Arms All-Russian Inspection Work - Chemistry Grade 11

Explanation of a model of all-Russian verification work

When familiarizing with a sample of verification work, it should be borne in mind that the tasks included in the sample do not reflect all the abilities and contents of the content that will be checked in the framework of the All-Russian verification work. A complete list of content and skill elements that can be checked in the work are given in the codifier of the content elements and requirements for graduates in the development of all-Russian inspection of chemistry. The purpose of the sample of the test work is to give an idea of \u200b\u200bthe structure of the All-Russian testing, the number and form of tasks, the level of their complexity.

Instructions for performing work

Verification work includes 15 tasks. For chemistry work, 1 hour 30 minutes is given (90 minutes).
Make answers in the text of work according to the instructions for tasks. In the case of the record of an incorrect answer, cross it and write down the new one.
When performing work, it is allowed to use the following additional materials:
- Periodic system of chemical elements D.I. Mendeleev;
- Table of solubility of salts, acids and bases in water;
- electrochemical series of metals voltage;
- unprogrammed calculator.
When executing tasks, you can use a draft. Entries in Chernovik are not checked and evaluated.
We advise you to perform tasks in the order in which they are given. To save time, skip the task that cannot be executed immediately and go to the next one. If, after performing all the work, you will have time, you can return to missing tasks.
The points you received for completed tasks are summed up. Try to perform as much tasks as possible and score the largest number of points.
We wish you success!

1. From the course of chemistry you know the following ways of separation of mixtures: settling, filtering, distillation (distillation), action magnet, evaporation, crystallization. Figures 1-3 present examples of using some of the listed methods.

Which of the mentioned methods of separation of mixtures can be applied to cleanse:
1) flour from iron sawdust in it;
2) Water from inorganic salts dissolved in it?
Write the figure in the table and the name of the corresponding method of separating the mixture.

iron sawdust attract a magnet

when distillation after condensation of water vapor in a vessel, salt crystals remain

2. Figure shows the model of the electronic structure of the atom of some chemicalelement.

Based on the analysis of the proposed model, follow these tasks:
1) determine the chemical element whose atom has such an electronic structure;
2) Specify the number number and number of the group in the periodic system of chemical elements D.I. Mendeleev, in which this element is located;
3) Determine, the metals or nonmetallax include a simple substance that forms this chemical element.
Answers write to the table.
Answer:

N; 2; 5 (or v); non-metal

to determine the chemical element, it is necessary to calculate the total number of electrons that we see in Figure (7)

taking the Mendeleev table, we can easily determine the element (the resulting number of electrons is equal to the atomic number of the element) (N-nitrogen)

after that, we define the number of the group (vertical column) (5) and the nature of this element (non-metall)

3. Periodic system of chemical elements D.I. Mendeleev - a rich storage of information on chemical elements, their properties and properties of their compounds, about the patterns of changes in these properties, on the methods of obtaining substances, as well as on finding them in nature. For example, it is known that with an increase in the sequence number of the chemical element in the periods of atoms, the atoms are reduced, and groups increase.
Considering these patterns, position the following elements in the order of increasing the radii of atoms: N, C, Al, Si. Record the designations of the elements in the desired sequence.

Answer: ____________________________

N → C → Si → Al

4. The table below lists the characteristic properties of substances that have a molecular and ion structure.

Using this information, determine which structure has a nitrogen substance N2 and the sodium salt NaCl. Write down the answer in the allotted place:

1) nitrogen N2 ____________________________________________________________
2) Salt NaCl ___________________________________________________

nitrogen N2 is a molecular structure;
Salt NaCl - ionic structure

5. Complex inorganic substances can be conditionally distributed, that is, classify, in four groups, as shown in the diagram. In this scheme for each of the four groups, enter the missed groups of groups or chemical formulas (for one example of formulas) belonging to this group.

The names of the groups are recorded: bases, salts;
The formulas of substances of the relevant groups are recorded.

Cao, Bases, HCl, Salt

Read the following text and perform tasks 6-8.

The food industry uses a dietary supplement E526, which is calcium hydroxide Ca (OH) 2. It finds use in production: fruit juices, baby food, pickled cucumbers, food salts, confectionery and sweets.
Obtaining calcium hydroxide on an industrial scale by mixing calcium oxide with waterThis process is called quenching.
Calcium hydroxide was widely used in the production of such building materials such as Belil, plaster and gypsum solutions. This is due to its ability. interact with CO2 carbon dioxidecontained in the air. This calcium hydroxide solution is used to measure the quantitative content of carbon dioxide in the air.
The useful feature of calcium hydroxide is its ability to act as a flocculant, cleaning waste water from suspended and colloidal particles (including iron salts). It is also used to increase the pH of water, since natural water contains substances (for example, acid) Corrosive in sanitary pipes.

1. Make a molecular equation of calcium hydroxide reaction, which
Mentioned in the text.

2. Explain why this process is called a quenching.
Answer:__________________________________________________________________________

________________________________________________________________________________

1) Cao + H 2 O \u003d Ca (OH) 2
2) when the interaction of calcium oxide with water highlights a large
The amount of heat, so water boils and hits, as when it gets into a split coal, when the fire is quenched with water (or "this process is called a quenching, because the exhaust lime is formed")

1. Make a molecular reaction equation between calcium hydroxide and carbon dioxide
Gas, which was mentioned in the text.
Answer:__________________________________________________________________________

2. Explain what particular features of this reaction can be used to detect it.
Carbon dioxide in the air.
Answer:__________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________

1) Ca (OH) 2 + CO 2 \u003d Caco 3 ↓ + H 2 O
2) As a result of this reaction, an insoluble substance is formed - calcium carbonate, observed the initial solution is observed, which allows you to judge the presence of carbon dioxide in the air (high-quality
Reaction to CO 2)

1. Make the abbreviated ion equation mentioned in the text of the reaction between
Calcium hydroxide and hydrochloric acid.
Answer:__________________________________________________________________________

2. Explain why this reaction is used to increase the pH of water.
Answer:__________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________

1) OH - + H + \u003d H 2 O (Ca (OH) 2+ 2HCl \u003d CaCl2 + 2H2O)
2) The presence of acid in natural water determines the low pH of this water. Calcium hydroxide neutralizes acid, and pH values \u200b\u200bincrease

scale pH exists from 0-14. From 0-6 - acidic medium, 7 - neutral medium, 8-14 - Alkaline medium

9. Dana scheme of the oxidation reaction.

H 2 S + Fe 2 O 3 → FES + S + H 2 O

1. Make an electronic balance of this reaction.
Answer:__________________________________________________________________________

2. Specify the oxidizer and reducing agent.
Answer:__________________________________________________________________________

3. Arrange the coefficients in the reaction equation.
Answer:__________________________________________________________________________

1) Electronic balance compiled:

2fe +3 + 2 → 2fe +2	2		1
		2
S -2 - 2ē → s 0	2		1

2) It is indicated that sulfur to the degree of oxidation -2 (or H 2 S) is a reducing agent, and iron to the degree of oxidation is +3 (or Fe 2 O 3) - the oxidizing agent;
3) The reaction equation is drawn up:
3H 2 S + Fe 2 O 3 \u003d 2FES + S + 3H 2 O

10. Dana scheme of transformations:

Fe → FECL 2 → Fe (NO 3) 2 → Fe (OH) 2

Write the molecular reactions equations with which you can implement
Specified transformations.
1) _________________________________________________________________________
2) _________________________________________________________________________
3) _________________________________________________________________________

The reaction equations corresponding to the transform scheme are written:
1) Fe + 2HCl \u003d FECL 2 + H 2
2) FECL 2 + 2AGNO 3 \u003d Fe (NO 3) 2 + 2AgCl
3) Fe (NO 3) 2 + 2KOH \u003d FE (OH) 2 + 2KNO 3
(Other, non-contrary to the condition of the task of the equation are allowed
reactions.)

11. Install the correspondence between the formula of the organic matter and class / groupTo which (s), this substance belongs: to each position indicated by the letter, select the appropriate position indicated by the number.

Write in the table selected numbers under the appropriate letters.
Answer:

BUT	B.	IN

1. C3H8 - CNH2N + 2 - Alcan
2. C3H6 - CNH2N- alkene
3. C2H6O - CNH2N + 2O- alcohol

12. In the proposed schemes of chemical reactions, insert the formulas of the missed substances and dispel the coefficients.

1) C 2 H 6 + ............... .. ... → C 2 H 5 CL + HCl
2) C 3 H 6 + ............... .. ... → CO 2 + H 2 O

1) C 2 H 6 + Cl 2 → C 2 H 5 CL + HCl
2) 2C 3 H 6 + 9O 2 → 6CO 2 + 6H 2 O
(Fractional coefficients are possible.)

13. Propane burns with low emissions of toxic substances into the atmosphere, Therefore, it is used as a source of energy in many areas, for example, gas lighters and during the heating of country houses.
What volume of carbon dioxide (N.U.) is formed with full combustion of 4.4 g of propane?
Write down a detailed solution of the problem.
Answer:__________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________

1) The propane burning reaction equation is compiled:
C 3 H 8 + 5O 2 → 3SO 2 + 4N 2
2) N (C 3 H 8) \u003d 4.4 / 44 \u003d 0.1 mol
n (CO 2) \u003d 3N (C 3 H 8) \u003d 0.3 mol
3) V (O 2) \u003d 0.3 · 22,4 \u003d 6.72 l

14. Isopropyl alcohol is used as a universal solvent: it is part of household chemicals, perfume and cosmetic products, glassy liquids for cars. In accordance with the scheme below, make the equation of reactions to obtain this alcohol. When writing the reaction equations, use structural formulas for organic substances.

1) _______________________________________________________
2) _______________________________________________________
3) _______________________________________________________

The reaction equations corresponding to the scheme are written:

(Other, non-contrary to the condition of the task of the reaction equation, are allowed.)

15. The physiological solution in medicine is called a 0.9% solution of sodium chloride in water. Calculate the mass of sodium chloride and mass of water, which is necessary for the preparation of 500 g of saline. Write down a detailed solution of the problem.
Answer:__________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________

1) M (NaCl) \u003d 4.5 g
2) M (water) \u003d 495.5 g

m (p-ra) \u003d 500g m (salts) \u003d x

x / 500 * 100% \u003d 0.9%

m (Coli) \u003d 500 * (0.9 / 100) \u003d 4.5 g

© 2017 Federal Service for Supervision in the field of education and science of the Russian Federation

On April 27, 2017, the All-Russian inspection works of HDP in chemistry in grade 11 were held for the first time.

Official website of the enterprise (Statgrad) - Vpr.Statgrad.org.

Variants of Arms in Chemistry 11 class 2017

№	Download Answers (evaluation criteria)
Option 11.	answers
Option 12.	answers
Option 13.	answers
Option 14.	answers
Option 15.	variant 15 Otvet.
Option 16.	variant 16 Otvet.
Option 17.	variant 17 Otvet.
Option 18.	variant 18 Otvet.

To familiarize yourself with exemplary work options on the official website of the FIPI, demonstration options are posted with answers and descriptions.

Samples of Arms in Chemistry 11 Class 2017 (demo version)

Verification work includes 15 tasks. For chemistry work, 1 hour 30 minutes is given (90 minutes).

When performing work, it is allowed to use the following additional materials:

- Periodic system of chemical elements D.I. Mendeleev;

- Table of solubility of salts, acids and bases in water;

- electrochemical series of metals voltage;

- unprogrammed calculator.

The structure and content of the All-Russian verification work of the Armediary for Chemistry

Each variant of HPP contains 15 tasks of various types and levels of complexity. The variants presents the tasks of various formats.

These tasks have differences on the required response entry form. For example, the answer may be: a sequence of numbers, symbols; the words; formulas of substances; Reaction equations.

The paper contains 4 tasks of the elevated level of complexity (their sequence numbers: 9, 10, 13, 14). These tasks are more complex, since their implementation implies the integrated application of the following skills:

- draw up the equations of reactions confirming the properties of substances and / or the relationship of various classes of substances, and the electronic balance of the redox reaction;

Explain the conditionality of the properties and methods of obtaining substances by their composition and structure;

- Simulate a chemical experiment based on its description.

When executing tasks, you can use a draft. Entries in Chernovik are not checked and evaluated.

PRP. Chemistry. Grade 11. 10 options for typical tasks. Drozdov A.A.

M.: 20. 1 7. - 9 6 s.

This manual fully complies with the Federal State Educational Standard (second generation). The book contains 10 options for typical tasks of the All-Russian test work (UPR) in chemistry for students of 11 classes. The collection is designed for students of 11 classes, teachers and methodologists using typical tasks to prepare for all-Russian verification work in chemistry.

Format: PDF.

The size: 3.4 MB

Watch, download:drive.google

Instructions for the performance of work, 4
Option 1 5.
Option 2 12.
Option 3 19.
Option 4 26.
Option 5 33.
Option 6 40.
Option 7 47.
Option 8 54.
Option 9 61.
Option 10 68.
Verification system for testing 75
Answers 76.
Appendices 93.

Verification work includes 15 tasks. For chemistry work, 1 hour 30 minutes is given (90 minutes).
Make answers in the text of work according to the instructions for tasks. In the case of the record of an incorrect answer, cross it and write down the new one.
When performing work, it is allowed to use the following additional materials:
- Periodic system of chemical elements D.I. Mendeleev;
- Table of solubility of salts, acids and bases in water;
- electrochemical series of metals voltage;
- unprogrammed calculator.
When executing tasks, you can use a draft. Entries in Chernovik are not checked and evaluated.
We advise you to perform tasks in the order in which they are given. To save time, skip the task that cannot be executed immediately and go to the next one. If, after performing all the work, you will have time, you can return to missing tasks.
The points you received for completed tasks are summed up. Try to perform as much tasks as possible and score the largest number of points.

Structure

Each version of the facilities in chemistry contains 15 tasks of various types and levels of complexity.

The paper contains 4 tasks of the elevated level of complexity (their sequence numbers: 9, 10, 13, 14).

The tasks included in the work can be conditionally distributed in four-intensive blocks: "Theoretical Basics of Chemistry", "Inorganic Chemistry", "Organic Chemistry", "Methods of Knowledge in Chemistry. Experimental bases of chemistry. Chemistry and life. "

Explanation of assuming tasks

The correct execution of the task with the sequence number 3 is estimated at 1 point.

The correct execution of each other tasks of the base level of complexity is estimated as 2 points. In the case of one error or incomplete response, 1 point is set. The remaining answers are considered incorrect and evaluated in 0 points.

Evaluation of the tasks of an increased level of complexity is carried out on the basis of a single analysis of students' responses. The maximum score for the correctly performed task is 3 points. Tasks with a detailed response can be performed by students in different ways. Therefore, sample decisions given in the evaluation criteria should be considered only as one of the possible response options.

Secondary education

Line Ukk V. V. Lunina. Chemistry (10-11) (bases.)

Line Ukk V. V. Lunina. Chemistry (10-11) (y)

Line Ukk N. E. Kuznetsova. Chemistry (10-11) (bases.)

Line Ukk N. E. Kuznetsova. Chemistry (10-11) (coal.)

PRD in chemistry. Grade 11

Verification work includes 15 tasks. For chemistry work, 1 hour 30 minutes is given (90 minutes).

Record to tasks Record in the field allotted for them. In the case of the record of an incorrect answer, cross it and write down the new one.

When performing the work, it is allowed to use:

• Periodic system of chemical elements D.I. Mendeleev;
• table solubility salts, acids and bases in water;
• electrochemical row of metals voltages;
• unprogrammed calculator.

When executing tasks, you can use a draft. Entries in Chernovik are not checked and evaluated.

We advise you to perform tasks in the order in which they are given. To save time, skip the task that cannot be executed immediately and go to the next one. If, after performing all the work, you will have time, you can return to missing tasks.

The points you received for completed tasks are summed up. Try to perform as much tasks as possible and score the largest number of points.

We wish you success!

From the course of chemistry you know the following ways of separation of mixtures: cresting, filtering, distillation (distillation), action magnet, evaporation, crystallization.

In fig. 1-3 shows examples of using some of the listed methods.

Determine which of the methods of separating the mixtures depicted in the figure, can be applied to the separation:

1. cereals and iron sawdust in her;
2. waters and salts dissolved in it.

Write the figure in the table and the name of the corresponding method of separating the mixture.

Decision

1.1. The separation of the mixture of cereals and iron sawdust is based on the property of the iron attracting a magnet. Figure 3.

1.2. The separation of the mixture of water and dissolved salts occurs during distillation. Water when heated to boiling point evaporates and, coolant in a water refrigerator, flows into a pre-prepared vessel. Picture 1.

The figure shows the electron distribution scheme according to the energy levels of the atom of some chemical element.

Based on the proposed scheme, follow these tasks:

1. write down the symbol of the chemical element that corresponds to this atom model;
2. record the period number and number of the group in the periodic system of chemical elements D.I. Mendeleev, in which this element is located;
3. determine, there is a simple substance to metals or non-metals, which forms this element.

Answers write to the table.

Decision

The figure shows the scheme of the building of the atom:

Where shown a core having a certain positive charge(n.), and rotating around the kernel on electronic electronic layers. Based on this, they ask to call this item, write the number of the period and the group in which it is located. Let's figure out:

1. Electrons rotate on three electronic layers, it means that the element is in the third period.
2. 5 electrons rotate on the last electron layer, it means that the element is located in the 5th group.

Task 3.

Periodic system of chemical elements D.I. Mendeleev is a rich storage of information on chemical elements, their properties and properties of their compounds. For example, it is known that with an increase in the sequence number of the chemical element, the main character of oxide in periods is reduced, and in groups it increases.

Considering these patterns, position the following elements in order to enhance the base base of oxides: Na, Al, Mg, V. Write down the characters of the elements in the desired sequence.

Answer: ________

Decision

As is known, the sum of protons in the nucleus of the atom is equal to the sequence number of the element. But the number of protons does not specify us. Since the atom is an electronic particle particle, the number of protons (positively charged particles) in the atom core is equal to the number of electrons (negatively charged particles) rotating around the atomic core. The total number of electrons rotating around the kernel is 15 (2 + 8 + 5), therefore, the sequence number of the element is 15. Now it remains to look at the periodic system of chemical elements D. I. Mendeleev and find number 15. This is P (phosphorus). Since phosphorus on the last electronic layer of 5 electrons, it is nonmetall; Metals on the last layer have from 1 to 3 electrons.

There are 4 elements from the periodic Mendeleev system: Na, Al, Mg, B. It is necessary to position them so that the bases of the oxides formed by them increase. Answering this Question to this issue, it is necessary to remember how metal properties change in periodic periodic systems and groups.

In periods from left to right, metal properties decrease and non-metallic increases. Therefore, the basicity of oxides decreases.

In groups, the main subgroups metallic properties increase from top to bottom. Consequently, the basicity of their oxides increases in the same order.

Now let's look at the data of us elements. Two of them are in the third group; This is b and Al. Aluminum in the group is below boron, therefore, it has metallic properties expressed stronger than at the boron. Accordingly, the basicity of aluminum oxide is stronger.

Al, Na and Mg are located in the 3rd period. Since in the period from left to right, the metallic properties decrease, decrease both the main properties of their oxides. Considering all this, you can position these elements in the following order:

Task 4.

The table below presents some characteristics of covalent and ionic types of chemical bonds.

Using this information, determine the type of chemical bond: 1) in calcium chloride (CAC 2); 2) in hydrogen molecule (H 2).

1. In calcium chloride _____________
2. In hydrogen molecule _____________

Decision

In the following question, it is necessary to determine which type of chemical communication is characteristic of CaCl 2, and which for H 2. This table has a hint:

Using it, it can be determined that for CACL 2 is characterized by an ion type of communication, since it consists of a metal atom (Ca) and non-metal atoms (CL), and for H 2 covalent, non-polar, since this molecule consists of atoms of the same element - hydrogen.

Complex inorganic substances can be conditionally distributed, that is, classify, in four classes, as shown in the diagram. In this scheme, enter the missing names of two classes and two formulas that are representatives of the relevant classes.

Decision

The next task to check the knowledge of the basic classes of inorganic substances.

In the table it is necessary to fill empty cells. In the first two cases, the formulas of substances are given, they must be attributed to a certain class of substances; In the last two, on the contrary, write formulas of representatives of these classes.

CO 2 is a complex substance consisting of atoms of various elements. One of which is oxygen. It is in second place. This is oxide. The overall formula of oxides is RO, where R is a certain element.

RBOH - refers to the classroom. General for all bases - the presence of a group of it, which is connected to the metal (the exception is NH 4 OH, where it is connected to the NH 4 group).

Acids are complex substances consisting of hydrogen atoms and an acid residue.

Therefore, the formulas of all acids begin with hydrogen atoms, and the acid residue is underway. For example: HCl, H 2 SO 4, HNO 3, etc.

And last, write a salt formula. Salts are complex substances consisting of metal atoms and an acid residue, for example NaCl, K 2 SO 4.

To execute the tasks 6-8, use the information contained in this text.

Phosphorus oxide (V) (P 2 O 5) is formed when the phosphorus is burning in air and is a white powder. This substance is very active and with the release of a large amount of heat reacts with water, so it is used as a gas dryer and liquids, watering means in organic synthesis.

The product of phosphorus oxide (V) reaction with water is phosphoric acid (H 3 PO 4). This acid exhibits all the common properties of acids, for example, interacts with the bases. Such reactions are called neutralization reactions.

Salts of phosphoric acid, such as sodium phosphate (Na 3 PO 4), are most widely used. They are injected into detergents and washing powders are used to reduce the rigidity of water. At the same time, the release of excess phosphates with wastewater in reservoirs contributes to the rapid development of algae (water flowering), which causes the need to carefully monitor the content of phosphates in the waste and natural waters. To detect phosphate ion, you can use a reaction with silver nitrate (AGNO 3), which is accompanied by the formation of the yellow sediment of silver phosphate (AG 3 PO 4)

Task 6.

1) Make an oxygen phosphorus reaction equation.

Answer: ________

2) On what property of phosphorus oxide (V) is it used to use it as a drying agent?

Answer: ________

Decision

In this task, it is necessary to make an oxygen phosphorus response equation and answer the question why the product of this reaction is used as a drying reagent.

We write the reaction equation and set the coefficients: 4 P. + 5 O. 2 = 2 P. 2 O. 5

Phosphorus oxide is used as a drying reagent for the ability to take water from substances.

Task 7.

1) Make a molecular equation of reaction between phosphoric acid and sodium hydroxide.

Answer: ________

2) Specify to which type of reactions (compounds, decomposition, substitution, exchange) refers to the interaction of phosphoric acid with sodium hydroxide.

Answer: ________

Decision

In the seventh task, it is necessary to draw up the equation of the reaction between phosphoric acid and sodium hydroxide. In order to do this, it is necessary to recall that this reaction refers to the exchange reactions when complex substances are exchanged components.

H. 3 PO. 4 + 3 Naoh. = Na. 3 PO. 4 + 3 H. 2 O.

Here we see that hydrogen and sodium in reaction products have changed in places.

Task 8.

1) Conduct an abbreviated ionic equation of reaction between sodium phosphate solutions (Na 3 PO 4) and silver nitrate.

Answer: ________

2) Specify the sign of the flow of this reaction.

Answer: ________

Decision

We will write the reaction equation in the abbreviated ion form between sodium phosphate solutions and silver nitrate.

In my opinion, it is first necessary to write the equation of the reaction in a molecular form, then place the coefficients and determine which of the substances leaves from the reaction medium, that is, it falls into a precipitate, it is released in the form of a gas or forms a smallssaging substance (for example, water). It will help us in this solubility table.

Na. 3 PO. 4 + 3 Agno. 3 = AG 3 PO. 4 + 3 Nano. 3

The arrow, standing near the phosphate of silver downward, indicates that this compound is not soluble in water and falls in the form of a precipitate, therefore it is not subjected to dissociation and in the ionic equations of the reaction is recorded as a molecule. We write the full ion equation of this reaction:

Now cross out the ions that moved from the left part of the equation to the right, without changing their charge:

3NA + + PO 4 3- + 3AG + 3NO 3 - \u003d AG 3 PO 4 + 3NA + + 3NO 3 -

All that is not crossed out, we will write down in the abbreviated ion equation:

PO 4 3- + 3AG + = AG 3 PO 4

Task 9.

Dana scheme of an oxidation reaction.

Mn (OH) 2 + Kbro 3 → MnO 2 + kv + H 2 O

1. Make an electronic balance of this reaction.

Answer: ________

2. Specify the oxidizer and reducing agent.

Answer: ________

3. Arrange the coefficients in the reaction equation.

Answer: ________

Decision

The following task is proposed to explain the redox process.

Mn (OH) 2 + KBRO 3 → MNO 2 + KBR + H 2 O

In order to do this, we will write to the symbol of each element its degree of oxidation in this connection. Do not forget that in the amount, all the degrees of oxidation of the substance are zero, as they are electronic. The degree of oxidation of atoms and molecules consisting of the same substance is also zero.

Mn 2+ (O 2- H +) 2 + K + BR 5+ O 3 2- → Mn 4+ O 2 2- + K + BR - + H 2 + O 2 -

Mn 2+. (O 2- H +) 2 + K + BR 5+ O 3 2- → Mn 4+O 2 2- + K + BR - + H 2 + O 2 -

Mn 2+ -2E → Mn 4+ The process of electron recoil - oxidation. At the same time, the element in the process of reaction increases the degree of oxidation. This element is a reducing agent, it restores the bromine.

Br 5+ + 6e → BR - electron acceptance process - recovery. In this case, the element in the reaction process the degree of oxidation decreases. This element is an oxidizing agent, it oxidizes the manganese.

The oxidizing agent is a substance that the electrons receives and is restored (the degree of oxidation of the element is reduced).

The reducing agent is a substance that the electrons gives it to and is oxidized (the degree of oxidation of the element decreases). In school, this is written as follows.

The figure 6, which stands after the first vertical line, is the lowest total multiple digit 2 and 6 - the numbers of the removable electrons with the reducing agent and the electrons received by the oxidizing agent. We divide this figure to the number of removable electrons with a reducing agent and obtain the number 3, it is placed after the second vertical feature and is a coefficient in the equation of the redox reaction, which is placed before the reducing agent, that is, manganese. Next, the number 6 is divided by number 6 - the number of accepted electrons by oxidizing agent. We obtain the figure 1. This is a coefficient that is placed in the equation of the redox reaction in front of the oxidizing agent, that is, bromo. We enter the coefficients into the abbreviated equation, and then transferred to the main equation.

3mn (OH) 2 + KBRO 3 → 3MNO 2 + KBR + 3H 2 O

If necessary, we arrange other coefficients with the calculation so that the number of atoms of the same item is the same. At the end, we check the number of oxygen atoms before and after the reaction. If their number turns out to be equal, it means that we did everything right. In this case, it is necessary to put the coefficient 3 before water.

Dana scheme of transformations:

Cu → Cucl 2 → Cu (OH) 2 → CU (NO 3) 2

Write the molecular equations of reactions with which you can implement the transformations.

Decision

We solve the scheme of transformations:

Cu.→ Cucl 2 → Cu.(Oh.) 2 → Cu.(No. 3 ) 2

1) Cu. + Cl. 2 = Cucl 2 - I care that copper with hydrochloric acid does not interact, as it stands in a row of stresses of metals after hydrogen. Therefore, one of the main reactions. Interaction directly with chlorine.

2) Cucl 2 + 2 Naoh. = Cu.(Oh.) 2 + 2 NaCl- Exchange exchange.

3) Cu.(Oh.) 2 + 2 Hno. 3 = Cu.(No. 3 ) 2 + 2 H. 2 O.- Hydroxide of copper - precipitate, so to obtain from it the copper nitrate of nitric acid salts will not fit.

Set the correspondence between the title of the organic matter and the class / group to which this substance belongs: to each position indicated by the letter, select the corresponding position indicated by the number.

Write in the table selected numbers under the appropriate letters.

1. Methanol is alcohol. Names of monohydric alcohols ends on-ol, so A2..

2. Acetylene is an unforeseen hydrocarbon. This is the trivial name here. On systematic nomenclature it is called ethin.Choose B4..

3. Glucose is a carbohydrate, monosaccharide. Therefore, choose IN 1.

In the proposed schemes of chemical reactions, insert the formulas of the missed substances and lay the coefficients where it is necessary.

1) C 6 H 6 + BR 2		C 6 H 5 -BR + ...

2) CH 3 CHO + ... → CH 3 CH 3 OH

Decision

It is necessary to insert the formulas of the missed substances and, if necessary, arrange the coefficients:

1) C 6 H 6 + BR 2 ⎯albr 3 → C 6 H 5 -BR + HBr for benzene and its homologues are characterized by replacement reaction, therefore, in this reaction, the hydrogen atom is replaced in the benzene and the bromobenzene is obtained.

2) CH 3 Cho + H 2 → CN 3 С 2 OH Recovery of the recovery of acetaldehyde to ethyl alcohol.

Acetic acid is widely used in the chemical and food industry. Aqueous solutions of acetic acid (food additive E260) are used in household cooking, in preservation, as well as for the preparation of medicinal and fragrant substances. The latter include numerous acetic esters of acetic acid, for example, propylacetate.

Calculate how many grams of propylacetate (CH 3 SOAP 3 H 7) can be obtained as a result of the reaction of 300 g of acetic acid (CH 3 of the Soam) with propanol-1 (C 3 H 7) at a 100% practical output. Record the equation of the flowing reaction and the detailed solution of the problem.

Answer: ________

A task. We write a brief condition of the task:

m (CH 3 coa 3 H 7) \u003d?

1. In the condition of the problem, it is said that acetic acid entered into a reaction by weighing 300 g. We define the number of moles in 300 g. To do this, we use the magic triangle, where N is the number of moles.

We substitute the numbers: n \u003d 300 g .: 60 g / mol \u003d 5 mol. Thus, acetic acid entered a reaction with a propyl alcohol with the amount of substance 5 mol. Next, we define how many moles of CH 3 coa 3 H 7 is formed from 5 mole CH 3 of the coxy. According to the reaction equation, the acetic acid reacts in the amount of 1 mol, and the ether is also formed 1 mol, since there are no coefficients in the reaction equation. Therefore, if you take the acid with the amount of 5 mol, then the ether will also get 5 mol. Since they react in a 1: 1 ratio.

Well, it remains to calculate the mass of 5 mol ether, using this triangle.

Substituting the numbers, we get: 5 mol · 102 g / mol \u003d 510

Answer: Ether mass \u003d 510

Acetylene is used as fuel with gas welding and cutting of metals, as well as raw materials for the production of vinyl chloride and other organic substances. In accordance with the scheme below, make the equations of reactions characteristic of acetylene. When writing the reaction equations, use structural formulas for organic substances.

Decision

Make the transformations characteristic of acetylene, according to the scheme.

I would like to say that acetylene is an unforeseen hydrocarbon having 2 π-bonds between carbon atoms, therefore, it is characterized by the reaction of attachment, oxidation, polymerization at the site of the π-bond break. Reactions can go in two stages.

Ringer's solution is widely used in medicine as a water-salt regulator, plasma substitute and other blood components. For its preparation in 1 l of distilled water, 8.6 g of sodium chloride is dissolved, 0.33 g of calcium chloride and 0.3 g of potassium chloride. Calculate the mass fraction of sodium chloride and calcium chloride in the resulting solution. Write down a detailed solution of the problem.

Answer: ________

Decision

To solve this task, we write its brief condition:

m (H 2 O) \u003d 1000 m (CaCl 2) \u003d 0.33 m (KCl) \u003d 0.3 g. m (NaCl) \u003d 8.6	Since the density of water is equal to one, 1 liter of water will have a mass equal to 1000 grams. Further, to find a mass fraction in percent of the solution, we use the magic triangle, m (V-BA) - mass of substance; m (p-ra) - mass of solution; ω is a mass fraction of a substance in percent in this solution. We derive the formula for finding ω% in solution. It will have the following form:
Ω% (P-RA NaCl)

In order to immediately move to finding a mass fraction in the percentage of NACI solution, we must know two other values, that is, the mass of the substance and the mass of the solution. The mass of the substance is known to us from the condition of the problem, and the mass of the solution should be found. The mass of the solution is equal to the mass of water plus the masses of all salts dissolved in water. The formula for calculating is simple: M (V-BA) \u003d M (H 2 O) + M (NaCl) + M (CaCl 2) + M (KCl), folding all values, we get: 1000 g. + 8.6 g + 0.3 g. + 0.33 g. \u003d 1009.23. This will be the mass of the whole solution.

Now we find a mass fraction of NaCl in solution:

Similarly, we calculate the mass of calcium chloride:

We substitute the numbers and get:

Answer: ω% in the solution NaCl \u003d 0.85%; Ω% in solution CaCl 2 \u003d 0.033%.