Genetic series of sodium metal. Genetic relationship between the main classes of inorganic substances

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 seem to be transferred from a simple substance to a more complex one, 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 the 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

Genetic series of metals and their compounds

Each such row consists of a metal, its basic oxide, a base, and any salt of the same metal:

To go from metals to basic oxides in all these series, reactions of combination with oxygen are used, for example:

2Ca + O 2 \u003d 2CaO; 2Mg + O 2 \u003d 2MgO;

The transition from basic oxides to bases in the first two rows is carried out by the hydration reaction known to you, for example:

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

As for the last two rows, the MgO and FeO oxides contained in them do not react with water. In such cases, to obtain bases, these oxides are first converted into salts, and then they are converted into bases. Therefore, for example, to carry out the transition from MgO oxide to Mg (OH) 2 hydroxide, successive reactions are used:

MgO + H 2 SO 4 \u003d MgSO 4 + H 2 O; MgSO 4 + 2NaOH \u003d Mg (OH) 2 ↓ + Na 2 SO 4.

The transitions from bases to salts are carried out by reactions already known to you. So, soluble bases (alkalis), which are in the first two rows, are converted into salts under the action of acids, acid oxides or salts. Insoluble bases from the last two rows form salts under the action of acids.

Genetic series of non-metals and their compounds.

Each such series consists of a non-metal, an acid oxide, the corresponding acid, and a salt containing the anions of this acid:

To go from non-metals to acidic oxides, in all these series, reactions of combination with oxygen are used, for example:

4P + 5O 2 \u003d 2 P 2 O 5; Si + O 2 \u003d SiO 2;

The transition from acid oxides to acids in the first three rows is carried out by the hydration reaction known to you, for example:

P 2 O 5 + 3H 2 O \u003d 2 H 3 PO 4.

However, you know that the SiO 2 oxide contained in the last row does not react with water. In this case, it is first converted into the corresponding salt, from which the desired acid is then obtained:

SiO 2 + 2KOH = K 2 SiO 3 + H 2 O; K 2 SiO 3 + 2HСl \u003d 2KCl + H 2 SiO 3 ↓.

The transitions from acids to salts can be carried out by reactions known to you with basic oxides, bases, or with salts.

It should be remembered:

Substances of the same genetic series do not react with each other.

Substances of genetic series of different types react with each other. The products of such reactions are always salts (Fig. 5):

Rice. 5. Scheme of the relationship of substances of different genetic series.

This scheme displays the relationship between different classes of inorganic compounds and explains the variety of chemical reactions between them.

Topic task:

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

1. Na → Na 2 O → NaOH → Na 2 CO 3 → Na 2 SO 4 → NaOH;

2. P → P 2 O 5 → H 3 PO 4 → K 3 PO 4 → Ca 3 (PO 4) 2 → CaSO 4;

3. Ca → CaO → Ca(OH) 2 → CaCl 2 → CaCO 3 → CaO;

4. S → SO 2 → H 2 SO 3 → K 2 SO 3 → H 2 SO 3 → BaSO 3;

5. Zn → ZnO → ZnCl 2 → Zn(OH) 2 → ZnSO 4 → Zn(OH) 2;

6. C → CO 2 → H 2 CO 3 → K 2 CO 3 → H 2 CO 3 → CaCO 3;

7. Al → Al 2 (SO 4) 3 → Al(OH) 3 → Al 2 O 3 → AlCl 3;

8. Fe → FeCl 2 → FeSO 4 → Fe(OH) 2 → FeO → Fe 3 (PO 4) 2;

9. Si → SiO 2 → H 2 SiO 3 → Na 2 SiO 3 → H 2 SiO 3 → SiO 2;

10. Mg → MgCl 2 → Mg(OH) 2 → MgSO 4 → MgCO 3 → MgO;

11. K → KOH → K 2 CO 3 → KCl → K 2 SO 4 → KOH;

12. S → SO 2 → CaSO 3 → H 2 SO 3 → SO 2 → Na 2 SO 3;

13. S → H 2 S → Na 2 S → H 2 S → SO 2 → K 2 SO 3;

14. Cl 2 → HCl → AlCl 3 → KCl → HCl → H 2 CO 3 → CaCO 3;

15. FeO → Fe(OH) 2 → FeSO 4 → FeCl 2 → Fe(OH) 2 → FeO;

16. CO 2 → K 2 CO 3 → CaCO 3 → CO 2 → BaCO 3 → H 2 CO 3;

17. K 2 O → K 2 SO 4 → KOH → KCl → K 2 SO 4 → KNO 3;

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

19. Al 2 O 3 → AlCl 3 → Al(OH) 3 → Al(NO 3) 3 → Al 2 (SO 4) 3 → AlCl 3;

20. SO 3 → H 2 SO 4 → FeSO 4 → Na 2 SO 4 → NaCl → HCl;

21. KOH → KCl → K 2 SO 4 → KOH → Zn(OH) 2 → ZnO;

22. Fe(OH) 2 → FeCl 2 → Fe(OH) 2 → FeSO 4 → Fe(NO 3) 2 → Fe;

23. Mg(OH) 2 → MgO → Mg(NO 3) 2 → MgSO 4 → Mg(OH) 2 → MgCl 2;

24. Al(OH) 3 → Al 2 O 3 → Al(NO 3) 3 → Al 2 (SO 4) 3 → AlCl 3 → Al(OH) 3;

25. H 2 SO 4 → MgSO 4 → Na 2 SO 4 → NaOH → NaNO 3 → HNO 3;

26. HNO 3 → Ca(NO 3) 2 → CaCO 3 → CaCl 2 → HCl → AlCl 3;

27. CuСO 3 → Cu(NO 3) 2 → Cu(OH) 2 → CuO → CuSO 4 → Cu;

28. MgSO 4 → MgCl 2 → Mg(OH) 2 → MgO → Mg(NO 3) 2 → MgCO 3;

29. K 2 S → H 2 S → Na 2 S → H 2 S → SO 2 → K 2 SO 3;

30. ZnSO 4 → Zn(OH) 2 → ZnCl 2 → HCl → AlCl 3 → Al(OH) 3;

31. Na 2 CO 3 → Na 2 SO 4 → NaOH → Cu(OH) 2 → H 2 O → HNO 3;

The material world in which we live and of which we are a tiny part is one and at the same time infinitely diverse. The unity and diversity of the chemical substances of this world is most clearly manifested in the genetic connection of substances, which is reflected in the so-called genetic series. Let us highlight the most characteristic features of such series.

1. All substances of this series must be formed by one chemical element. For example, a series written using the following formulas:

2. Substances formed by the same element must belong to different classes, i.e., reflect different forms of its existence.

3. Substances that form the genetic series of one element must be connected by mutual transformations. On this basis, one can distinguish between complete and incomplete genetic series.

For example, the above genetic series of bromine will be incomplete, incomplete. And here is the next row:

can already be considered as complete: it began with the simple substance bromine and ended with it.

Summarizing the above, we can give the following definition of the genetic series.

genetic series- this is a number of substances - representatives of different classes, which are compounds of one chemical element, connected by mutual transformations and reflecting the common origin of these substances or their genesis.

genetic connection- the concept is more general than the genetic series, which is, albeit a vivid, but particular manifestation of this connection, which is realized in any mutual transformations of substances. Then, obviously, the first given series of substances also fits this definition.

There are three types of genetic series:

The richest series of metals, which exhibit different degrees of oxidation. As an example, consider the genetic series of iron with oxidation states +2 and +3:

Recall that for the oxidation of iron to iron (II) chloride, you need to take a weaker oxidizing agent than to obtain iron (III) chloride:

Similarly to the metal series, the non-metal series with different oxidation states is richer in bonds, for example, the genetic series of sulfur with oxidation states +4 and +6:

Difficulty can cause only the last transition. Follow the rule: in order to obtain a simple substance from an oxidized compound of an element, you need to take its most reduced compound for this purpose, for example, the volatile hydrogen compound of a non-metal. In our case:

By this reaction, sulfur is formed from volcanic gases in nature.

Similarly for chlorine:

3. The genetic series of the metal, which corresponds to the amphoteric oxide and hydroxide,it is very rich in bonds, because, depending on the conditions, they exhibit either acidic or basic properties.

For example, consider the genetic series of zinc:

Genetic relationship between classes of inorganic substances

Reactions between representatives of different genetic series are characteristic. Substances from the same genetic series, as a rule, do not interact.

For example:
1. metal + non-metal = salt

Hg + S = HgS

2Al + 3I 2 = 2AlI 3

2. basic oxide + acid oxide = salt

Li 2 O + CO 2 \u003d Li 2 CO 3

CaO + SiO 2 \u003d CaSiO 3

3. base + acid = salt

Cu(OH) 2 + 2HCl \u003d CuCl 2 + 2H 2 O

FeCl 3 + 3HNO 3 \u003d Fe (NO 3) 3 + 3HCl

salt acid salt acid

4. metal - basic oxide

2Ca + O 2 \u003d 2CaO

4Li + O 2 \u003d 2Li 2 O

5. non-metal - acid oxide

S + O 2 \u003d SO 2

4As + 5O 2 \u003d 2As 2 O 5

6. basic oxide - base

BaO + H 2 O \u003d Ba (OH) 2

Li 2 O + H 2 O \u003d 2LiOH

7. acid oxide - acid

P 2 O 5 + 3H 2 O \u003d 2H 3 PO 4

SO 3 + H 2 O \u003d H 2 SO 4

9 cells Lesson number 47 Topic: "Genetic relationship of 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 of 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 carry out reflection of one's activity, to act according to the algorithm, to be able to compose the algorithm of a 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 a common product and result.

Lesson type:

for the didactic purpose: a lesson in 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

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 draw up simple schemes of 2 series: 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 of 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

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 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 a)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 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

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 nice 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")

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

Appendix 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.