Determine, atoms of which of those specified in the number of elements are mainly containing one unpaired electron.
Record in the response field of the selected items.
Answer:

Answer: 23.
Explanation:
We write the electronic formula for each of these chemical elements and depicting the electronic graphic formula of the last electronic level:
1) S: 1S 2 2S 2 2P 6 3S 2 3P 4

2) NA: 1S 2 2S 2 2P 6 3S 1

3) AL: 1S 2 2S 2 2P 6 3S 2 3P 1

4) Si: 1S 2 2S 2 2P 6 3S 2 3P 2

5) MG: 1S 2 2S 2 2P 6 3S 2

From those specified in a number of chemical elements, select three metal elements. Place the selected items in order of increasing reducing properties.

Write down in the response field of the selected items in the desired sequence.

Answer: 352.
Explanation:
In the main subgroups of the Mendeleev table, the metals are located under the diagonal of Bor-Astat, as well as in side subgroups. Thus, the metals from the specified list include Na, Al and Mg.
Metal and, therefore, the reduction properties of the elements increase when moving to the left at the period and down the subgroup.
Thus, the metal properties of the metals listed above are increasing in the series Al, Mg, Na

From among those indicated in a number of items, select two elements that in compound with oxygen exhibit the degree of oxidation +4.

Record in the response field of the selected items.

Answer: 14.
Explanation:
The main degrees of oxidation of elements from the submitted list in complex substances:
Sulfur - "-2", "+4" and "+6"
Sodium Na - "+1" (only)
Aluminum Al - "+3" (only)
Si - "-4", "+4"
Magnesium MG - "+2" (only)

From the proposed list of substances, select two substances in which ionic chemical bond is present.

Answer: 12.

Explanation:

It is possible to determine the presence of an ionic type of communication in compound in the overwhelming majority of cases, it is possible for the composition of its structural units at the same time the atoms of typical metal and the non-metal atoms are included.

Based on this criterion, the ion type of communication takes place in KCL and KNO 3 connections.

In addition to the above feature, the presence of ion communication in the compound can be said if the composition of its structural unit contains ammonium cation (NH 4 + ) Or its organic analogs - Alkylammonium cations RNH 3 + , Dialkylamonia R. 2 NH 2 + , trialkilammonium R. 3 NH +. and tetraalklammonium R. 4 N +. where R is some hydrocarbon radical. For example, the ion type of communication takes place in the connection (CH 3 ) 4 NCL between the cation (CH 3) 4 + and CL chloride ion -.

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

Answer: 241.

Explanation:

N 2 O 3 - Nemetal oxide. All non-metallic oxides besides N 2 O, NO, SiO and CO are acidic.

Al 2 O 3 - metal oxide into oxidation degree +3. Metal oxides into oxidation degree + 3, + 4, as well as BEO, ZNO, SNO and PBO, are amphoter.

HCLO 4 is a typical source representative, because During dissociation in aqueous solution, only n + cations are formed from the cations:

HCLO 4 \u003d H + + CLO 4 -

From the proposed list of substances, select two substances with each of which is interacting with zinc.

1) nitric acid (r-p)

2) iron hydroxide (II)

3) magnesium sulfate (rr)

4) sodium hydroxide (r-p)

5) Aluminum chloride (R-R)

Write down in the response field of the selected substances.

Answer: 14.

Explanation:

1) nitric acid is a strong oxidizing agent and reacts with all metals besides platinum and gold.

2) iron hydroxide (LL) is an insoluble base. With insoluble hydroxides, the metals do not react at all, and with soluble (alkalis), only three metal react - BE, Zn, Al.

3) magnesium sulfate - salt of more active metal than zinc, and therefore the reaction does not flow.

4) sodium hydroxide - alkali (soluble metal hydroxide). Alits from metals only be, zn, al.

5) AlCl 3 - salt more active than zinc metal, i.e. The reaction is impossible.

From the proposed list of substances, select two oxide that react with water.

Write down in the response field of the selected substances.

Answer: 14.

Explanation:

From oxides with water, only alkaline and alkaline earth metal oxides react, as well as all acid oxides except SiO 2.

Thus, the options for answers 1 and 4 are suitable:

Bao + H 2 O \u003d BA (OH) 2

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

1) Bromomopod

3) sodium nitrate

4) Sulfur Oxide (IV)

5) aluminum chloride

Write in the table selected numbers under the appropriate letters.

Answer: 52.

Explanation:

Salts among these substances are only sodium nitrate and aluminum chloride. All nitrates, as well as sodium soluble salts, due to the precipitate, sodium nitrate cannot give in principle with any of the reagents. Therefore, salt X can only be aluminum chloride.

A common error among the exams in chemistry is not understanding that in an aqueous solution of ammonia forms a weak base - ammonium hydroxide due to the reaction flow:

NH 3 + H 2 O<=> NH 4 Oh.

In this regard, an aqueous solution of ammonia gives a precipitate when mixed with solutions of metals salts forming insoluble hydroxides:

3NH 3 + 3H 2 O + AlCl 3 \u003d Al (OH) 3 + 3NH 4 Cl

In a given scheme of transformations

CU X\u003e CUCL 2 Y\u003e CUI

substances X and Y are:

Answer: 35.

Explanation:

Copper - metal, located in a row of activity to the right of hydrogen, i.e. Does not react with acids (except H 2 SO 4 (conc.) and HNO 3). Thus, the formation of copper chloride (LL) is possible in our case only when reactions with chlorine:

Cu + Cl 2 \u003d CUCL 2

Iodide ions (I -) cannot coexist in one solution with bivalent copper ions, because oxidize them:

Cu 2+ + 3i - \u003d Cui + I 2

Set the correspondence between the reaction equation and the substance-oxidizing agent in this reaction: to each position indicated by the letter, select the corresponding position indicated by the number.

Reaction equation A) h 2 + 2li \u003d 2lih B) N 2 H 4 + H 2 \u003d 2NH 3 C) n 2 o + h 2 \u003d n 2 + h 2 o D) N 2 H 4 + 2N 2 O \u003d 3N 2 + 2H 2 O

OXIDIZING AGENT

Write in the table selected numbers under the appropriate letters.

Answer: 1433.
Explanation:
The oxidizer in the reaction is the substance that contains an element that reduces its oxidation degree

Install the correspondence between the formula of the substance and reagents, with each of which this substance can interact: to each position indicated by the letter, select the corresponding position indicated by the number.

Formula of substances	Reagents
A) Cu (NO 3) 2	1) NaOH, MG, BA (OH) 2 2) HCl, Liohn, H 2 SO 4 (R-P) 3) BACL 2, PB (NO 3) 2, s 4) CH 3 COOH, KOH, FES 5) O 2, BR 2, HNO 3

Write in the table selected numbers under the appropriate letters.

Answer: 1215.

Explanation:

A) Cu (NO 3) 2 + NaOH and CU (NO 3) 2 + Ba (OH) 2 - similar interactions. Salt with metal hydroxide reacts if the source substances are soluble, and in the products there is a precipitate, gas or a slightly subsorative substance. And for the first and for the second reaction, both requirements are performed:

Cu (NO 3) 2 + 2NAOH \u003d 2Nano 3 + Cu (OH) 2 ↓

Cu (NO 3) 2 + BA (OH) 2 \u003d Na (NO 3) 2 + Cu (OH) 2 ↓

Cu (NO 3) 2 + Mg - salt with metal reacts if the free metal is more active in the composition of the salt. Magnesium in a row of activity is located left copper, which indicates its greater activity, therefore, the reaction proceeds:

Cu (NO 3) 2 + Mg \u003d Mg (NO 3) 2 + Cu

B) Al (OH) 3 - metal hydroxide into the degree of oxidation +3. Metal hydroxides in the degree of oxidation + 3, + 4, as well as the exceptions of BE hydroxides BE (OH) 2 and Zn (OH) 2, are amphoter.

By definition, the amphoteric hydroxides are called those that react with alkali and almost all soluble acids. For this reason, you can immediately conclude that the answer option is suitable:

Al (OH) 3 + 3HCl \u003d AlCl 3 + 3H 2 O

Al (OH) 3 + LiOh (p-p) \u003d Li or Al (OH) 3 + LiOh (TV.) \u003d To \u003d\u003e Lialo 2 + 2H 2 O

2AL (OH) 3 + 3H 2 SO 4 \u003d Al 2 (SO 4) 3 + 6H 2 O

C) ZnCl 2 + NaOH and ZnCl 2 + Ba (OH) 2 - the interaction of the type "salt + hydroxide metal". The explanation is given in P.A.

ZnCl 2 + 2NAOH \u003d Zn (OH) 2 + 2NACL

ZnCl 2 + Ba (OH) 2 \u003d Zn (OH) 2 + BaCl 2

It should be noted that with an excess of NaOH and BA (OH) 2:

ZnCl 2 + 4NAOH \u003d Na 2 + 2NACL

ZnCl 2 + 2BA (OH) 2 \u003d BA + BACL 2

D) br 2, o 2 - strong oxidizers. Metals do not react only with silver, platinum, gold:

CU + BR 2 t ° \u003e Cubr 2.

2CU + O 2 t ° \u003e 2Cuo.

HNO 3 - acid with strong oxidative properties, because oxidizes not by hydrogen cations, but by the acid-forming element - nitrogen N +5. Reacts with all metals besides platinum and gold:

4HNO 3 (conc.) + Cu \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O

8HNO 3 (RSS) + 3CU \u003d 3CU (NO 3) 2 + 2NO + 4H 2 O

Set the correspondence between the general formula of the homologous series and the name of the substance belonging to this row: 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.

Answer: 231.

Explanation:

From the proposed list of substances, select two substances that are isomers of cyclopentane.

1) 2-methylbutan

2) 1,2-dimethylcyclopropane

3) Penten-2

4) hexen-2

5) Cyclopenten

Write down in the response field of the selected substances.

Answer: 23.
Explanation:
Cyclopentane has a molecular formula C 5 H 10. We write the structural and molecular formulas listed in the condition

Name of substance	Structural formula	Molecular formula
cyclopentane		C 5 H 10
2-methylbutan		C 5 H 12
1,2-dimethylcyclopropan		C 5 H 10
penten-2		C 5 H 10
hexen-2.		C 6 H 12
cyclopenten		C 5 H 8

From the proposed list of substances, select two substances, each of which reacts with a solution of potassium permanganate.

1) methylbenzene.

2) Cyclohexane

3) methylpropan

Write down in the response field of the selected substances.

Answer: 15.

Explanation:

From hydrocarbons with an aqueous solution of potassium permanganate, those containing in their structural formula C \u003d C or C≡C of communication, as well as benzene homologs (except for benzene itself).
This is suitable methyl benzene and styrene.

From the proposed list of substances, select two substances with which phenol interacts.

1) hydrochloric acid

2) sodium hydroxide

4) nitric acid

5) sodium sulfate

Write down in the response field of the selected substances.

Answer: 24.

Explanation:

Phenol has weak acid properties, pronounced more brightly than alcohols. For this reason, phenols are unlike alcohols with alkalis:

C 6 H 5 OH + NaOH \u003d C 6 H 5 ONA + H 2 O

Phenol contains in its molecule a hydroxyl group directly attached to the benzene ring. The hydroxy group is the first-kind orientant, that is, facilitates reactions of substitution in ortho and para-positions:

From the proposed list of substances, select two substances that are subjected to hydrolysis.

1) Glucose

2) Sakhares

3) fructose

5) Stachmal

Write down in the response field of the selected substances.

Answer: 25.

Explanation:

All listed substances are carbohydrates. From hydrolysis of hydrolysis is not subject to monosaccharides. Glucose, fructose and ribosis are monosaccharides, sucrose - disaccharide, and starch - polysaccharide. Therefore, the hydrolysis is subjected from the specified list of sucrose and starch.

The following scheme of the transformation of substances is given:

1,2-Diberomethane → X → Brometan → Y → Ethyl formate

Determine which of these substances are substances X and Y.

2) Ethanal

4) Chlorhetan

5) Acetylene

Record into the table number selected substances under the appropriate letters.

Answer: 31.

Explanation:

Set the correspondence between the title of the starting material and the product, which is preferably formed when the interaction of this substance with bromine: 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: 2134.

Explanation:

Replacement at a secondary carbon atom flows to a greater extent than during primary. Thus, the main product of propane bromination is 2-bromopropane, and not 1-bromopropane:

Cyclohexane - cycloalkan with a cycle size of more than 4 carbon atoms. Cycloalkanes with a cycle size of more than 4 carbon atoms when interacting with halogens, react substitution with the cycle preservation:

Cyclopropin and cyclobutane - cycloalkanes with a minimum cycle size preferably enter the attachment reaction, accompanied by a cycle break:

The substitution of hydrogen atoms with a tertiary carbon atom occurs to a greater extent than when secondary and primary. Thus, the bromination of isobutan proceeds mainly as follows:

Install the correspondence between the reaction scheme and the organic matter, which is the product of this reaction: 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.

Answer: 6134.

Explanation:

The heating of aldehydes with freshly fired copper hydroxide leads to the oxidation of the aldehyde group to carboxyl:

Aldehydes and ketones are reduced by hydrogen in the presence of nickel, platinum or palladium to alcohols:

Primary and secondary alcohols are oxidized by the hot Cuo to aldehydes and ketones, respectively:

Under the action of concentrated sulfuric acid on ethanol, when heated, two different products are possible. When heated to a temperature below, 140 OS is preferably intermolecular dehydration with the formation of diethyl ether, and when heated, more than 140 ° C is intramolecular, as a result of which ethylene is formed:

From the proposed list of substances, select two substances, the response of the thermal decomposition of which is the redox.

1) aluminum nitrate

2) potassium bicarbonate

3) aluminum hydroxide

4) Ammonium carbonate

5) ammonium nitrate

Write down in the response field of the selected substances.

Answer: 15.

Explanation:

Oxidative and restores are called such reactions as a result of which chemical one or more chemical elements change their degree of oxidation.

The decomposition reactions are absolutely all nitrates relate to redox. Nitrates of metals from Mg to Cu inclusive decompose to metal oxide, nitrogen dioxide and molecular oxygen:

All metal bicarbonates decompose already with insignificant heating (60 o C) to metal carbonate, carbon dioxide and water. At the same time, changes in oxidation degrees does not occur:

Insoluble oxides decompose when heated. The reaction is not oxidative and reducing. No chemical element of the degree of oxidation as a result of it does not change:

Ammonium carbonate decomposes when heated on carbon dioxide, water and ammonia. The reaction is not redox:

Ammonium nitrate decomposes on nitrogen oxide (I) and water. The reaction refers to ORP:

From the proposed list, select two external exposure, which lead to an increase in the reaction rate of nitrogen with hydrogen.

1) temperature decrease

2) Increased pressure in the system

5) the use of an inhibitor

Write down in the response field of the number of selected external influences.

Answer: 24.

Explanation:

1) temperature decrease:

The speed of any reaction when the temperature decreases decreases

2) Increased pressure in the system:

Increased pressure increases the speed of any reaction in which at least one gaseous substance takes part.

3) decrease in hydrogen concentration

Reducing the concentration always reduces the reaction rate

4) an increase in nitrogen concentration

An increase in the concentration of reagents always increases the reaction rate

5) the use of an inhibitor

Inhibitors call substances that slow down the reaction rate.

Install the correspondence between the formula of the substance and the electrolysis products of the aqueous solution of this substance on the inert electrodes: 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: 5251.

Explanation:

A) Nabr → Na + + Br -

Cathiode among themselves compete Na + cations and water molecules.

2H 2 O + 2E - → H 2 + 2OH -

2CL - -2E → Cl 2

B) Mg (NO 3) 2 → MG 2+ + 2NO 3 -

Cathiode among themselves compete mg 2+ cations and water molecules.

Alkali metal cations, as well as magnesium and aluminum are not capable of recovering under aqueous solution conditions due to high activity. For this reason, water molecules are restored instead of them in accordance with equation:

2H 2 O + 2E - → H 2 + 2OH -

Anions NO 3 and water molecules compete with each other.

2H 2 O - 4E - → O 2 + 4H +

Thus, response 2 (hydrogen and oxygen) is suitable.

C) ALCL 3 → AL 3+ + 3CL -

Alkali metal cations, as well as magnesium and aluminum are not capable of recovering under aqueous solution conditions due to high activity. For this reason, water molecules are restored instead of them in accordance with equation:

2H 2 O + 2E - → H 2 + 2OH -

The anions cl - and water molecules compete with each other.

Anions consisting of one chemical element (except F -) win competition in water molecules for oxidation on an anode:

2CL - -2E → Cl 2

This suits the answer option 5 (hydrogen and halogen).

D) Cuso 4 → Cu 2+ + SO 4 2-

Metal cations The right of hydrogen in a row of activity is easily restored under an aqueous solution:

Cu 2+ + 2e → Cu 0

Acid residues containing acid-forming element in the highest oxidation, lose competition of water molecules for oxidation on the anode:

2H 2 O - 4E - → O 2 + 4H +

This suits the option of response 1 (oxygen and metal).

Set the correspondence between the name of the salt and the aqueous solution of this salt: 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.

Answer: 3312.

Explanation:

A) iron sulfate (III) - Fe 2 (SO 4) 3

it is formed by a weak "base" Fe (OH) 3 and a strong acid H 2 SO 4. Conclusion - sour Wednesday

B) chromium chloride (III) - CRCL 3

educated weak "base" CR (OH) 3 and a strong acid HCl. Conclusion - sour Wednesday

C) sodium sulfate - Na 2 SO 4

Formed by a strong base of NaOH and a strong acid H 2 SO 4. Conclusion - Neutral medium

D) sodium sulphide - Na 2 S

Educated by a strong base of NaOH and weak acid H 2 S. Conclusion - alkaline medium.

Set the correspondence between the way of exposure to the equilibrium system

Co (g) + Cl 2 (g) COCL 2 (g) + Q

and the direction of chemical equilibrium displacement as a result of this impact: 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: 3113.

Explanation:

The displacement of equilibrium with external exposure to the system occurs in such a way as to minimize the effect of this external influence (the principle of Le Chatel).

A) an increase in the concentration of CO leads to a displacement of equilibrium towards a direct reaction, since as a result of it, the amount of CO decreases.

B) The temperature increase will shift the equilibrium towards the endothermic reaction. Since the direct reaction is exothermic (+ q), then the equilibrium will be shifted towards the reverse reaction.

C) reduction of pressure will shift equilibrium towards the reaction as a result of which an increase in the amount of gases occurs. As a result of the reverse reaction, more gases are formed than as a result of direct. Thus, the equilibrium will shift towards the reverse reaction.

D) an increase in chlorine concentration leads to a displacement of equilibrium towards a direct reaction, since as a result of it, the amount of chlorine is reduced.

Set the correspondence between the two substances and the reagent, with which you can distinguish between these substances: to each position indicated by the letter, select the appropriate position indicated by the number.

Substances A) FESO 4 and FECL 2 B) Na 3 PO 4 and Na 2 SO 4 C) KOH and CA (OH) 2 D) Kon and KCL

REAGENT

Write in the table selected numbers under the appropriate letters.

Answer: 3454.

Explanation:

You can distinguish two substances with the help of the third only if these two substances interact with it differently, and, most importantly, these differences are extremely distinguishable.

A) FESO 4 and FECL 2 solutions can be distinguished using a barium nitrate solution. In the case of FESO 4, the formation of a white precipitate of barium sulfate:

FESO 4 + BACL 2 \u003d BASO 4 ↓ + FECL 2

In the case of FECL 2, there is no visible signs of interaction, since the reaction does not proceed.

B) Na 3 PO 4 and Na 2 SO 4 solutions can be distinguished using the MGCl 2 solution. The solution of Na 2 SO 4 does not enter the reaction, and in the case of Na 3 PO 4, a white precipitate of magnesium phosphate falls:

2NA 3 PO 4 + 3MGCl 2 \u003d Mg 3 (PO 4) 2 ↓ + 6NACL

C) KOH and CA (OH) 2 solutions can be distinguished by the solution of Na 2 CO 3. KOH with Na 2 CO 3 does not react, and Ca (OH) 2 gives with Na 2 CO 3 white calcium carbonate precipitate:

Ca (OH) 2 + Na 2 CO 3 \u003d Caco 3 ↓ + 2NAOH

D) Kon and KCL solutions can be distinguished by the MGCl 2 solution. KCl with MgCl 2 does not react, and the mixing of solutions of Kon and MgCl 2 leads to the formation of a white precipitate of magnesium hydroxide:

MgCl 2 + 2Kone \u003d Mg (OH) 2 ↓ + 2KCL

Install the correspondence between the substance and the application area: 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: 2331.
Explanation:
Ammonia - used in the production of nitrogen fertilizers. In particular, ammonia is the raw material for the production of nitric acid, which in turn receives fertilizers - sodium, potassium and ammonium nitrate (Nano 3, Kno 3, NH 4 NO 3).
Carbon tetrachloride and acetone are used as solvents.
Ethylene is used to obtain high molecular weight compounds (polymers), namely polyethylene.

The answer to the tasks 27-29 is the number. Record this number in the response field in the text of the work, while observing the specified degree of accuracy. Then transfer this number to the response form No. 1 to the right of the number of the corresponding task, starting from the first cell. Each character is written in a separate cell in accordance with the samples given in the form. Units of measurement of physical quantities are not needed. To the reaction, the thermochemical equation of which MGO (TV.) + CO 2 (g) → MGCO 3 (TV.) + 102 kJ, 88 g of carbon dioxide entered. What amount of heat is highlighted at the same time? (Record the number up to the integer.) Answer: ___________________________ KJ. Answer: 204. Explanation: Calculate the amount of carbon dioxide substance: n (CO 2) \u003d N (CO 2) / M (CO 2) \u003d 88/44 \u003d 2 mole, According to the reaction equation, the interaction of 1 mol CO 2 with magnesium oxide is released 102 kJ. In our case, the amount of carbon dioxide is 2 mol. Denote by the amount of heat released at the same time as X KJ, you can write down the following proportion: 1 mol CO 2 - 102 kJ 2 mol CO 2 - X KJ Consequently, the equation is true: 1 ∙ x \u003d 2 ∙ 102 Thus, the amount of heat that is extended with the participation in the reaction with magnesium oxide 88 g of carbon dioxide is 204 kJ. Determine the mass of zinc, which reacts with hydrochloric acid to produce 2.24 liters (N.O.) of hydrogen. (Record the number up to the tenths.) Answer: ___________________________ Answer: 6.5 Explanation: We write the reaction equation: Zn + 2HCl \u003d ZnCl 2 + H 2 Calculate the amount of hydrogen substance: n (H 2) \u003d V (H 2) / V m \u003d 2.24 / 22.4 \u003d 0.1 mol. Since there are equal coefficients in the reaction equation in front of zinc and hydrogen, this means that the amounts of zinc substances entered into the reaction and hydrogen formed as a result of it are also equal, i.e. n (zn) \u003d n (H 2) \u003d 0.1 mol, therefore: m (zn) \u003d n (zn) ∙ m (zn) \u003d 0.1 ∙ 65 \u003d 6.5 g Do not forget to transfer all the answers to the answer blank number 1 in accordance with the instructions for performing the work. C 6 H 5 COOH + CH 3 OH \u003d C 6 H 5 Coch 3 + H 2 O Sodium bicarbonate weighing 43.34 g was performed to constant mass. The residue was dissolved in the excess of hydrochloric acid. The resulting gas was missed through 100 g of a 10% sodium hydroxide solution. Determine the composition and mass of the resulting salt, its mass fraction in solution. In response, write down the reaction equations that are specified in the task condition, and give all the necessary calculations (specify the units of measurement of the desired physical quantities). Answer: Explanation: Sodium bicarbonate when heated is decomposed in accordance with the equation: 2NAHCO 3 → Na 2 CO 3 + CO 2 + H 2 O (I) The resulting solid residue obviously consists only of sodium carbonate. When the sodium carbonate is dissolved in hydrochloric acid, the following reaction flows: Na 2 CO 3 + 2HCl → 2NACL + CO 2 + H 2 O (II) Calculated the amount of substance of sodium bicarbonate and sodium carbonate: n (NaHCO 3) \u003d M (NaHCO 3) / M (NaHCO 3) \u003d 43.34 g / 84 g / mol ≈ 0.516 mol, hence, n (Na 2 CO 3) \u003d 0.516 mol / 2 \u003d 0.258 mol. Calculate the amount of carbon dioxide formed by reaction (II): n (CO 2) \u003d N (Na 2 CO 3) \u003d 0.258 mol. We calculate the mass of pure sodium hydroxide and its amount of the substance: m (NaOH) \u003d m p-ra (NaOH) ∙ Ω (NaOH) / 100% \u003d 100 g ∙ 10% / 100% \u003d 10 g; n (NaOH) \u003d M (NaOH) / M (NaOH) \u003d 10/40 \u003d 0.25 mol. The interaction of carbon dioxide with sodium hydroxide, depending on their proportions, may occur in accordance with two different equations: 2NAOH + CO 2 \u003d Na 2 CO 3 + H 2 O (with an excess of alkali) NaOH + CO 2 \u003d NaHCO 3 (with an excess of carbon dioxide) From the presented equations, it follows that only the average salt is obtained with the ratio N (NaOH) / N (CO 2) ≥2, and only acid, with a ratio N (NaOH) / N (CO 2) ≤ 1. According to calculations ν (CO 2)\u003e ν (NaOH), therefore: n (NaOH) / N (CO 2) ≤ 1 Those. The interaction of carbon dioxide with sodium hydroxide occurs solely to the formation of acid salt, i.e. In accordance with equation: NaOH + CO 2 \u003d NaHCO 3 (III) Calculation is carried out on shortness of alkali. By the reaction equation (III): n (NaHCO 3) \u003d n (NaOH) \u003d 0.25 mol, therefore: m (NaHCO 3) \u003d 0.25 mol ∙ 84 g / mol \u003d 21 g The mass of the resulting solution will fold from the mass of the alkali solution and the mass of carbon dioxide absorbed by it. From the reaction equation, it follows that reacted, i.e. Miselid only 0.25 mol CO 2 of 0.258 mol. Then the mass of the absorbed CO 2 is: m (CO 2) \u003d 0.25 mol ∙ 44 g / mol \u003d 11 g Then, the solution of the solution is equal to: m (p-ra) \u003d m (p-ra NaOH) + m (CO 2) \u003d 100 g + 11 g \u003d 111 g, a mass fraction of sodium bicarbonate in the solution will thus be equal to: ω (NaHCO 3) \u003d 21 g / 111 g ∙ 100% ≈ 18.92%. With a combustion of 16.2 g of the organic substance of the non-cyclic structure, 26.88 liters (N.O.) of carbon dioxide and 16.2 g of water were obtained. It is known that 1 mol of this organic matter in the presence of the catalyst attaches only 1 mol of water and this substance does not react with the ammonia solution of silver oxide. On the basis of these conditions of the problem: 1) Calculates needed to establish the molecular formula of the organic matter; 2) write down the molecular formula of the organic matter; 3) make a structural formula of an organic matter, which uniquely reflects the order of communication of atoms in its molecule; 4) Write an organic substance hydration reaction equation. Answer: Explanation: 1) To determine the elemental composition, calculate the amount of carbon dioxide, water, and then the masses of the elements included in them: n (CO 2) \u003d 26.88 l / 22.4 l / mol \u003d 1.2 mol; n (CO 2) \u003d n (C) \u003d 1.2 mol; M (C) \u003d 1.2 mol ∙ 12 g / mol \u003d 14.4 g. n (H 2 O) \u003d 16.2 g / 18 g / mol \u003d 0.9 mol; n (h) \u003d 0.9 mol ∙ 2 \u003d 1.8 mol; M (H) \u003d 1.8 g m (Org. V-Ba) \u003d M (C) + M (H) \u003d 16.2 g, therefore, there is no oxygen in the organic substance. The general formula of the organic compound is C x H y. x: y \u003d ν (c): ν (H) \u003d 1,2: 1, 8 \u003d 1: 1,5 \u003d 2: 3 \u003d 4: 6 Thus, the simplest formula of the substance C 4 H 6. The true formula can substance can coincide with the simplest, and may differ from it for an integer time. Those. Be, for example, with 8 H 12, with 12 H 18, etc. The condition states that the hydrocarbon is non-cyclic and one of its molecule can attach only one water molecule. This is possible if there is only one multiple communication (double or triple) in the structural formula of substance. Since the desired hydrocarbon is non-cyclic, it is obvious that one multiple bond can only be for a substance with a C 4 H 6 formula. In the case of other hydrocarbons with a greater molecular weight, the number of multiple connections is more than one everywhere. Thus, the molecular formula of the substance C 4 H 6 coincides with the simplest. 2) The molecular formula of the organic matter is C 4 H 6. 3) Alkins interact from hydrocarbons with ammonia solution of silver oxide, in which the triple bond is located at the end of the molecule. In order for the interaction with ammonia solution of silver oxide, alkin of the C 4 H 6 composition should be the following structure: CH 3 -C≡C-CH 3 4) Alkin hydration proceeds in the presence of bivalent mercury salts:

EGE 2017 Chemistry Typical Test Quests Medvedev

M.: 2017. - 120 s.

Typical test tasks for chemistry contain 10 options for sets of tasks drawn up taking into account all the features and requirements of the Unified State Exam in 2017. The purpose of the manual is to provide readers information about the structure and content of the Kim 2017 in chemistry, the degree of difficulty of tasks. In the collection there are answers to all test options and solutions all the tasks of one of the options are given. In addition, samples of the forms used on the USE are given to record responses and solutions. The author of the tasks is a leading scientist, a teacher and a methodologist who takes direct participation in the development of control measuring materials of the EGE. The manual is intended for teachers to prepare students for the chemistry exam, as well as high school students and graduates - for self-preparation and self-control.

Format: PDF.

The size: 1.5 MB

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CONTENT
Preface 4.
Instructions for performing work 5
Option 1 8.
Part 1 8.
Part 2, 15
Option 2 17.
Part 1 17.
Part 2 24.
Option 3 26.
Part 1 26.
Part 2 33.
Option 4 35.
Part 1 35.
Part 2 41.
Option 5 43.
Part 1 43.
Part 2 49.
Option 6 51.
Part 1 51.
Part 2 57.
Option 7 59.
Part 1 59.
Part 2 65.
Option 8 67.
Part 1 67.
Part 2 73.
Option 9 75.
Part 1 75.
Part 2 81.
Option 10 83.
Part 1 83.
Part 2 89.
Answers and Solutions 91
Answers to the tasks of part 1 91
Solutions and answers to the tasks of part 2 93
Solution of the tasks of option 10 99
Part 1 99.
Part 2 113.

The present tutorial is a collection of tasks for preparing for the delivery of a single state exam (USE) for chemistry, which is both the final exam for secondary school and the entrance exam in the university. The structure of the manual reflects the current requirements for the procedure for passing the exam in chemistry, which will allow you to better prepare for new forms of graduation certification and admission to universities.
The manual consists of 10 options for assignments, which in form and content are approximate to the demoralization of the USE and do not go beyond the content of the chemistry course, normatively determined by the federal component of the state standard of general education. Chemistry (order of the Ministry of Education No. 1089 of 05.03.2004).
The level of presentation of the consideration of the educational material in the tasks is related to the requirements of the State Standard to the preparation of graduates of the average (full) school in chemistry.
In the control measuring materials of the Unified State Exam, the tasks of three types are used:
- tasks of the basic level of complexity with a brief answer,
- Quests of an increased level of difficulty with a brief answer,
- Tasks of a high level of difficulty with a detailed answer.
Each version of the examination work is built according to the Unified Plan. The work consists of two parts that include total 34 tasks. Part 1 contains 29 tasks with a brief response, including 20 tasks of the base level of complexity and 9 tasks of the increased level of complexity. Part 2 contains 5 high-level tasks, with an expanded response (tasks under the numbers 30-34).
In the tasks of a high level of complexity, the text of the decision is written on a special form. The tasks of this type are the bulk of the written work on chemistry on the entrance exams in universities.

Tips for preparation for the exam in chemistry on site site

How to competently pass the exam (and OGE) in chemistry? If the time is just 2 months, and you are not ready? And do not be friends with chemistry ...

It offers tests with answers on each topic and a task, passing which you can explore the basic principles, patterns and the theory that occur in the exam in chemistry. Our allow you to find answers to most questions encountered in the exam in chemistry, and our tests allow us to consolidate the material, find weak places, and work out the material.

All you need is the Internet, stationery, time and site. It is best to start a separate notebook for formulas / solutions / marks and dictionary of trivial titles of connections.

1. From the very beginning you need to evaluate your current level and then the number of points you need is worth passing for this. If everything is very bad, and you need excellent performance - congratulations, even now is not all lost. Suppress yourself for successful surrender can be soothing and without the help of a tutor.
   Decide with the minimum number of balls you want to dial, it will allow you to understand how many tasks you must solve exactly to get the score you need.
   Naturally, consider that everything can go not so smoothly and decide as much as possible tasks, and better everything. The minimum that you have determined for yourself - you must solve perfectly.
2. Let us turn to the practical part of the contract on the decision.
   The most effective way is the next one. Choose only the exam you are interested in and solve the corresponding test. About 20 solvable tasks guarantee the meeting of all types of tasks. As soon as you start feeling that every task you see you know how to solve from beginning to end - proceed to the next task. If you do not know how to solve any task - use the search on our site. The solution on our site is almost always, otherwise you just write the tutor by clicking on the icon in the lower left corner - it is free.
3. In parallel, repeat the third item for all of our site, starting with.
4. When the first part is given to you at least at the average level - start deciding. If one of the tasks is poorly amenable, and you are mistaken in its execution, then return to tests on this task or the appropriate topic with tests.
5. Part 2. If you have a tutor - focus with him on the study of this part. (provided that you are able to solve the rest at least 70%). If you started part 2, then the passage point you must recruit without any problems in 100% of cases. If this does not happen, it is better to stay on the first part. When you are ready for part 2, we recommend getting a separate notebook, where you will record only solutions of part 2. The key to success is the dashing as much as possible tasks as in part 1.