Genetic connection A genetic series is a chain that unites substances of different classes, which are compounds of one element and are connected by interconversions. Genetic connection of metals, non-metals and their compounds

Topic: GENETIC RELATION BETWEEN metals and non-metals and their compounds. 9th grade.

Objectives: educational: to consolidate the concepts of "genetic series", "genetic connection"; to teach how to compose the genetic series of elements (metals and non-metals), to compose equations of reactions corresponding to the genetic series; check how knowledge is learned about chemical properties oxides, acids, salts, bases; developing: develop the ability to analyze, compare, generalize and draw conclusions, draw equations chemical reactions; educational: to promote the formation of a scientific worldview.

Ensuring the lesson: tables " Periodic system”, “Table of solubility”, “Series of activity of metals”, instructions for students, tasks for testing knowledge.

Work progress: 1) Org. moment

2) Checking d / z

3) Learning new material

4) Fixing

5) D/Z

1) Org. moment. Greetings.

2) Checking d / z.

Genetic links are links between different classes based on their mutual transformations.
Knowing the classes of inorganic substances, it is possible to compose the genetic series of metals and non-metals. These rows are based on the same element.

Among metals, two types of series can be distinguished:

1 . A genetic series in which alkali acts as a base. This series can be represented using the following transformations:

metal→basic oxide→alkali→salt

For example, K→K 2 O→KOH→KCl

2 . A genetic series, where an insoluble base acts as a base, then the series can be represented by a chain of transformations:

metal→basic oxide→salt→insoluble base→

→basic oxide→metal

For example, Cu→CuO→CuCl 2 →Cu(OH) 2 →CuO→Cu

1 . The genetic series of non-metals, where a soluble acid acts as a link in the series. The chain of transformations can be represented as follows:

nonmetal→acidic oxide→soluble acid→salt

For example, P→P 2 O 5 →H 3 PO 4 →Na 3 PO 4

2 . The genetic series of non-metals, where an insoluble acid acts as a link in the series:

nonmetal→acidic oxide→salt→acid→

→acid oxide→non-metal

For example,Si→ SiO 2 → Na 2 SiO 3 → H 2 SiO 3 → SiO 2 → Si

Frontal discussion on:

What is a genetic link? Genetic links are links between different classes based on their mutual transformations. What is a genetic series?

Genetic series - a number of substances - representatives of different classes, which are compounds of the same chemical element, connected by mutual transformations and reflecting the transformations of these substances. These rows are based on the same element.

What types of genetic series are usually distinguished? Among metals, two types of series can be distinguished:

a) A genetic series in which alkali acts as a base. This series can be represented using the following transformations:

metal → basic oxide → alkali → salt

for example, the potassium genetic series K → K 2 O → KOH → KCl

b) A genetic series, where an insoluble base acts as a base, then the series can be represented by a chain of transformations:

metal → basic oxide → salt → insoluble base → basic oxide → metal

e.g.: Cu → CuO → CuCl 2 → Cu(OH) 2 → CuO → Cu

Among non-metals, two types of series can also be distinguished:

a) The genetic series of non-metals, where a soluble acid acts as a link in the series. The chain of transformations can be represented as follows: non-metal → acid oxide → soluble acid → salt.

For example: P → P 2 O 5 → H 3 PO 4 →Na 3 PO 4

b) The genetic series of non-metals, where an insoluble acid acts as a link in the series: non-metal → acid oxide → salt → acid → acid oxide → non-metal

For example: Si → SiO 2 → Na 2 SiO 3 → H 2 SiO 3 → SiO 2 → Si

Completing tasks by options:

1. Choose the formulas of oxides in your version, explain your choice, based on knowledge of the characteristics of the composition this class connections. Name them.

2. In the column of formulas of your option, find the formulas of acids and explain your choice based on the analysis of the composition of these compounds.

3. Determine the valency of acid residues in the composition of acids.

4. Choose salt formulas and name them.

5. Make formulas of salts that can be formed by magnesium and acids of your choice. Write them down, name them.

6. In the formula column of your option, find the base formulas and explain your choice based on the analysis of the composition of these compounds.

7. In your version, select the formulas of substances with which a solution of phosphoric acid (hydrochloric, sulfuric) can react. Write the appropriate reaction equations.

9. Among the formulas of your option, select the formulas of substances that can interact with each other. Write the appropriate reaction equations.

10. Make a chain of genetic links not organic compounds, which will include a substance whose formula is given in your version at number one.

Option 1

Option 2

CaO

HNO 3

Fe(OH) 3

N 2 O

Zn(NO 3 ) 2

Cr(OH) 3

H 2 SO 3

H 2 S

PbO

LiOH

Ag 3 PO 4

P 2 O 5

NaOH

ZnO

CO 2

BaCl 2

HCl

H 2 CO 3

H 2 SO 4

CuSO 4

From these substances, make a genetic series using all the formulas. Write the reaction equations with which you can carry out this chain of transformations:

I option: ZnSO 4, Zn, ZnO, Zn, Zn(OH) 2 : IIoption:Na 2 SO 4, NaOH, Na, Na 2 O 2 , Na 2 O

4) Fixing1.Al→ Al 2 O 3 → AlCl 3 → Al( Oh) 3 → Al 2 O 3

2. P→ P 2 O 5 → H 3 PO 4 → Na 3 PO 4 → Ca 3 ( PO 4 ) 2

3. Zn→ZnCl 2 →Zn(OH) 2 →ZnO→Zn(NO 3 ) 2

4.Cu→CuO→CuCl 2 →Cu(OH) 2 →CuO→Cu

5.N 2 O 5 →HNO 3 →Fe(NO 3 ) 2 →Fe(OH) 2 →FeS→FeSO 4

5)Homework: chart gradual transition from calcium to calcium carbonate and prepare a report on the medical use of any salt (using supplementary literature).

Instructions for students in the correspondence course "General Chemistry for Grade 12" 1. Category of students: the materials of this presentation are provided to the student for self-study topics "Substances and their properties", from the course general chemistry 12th grade. 2. Course content: includes 5 topic presentations. Each learning topic contains a clear structure educational material on a specific topic, the last slide is a control test - tasks for self-control. 3. Duration of study for this course: from one week to two months (determined individually). 4. Knowledge control: the student provides a progress report test items- a sheet with options for tasks, indicating the topic. 5. Evaluation of the result: "3" - 50% of tasks completed, "4" - 75%, "5"% of tasks. 6. Learning outcome: pass (fail) of the studied topic.

Reaction equations: 1. 2Cu + o 2 2CuO copper (II) oxide 2. CuO + 2 HCl CuCl 2 + H 2 O copper (II) chloride 3. CuCl NaOH Cu (OH) Na Cl copper (II) hydroxide 4. Cu (OH) 2 + H 2 SO 4 CuSO 4 + 2H 2 O copper (II) sulfate

Genetic series of organic compounds. If the basis of the genetic series is not organic chemistry are substances formed by one chemical element, then the basis of the genetic series in organic chemistry is made up of substances with the same number carbon atoms in a molecule.

Reaction scheme: Each number above the arrow corresponds to a specific reaction equation: ethanal ethanol ethene ethane chloroethane ethine Acetic (ethanoic) acid

Reaction equations: 1. C 2 H 5 Cl + H 2 O C 2 H 5 OH + HCl 2. C 2 H 5 OH + O CH 3 CH O + H 2 O 3. CH 3 CH O + H 2 C 2 H 5 OH 4. C 2 H 5 OH + HCl C 2 H 5 Cl + H 2 O 5. C 2 H 5 Cl C 2 H 4 + HCl 6. C 2 H 4 C 2 H 2 + H 2 7. C 2 H 2 + H 2 O CH 3 CH O 8. CH 3 CH O + Ag 2 O CH 3 COOH + Ag

This lesson is devoted to the generalization and systematization of knowledge on the topic "Classes of inorganic substances." The teacher will tell you how you can get a substance of another class from substances of one class. The acquired knowledge and skills will be useful for compiling reaction equations for chains of transformations.

In the course of chemical reactions, a chemical element does not disappear; atoms pass from one substance to another. The atoms of a chemical element are, as it were, transferred from a simple substance to more complex ones, and vice versa. Thus, the so-called genetic series arise, starting with a simple substance - metal or non-metal - and ending with salt.

Let me remind you that the composition of salts includes metals and acid residues. So, the genetic series of a metal might look like this:

A basic oxide can be obtained from a metal as a result of a compound reaction with oxygen, a basic oxide, when interacting with water, gives a base (only if this base is an alkali), a salt can be obtained from a base as a result of an exchange reaction with an acid, salt or acid oxide.

Please note that this genetic series is only suitable for metals whose hydroxides are alkalis.

Let us write down the reaction equations corresponding to the transformations of lithium in its genetic series:

Li → Li 2 O → LiOH → Li 2 SO 4

As you know, metals, when interacting with oxygen, usually form oxides. When oxidized by atmospheric oxygen, lithium forms lithium oxide:

4Li + O 2 = 2Li 2 O

Lithium oxide, interacting with water, forms lithium hydroxide - a water-soluble base (alkali):

Li 2 O + H 2 O \u003d 2LiOH

Lithium sulfate can be obtained from lithium in several ways, for example, as a result of a neutralization reaction with sulfuric acid:

2. Chemical information network ().

Homework

1. p. 130-131 №№ 2,4 from Workbook in chemistry: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry. Grade 8” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006.

2. p.204 Nos. 2, 4 from the textbook P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova "Chemistry: 8th grade", 2013

9 cells Lesson #47 Topic: " genetic connection Me, NeMe and their compounds".

Goals and objectives of the lesson:

Understand the concept of genetic connection.

Learn how to make genetic series of metals and non-metals.

Based on the knowledge of students about the main classes of inorganic substances, bring them to the concept of "genetic connection" and the genetic series of metal and non-metal;

To consolidate knowledge about the nomenclature and properties of substances belonging to different classes;

Develop skills to highlight the main thing, compare and generalize; identify and establish relationships;

Develop ideas about the cause-and-effect relationships of phenomena.

Restore in memory the concepts of simple and complex matter, of metals and non-metals, of the main classes inorganic compounds;

To form knowledge about the genetic relationship and the genetic series, learn how to compose the genetic series of metals and non-metals.

Develop the ability to generalize facts, build analogies and draw conclusions;

Continue developing a culture of communication, the ability to express one's views and judgments.

Cultivate a sense of responsibility for the acquired knowledge.

Planned results:

Know definitions and classification of inorganic substances.

Be able to classify inorganic substances by composition and properties; make up the genetic series of metal and non-metal;

illustrate the genetic relationship between the main classes of inorganic compounds with the equations of chemical reactions.

Competencies:

cognitive skills : systematize and classify information from written and oral sources.

Activity skills : to reflect on one’s activities, act according to the algorithm, be able to compose an algorithm on one’s own new activity, amenable to algorithmization; understand the language of diagrams.

Communication skills : build communication with other people - conduct a dialogue in pairs, take into account the similarities and differences in positions, interact with partners to obtain common product and result.

Lesson type:

on didactic purpose: a lesson on updating knowledge;

according to the method of organization: generalizing with the assimilation of new knowledge (combined lesson).

During the classes

I. Organizational moment.

II. Updating the basic knowledge and methods of action of students.

Lesson motto:"The only way,
leading to knowledge is activity” (B. Shaw). slide 1

At the first stage of the lesson, I update the basic knowledge that is necessary to solve the problem. This prepares students for the perception of the problem. I conduct the work in an entertaining way. I conduct a “brainstorming” on the topic: “Main classes of inorganic compounds” Work on cards

Task 1. “Third extra” slide 2

The students were given cards with three formulas written on them, and one of them was superfluous.

Students identify the extra formula and explain why it is superfluous.

Answers: MgO, Na 2 SO 4, H 2 S slide 3

Task 2. “Name and choose us” (“Name us”) slide 4

	nonmetals				hydroxides	Anoxic acids

Give the name of the selected substance (“4-5” write down the answers with formulas, “3” with words).

(Students work in pairs, wishing at the blackboard. (“4-5” write down answers in formulas, “3” in words).

Answers: slide 5

1. copper, magnesium;

4. phosphoric;

5. magnesium carbonate, sodium sulfate

7. salt

III. Learning new material.

1. Determining the topic of the lesson together with students.

As a result of chemical transformations, substances of one class are transformed into substances of another: an oxide is formed from a simple substance, an acid is formed from an oxide, and a salt is formed from an acid. In other words, the classes of compounds you have studied are interconnected. Let's distribute the substances into classes, according to the complexity of the composition, starting from a simple substance, according to our scheme.

Students express their versions, thanks to which we compose simple circuits 2 rows: metals and non-metals. Scheme of genetic series.

I draw students' attention to the fact that each chain has something in common - these are the chemical elements metal and non-metal, which pass from one substance to another (as if by inheritance).

(for strong students) CaO, P 2 O 5, MgO, P, H 3 PO 4, Ca, Na 3 PO 4, Ca (OH) 2, NaOH, CaCO 3, H 2 SO 4

(For weak students) CaO, CO 2 , C, H 2 CO 3 , Ca, Ca(OH) 2 , CaCO 3 slide 6

Answers: slide 7

P P2O5 H3PO4 Na3 PO4

Ca CaO Ca(OH)2 CaCO3

What is the name of the carrier of hereditary information in biology? (Gene).

What element do you think will be the “gene” for each chain? (metal and non-metal).

Therefore, such chains or series are called genetic. The topic of our lesson is "Genetic connection of Me and NeMe" slide 8. Open your notebook and write down the date and topic of the lesson. What do you think is the purpose of our lesson? To get acquainted with the concept of "genetic connection". Learn to compose the genetic series of metals and non-metals.

2. Let's define the genetic link.

genetic connection - called the connection between substances of different classes, based on their mutual transformations and reflecting the unity of their origin. slide 9,10

Features that characterize the genetic series: slide 11

1. Substances of different classes;

2. Different substances formed by one chemical element, i.e. represent different forms the existence of one element;

3. Different substances of one chemical element are connected by mutual transformations.

3. Consider examples of the genetic relationship of Me.

2. A genetic series, where an insoluble base acts as a base, then the series can be represented by a chain of transformations: slide 12

metal→basic oxide→salt→insoluble base→basic oxide→metal

For example, Cu→CuO→CuCl2→Cu(OH)2→CuO
1. 2 Cu + O 2 → 2 CuO 2. CuO + 2HCI → CuCI 2 3. CuCI 2 + 2NaOH → Cu (OH) 2 + 2NaCI

4. Cu (OH) 2 CuO + H 2 O

4. Consider examples of the genetic connection of NeMe.

Among non-metals, two types of series can also be distinguished: slide 13

2. The genetic series of non-metals, where a soluble acid acts as a link in the series. The chain of transformations can be represented as follows: non-metal → acid oxide → soluble acid → salt For example, P → P 2 O 5 → H 3 PO 4 → Ca 3 (PO 4) 2
1. 4P + 5O 2 → 2P 2 O 5 2. P 2 O 5 + H 2 O → 2H 3 PO 4 3. 2H 3 PO 4 +3 Ca (OH) 2 → Ca 3 (PO 4) 2 +6 H 2 O

5. Compilation of the genetic series. Slide 14

1. A genetic series in which alkali acts as a base. This series can be represented using the following transformations: metal → basic oxide → alkali → salt

O 2, + H 2 O, + HCI

4K + O 2 \u003d 2K 2 O K 2 O + H 2 O \u003d 2KOH KOH + HCI \u003d KCl slide 15

2. The genetic series of non-metals, where an insoluble acid acts as a link in the series:

nonmetal→acid oxide→salt→acid→acid oxide→nonmetal

For example, Si→SiO 2 →Na 2 SiO 3 →H 2 SiO 3 →SiO 2 →Si (make equations yourself, who works "4-5"). Self-test. All equations are correct "5", one error "4", two errors "3".

5. Performing differential exercises (self-examination). slide 15

Si + O 2 \u003d SiO 2 SiO 2 + 2NaOH \u003d Na 2 SiO 3 + H 2 O Na 2 SiO 3 + 2НCI \u003d H 2 SiO 3 + 2NaCI H 2 SiO 3 \u003d SiO 2 + H 2 O

SiO 2 +2Mg \u003d Si + 2MgO

1. Carry out transformations according to the scheme. (task "4-5")

Task 1. In the figure, connect the formulas of substances with lines in accordance with their location in the genetic series of aluminum. Write reaction equations. slide 16

Self-test.

4AI + 3O 2 \u003d 2AI 2 O 3 AI 2 O 3 + 6HCI \u003d 2AICI 3 + 3H 2 O AICI 3 + 3NaOH \u003d AI (OH) 3 + 3NaCI

AI(OH) 3 \u003d AI 2 O 3 + H 2 O slide 17

Task 2. "Hit the target." Select the formulas of the substances that make up the genetic series of calcium. Write the reaction equations for these transformations. Slide 18

Self-test.

2Ca + O 2 \u003d 2CaO CaO + H 2 O \u003d Ca (OH) 2 Ca (OH) 2 +2 HCI \u003d CaCI 2 + 2 H 2 O CaCI 2 + 2AgNO 3 \u003d Ca (NO 3) 2 + 2AgCI slide 19

2. Carry out the task according to the scheme. Write the reaction equations for these transformations.

O 2 + H 2 O + NaOH

S SO 2 H 2 SO 3 Na 2 SO 3 or light version

S + O 2 \u003d SO 2 + H 2 O \u003d H 2 SO 3 + NaOH \u003d

SO 2 + H 2 O \u003d H 2 SO 3

H 2 SO 3 + 2NaOH \u003d Na 2 SO 3 + 2H 2 O

IV. AnchoringZUN

Option 1.

Part A.

1. The genetic series of the metal is: a) substances that form a series based on one metal

but)CO 2 b) CO c) CaO d) O 2

3. Determine the substance "Y" from the transformation scheme: Na → Y→NaOH but)Na 2 O b) Na 2 O 2 c) H 2 O d) Na

4. In the transformation scheme: CuCl 2 → A → B → Cu by the formulas intermediate products A and B are: a) CuO and Cu(OH) 2 b) CuSO 4 and Cu(OH) 2 c) CuCO 3 and Cu(OH) 2 G)Cu(Oh) 2 AndCuO

5. The end product in the chain of transformations based on carbon compounds CO 2 → X 1 → X 2 → NaOH a) sodium carbonate b) sodium bicarbonate c) sodium carbide d) sodium acetate

E → E 2 O 5 → H 3 EO 4 → Na 3 EO 4 a) N b) Mn in)P d)Cl

Part B.

Fe + Cl 2 A) FeCl 2

Fe + HCl B) FeCl 3

FeO + HCl B) FeCl 2 + H 2

Fe 2 O 3 + HCl D) FeCl 3 + H 2

E) FeCl 2 + H 2 O

E) FeCl 3 + H 2 O

1 B, 2 A, 3D, 4E

a) potassium hydroxide (solution) b) iron c) barium nitrate (solution) d) aluminum oxide

e) carbon monoxide (II) f) sodium phosphate (solution)

Part C.

1. Implement the scheme of transformation of substances: Fe → FeO → FeCI 2 → Fe (OH) 2 → FeSO 4

2Fe + O 2 \u003d 2FeO FeO + 2HCI \u003d FeCI 2 + H 2 O FeCI 2 + 2NaOH \u003d Fe (OH) 2 + 2NaCI

Fe(OH) 2 + H 2 SO 4= FeSO 4 +2 H 2 O

Option 2.

Part A. (questions with one correct answer)

b) substances forming a series based on one non-metal c) substances forming a series based on a metal or non-metal d) substances from different classes of substances connected by transformations

2. Determine the substance "X" from the transformation scheme: P → X → Ca 3 (PO 4) 2 but)P 2 O 5 b) P 2 O 3 c) CaO d) O 2

a) Ca b)CaO c) CO 2 d) H 2 O

4. In the conversion scheme: MgCl 2 → A → B → Mg, the formulas of intermediate products A and B are: a) MgO and Mg (OH) 2 b) MgSO 4 and Mg (OH) 2 c) MgCO 3 and Mg (OH) 2 G)mg(Oh) 2 AndMgO

CO 2 → X 1 → X 2 → NaOH a) sodium carbonate b) sodium bicarbonate

6. Element "E", participating in the chain of transformations:

Part B. (tasks with 2 or more the right options response)

1. Establish a correspondence between the formulas of the starting substances and the reaction products:

Formulas of starting substances Formulas of products

NaOH + CO 2 A) NaOH + H 2

Na + H 2 O B) NaHCO 3

NaOH + HCl D) NaCl + H 2 O

1B, 2V, 3 A, 4G

a) sodium hydroxide (solution) b) oxygen c) sodium chloride (solution) d) calcium oxide

e) potassium permanganate (crystalline) e) sulfuric acid

Part C. (with extended answer)

S + O 2 \u003d SO 2 2SO 2 + O 2 \u003d 2 SO 3 SO 3 + H 2 O \u003d H 2 SO 4 H 2 SO 4 + Ca (OH) 2 \u003d CaSO 4 +2 H 2 O

CaSO 4 + BaCI 2 \u003d BaSO 4 + CaCI 2

v.Resultslesson. Grading.

VI.D/Z p.215-216 prepare for project No. 3 Option 1 of task No. 2,4, 6, Option 2 of task No. 2,3, 6. slide 20

VII. Reflection.

Students write down on paper what they did well and what they didn't. What were the difficulties. And a wish to the teacher.

The lesson is over. Thank you all and have a good day. slide 21

If there is time.

A task
Once Yuh conducted experiments to measure the electrical conductivity of solutions of various salts. Chemistry beakers with solutions were on his laboratory table. KCl, BaCl 2 , K 2 CO 3 , Na 2 SO 4 and AgNO 3 . Each glass was neatly labeled. There was a parrot in the lab whose cage didn't lock very well. When Juh, absorbed in the experiment, looked back at the suspicious rustle, he was horrified to find that the parrot, grossly violating safety regulations, was trying to drink from a glass of BaCl 2 solution. Knowing that all soluble barium salts are extremely poisonous, Yuh quickly grabbed a glass with a different label from the table and forcibly poured the solution into the parrot's beak. The parrot was saved. What glass of solution was used to save the parrot?
Answer:
BaCl 2 + Na 2 SO 4 \u003d BaSO 4 (precipitate) + 2NaCl (barium sulfate is so slightly soluble that it cannot be poisonous, like some other barium salts).

Attachment 1

9 "B" class F.I.______________________________ (for weak students)

Task 1. “The third extra”.

(4 correct - "5", 3-"4", 2-"3", 1-"2")

	nonmetals				hydroxides	Anoxic acids

Students define the chosen class and select the appropriate substances from the provided handout.

copper, silicon oxide, hydrochloric, barium hydroxide, coal, magnesium, phosphoric, barium hydroxide, magnesium oxide, iron (III) hydroxide, magnesium carbonate, sodium sulfate.

("4-5" write down the answers with formulas, "3" with words).

12 answers "5", 11-10 - "4", 9-8 - "3", 7 or less - "2"

Task 3.

O 2, + H 2 O, + HCI

For example, K → K 2 O → KOH → KCl (make equations yourself, who works "3", one error "3", two errors "2").

Task 4. Perform the task according to the scheme. Write the reaction equations for these transformations.

O 2 + H 2 O + NaOH

S SO 2 H 2 SO 3 Na 2 SO 3

or light version

H 2 SO 3 + NaOH \u003d

Option 1.

Part A. (questions with one correct answer)

1. The genetic series of a metal is: a) substances that form a series based on one metal

b) substances forming a series based on one non-metal c) substances forming a series based on a metal or non-metal d) substances from different classes of substances connected by transformations

2. Determine the substance "X" from the transformation scheme: C → X → CaCO 3

a) CO 2 b) CO c) CaO d) O 2

3. Determine the substance "Y" from the transformation scheme: Na → Y→NaOH a) Na 2 O b) Na 2 O 2 c) H 2 O d) Na

4. In the transformation scheme: CuCl 2 → A → B → Cu, the formulas of intermediate products A and B are: a) CuO and Cu (OH) 2 b) CuSO 4 and Cu (OH) 2 c) CuCO 3 and Cu (OH) 2 g) Cu(OH) 2 and CuO

5. The end product in the chain of transformations based on carbon compounds CO 2 → X 1 → X 2 → NaOH a) sodium carbonate b) sodium bicarbonate c) sodium carbide d) sodium acetate

6. Element "E", participating in the chain of transformations: E → E 2 O 5 → H 3 EO 4 → Na 3 EO 4 a) N b) Mn c) P d) Cl

Part B. (tasks with 2 or more correct answers)

1. Establish a correspondence between the formulas of the starting substances and the reaction products:

Formulas of starting substances Formulas of products

Fe + Cl 2 A) FeCl 2

Fe + HCl B) FeCl 3

FeO + HCl B) FeCl 2 + H 2

Fe 2 O 3 + HCl D) FeCl 3 + H 2

E) FeCl 2 + H 2 O

E) FeCl 3 + H 2 O

2. A solution of copper sulfate (II) interacts:

a) potassium hydroxide (solution) b) iron c) barium nitrate (solution) d) aluminum oxide

e) carbon monoxide (II) f) sodium phosphate (solution)

Part C. (with extended answer)

1. Implement a scheme for the transformation of substances:

Fe → FeO → FeCI 2 → Fe(OH) 2 → FeSO 4

Annex 2

9 "B" class F.I.______________________________ (for a strong student)

Task 1. “The third extra”. Identify the redundant formula and explain why it is redundant.

(4 correct - "5", 3-"4", 2-"3", 1-"2")

Task 2. “Name and choose us” (“Name us”). Give the name of the selected substance, fill in the table.

Students define the chosen class and select the appropriate substances from the provided handout.

copper, silicon oxide, hydrochloric, barium hydroxide, coal, magnesium, phosphoric, barium hydroxide, magnesium oxide, iron (III) hydroxide, magnesium carbonate, sodium sulfate. ("4-5" write down the answers with formulas, "3" with words).

12 answers "5", 11-10 - "4", 9-8 - "3", 7 or less - "2"

Task 3.

Si→SiO 2 →Na 2 SiO 3 →H 2 SiO 3 →SiO 2 →Si (make equations yourself, who works "4-5"). Self-test. All equations are correct "5", one error "4", two errors "3".

Task 4. In the figure, connect the formulas of substances with lines in accordance with their location in the genetic series of aluminum. Write reaction equations. All equations are correct "5", one error "4", two errors "3".

Task 5. "Hit the target." Select the formulas of the substances that make up the genetic series of calcium. Write the reaction equations for these transformations. All equations are correct "5", one error "4", two errors "3".

Option 2.

Part A. (questions with one correct answer)

1. The genetic series of a non-metal is: a) substances that form a series based on one metal

b) substances forming a series based on one non-metal c) substances forming a series based on a metal or non-metal d) substances from different classes of substances connected by transformations

2. Determine the substance "X" from the transformation scheme: P → X → Ca 3 (PO 4) 2 a) P 2 O 5 b) P 2 O 3 c) CaO d) O 2

3. Determine the substance "Y" from the transformation scheme: Ca → Y→Ca(OH) 2

a) Ca b) CaO c) CO 2 d) H 2 O

4. In the conversion scheme: MgCl 2 → A → B → Mg, the formulas of intermediate products A and B are: a) MgO and Mg (OH) 2 b) MgSO 4 and Mg (OH) 2 c) MgCO 3 and Mg (OH) 2 g) Mg (OH) 2 and MgO

5. The end product in the chain of transformations based on carbon compounds:

CO 2 → X 1 → X 2 → NaOH a) sodium carbonate b) sodium bicarbonate

c) sodium carbide d) sodium acetate

6. Element "E", participating in the chain of transformations:

E → EO 2 → EO 3 → H 2 EO 4 → Na 2 EO 4 a) N b) S c) P d) Mg

Part B. (tasks with 2 or more correct answers)

1. Establish a correspondence between the formulas of the starting substances and the reaction products:

Formulas of starting substances Formulas of products

NaOH + CO 2 A) NaOH + H 2

NaOH + CO 2 B) Na 2 CO 3 + H 2 O

Na + H 2 O B) NaHCO 3

NaOH + HCl D) NaCl + H 2 O

2. Hydrochloric acid does not interact:

a) sodium hydroxide (solution) b) oxygen c) sodium chloride (solution) d) calcium oxide

e) potassium permanganate (crystalline) f) sulfuric acid

Part C. (with extended answer)

Implement the scheme of transformation of substances: S → SO 2 → SO 3 → H 2 SO 4 → CaSO 4 → BaSO 4

Annex 3

Answer sheet "4-5":

Task 1. MgO, Na 2 SO 4, H 2 S

Task 2.

1. copper, magnesium;

3. silicon oxide, magnesium oxide;

4. phosphoric,

5. magnesium carbonate, sulfate;

6. barium hydroxide, iron (III) hydroxide;

7. sodium hydrochloride

Task 3.

SiO 2 + 2NaOH \u003d Na 2 SiO 3 + H 2 O

Na 2 SiO 3 + 2НCI \u003d H 2 SiO 3 + 2NaCI

H 2 SiO 3 \u003d SiO 2 + H 2 O

SiO 2 +2Mg \u003d Si + 2MgO

Task 4.

4AI + 3O 2 \u003d 2AI 2 O 3

AI 2 O 3 + 6HCI \u003d 2AICI 3 + 3H 2 O

AICI 3 + 3NaOH \u003d AI (OH) 3 + 3NaCI

AI (OH) 3 \u003d AI 2 O 3 + H 2 O

Task 5.

CaO + H 2 O \u003d Ca (OH) 2

Ca (OH) 2 +2 HCI \u003d CaCI 2 + 2 H 2 O

CaCI 2 + 2AgNO 3 \u003d Ca (NO 3) 2 + 2AgCI

Self-assessment sheet.

Full name of the student	job number