Structure

Each version of VLOOKUP in chemistry contains 15 tasks of various types and difficulty levels.

The work contains 4 tasks of an increased level of complexity (their serial numbers: 9, 10, 13, 14).

The tasks included in the work can be conditionally divided into four meaningful blocks: "Theoretical foundations of chemistry", "Inorganic chemistry", " Organic chemistry"," Methods of cognition in chemistry. Experimental Foundations of Chemistry. Chemistry and Life ".

Assignment Grading Explanations

Correct performance of the task with serial number 3 is estimated at 1 point.

Correct execution of each of the other tasks basic level difficulty is estimated with a maximum of 2 points. In case of one error or incomplete answer, 1 point is given. The rest of the answer options are considered incorrect and are evaluated at 0 points.

Assessment of tasks of an increased level of complexity is carried out on the basis of an element-by-element analysis of students' answers. The maximum mark for a correctly completed task is 3 points. Students can complete tasks with a detailed answer in different ways. Therefore, the sample decisions given in the evaluation criteria should be considered only as one of the possible answers.

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

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

This allowance is fully consistent with the federal state educational standard (second generation). The book contains 10 options for typical tasks of the All-Russian verification work (VLOOKUP) in chemistry for 11th grade students. The collection is intended for 11th grade students, teachers and methodologists who use standard tasks to prepare for the All-Russian Chemistry Testing Work.

Format: pdf

The size: 3.4 MB

Watch, download:drive.google

Work instructions, 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
Scoring system for test work 75
76 responses
Appendices 93

Testing work includes 15 tasks. Chemistry work takes 1 hour 30 minutes (90 minutes).
Fill out the answers in the text of the work according to the instructions for the tasks. If you write down an incorrect answer, cross it out and write a new one next to it.
When performing work, it is allowed to use the following additional materials:
- Periodic system chemical elements D.I. Mendeleev;
- table of solubility of salts, acids and bases in water;
- electrochemical series of metal voltages;
- non-programmable calculator.
When completing assignments, you can use a draft. Draft entries will not be reviewed or graded.
We advise you to complete the tasks in the order in which they are given. To save time, skip a task that cannot be completed immediately and move on to the next. If, after completing all the work, you have time left, you can return to the missed tasks.
The points received by you for completed tasks are summed up. Try to complete as many tasks as possible and score the most points.

On April 27, 2017, for the first time, all-Russian verifications of the CDP in chemistry in 11 grades took place in the approbation mode.

Official site of VPR (StatGrad) - vpr.statgrad.org

VLOOKUP options for chemistry grade 11 2017

№	Download answers (evaluation criteria)
Option 11	the answers
Option 12	the answers
Option 13	the answers
Option 14	the answers
Option 15	variant 15 otvet
Option 16	variant 16 otvet
Option 17	variant 17 otvet
Option 18	variant 18 otvet

To get acquainted with the approximate options for work on the official website of the FIPI, posted demo options with answers and descriptions.

Samples of VLOOKUP in chemistry grade 11 2017 (demo version)

Testing work includes 15 tasks. Chemistry work takes 1 hour 30 minutes (90 minutes).

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

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

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

- electrochemical series of metal voltages;

- non-programmable calculator.

The structure and content of the all-Russian verification work of the VPR in chemistry

Each version of the VLOOKUP contains 15 tasks of various types and difficulty levels. The variants contain tasks of various formats.

These tasks differ in the required form of recording the answer. So, for example, the answer can be: a sequence of numbers, symbols; the words; formulas of substances; reaction equations.

The work contains 4 tasks of an increased level of complexity (their serial numbers: 9, 10, 13, 14). These tasks are more difficult, since their implementation involves the complex application of the following skills:

- to draw up reaction equations that confirm 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 completing assignments, you can use a draft. Draft entries will not be reviewed or graded.

The average general education

The line of the UMK V. V. Lunin. Chemistry (10-11) (basic)

The line of the UMK V. V. Lunin. Chemistry (10-11) (D)

UMK line N. E. Kuznetsova. Chemistry (10-11) (basic)

UMK line N. E. Kuznetsova. Chemistry (10-11) (in-depth)

CDF in chemistry. Grade 11

Testing work includes 15 tasks. Chemistry work takes 1 hour 30 minutes (90 minutes).

Write down the answers to the tasks in the space provided for them. If you write down an incorrect answer, cross it out and write a new one next to it.

When performing work, it is allowed to use:

• Periodic table of chemical elements D.I. Mendeleev;
• table of solubility of salts, acids and bases in water;
• electrochemical series of metal voltages;
• non-programmable calculator.

When completing assignments, you can use a draft. Draft entries will not be reviewed or graded.

We advise you to complete the tasks in the order in which they are given. To save time, skip a task that cannot be completed immediately and move on to the next. If, after completing all the work, you have time left, you can return to the missed tasks.

The points received by you for completed tasks are summed up. Try to complete as many tasks as possible and score the most points.

We wish you success!

From the course of chemistry, you know the following methods for separating mixtures: settling, filtration, distillation (distillation), magnet action, evaporation, crystallization.

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

Determine which of the methods of separation of mixtures shown in the figure can be used for separation:

1. cereals and iron filings trapped in it;
2. water and salts dissolved in it.

Write down the figure number and the name of the corresponding mixture separation method in the table.

Decision

1.1. The separation of a mixture of cereals and iron filings is based on the property of iron to be attracted by a magnet. Figure 3.

1.2. Separation of a mixture of water and dissolved salts occurs during distillation. When heated to the boiling point, water evaporates and, when cooled in a water refrigerator, flows into a previously prepared vessel. Picture 1.

The figure shows a diagram of the distribution of electrons over the energy levels of an atom of a certain chemical element.

Based on the proposed scheme, complete the following tasks:

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

Write down the answers in the table.

Decision

The figure shows a diagram of the structure of an atom:

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

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

Assignment 3

Periodic table of chemical elements D.I. Mendeleev - a rich repository of information about chemical elements, their properties and the properties of their compounds. So, for example, it is known that with an increase in the serial number of a chemical element, the basic character of the oxide decreases in periods, and increases in groups.

Taking into account these regularities, arrange the following elements in order of increasing the basicity of oxides: Na, Al, Mg, B. Write down the symbols of the elements in the required sequence.

Answer: ________

Decision

As you know, the sum of protons in the nucleus of an atom is equal to the ordinal number of the element. But the number of protons is not indicated to us. Since an atom is an electrically neutral particle, the number of protons (positively charged particles) in the nucleus of an atom is equal to the number of electrons (negatively charged particles) revolving around the nucleus of the atom. The total number of electrons revolving around the nucleus is 15 (2 + 8 + 5), therefore, the ordinal number of the element is 15. Now it remains to look into the periodic system of chemical elements of DI Mendeleev and find number 15. This is P (phosphorus). Since phosphorus has 5 electrons on the last electron layer, it is non-metal; metals in the last layer have 1 to 3 electrons.

4 elements from the periodic system of Mendeleev are given: Na, Al, Mg, B. They must be arranged so that the basicity of the oxides formed by them increases. Answering this VLOOKUP question, it is necessary to remember how the metallic properties change in the periods and groups of the periodic system.

In periods from left to right, metallic properties decrease and non-metallic properties increase. Consequently, the basicity of oxides also decreases.

In groups, 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 elements given to us. Two of them are in the third group; these are B and Al. Aluminum in the group is lower than boron, therefore, its metallic properties are more pronounced than boron. Accordingly, the basicity of aluminum oxide is more pronounced.

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

Assignment 4

The table below summarizes some of the characteristics of covalent and ionic bonding.

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

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

Decision

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

Using it, one can determine that CaCl 2 is characterized by an ionic type of bond, since it consists of a metal atom (Ca) and non-metal atoms (Cl), and for H 2 it is covalent non-polar, since this molecule consists of atoms of the same element - hydrogen.

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

Decision

The next task is to test the knowledge of the main classes of inorganic substances.

Empty cells must be filled in the table. In the first two cases, the formulas of substances are given, it is necessary to class them as belonging to a certain class of substances; in the latter two, on the contrary, write formulas for representatives of these classes.

CO 2 is a complex substance, composed of atoms of various elements. One of which is oxygen. He is in second place. It is an oxide. The general formula for oxides is RO, where R is a specific element.

RbOH - belongs to the class of bases. Common to all bases is the presence of the OH group, which is connected to the metal (the exception is NH 4 OH, where the OH group is connected to the NH 4 group).

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

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

Lastly, write the salt formula. Salts are complex substances consisting of metal atoms and an acidic residue, for example NaCl, K 2 SO 4.

To complete assignments 6-8, use the information contained in this text

Phosphorus (V) oxide (P 2 O 5) is formed by combustion of phosphorus in air and is a white powder. This substance is very active and reacts with water with the release of a large amount of heat, therefore it is used as a desiccant for gases and liquids, as a dehydrating agent in organic synthesis.

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

Salts of phosphoric acid, for example sodium phosphate (Na 3 PO 4), are widely used. They are added to detergents and washing powders, used to reduce the hardness of water. At the same time, the ingress of an excess amount of phosphates with wastewater into water bodies contributes to the rapid development of algae (water bloom), which necessitates careful control of the phosphate content in wastewater and natural waters. To detect the phosphate ion, you can use the reaction with silver nitrate (AgNO 3), which is accompanied by the formation of a yellow precipitate of silver phosphate (Ag 3 PO 4)

Assignment 6

1) Make the equation for the reaction of phosphorus with oxygen.

Answer: ________

2) What property of phosphorus (V) oxide is its use as a drying agent based on?

Answer: ________

Decision

In this task, it is necessary to draw up an equation for the reaction of phosphorus with oxygen and answer the question of why the product of this reaction is used as a drying agent.

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

Phosphorus oxide is used as a drying agent for its ability to remove water from substances.

Assignment 7

1) Write the molecular equation for the reaction between phosphoric acid and sodium hydroxide.

Answer: ________

2) Indicate what type of reactions (compound, decomposition, substitution, exchange) the interaction of phosphoric acid with sodium hydroxide belongs to.

Answer: ________

Decision

In the seventh task, it is necessary to draw up an equation for the reaction between phosphoric acid and sodium hydroxide. In order to do this, it is necessary to remember that this reaction refers to exchange reactions, when complex substances exchange their constituent parts.

H 3 PO 4 + 3 NaOH = Na 3 PO 4 + 3 H 2 O

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

Assignment 8

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

Answer: ________

2) Indicate the sign of this reaction.

Answer: ________

Decision

Let us write the reaction equation in an abbreviated ionic form between solutions of sodium phosphate and silver nitrate.

In my opinion, first it is necessary to write the reaction equation in molecular form, then arrange the coefficients and determine which of the substances leaves the reaction medium, that is, precipitates, escapes as a gas, or forms a low-dissociating substance (for example, water). The solubility table will help us with this.

Na 3 PO 4 + 3 AgNO 3 = Ag 3 PO 4 + 3 NaNO 3

The downward arrow next to silver phosphate indicates that this compound is insoluble in water and precipitates, therefore it does not undergo dissociation and is written in the ionic equations of the reaction as a molecule. Let's write the complete ionic equation for this reaction:

Now we delete the ions that have passed from the left side 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 -

Everything that is not crossed out, we write in the abbreviated ionic equation:

PO 4 3– + 3Ag + = Ag 3 PO 4

Assignment 9

The scheme of the redox reaction is given.

Mn (OH) 2 + KBrO 3 → MnO 2 + KBr + H 2 O

1. Make an electronic balance of this reaction.

Answer: ________

2. Specify the oxidizing agent and reducing agent.

Answer: ________

3. Place the coefficients in the reaction equation.

Answer: ________

Decision

The next task is to explain the redox process.

Mn (OH) 2 + KBrO 3 → MnO 2 + KBr + H 2 O

In order to do this, we write next to the symbol of each element its oxidation state in this compound. Do not forget that in total all the oxidation states of a substance are zero, since they are electrically neutral. The oxidation state 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 donating electrons - oxidation. In this case, the oxidation state of the element increases during the reaction. This element is a reducing agent; it reduces bromine.

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

An oxidizing agent is a substance that accepts electrons and is reduced at the same time (the oxidation state of an element decreases).

Reducing agent is a substance that gives up electrons and is oxidized at the same time (the oxidation state of an element decreases). At school, this is recorded as follows.

The number 6 after the first vertical bar is the smallest common multiple of the numbers 2 and 6 - the number of electrons donated by the reducing agent and electrons taken by the oxidizing agent. We divide this figure by the number of electrons donated by the reducing agent and we get the figure 3, it is placed after the second vertical line and is the coefficient in the equation of the redox reaction, which is placed before the reducing agent, that is, manganese. Further, the number 6 is divided by the number 6 - the number of electrons received by the oxidizer. We get the number 1. This is the coefficient that is put in the equation of the redox reaction before the oxidizing agent, that is, bromine. We enter the coefficients into the abbreviated equation, and then we transfer them to the main equation.

3Mn (OH) 2 + KBrO 3 → 3MnO 2 + KBr + 3H 2 O

If necessary, we arrange other coefficients so that the number of atoms of the same element is the same. Finally, we check the number of oxygen atoms before and after the reaction. If their number turns out to be equal, then we did everything right. In this case, it is necessary to put a coefficient of 3 in front of the water.

The scheme of transformations is given:

Cu → CuCl 2 → Cu (OH) 2 → Cu (NO 3) 2

Write down the molecular reaction equations that can be used to carry out the indicated transformations.

Decision

We solve the transformation scheme:

Cu→ CuCl 2 → Cu(OH) 2 → Cu(NO 3 ) 2

1) Cu + Cl 2 = CuCl 2 - I would like to draw your attention to the fact that copper does not interact with hydrochloric acid, since it is among the 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–Reaction of exchange.

3) Cu(OH) 2 + 2 HNO 3 = Cu(NO 3 ) 2 + 2 H 2 O- copper hydroxide is a precipitate, therefore, nitric acid salts are not suitable for obtaining copper nitrate from it.

Establish a correspondence between the name of the organic substance and the class / group to which this substance belongs: for each position indicated by a letter, select the corresponding position indicated by a number.

Write down the selected numbers in the table under the corresponding letters.

1. Methanol is alcohol. The names of monohydric alcohols end in -ol, so A2.

2. Acetylene is an unsaturated hydrocarbon. This is the trivial name given here. According to the systematic nomenclature, it is called ethin.We choose B4.

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

Insert the formulas of the missing substances into the proposed schemes of chemical reactions and place the coefficients where 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 missing 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, substitution reactions are characteristic, therefore in this reaction bromine replaces the hydrogen atom in benzene and bromobenzene is obtained.

2) CH 3 CHO + H 2 → CH 3 CH 2 OH The reaction of reducing acetaldehyde to ethyl alcohol.

Acetic acid is widely used in the chemical and food industries. Aqueous solutions of acetic acid (food additive E260) are used in household cooking, canning, as well as to obtain medicinal and aromatic substances. The latter include numerous acetic acid esters such as propyl acetate.

Calculate how many grams of propyl acetate (CH 3 COOC 3 H 7) can be obtained by reacting 300 g of acetic acid (CH 3 COOH) with propanol-1 (C 3 H 7 OH) at 100% practical yield. Write down the equation of the reaction and a detailed solution to the problem.

Answer: ________

Task. We write down a short statement of the problem:

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

1. In the condition of the problem it is said that acetic acid reacted with a mass of 300 g. Let us determine the number of moles in 300 g. To do this, we will use a magic triangle, where n is the number of moles.

Substitute the numbers: n \u003d 300 g: 60 g / mol \u003d 5 mol. Thus, acetic acid reacted with propyl alcohol in an amount of 5 mol. Next, we determine how many moles of CH 3 COOC 3 H 7 are formed from 5 moles of CH 3 COOH. According to the reaction equation, acetic acid reacts in an amount of 1 mol, and the ether is also formed 1 mol, since there are no coefficients in the reaction equation. Therefore, if we take an acid in the amount of 5 mol, then the ether will also turn out to be 5 mol. Since they react in a 1: 1 ratio.

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

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

Answer: ether mass \u003d 510 g.

Acetylene is used as a fuel for gas welding and metal cutting, as well as a raw material for the production of vinyl chloride and other organic substances. In accordance with the scheme below, draw up the reaction equations for acetylene. When writing reaction equations, use the structural formulas of organic substances.

Decision

Carry out the transformations characteristic of acetylene according to the given scheme.

I would like to say that acetylene is an unsaturated hydrocarbon that has 2 π-bonds between carbon atoms, therefore, it is characterized by addition, oxidation, and polymerization reactions at the site of breaking π-bonds. The reactions can proceed in two stages.

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

Answer: ________

Decision

To solve this problem, we write down its brief condition:

m (H 2 O) \u003d 1000 g. m (CaCl 2) \u003d 0.33 g. m (KCl) \u003d 0.3 g. m (NaCl) \u003d 8.6 g.

Since the density of water is equal to one, 1 liter of water will have a mass equal to 1000 grams. Next, to find the mass fraction in percent of the solution, we use the magic triangle, m (in-va) - the mass of the substance; m (solution) is the mass of the solution; ω - mass fraction of a substance in percent in a given solution. Let's derive the formula for finding ω% in solution. It will look like this:

ω% (solution NaCl)

In order to go directly to finding the mass fraction in percent of the NaCl solution, we must know two other values, that is, the mass of the substance and the mass of the solution. We know the mass of the substance 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 mass of all salts dissolved in water. The formula for the calculation is simple: m (in-va) \u003d m (H 2 O) + m (NaCl) + m (CaCl 2) + m (KCl), adding all the values, we get: 1000 g + 8.6 g. + 0.3 g + 0.33 g \u003d 1009.23 g. This will be the mass of the entire solution.

Now we find the mass fraction of NaCl in the solution:

Similarly, we calculate the mass of calcium chloride:

Substitute the numbers and get:

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

VPR All-Russian Testing Work- Chemistry Grade 11

Explanations for a sample of all-Russian verification work

When familiarizing yourself with the sample test work, it should be borne in mind that the tasks included in the sample do not reflect all the skills and content issues that will be tested as part of the All-Russian test work. A complete list of content elements and skills that can be tested in the work is given in the codifier of content elements and requirements for the level of training of graduates for the development of an all-Russian test work in chemistry. The purpose of the test work sample is to give an idea of \u200b\u200bthe structure of the All-Russian test work, the number and form of tasks, the level of their complexity.

Work instructions

Testing work includes 15 tasks. Chemistry work takes 1 hour 30 minutes (90 minutes).
Fill out the answers in the text of the work according to the instructions for the tasks. If you write down an incorrect answer, cross it out and write a new one next to it.
When performing work, it is allowed to use the following additional materials:
- Periodic table of chemical elements D.I. Mendeleev;
- table of solubility of salts, acids and bases in water;
- electrochemical series of metal voltages;
- non-programmable calculator.
When completing assignments, you can use a draft. Draft entries will not be reviewed or graded.
We advise you to complete the tasks in the order in which they are given. To save time, skip a task that cannot be completed immediately and move on to the next. If, after completing all the work, you have time left, you can return to the missed tasks.
The points received by you for completed tasks are summed up. Try to complete as many tasks as possible and score the most points.
We wish you success!

1. From the course of chemistry you know the following methods of separating mixtures: settling, filtration, distillation (distillation), magnet action, evaporation, crystallization. Figures 1–3 show examples of using some of these methods.

Which of the above methods of separating mixtures can be used for purification:
1) flour from iron filings trapped in it;
2) water from dissolved inorganic salts?
Write down the figure number and the name of the corresponding mixture separation method in the table.

iron filings are attracted by a magnet

during distillation after condensation of water vapor, salt crystals remain in the vessel

2. The figure shows a model of the electronic structure of an atom of some chemicalelement.

Based on the analysis of the proposed model, complete the following tasks:
1) determine the chemical element whose atom has such an electronic structure;
2) indicate the number of the period and the number of the group in the Periodic Table of Chemical Elements of D.I. Mendeleev, in which this element is located;
3) determine whether a simple substance that forms this chemical element belongs to metals or non-metals.
Write down the answers in the table.
Answer:

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

to determine a chemical element, one should calculate the total number of electrons, which we see in figure (7)

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

after that we determine the group number (vertical column) (5) and the nature of this element (non-metal)

3. Periodic table of chemical elements D.I. Mendeleev - a rich repository of information about chemical elements, their properties and the properties of their compounds, about the patterns of changes in these properties, about the methods of obtaining substances, as well as about their finding in nature. So, for example, it is known that with an increase in the ordinal number of a chemical element in periods, the radii of atoms decrease, and in groups they increase.
Considering these patterns, arrange the following elements in order of increasing atomic radii: N, C, Al, Si. Write down 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 ionic structure.

Using this information, determine the structure of the substances nitrogen N2 and sodium chloride. Write your answer in the space provided:

1) nitrogen N2 ________________________________________________________________
2) table salt NaCl ___________________________________________________

nitrogen N2 - molecular structure;
table salt NaCl - ionic structure

5. Complex inorganic substances can be conditionally distributed, that is, classified, into four groups, as shown in the diagram. In this diagram, for each of the four groups, write the missing names of the groups or chemical formulas of substances (one example of the formulas) belonging to this group.

The names of the groups are recorded: bases, salts;
the formulas of the substances of the corresponding groups are written

CaO, bases, HCl, salts

Read the following text and complete assignments 6–8.

In the food industry, the food additive E526 is used, which is calcium hydroxide Ca (OH) 2. It finds application in the production of: fruit juices, baby food, pickled cucumbers, table salt, confectionery and sweets.
Production of calcium hydroxide on an industrial scale is possible by mixing calcium oxide with waterThis process is called quenching.
Calcium hydroxide is widely used in the production of building materials such as whitewash, plaster and gypsum solutions. This is due to his ability interact with carbon dioxide CO2contained in the air. The same property of calcium hydroxide solution is used to measure the quantitative content of carbon dioxide in the air.
A useful property of calcium hydroxide is its ability to act as a flocculant, cleaning wastewater from suspended and colloidal particles (including iron salts). It is also used to raise the pH of water, as natural water contains substances (e.g. acid) that cause corrosion in plumbing pipes.

1. Make a molecular equation for the reaction of obtaining calcium hydroxide, which
mentioned in the text.

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

________________________________________________________________________________

1) CaO + H 2 O \u003d Ca (OH) 2
2) When calcium oxide interacts with water, a large
the amount of heat, so the water boils and hisses, as if it hits a hot coal, when the fire is extinguished with water (or "this process is called extinguishing, because as a result slaked lime is formed")

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

2. Explain what features of this reaction make it possible to use it to detect
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, turbidity of the initial solution is observed, which makes it possible to judge the presence of carbon dioxide in the air (qualitative
reaction to CO 2)

1. Make an abbreviated ionic equation for the reaction mentioned in the text between
calcium hydroxide and hydrochloric acid.
Answer:__________________________________________________________________________

2. Explain why this reaction is used to raise 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 causes low pH values \u200b\u200bof this water. Calcium hydroxide neutralizes acid and pH values \u200b\u200brise

the pH scale is from 0-14. from 0-6 - acidic environment, 7- neutral environment, 8-14 - alkaline environment

9. The scheme of the redox reaction is given.

H 2 S + Fe 2 O 3 → FeS + S + H 2 O

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

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

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

1) An electronic balance has been drawn up:

2Fe +3 + 2ē → 2Fe +2	2		1
		2
S -2 - 2ē → S 0	2		1

2) It is indicated that sulfur in the oxidation state –2 (or H 2 S) is a reducing agent, and iron in the oxidation state +3 (or Fe 2 O 3) is an oxidizing agent;
3) The reaction equation is composed:
3H 2 S + Fe 2 O 3 \u003d 2FeS + S + 3H 2 O

10. The scheme of transformations is given:

Fe → FeCl 2 → Fe (NO 3) 2 → Fe (OH) 2

Write the molecular reaction equations that can be used to carry out
indicated transformations.
1) _________________________________________________________________________
2) _________________________________________________________________________
3) _________________________________________________________________________

The reaction equations are written corresponding to the transformation scheme:
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 are allowed that do not contradict the condition for setting the equation
reactions.)

11. Establish a correspondence between the formula of organic matter and the class / groupto which the substance belongs: for each letter-marked position, select the corresponding number-marked position.

Write down the selected numbers in the table under the corresponding letters.
Answer:

AND	B	AT

1. C3H8 - CnH2n + 2 - alkane
2. C3H6 - CnH2n- alkene
3. C2H6O - CnH2n + 2O- alcohol

12. Insert the formulas of the missing substances into the proposed schemes of chemical reactions and place the coefficients.

1) С 2 Н 6 + …………… ..… → С 2 Н 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 odds are possible.)

13. Propane burns with low emission of toxic substances into the atmosphere, therefore, it is used as an energy source in many areas, such as gas lighters and heating country houses.
What is the volume of carbon dioxide (n.u.) produced by the complete combustion of 4.4 g of propane?
Write down a detailed solution to the problem.
Answer:__________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________

1) The equation for the reaction of propane combustion is compiled:
C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O
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 a part of household chemicals, perfumery and cosmetic products, windshield washer fluids for cars. In accordance with the scheme below, compose the reaction equations for the production of this alcohol. When writing reaction equations, use the structural formulas of organic substances.

1) _______________________________________________________
2) _______________________________________________________
3) _______________________________________________________

The reaction equations are written, corresponding to the scheme:

(Others are allowed that do not contradict the condition for setting the reaction equation.)

15. Physiological solution in medicine is a 0.9% solution of sodium chloride in water. Calculate the mass of sodium chloride and mass of water required to prepare 500 g of saline. Write down a detailed solution to the problem.
Answer:__________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________

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

m (solution) \u003d 500g m (salt) \u003d x

x / 500 * 100% \u003d 0.9%

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

© 2017 Federal Service for Supervision in Education and Science of the Russian Federation