containing 2.96 g of slaked lime. Determine the composition of the initial gas mixture (in% by volume), as well as the composition and mass of the sediment formed.
Show how alkaline earth metal hydroxides can be obtained. What is the strongest alkali hydroxide? Indicate the technical names for calcium and barium hydroxides.
In the industrial production of copper sulfate, copper scrap is oxidized when heated by atmospheric oxygen and the resulting copper (II) oxide dissolves in sulfuric acid. Calculate the consumption of copper and 80% H 2 SO 4 per ton of CuSO 4 5H 2 O if the product yield is 75%.
Write down the reaction equations: a) magnesium iodide with bromine; b) magnesium with a solution of hydrobromic acid. Indicate which element in each case is an oxidizing agent and which reducing agent, show the electron transition.
In one collective farm, in addition to manure, the following masses of mineral fertilizers per hectare were applied for potatoes: granular superphosphate containing 12.5% \u200b\u200bof the assimilable P 2 O 5 - 0.15 tons, ammonium nitrate - 0.1 tons and potassium chloride containing 90% KCl - 0.1 t. Calculate what masses of calcium hydrogen phosphate, nitrogen and potassium oxide this corresponds to.
The reaction proceeds according to the equation: 2SO 2 (g) + O 2 (g) ⇄ 2SO 3 (g) + 284.2 kJ. By changing what parameters can the equilibrium shift towards the formation of sulfur (VI) oxide?
Write down the equations of all reactions that can occur when lithium and sodium metal are stored in air.
Will sodium ions remain: a) when sodium hydroxide interacts with hydrochloric acid; b) when sodium hydroxide interacts with copper (II) chloride. Write down the equations of the corresponding reactions.
Write the reaction equations with which you can carry out the following transformations:
Option 4
Explain how aqueous solutions of amphoteric bases dissociate into ions. Give examples of such bases and show their dissociation into ions.
Indicate which one reagent can be used to recognize solutions of three substances: potassium chloride, aluminum chloride and magnesium chloride.
The laboratory contains iron, hydrochloric acid, caustic soda, calcium carbonate, copper (II) oxide. Is it possible to obtain 12 new inorganic substances if we use these reagents and their interaction products as starting materials? Write down the equations of the corresponding reactions.
{!LANG-fa8191ac9c2e3064dee15a0881dc74f2!}
{!LANG-1fd58b71161dcb0dd32d2a898fe41a57!}
{!LANG-06d3704f8cc5bfe6fa2571e77e87c696!}
{!LANG-d383f6925a88c28de4ba43c8e6e95cc8!}
{!LANG-10a6cddb6a245bd24d0051eeb7f125a2!}
{!LANG-8477b2f796d08e06935b0176c7520c56!}
{!LANG-2065d5fbe160c98352e8de559f9b1688!}
{!LANG-80882494f944b9d81e2d1e2b67049ad8!}
Write the reaction equations with which you can carry out the following transformations:
{!LANG-95e621727ab0034b8a10d273c7b79ded!}
{!LANG-3d721e315fc72c5011570865edbd8349!}
{!LANG-4572b4a80d7930ed03a855a3f57e8d40!}
{!LANG-9901b8ff7a34dbebfb448c1a122aaf7e!}
{!LANG-3b4ea6734620acdd1ca16a9d414f9262!}
{!LANG-5a2fdd635558ba1a6884a27e44dbf694!}
{!LANG-4b8c777ff3892dada912c97c548b5f3b!} 
{!LANG-7cd8e6992762353ce2178eba511574a7!} 
Decision:
{!LANG-ee1d4657104c3c5849aede7632f61e37!}
{!LANG-a880d8bf6a4a3f256338b86511275e28!}
{!LANG-cf31bb670359610b397ed876ee48b62c!}
{!LANG-01c4c4eb046ae1491c3a4e6511440468!}
{!LANG-fc3058fbb667c0326a62a87a75bb864d!} 
{!LANG-0a170f0da04921d2150e3982751c52c1!}
Decision:
{!LANG-110a60651d648b4eb94b810895416c85!}
{!LANG-78cda9fa388edcac6a07f2d30f3feaf3!}
{!LANG-ff5e357741e74e1a1f5f5e34117db90c!} 
{!LANG-7a49dd3b7238f5bdd83c2afca8c32ce9!}
{!LANG-7fa9eec4eb206427ec3ad2cffbdd755a!}
{!LANG-a3a194b92694fbe4fd8f782b5d459ef2!}
{!LANG-0f9e00eb4d1a7a6b81baec4499569681!}
{!LANG-efc16a6598e8234c646bd15fedf717e5!}
{!LANG-c9a47e6f02e157bca3b2677cd5ef7ec3!}
{!LANG-84b4dfd5a2111ff38a682941df111c84!}
{!LANG-e8df70e59270fd4c054fb18d64b1cee9!}
{!LANG-38922f86386289dd2e79c169e09fb0ee!} 
.
{!LANG-71e26536b2ddca18ab93b1341d22423b!}
Decision:
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{!LANG-abb5a45805965f51464e3d66e152651e!}
{!LANG-b356a49c68a5d4d17700033cc54194eb!}
{!LANG-e32b3b0bda660d4165145a02be907253!}
{!LANG-aa759dac6ade1975e4284d4d4def607a!}
{!LANG-37c40bc575e614f5e87b5b78682f742d!}
{!LANG-002ae18e040b0cc60ff0ee3c397a1561!}
{!LANG-753e7e89a76877ad78866aff08c39d97!}
{!LANG-5dfa9639ea185427745f50563d1c8d84!}
{!LANG-e92241b14fe4303f39e1c18af8a009c0!}
.
Decision:
{!LANG-5da64971155146565606c11f3843d004!} 
{!LANG-96a9177d266e6533fe16478ac1736f42!} 
{!LANG-ceb7d137351a85e73aab0202293a16a6!}
.
{!LANG-59c6382efb7190cb00d32a4deabf762e!}
{!LANG-99f9a69aa4531dd314f462f742f36610!}
{!LANG-37e2e075fc2de955fbdceb2d85540284!}
{!LANG-425a3e9d228952d6bf6ae19e67d8ade4!}{!LANG-eed6c7f27ef227792e6791c66ab965f6!}
{!LANG-69b64623f86def16ce17d454b8be41ae!}{!LANG-cf0c69088beeb4853a707f23a0455095!}
{!LANG-370751da74d8145ea4153e218746f10f!}{!LANG-c171cee3bf60fe5b4135a7289445db28!}
{!LANG-c18eafc2a4a0a61edd295a631d5b843b!}{!LANG-0b710ebcacaa257d7b5ceab4ba0f924c!}
{!LANG-7aa42d9d4ea52e8a68340909921225c2!}
{!LANG-425a3e9d228952d6bf6ae19e67d8ade4!}{!LANG-deb61dfa37db3cb941c2e76e64e49df6!}
{!LANG-425a3e9d228952d6bf6ae19e67d8ade4!}{!LANG-42ac52f73453b611dab3241653e7a659!}
{!LANG-69b64623f86def16ce17d454b8be41ae!}{!LANG-6cd225ada7853a96aa439503324bda26!}
Decision:
{!LANG-224f5d3fffc213cc848b5f2fe01877b4!}
{!LANG-5ce3af2a7047594c1a809a88765b1bec!}
{!LANG-fe13119fb084fe8bbf5fe3ab7cc89b3b!}{!LANG-58aa919388da4127c5ed11287ca34392!}
{!LANG-537ac3fd61e64b26ee8f05ad7d15aafa!}{!LANG-cff9874ad6c1bc251ee3e646bef1b24f!}
{!LANG-d849a97bff355053ba5e16980b3d3b63!}
{!LANG-7e2c222c520c72be15f7120c607fd86a!}
{!LANG-4ecb828eda33fac4e973b8623c308a27!} x{!LANG-60440bc1e9ec7cbd58466b5d038e23d9!} x{!LANG-c3626383e1d59e652391cad52dd6d76a!} 
{!LANG-dda004641b52a6842f6947fc174856b7!} 
{!LANG-78cd242058dac5f5d64e6b4df9bca2fa!} {!LANG-12978cfd7293225b6316c3fbc196c477!}
{!LANG-12be92ad1553ebe80cfbdba8f90cdacc!}
{!LANG-1f140e05e6fb4a1b5340b5cb1f546818!}{!LANG-bc384d1f72b89499ec10e89ac88ccbfa!} {!LANG-c246a467e2c14c44e50fd0d0bdd87d19!}{!LANG-a37e7c309930d5ea2fd5f4fa2534750e!} « {!LANG-ef540fb8c7ad0867e3f2d670915e6d23!}{!LANG-12978cfd7293225b6316c3fbc196c477!}{!LANG-01df403fc8ff411f35cda17dedb632eb!} for{!LANG-ee2eabaac7e6d3cba934586fa8894088!} {!LANG-85179979d0ed7f3c555cfb1a7172e8a6!}{!LANG-73365e38bf2b65234d77914e34d9fae0!} {!LANG-ab540ccbfb602574efd8a30ddddd0102!}for{!LANG-934ea188e5f58b8a59d80f08a039813d!}{!LANG-8f18781685d0d2d132d9cb585362f5e6!} « {!LANG-ef540fb8c7ad0867e3f2d670915e6d23!}{!LANG-12978cfd7293225b6316c3fbc196c477!}{!LANG-2088b05cc62b1eee542fbf4890737ee9!}
{!LANG-585a01fc86c2a1e5fde762e0367b4c56!}
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{!LANG-ef843f28ff9aa88dce98aa3c1e95160f!}
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