Presentation preparation for the exam in chemistry. Ege in chemistry. Both aniline and dimethylamine react with

06.10.2021 Sport

Dmitry Yurievich Dobrotin, Senior Researcher Laboratory of Didactics of Chemistry, ISMO RAO, Ph.D. site

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The structure of school chemistry education

1. Propaedeutic stage (grades 1 - 7) 2. Basic (basic) training of students in chemistry (grades 8 - 9) 3. Profile training of students in chemistry (grades 10 - 11)

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propaedeutic stage

The world around us (grades 1-4) Natural history (grade 5) Natural science (grades 5, 6, 7) Physics and Chemistry (grades 5-6) Introduction to chemistry (grade 7)

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book to read

Designed for children 6-12 years old

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Main ideas of the book

How is the world? Methods of knowing nature Substances around us Reactions around us Household chemical literacy

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Main school

Examinations in a new format Thematic examinations in chemistry for grade 8

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Topics of tests included in the manual

1. Initial chemical concepts 2. Oxygen. Oxides 3. Hydrogen. Water. Solutions 4. Main classes of inorganic compounds 5. Periodic law and periodic system chemical elements D.I. Mendeleev. Chemical bond 6. Final test

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State final certification of graduates in a new form in chemistry

On the new form of state final certification in grade 9 in chemistry Recommendations for preparing for the GIA in chemistry Tasks for independent work on content blocks Training options Answers to tasks Codifier

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High school

USE. Universal materials for preparing students Unified State Exam Recommendations for preparing for the Unified State Examination in Chemistry Assignments for independent work on content blocks Training options Answers to assignments Demonstration plan Examination plan for the Unified State Examination

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Manuals of the publishing house "Intellect-Center"

Designed for use at all stages of the chemical training of students The tasks included in the manuals are aimed at testing the knowledge and skills, the mastery of which is provided for by the Mandatory minimum and the Requirements for the level of graduates' training Can serve as an addition to any teaching materials in chemistry as didactic materials Allows you to work out material at various levels Difficulties Reflect the logic of reasoning when performing tasks of various types and levels of complexity Allow you to build an individual trajectory when working out the material in preparation for control measures, including the GIA and the Unified State Examination

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The structure of part C of KIMs in chemistry in 2006 C1. Redox reactions C2. Compilation of possible equations of reactions C3. Chains of transformations in organic chemistry С4. Calculation tasks on the topics: "Solutions", "Calculations according to the reaction equation" C5. The task of deriving the formula of a substance

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Redox reactions The basic rule that is used in determining the products of OVR is that one element lowers its oxidation state, another element increases its oxidation state. The reaction products depend on the environment. In an acidic environment, the change in the degree of oxidation occurs more deeply.

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Redox reactions C1. Using the electron balance method, write the reaction equation: P + HClO3 + ...= HCl + ... Determine the oxidizing agent and reducing agent. DECISION: Let's determine the oxidation states and establish which element will lower the oxidation state, and which will increase it. P0 + H Cl + 5O3 + ... = HCl-1 + ... In this equation, Cl + 5 (oxidizing agent) turns into Cl-1, therefore, P0 (reducing agent) will increase its oxidation state to P + 5 Electron transition: P0-5e- = P+5 6 Cl+5+6e- = Cl-1 5 The equation is: 6P + 5HClO3 + 9H2O = 5HCl + 6H3PO4

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Redox reactions C1. Using the electron balance method, formulate the equation: Al + K2Cr2O7 + … = … + Cr2(SO4)3 + K2SO4 + H2O SOLUTION: Al0 + K2Cr2+6O7 + … = … + Cr2+3(SO4)3 + K2SO4 + H2O Al0 increases oxidation state to +3, Cr+6 lowers the oxidation state to Cr+3. Sulfuric acid acts as a medium. Electron transfer: Al0 - 3e- = Al+3 2 reducing agent 2Cr+6 + 6e- = 2Cr+3 1 oxidizing agent

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Compilation of possible equations of reactions C2. Are given aqueous solutions: sodium sulfide, hydrogen sulfide, aluminum chloride and chlorine. Write the equations of four possible reactions between these substances. SOLUTION: Na2S + H2S = 2NaHS H2S + 4Cl2 + 4H2O = 8HCl + H2SO4 Na2S + Cl2 = 2NaCl + S 2AlCl3 + 3Na2S + 6H2O = 2Al(OH)3 + 6NaCl + 3H2S

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Compilation of possible equations of reactions C2. Aqueous solutions are given: potassium hexahydroxoaluminate K3, aluminum chloride, hydrogen sulfide and rubidium hydroxide. Write the equations for the four possible reactions between these substances. SOLUTION: K3 + 3H2S = 3K2S + 2Al(OH)3 + 6H2O H2S + 2RbOH = Rb2S + 2H2O or H2S + RbOH = RbHS + H2O AlCl3 + 3RbOH = 3RbCl + Al(OH)3 or AlCl3 + 4RbOH = 3RbCl + Rb 2AlCl3+ 3H2S = Al2S3 + 6HCl Al2S3 + 6H2O = 2Al(OH)3 + 3H2S

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Compilation of possible equations of reactions C2. Aqueous solutions are given: sodium hexahydroxochromate Na3, iron (III) chloride, sulfur dioxide and hydrogen peroxide. Write the equations for the four possible reactions between these substances. SO2 + H2O2 = H2SO4 2FeCl3 + SO2 + 2H2O = 2FeCl2 + H2SO4 + 2HCl 2Na3 +3H2O2 = 2Na2CrO4 + 2NaOH + 8H2O Na3 + FeCl3 = Fe(OH)3 + Cr(OH)3 + 3NaCl

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Compilation of possible reaction equations C2. Aqueous solutions are given: potassium hexahydroxoaluminate K3, chromium (III) chloride, potassium carbonate, and carbonic acid. Write the equations for the four possible reactions between these substances. SOLUTION: K2CO3 + H2CO3 = 2KHCO3 2CrCl3 + 3K2CO3 + 3H2O = 2Cr(OH)3 + 3CO2 + 6KCl K3 + CrCl3 = Cr(OH)3 + Al(OH)3 + 3KCl 2K3 + 3H2CO3 = 3K2CO3 + 2Al(OH)3 + 6H2O

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The relationship of organic substances C3. Write the reaction equations that can be used to carry out the transformations: H2SO4(conc.),t Br2 H2O, Hg2+ H2, cat. Propanol-1 X1 X2 propyne X3 X4 SOLUTION: CH3–CH2–CH2–OH CH3–CH=CH2 + H2O CH3–CH=CH2 + Br2 CH3–CHBr–CH2Br CH3–CHBr–CH2Br + 2KOH 2KBr + 2H2O + CH3–CCH CH3–CCH + HOH CH3–CO–CH3 CH3–CO–CH3 + H2 CH3–CHOH–CH3

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The relationship of organic substances C3. Write the reaction equations with which you can carry out the transformations: KOH (alcohol), tt, Cact CH3Cl, AlCl3 KMnO4, H2SO4 C2H4 C2H4Cl2 X1 C6H6 X2 C6H5COOH SOLUTION: C2H4 + Cl2 C2H4Cl2 C2H4Cl2 + 2KOH 2KCl + 2H2O + C2H2 3C2H2 C6H6 C6H6 + CH3Cl C6H5CH3 + HCl 5C6H5CH3 + 6KMnO4 + 9H2SO4 5C6H5COOH + 3K2SO4 + 6MnSO4 + 14H2O

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The relationship of organic substances C3. Write the reaction equations that can be used to carry out the transformations: H2O, Hg2+ KMnO4, H2SO4 Cl2, P ex. NH3 C2H2 X1 CH3COOH X2 X3 N2 SOLUTION: С2H2 + HOH CH3COH 5CH3COH + 2KMnO4 + 3H2SO4 5CH3COOH + K2SO4 + 2MnSO4 + 3H2O CH3COOH + Cl2 CH2ClCOOH + HCl CH2ClCOOH + 2NH3 CH2NH2COOH + NH4OH + H2NH2COOH + HNO2 CH + HNO2

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Calculations according to the equations of reactions С4. For the oxidation of a certain amount of sulfur, such a volume of oxygen (n.o.) was required, which is formed during the decomposition of 215 g of potassium chlorate containing 5% oxygen-free impurities. Determine the mass of sulfur that reacted and the volume of the resulting gaseous oxidation product. SOLUTION: 2KClO3 = 2KCl + 3O2 m(KClO3) = 215 ∙ 0.95 = 204.24 g; n(KClO3) = 1.667 mol According to the reaction equation, we calculate the amount of oxygen substance. n(O2) = 2.5 mol S + O2 = SO2 According to the reaction equation n(S) = n(SO2) = n(O2) = 2.5 mol m(S) = 2.5 mol ∙ 32 g/mol \u003d 8 g V (SO2) \u003d 2.5 mol ∙ 22.4 l / mol \u003d 56 l ANSWER: 56 l SO2; 8g S.

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Calculations by reaction equations C4. Examples of tasks: a) For the oxidation of a certain amount of ammonia, such a volume of oxygen (n.o.) was required, which is formed during the decomposition of 245.6 g of potassium permanganate, containing 3.5% of an oxygen-free impurity. Determine the mass of ammonia that reacted and the volume of the resulting nitrogen-containing oxidation product (n.o.). b) For the oxidation of a certain amount of sulfur, such a volume of oxygen (n.o.) was required, which is formed during the decomposition of 330.9 g of potassium permanganate, containing 4.5% of an oxygen-free impurity. Determine the mass of sulfur that reacted and the volume of the resulting oxidation product.

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Calculations on the topic "Solutions" С4. 100 ml of 15% potassium hydroxide solution (density 1.10 g/ml) and 150 ml of 10% hydrochloric acid solution (density 1.05 g/ml) were mixed. Determine the environment of the resulting solution and the mass fraction of potassium chloride in it. SOLUTION: KOH + HCl = KCl + H2O Let's find the mass of each solution and the mass of the substance. mr-ra (KOH) \u003d 100 ∙ 1.1 \u003d 110 g; m(KOH) = 110 ∙ 0.15 = 16.5 g mp-ra (HCl) = 150 ∙ 1.05 = 157.5 g; m(HCl) = 157.5 ∙ 0.1 = 15.75 g Let's find the amount of substance for each compound. n(KOH) = 0.294 mol; n(HCl) = 0.43 mol; n(KCl) = 0.294 mol Acid medium Find the mass of salt m(KCl) = 21.9 g Calculate the mass of the total solution: m(p-ra) = 110 + 157.5 = 267.5 g Determine the mass fraction of potassium chloride in the solution :=21.9/267.5=0.08

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Tasks for the derivation of the formula of the substance C5. The combustion of gaseous organic matter that does not contain oxygen released 2.24 L (n.o.s.) carbon dioxide, 1.8 g of water and 3.65 g of hydrogen chloride. Set the molecular formula of the burned substance. Solution: Set the amount of substance carbon, hydrogen and chlorine. 1) 22.4 l CO2 - 1 moln (C) \u003d 0, 1 mol 2.24 l CO2 - x mol 2) 36.5 g HCl - 35.5 g Clz (Cl) \u003d 3.55 g Cl; n (Cl) \u003d 0.1 mol 3.65 g HCl - z g Cl 3) 36.5 g HCl - 1 g H y (H) \u003d 0.1 g; n(H) \u003d 0.1 mol 3.65 g HCl - y g H 18 g H2O - 2 g H y1 \u003d 0.2 g; n1(H)= 0.2 mol 1.8 g H2O – y1 g H The total amount of hydrogen substance is 0.3 mol. x:y:z = 0.1: 0.3: 0.1. Formula CH3Cl.

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Calculations on the topic "Solutions" C5. Sample tasks: a) Set the formula of an inorganic compound containing 20% magnesium, 53.33% oxygen and 26.67% of some element. b) With the complete combustion of gaseous organic matter that does not contain oxygen, 4.48 l (n.o.) of carbon dioxide, 1.8 g of water and 4 g of hydrogen fluoride were released. Set the formula for the burnt compound. c) The thermal decomposition of an unknown substance weighing 49 g released 13.44 l (n.o.) of oxygen and left a solid containing 52.35% potassium and 47.65% chlorine. Determine the formula of the substance.

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MBOU "Pochaevskaya secondary school" Master class on solving problems of Part II in preparing students for the exam in chemistry. Prepared by: chemistry teacher Kasilova Elena Ivanovna 2012

Work plan The role of elective courses in preparing for the exam Redox reactions in tasks of level C Tasks Part - 2 Chains of transformations in organic chemistry Solving problems by chemical equations Solving problems for finding formulas of substances

The program of the elective course "redox reactions" HNO 3 H 2 O 2 H 2 SO 4

"redox reactions in organic chemistry"

Scheme for KMnO 4 H 2 O

Using the electron balance method, write an equation for the reaction. Specify the oxidizing agent and reducing agent: As 2 O 3 + ... + KMnO 4 + H 2 O \u003d H 3 AsO 4 + K 2 SO 4 + ... Set the oxidation states: As +3 2 O 3 + ... + KMn +7 O 4 + H 2 O \u003d H 3 As +5 O 4 + K 2 SO 4 + ... Compose the electronic balance and put down the main coefficients in the reaction equation: Mn +7 + 5 e - Mn +2 | 2 oxidizing agent 2 As +3 - 4 e - 2 As 2 +5 |5 reducing agent 5As 2 O 3 + H 2 SO 4 + 4KMnO 4 + H 2 O \u003d 10H 3 AsO 4 + 2K 2 SO 4 + 4MnSO 4 5As 2 O 3 (reducing agent) + 6 H 2 SO 4 + 4KMnO 4 (oxidizer) + 9 H 2 O \u003d 10H 3 AsO 4 + 2K 2 SO 4 + 4MnSO 4.

Using the electron balance method, write an equation for the reaction. Indicate the oxidizing agent and reducing agent: K 2 S + H 2 O + KBrO 4  S + KBr + ... 1. Determine the oxidation states of the elements to understand whether all the key substances are indicated in the scheme: K 2 S - 2 + H 2 O + KBr +7 O 4  S 0 + KBr - 1 + ... The oxidation state was changed by sulfur and bromine. So, all the key substances are indicated. 2. Compile the electronic balance and begin to equalize the scheme: S - 2 - 2 e  S 0 4 Br +7 +8 e  Br - 1 1 4K 2 S - 2 + H 2 O +1 KBr +7 O 4  4S 0 +1 KBr - 1 + ... 3. As you can see, there are 8 "extra" potassium atoms on the left side, therefore, KOH 4K 2 S - 2 + H 2 O +1 KBr +7 O 4  4S 0 + 1 KBr - 1 + 8 KOH 4. Now we will equalize hydrogen: 4K 2 S - 2 + 4H 2 O +1 KBr +7 O 4  4S 0 +1 KBr - 1 + 8 KOH 5. Let's check for oxygen. Oxygen left and right 8 each. The final solution looks like this: 4K 2 S - 2 + 4H 2 O + KBr +7 O 4  4S 0 + KBr - 1 + 8 KOH S - 2 - 2 e  S 0 4 Br +7 +8 e  Br - 1 1 K 2 S - reducing agent due to S - 2 KBr +7 O 4 - oxidizing agent due to Br +7

Using the electron balance method, write an equation for the reaction. Indicate the oxidizing agent and reducing agent: KNO 2 + ... + H 2 O  MnO 2 + KOH + ... 1. Let's determine the oxidation states of the elements in order to understand whether all the key substances are indicated in the scheme: KN + 3 O 2 + ... + H 2 O  Mn +4 O 2 + KOH + ... Potassium, hydrogen and oxygen did not change their oxidation state, there is no manganese on the left side of the diagram, and nitrogen on the right side. These elements changed their oxidation state. 2. Let's determine the role of manganese in this reaction: most likely, manganese in this reaction acts as an oxidizing agent, in a neutral environment, manganese compounds +7 and +6, acting as oxidizing agents, turn into MnO 2. Let us choose KMnO 4 as the oxidizing agent. Then nitrite will be a reducing agent, oxidized to nitrate. 3. Let's write down the reaction scheme with the selected substances and draw up an electronic balance: KN + 3 O 2 + KMn + 7 O 4 + H 2 O  Mn + 4 O 2 + KOH + KN + 5 O 3 N + 3 - 2 e  N +5 3 Mn +7 + 3 e  Mn +4 2 4. Equalize the scheme: 3 KN +3 O 2 + 2 KMn +7 O 4 + H 2 O  2 Mn +4 O 2 + 2 KOH + 3 KN + 5 O 3 KNO 2 - reducing agent due to N + 3 KMnO 4 - oxidizing agent due to Mn +7

Using the electron balance method, write an equation for the reaction. Specify the oxidizing agent and reducing agent: Na 2 O 2 +...+ KMnO 4 = O 2 + MnSO 4 + Na 2 SO 4 + K 2 SO 4 +… Set the oxidation state: Na 2 O 2 -1 +...+ KMn +7 O 4 \u003d O 2 0 + Mn +2 SO 4 + Na 2 SO 4 + K 2 SO 4 + ... Na 2 O 2 (reducing agent) + H 2 SO 4 + KMnO 4 (oxidizing agent) \u003d O 2 + MnSO 4 + Na 2 SO 4 + K 2 SO 4 + H 2 O. We compose the electronic balance and put down the main coefficients in the reaction equation: Mn + 7 + 5 e - Mn + 2 | 2 oxidizing agent 2 O -1 - 2 e - O 2 0 | 5 reducing agent 5Na 2 O 2 + H 2 SO 4 + 2KMnO 4 \u003d 5O 2 + 2MnSO 4 + 5Na 2 SO 4 + K 2 SO 4 + H 2 O. 5Na 2 O 2 (reductant) + 8H 2 SO 4 + 2KMnO 4 (oxidizing agent) \u003d 5O 2 + 2MnSO 4 + 5Na 2 SO 4 + K 2 SO 4 + 8H 2 O.

Using the electron balance method, write an equation for the reaction. Specify the oxidizing agent and reducing agent: KI +…+ KMnO 4 = I 2 …+ K 2 SO 4 + H 2 O. Set the oxidation state: KI -1 +…+ KMn +7 O 4 = I 2 0 …+ K 2 SO 4 + H 2 O. KI (reductant) + H 2 SO 4 + KMnO 4 (oxidizing agent) \u003d I 2 + MnSO 4 + K 2 SO 4 + H 2 O. Compose the electronic balance and put down the main coefficients in the reaction equation: Mn +7 + 5e - Mn +2 | 2 oxidizing agent 2 I -1 - 2e - I 2 0 | 5 reducing agent 10 KI (reducing agent) + 8 H 2 SO 4 + 2KMnO 4 (oxidizing agent) \u003d 5 I 2 + 2MnSO 4 + 6K 2 SO 4+8H2O.

Given: chromium(III) chloride, potash, hydrobromic acid. lithium hydroxide. Write an equation for four possible reactions between these substances. 2CrCl 3 + 3K 2 CO 3 + 3H 2 O \u003d 2Cr (OH) 3 + 3CO 2 + 6KCl. CrCl 3 + 3LiOH = Cr(OH) 3 + 3LiCl K 2 CO 3 + 2HBr = 2KBr + CO 2 + H 2 O HBr + LiOH = LiBr + H 2 O

Given: Potassium tetrahydroxoaluminate, iron (III) chloride, chlorine, potassium carbonate 1) 3K+ FeCl 3 = Fe(OH) 3 +3Al(OH) 3 +3KCl (properties of complex salts) 2) K [ Al (OH) 4 ] = KAlO 2 +2 H 2 O (properties of complex salts) 3) 2FeCl 3 + 3K 2 CO 3 +3 H 2 O \u003d \u003d 6KCl + 2Fe (OH) 3 + 3CO 2 (hydrolysis) 4) K 2 CO 3 + Cl 2 + H 2 O \u003d KClO + 2KHCO 3 + KCl (hydrolysis)

Given: magnesium bromide, sodium carbonate, hydrochloric acid, lithium hydroxide. Write an equation for four possible reactions between these substances. MgBr 2 + Na 2 C O 3 \u003d MgCO 3 + 2NaBr. MgBr 2 + 2LiOH + Mg(OH) 2 + 2 LiBr Na 2 CO 3 +2 HCl = 2NaCl + H 2 O + CO 2 Hl + LiOH = LiCl + H 2 O

AlCl3 1) C 6 H 6 + C 2 H 5 Cl → C 6 H 5 - CH 2 CH 3 + HCl alkylation of benzene, obtaining benzene homologues h ν 2) C 6 H 5 CH 2 CH 3 + Br 2 → C 6 H 5 CH - CH 3 + HBr radical substitution reaction | Br C -3 C 2 H 5 Cl, AlCl 3 Br 2, hy KOH + H 2 O H 2 SO 4, t 0 KMnO 4, H + Benzene X 1 X 2 X 3 X 4 X5

3) C 6 H 5 CHBrCH 3 + KOH → C 6 H 5 CHOH-CH 3 + KBr obtaining alcohols H2SO4 (conc), t ° 4) C 6 H 5 CHOHCH 3 → C 6 H 5 CH \u003d CH 2 + H 2 O dehydration of alcohols in the presence of concentrated sulfuric acid. 5) C 6 H 5 CH \u003d CH 2 + 2KMnO 4 + 3H 2 SO 4 → C 6 H 5 COOH + CO 2 + K 2 SO 4 + 2 MnSO 4 + 4 H 2 O oxidation of vinylbenzene in the presence of potassium permanganate in an acidic medium , redox reaction.

electrolyzes CH 3 COONa + 2H 2 O → C 2 H 6 + 2NaHCO 3 + H 2 production of alkanes by electrolysis of salts of carboxylic acids light 2) C 2 H 6 + Cl 2 → C 2 H 5 Cl + HCl Chemical properties alkanes, substitution reaction A sodium cetate → ethane → X → ethanol → diethyl ether ↓ CO 2

H 2 O 3) C 2 H 5 Cl + NaOH → C 2 H 5 OH + NaCl obtaining alcohols from halogen derivatives H 2 SO 4k, t

Formulas for solving problems M (CxHy) = D (H 2) M (H 2)

Solution. Calculate the mass of impurities and the mass of magnesium carbonate. m (approx.) \u003d 0.08 * 400 \u003d 32g; m (MgCO 3) = 400 - 32 = 368g; Let's calculate the amount of substance MgCO 3 ; n (MgCO 3) \u003d 368/84 \u003d 4, 38 mol. We compose the chemical reaction equation: MgCO 3 + 2 HNO 3 \u003d Mg (NO 3) 2 + CO 2 + H 2 O. It follows from the reaction equation that n (CO 2) \u003d n (MgCO 3). Determine V (CO 2): V (CO 2) \u003d 4.38 * 22, 4 \u003d 98.1 l. After rounding - 98 l. Answer 98 l. CO 2 A sample of magnesium carbonate weighing 400 g, containing in its composition 8% sulfate impurities, was treated with an excess of nitric acid. What volume of gas will be released? (Round your answer to the nearest whole number.)

General formulas of organic compounds Hydrocarbons Alkanes - C n H 2n +2 Alkenes - C n H 2n Cycloalkanes - C n H 2n Alkynes - C n H 2n-2 Alkadienes - C n H 2n-2 Arenes - C n H 2n-6

Oxygen-containing organic substances Alcohols - CnH 2 n + 1OH, CnH 2 n + 2O Ethers ROR (R \u003d CnH 2 n + 1) Carboxylic acids R-COOH Esters - RC OOR 1 Phenols - C n H (2n-6) - m (OH) m Aldehydes - C n H 2n + 1 - C (O) H Ketones - R - C - R 1 R, R 1 \u003d C n H 2n + 1 ‖ O Carbohydrates - C m (H 2 O) n Amines - R-NH 2, R 1 -NH-R 2, R 1 - N - R 2 | R 3 Amino acids- C n H 2n (NH 2) COOH

Problem 10 g of copper oxide (II) went to the oxidation of saturated monohydric alcohol. As a result of the reaction, aldehyde weighing 10.75 g, copper and water were obtained. Determine the molecular formula of the original alcohol.

Task (solution) Response elements: The general equation of the reaction is given: С n Н 2 n +2 О + С u О -> С n Н 2 n О + Сu + Н 2 О CuO) \u003d n (C n H 2n O) 10 \u003d 10.75, from where n \u003d 5 80 14 n +16 The alcohol formula is determined: C 5 H 11 OH

During the combustion of the amine, 0.448 l (n.o.) of carbon dioxide, 0.495 g of water and 0.056 l of nitrogen were released. Set the molecular formula of this amine. Task

Solution Response elements: the numbers of moles of carbon, hydrogen and nitrogen atoms in a sample of amine were calculated: n (C) \u003d n (CO 2) \u003d 0.448 / 22.4 \u003d 0.02 mol n (H) \u003d 2 n (H 2 O) = 2 . 0.495/18 \u003d 0.055 mol n (N) \u003d 2 n (N 2) \u003d 2. 0.056 / 22.4 \u003d 0.005 mol 2) the ratio of atoms C: H: N \u003d 0.02: 0.055: 0.005 \u003d 4: 11: 1 was established, the true molecular formula of the amine was determined: C 4 H 11 N

STAGES OF PREPARING STUDENTS FOR THE USE IN CHEMISTRY s / pp / s The content of the stage period 1 Formation: interest in the subject and motivation for its study; solid basic knowledge; the ability to work independently with literature, to systematically engage in problem solving; ability to work with tests various types. 8-9 grade

STAGES OF PREPARING STUDENTS FOR THE USE IN CHEMISTRY s / pp / n Content of the stage period 2 Profile orientation of students Determination of the range of subjects that require preparation for the USE Formation of a group of students who need preparation for the USE in chemistry grades 9-10

STAGES OF PREPARING STUDENTS FOR THE USE IN CHEMISTRY s / pp / s Content of stage period 3 In-depth preparation of a group of students in chemistry: a differentiated approach in the classroom elective courses individual consultations class

STAGES OF PREPARING STUDENTS FOR THE USE IN CHEMISTRY s / pp / n Content of the stage period 4 Direct preparation for the exam: familiarity with the structure of KIM in chemistry, regulatory documents; selection of particularly complex topics, selection of tasks of different levels of complexity on these topics; study of material on topics, while: - repetition of the theory; - independent work with USE assignments related to this topic; - analysis of all misunderstood and unresolved. Grade 11

WORKING WITH PARENTS The purpose of the work: awareness by a certain part of them of the fact that in order to achieve their life goals, their children need in-depth training in the natural sciences. federal standard the basic level does not provide for the purpose of preparing students for entering a university in this area. The student choosing this exam must plan for a large amount of independent work on the subject. To draw the attention of parents to the role of their help in the preparation of their children. The teacher should also talk about the help that he is able to provide to students in their work (materials, consultations).

DIFFERENTIATED APPROACH IN THE LESSONS Preparing a small group of students for the USE. The teacher should prepare assignments high level, tasks in the format of the exam. The teacher offers additional free homework. Involvement of these students as teacher assistants, tutors in the organization of mutual learning and mutual control in the educational process.

INDEPENDENT WORK OF STUDENTS Part of the tests for independent work can be given to students with ready-made answers. Performing such tests, the student checks his answers with the "key", notes the mistakes made. Then he must analyze them. A special icon marks mistakes made due to inattention, a special one - those that were corrected with the help of the manual, a special one - those that the student could not understand. The student shows the results of this analysis to the teacher at consultation interviews or at any convenient time. There may be control tests that are checked directly at the consultations.

METHODOLOGY FOR USING TESKS IN THE LESSONS For successful work on the USE and other types of control in a similar format, students must know the main types of test tasks, navigate their structure, and understand in what form they need to give their answer. In a test, especially a short one, there is always the possibility of getting an undeserved mark by guessing. When using them, oral and written speech student.

METHODOLOGY FOR USING TESKS IN THE LESSONS Reception of work with tests - "Answer with comments": The student, orally or in writing, completing the test task, not only indicates the correct answer, but also comments on it, gives the motivation for his choice. Motivation can be a verbal explanation based on the properties of a certain class of substances, the formulation of a definition, a rule, a law, the formulation of a reaction equation, the solution of a calculation problem ...

FEATURES OF PREPARING STRONG STUDENTS The purpose of preparing for the USE for strong students is to master the skills to perform the most difficult tasks of part C. However, they must regularly train themselves on completing tasks of parts A and B. At the same time, they are tasked with: a) to minimize the mistakes made by carelessness; b) identify those topics of the chemistry course where mistakes are made due to ignorance, study these sections, work out the acquired knowledge on thematic tests.

PREPARATION OF WEAK AND AVERAGE PUPILS It is focused on the successful passing of the Unified State Examination at least for a satisfactory mark. For this, it is enough to confidently complete part A of the test materials (30 primary points that can be scored in part A approximately correspond to 50 points on the USE scale). Preparation should be carried out according to the most characteristic properties of substances, the most common types of reactions, the simplest and most often used algorithms for solving problems.

EXAM STRATEGY Rules of the first circle: Mark what is right and skip the difficult. (45 min.) Rules of the second circle: Check what is done, discard the wrong, use the “three cheat sheets” and the laws of chemistry. (95 min.) Rule of the third circle: Check what is done and guess what is not found .

Positive experience in preparing for the Unified State Examination and the OGE in chemistry. Prepared by: chemistry teacher MBOU secondary school No. 33 I.E. Karabak

1 step Preparatory work.

Using tests from grade 8.

Development of final examinations in the format of the OGE, Unified State Examination Development of elective courses aimed at preparing for the final certification in a new form.