1 of 20

Presentation on the topic:

slide number 1

Description of the slide:

slide number 2

Description of the slide:

The structure of part C of KIMs in chemistry in 2006 C1. Redox reactions C2. Drafting possible equations C3 reactions. Chains of transformations into organic chemistry C4. Calculation tasks on the topics: "Solutions", "Calculations according to the reaction equation" С5. The task of deriving the formula of a substance

slide number 3

Description of the slide:

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

slide number 4

Description of the slide:

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

slide number 5

Description of the slide:

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. Acts as a medium sulfuric acid. Electron transfer: Al0 - 3e- = Al+3 2 reducing agent 2Cr+6 + 6e- = 2Cr+3 1 oxidizing agent

slide number 6

Description of the slide:

slide number 7

Description of the slide:

Compilation of possible equations of reactions C2. Are given aqueous solutions: sodium sulfide, hydrogen sulfide, aluminum chloride and chlorine. Write the equations for the 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

slide number 8

Description of the slide:

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

slide number 9

Description of the slide:

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

slide number 10

Description of the slide:

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

slide number 11

Description of the slide:

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

slide number 12

Description of the slide:

The relationship of organic substances C3. Write the reaction equations with which you can carry out the transformations: KOH (alcohol), t t, Cact CH3Cl, AlCl3 KMnO4, H2SO4 C2H4 C2H4Cl2 X1 C6H6 X2 C6H5COOH SOLUTION: C2H4 + Cl2 C2H4Cl2 C2H4Cl2 + 2KOH 2KCl + 2H2O + C2H2 3C2H2 C6H 6 C6H6+ CH3Cl C6H5CH3 + HCl 5C6H5CH3 + 6KMnO4 + 9H2SO4 5C6H5COOH + 3K2SO4 + 6MnSO4 + 14H2O

slide number 13

Description of the slide:

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 + NH4Cl CH2NH2CO OH + HNO2 N2 + CH2OH–COOH + H2O

slide number 14

Description of the slide:

slide number 15

Description of the slide:

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.

slide number 16

Description of the slide:

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.

slide number 17

Description of the slide:

Calculations on the topic "Solutions" С4. Mixed 100 ml of 15% potassium hydroxide solution (density 1.10 g/ml) and 150 ml of 10% solution of hydrochloric acid(density 1.05 g/ml). 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

slide number 18

Description of the slide:

slide number 19

Description of the slide:

Tasks for the derivation of the formula of the substance C5. When burning gaseous organic matter, containing no oxygen, 2.24 l (n.o.) of carbon dioxide, 1.8 g of water and 3.65 g of hydrogen chloride were released. Install molecular formula 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.

slide number 20

Description of the slide:

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) During the thermal decomposition of an unknown substance weighing 49 g, 13.44 liters (N.O.) of oxygen were released and remained solid containing 52.35% potassium and 47.65% chlorine. Determine the formula of the substance.

slide 1

slide 2

slide 3

slide 4

slide 5

slide 6

Slide 7

Slide 8

Slide 9

slide 10

slide 11

slide 12

slide 13

slide 14

slide 15

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

slide 2

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)

slide 3

propaedeutic stage

The world(grades 1-4) Natural History (grade 5) Natural Science (grades 5, 6, 7) Physics and Chemistry (grades 5-6) Introduction to Chemistry (grade 7)

slide 4

book to read

Designed for children 6-12 years old

slide 5

Main ideas of the book

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

slide 6

Main school

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

Slide 7

Topics of tests included in the manual

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

Slide 8

State final certification of graduates in a new form in chemistry

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

Slide 9

High school

USE. Universal materials for preparing students The 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 version of the exam

slide 11

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 complexity Reflect the logic of reasoning when performing tasks of various types and levels of complexity Allow to build individual trajectory when working out the material in preparation for control measures, including the GIA and the Unified State Examination

View all slides

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 of 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 consultation class

STAGES OF PREPARING STUDENTS FOR THE USE IN CHEMISTRY s / pp / s Content of the stage period 4 Direct preparation for the exam: acquaintance with the structure of KIM in chemistry, regulatory documents; highlighting difficult topics, selection of tasks different levels difficulties 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 the life goals of their children, they need in-depth training in the natural sciences. federal standard 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 time. convenient time. There may be control tests, which 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 test items, navigate in their structure, understand in what form you need to give your 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 language student.

METHODOLOGY FOR USING TESKS IN THE LESSONS Receiving work with tests - “Answer with comments”: The student, performing the test task orally or in writing, 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 goal of preparing for the USE for strong students is to master the skills to complete 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.

TRAINING FOR WEAK AND INTERMEDIATE STUDENTS Focused on successful passing the exam at least to a satisfactory level. To do this, it is enough to confidently complete part A of the test materials (30 primary points that can be scored in part A correspond approximately to 50 points in USE scale). Training 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 .

Bratyakova S.B.

Preparation for the exam in chemistry Tasks part 1, part 2

Bratyakova S.B.

Goals:

• Get to know the material examination work in chemistry.
• Check your preparation for the exam.
• Expand the experience of performing test tasks in chemistry

The presentation is presented in two parts:

• To select the correct answer to the question of part 1, move the cursor over the selected answer and click on the left mouse button, if the answer is correct, the word “True” will appear, if the answer is incorrect, “False”

Answers and solutions for tasks of part 2 will appear by clicking

Mendeleev table

Solubility table

Bratyakova S.B.

Are the following judgments about the synthesis of methanol and ammonia in industry correct?

A. Reactions for the synthesis of methanol and ammonia are exothermic. B. In the production of methanol and ammonia, a circulating

1) both statements are correct

Right

2) both judgments are wrong

Wrong

3) only B is correct

Wrong

4) only A is correct

Wrong

Bratyakova S.B.

What mass of calcium oxide should be taken to prepare 200 g of calcium hydroxide solution with a mass fraction of 3.7%?

1) 2.8 g

Wrong

2) 4.2 g

Wrong

Right

3) 5.6 g

4) 8.4 g

Wrong

Bratyakova S.B.

As a result of the reaction, the thermochemical equation of which

2C 2 H 2 +5O 2 = 4CO 2 + 2H 2 O + 2610 kJ.

652.5 kJ of heat was released.

The volume of burned acetylene is

1) 33.6 l

Wrong

2) 22.4 l

Wrong

Right

3) 11.2 l

4) 5.6 l

Wrong

Bratyakova S.B.

How many liters of oxygen will be required for the catalytic oxidation of 100 liters of sulfur oxide (IV) to sulfur oxide (VI)? The volumes of gases are measured under the same conditions

Right

Wrong

Wrong

Wrong

Establish a correspondence between the name of the substance and the class (group) of inorganic compounds to which it belongs

Bratyakova S.B.

name of the substance

connection classes

• A) CH 3 OCH 3
• B) HOCH 2 CH 2 CH 3
• B) HCOOCH 2 CH 3
• D) CH 3 C(OH) 2 CH 3

• 1) alkene
• 2) monohydric alcohol
• 3) polyhydric alcohol
• 4) simple ether
• 5) ester

Wrong

Right

Wrong

Wrong

Establish a correspondence between the name of the substance and the oxidation state of carbon

Bratyakova S.B.

substances oxidation state C

A) methanol 1) –4

B) methanoic acid 2) -3

B) methane 3) –2

D) ethane 4) 0

Wrong

Wrong

Wrong

Right

Establish a correspondence between the substance and the product on the anode during the electrolysis of its solution

Bratyakova S.B.

salt formula product at the anode

A) Na 2 CO 3 1) H 2 S

B) Na 2 S 2) Cu

B) NaNO 3 3) O 2

D) CuCl 2 4)Cl 2

6) CO 2

Wrong

Right

Wrong

Wrong

Establish a correspondence between the formula of salt and its ability to hydrolyze.

Bratyakova S.B.

relation to hydrolysis

1) by cation

salt formula

1) Al 2 S 3

2) by anion

3) by cation and anion

2) K 2 SO 3

4) does not undergo hydrolysis

3)CrCl 3

4) K 2 SO 4

Wrong

Right

Wrong

Wrong

Establish a correspondence between the reactant and the reaction product

Bratyakova S.B.

Products

1) CaCO 3

2) CaCO 3 + H 2 ABOUT

3) Ca(HCO 3 ) 2

4) Ca(HCO 3 ) 2 + H 2 ABOUT

5) CO + H 2

6) H 2 CO 3

Reagents

A) CO 2 + H 2 ABOUT

B) CO 2 + CaO

B) CO 2 + Ca(OH) 2

D) CO 2 (excess) + Ca(OH) 2

Wrong

Wrong

Wrong

Right

Bratyakova S.B.

Establish a correspondence between two substances and a reagent with which you can distinguish between these substances:

Substances Reactive

A) butanone-2 and ethanol 1) HCl (solution)

B) hexene-1 and ethyl formate 2) FeCl 3 (water)

C) aniline and phenol 3) Br 2 (water)

D) benzene and aniline 4) Na

5) HNO 3

Wrong

Right

Wrong

Wrong

For both benzene and styrene,

Bratyakova S.B.

1) the presence in the molecule of a conjugated electronic system

2) sp 3 -hybridization of carbon atoms

3) interaction with hydrogen in the presence of a catalyst

4) discoloration of bromine water

5) combustibility

6) good solubility in water

Wrong

Wrong

Right

Wrong

methanol reacts with

Bratyakova S.B.

• CH 3 COOH
• NaOH
• Ag 2 O(NH 3 )
• Br 2 (H 2 O)
• KMnO 4 (H+)

Wrong

Right

Wrong

Wrong

Both aniline and dimethylamine react with

Bratyakova S.B.

1 ) water

2) bromoethane

3) sulfuric acid

4) hydrogen bromide

5) sodium hydroxide

6) potassium permanganate solution

Wrong

Wrong

Right

Wrong

Bratyakova S.B.

30. Using the electronic balance method, write the reaction equation:

NaNO 2 +NaI+...=NO+...+Na 2 SO 4 +...

Determine the oxidizing agent and reducing agent.

Answer: 2NaNO 2 +2NaI+2H 2 SO 4 =2NO+I 2 +2Na 2 SO 4 +2H 2 O

N +3 + e - -N +2 oxidizer

2I - + 2e - -I 2 reducing agent

31. Copper was dissolved in concentrated sulfuric acid. The resulting salt was treated with an excess of sodium hydroxide solution.

The resulting blue precipitate was filtered off and calcined. Then, the resulting substance was calcined with coal.

Write the equations for the four described reactions .

Bratyakova S.B.

Answer: 1) Сu+2H 2 SO 4 conc. = CuSO 4 + SO 2 + 2H 2 O

2) CuSO 4 + 2NaOH = Na 2 SO 4 + Cu(OH) 2

3) Cu(OH) 2 = CuO + H 2 O

4) 2CuO + C = 2Cu + CO 2

32. Write the reaction equations that can be used to carry out the following transformations .

Bratyakova S.B.

(+H 2 O) 1500 0 With Sakt. t 0 (+Br 2 ), AlBr 3 (+NH 3 ), t 0 , p

Al 4 C 3 → X 1 → X 2 → X 3 → X 4 → X 5

Al 4 C 3 + 12H 2 O → 3CH 4 + 4Al(OH) 3

2CH 4 → C 2 H 2 + 3H 2

3C 2 H 2 →C 6 H 6

C 6 H 6 +Br 2 → C 6 H 5 Br+HBr

C 6 H 5 Br+3NH 3 → C 6 H 2 Br(NH 2 ) 3 + 3H 2

33. Hydrogen sulfide, released during the interaction of an excess of concentrated sulfuric acid with 1.44 g of magnesium, was passed through 160 g of a 1.5% bromine solution. Determine the mass of the precipitate

Bratyakova S.B.

1) 5H 2 SO 4 conc. + 4Mg = 4MgSO 4 + H 2 S+4H 2 O

Br 2 + H 2 S=2HBr+S

2) n(Mg) = 1.44/ 24 = 0.06 (mol)

n(H 2 S)=0.015 (mol)

n(Br 2 ) = 160⋅0.015/160=0.015 (mol) (both substances react completely)

3)n(H 2 S)=n(Br 2 )=n(S)=0.015 (mol)

m(S)=32g/mol 0.015 mol=0.48g

Answer: sediment weight = 0.48 g.

34. During the interaction of 11.6 g of saturated aldehyde with an excess of copper (II) hydroxide, a precipitate of 28.8 g was formed during heating. Derive the molecular formula of the aldehyde. write down structural formula and reaction equation

Bratyakova S.B.

1) C n H 2n O + Cu(OH) 2 = C n H 2n+1 O 2 + H 2 O + Cu 2 O

2) n(Cu 2 O) \u003d 28.8 / 144 \u003d 0.2 (mol)

n(Cu 2 O) = n(C n H 2n O) = 0.2 mol

M(S n H 2n O) \u003d 11.6 / 0.2 \u003d 58 g / mol

3)M(S n H 2n O) = 12n + 2n + 16 = 58

14n + 16 = 58 , 14n = 42 , n = 3

Molecular formula - C 3 H 6 O - propanal

Bratyakova S.B.

Bratyakova S.B.

Bratyakova S.B.

Literature:

1 . open bank USE assignments http://www.fipi.ru/os11/xmodules/qprint/afrms.php?proj =

2. http://chem.reshuege.ru/test?theme=20

3. http://www.otbet.ru/ege/demo-varianty-ege/demo-himiya/

4. O.S. Gabrielyan. Chemistry grade 10. Bustard. M. 2012

5. O.S. Gabrielyan. Chemistry grade 11. Bustard. M. 2012

6. Template Alekseeva L.A.