Exam in chemistry preparation of tasks. Exam in chemistry
Determine which atoms of the elements indicated in the series contain one unpaired electron in the ground state.
Write down the numbers of the selected elements in the answer field.
Answer:
Answer: 23
Explanation:
Let's write down the electronic formula for each of the indicated chemical elements and depict the electron-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 the chemical elements indicated in the series, select three metal elements. Arrange the selected elements in order of increasing reducing properties.
Write down the numbers of the selected elements in the required sequence in the answer field.
Answer: 352
Explanation:
In the main subgroups of the periodic table, metals are located under the boron-astatine diagonal, as well as in secondary subgroups. Thus, the metals from this list include Na, Al and Mg.
The metallic and, therefore, reducing properties of the elements increase when moving to the left along the period and down the subgroup.
Thus, the metallic properties of the metals listed above increase in the order Al, Mg, Na
From among the elements indicated in the series, select two elements that, when combined with oxygen, exhibit an oxidation state of +4.
Write down the numbers of the selected elements in the answer field.
Answer: 14
Explanation:
The main oxidation states of elements from the presented list in complex substances:
Sulfur – “-2”, “+4” and “+6”
Sodium Na – “+1” (single)
Aluminum Al – “+3” (single)
Silicon Si – “-4”, “+4”
Magnesium Mg – “+2” (single)
From the proposed list of substances, select two substances in which an ionic chemical bond is present.
Answer: 12
Explanation:
In the vast majority of cases, the presence of an ionic type of bond in a compound can be determined by the fact that its structural units simultaneously include atoms of a typical metal and atoms of a non-metal.
Based on this criterion, the ionic type of bond occurs in the compounds KCl and KNO 3.
In addition to the above characteristic, the presence of an ionic bond in a compound can be said if its structural unit contains an ammonium cation (NH 4 + ) or its organic analogues - alkylammonium cations RNH 3 + , dialkylamonium R 2NH2+ , trialkylammonium R 3NH+ and tetraalkylammonium R 4N+ , where R is some hydrocarbon radical. For example, the ionic type of bond occurs in the compound (CH 3 ) 4 NCl between the cation (CH 3 ) 4 + and chloride ion Cl − .
Establish a correspondence between the formula of a 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.
Answer: 241
Explanation:
N 2 O 3 is a non-metal oxide. All non-metal oxides except N 2 O, NO, SiO and CO are acidic.
Al 2 O 3 is a metal oxide in the oxidation state +3. Metal oxides in the oxidation state +3, +4, as well as BeO, ZnO, SnO and PbO, are amphoteric.
HClO 4 is a typical representative of acids, because upon dissociation in an aqueous solution, only H + cations are formed from cations:
HClO 4 = H + + ClO 4 —
From the proposed list of substances, select two substances, with each of which zinc interacts.
1) nitric acid (solution)
2) iron(II) hydroxide
3) magnesium sulfate (solution)
4) sodium hydroxide (solution)
5) aluminum chloride (solution)
Write down the numbers of the selected substances in the answer field.
Answer: 14
Explanation:
1) Nitric acid is a strong oxidizing agent and reacts with all metals except platinum and gold.
2) Iron hydroxide (ll) is an insoluble base. Metals do not react with insoluble hydroxides at all, and only three metals react with soluble (alkalis) - Be, Zn, Al.
3) Magnesium sulfate is a salt of a more active metal than zinc, and therefore the reaction does not proceed.
4) Sodium hydroxide - alkali (soluble metal hydroxide). Only Be, Zn, Al work with metal alkalis.
5) AlCl 3 – a salt of a metal more active than zinc, i.e. reaction is impossible.
From the proposed list of substances, select two oxides that react with water.
Write down the numbers of the selected substances in the answer field.
Answer: 14
Explanation:
Of the oxides, only oxides of alkali and alkaline earth metals, as well as all acidic oxides except SiO 2, react with water.
Thus, answer options 1 and 4 are suitable:
BaO + H 2 O = Ba(OH) 2
SO 3 + H 2 O = H 2 SO 4
1) hydrogen bromide
3) sodium nitrate
4) sulfur oxide(IV)
5) aluminum chloride
Write down the selected numbers in the table under the corresponding letters.
Answer: 52
Explanation:
The only salts among these substances are sodium nitrate and aluminum chloride. All nitrates, like sodium salts, are soluble, and therefore sodium nitrate cannot form a precipitate in principle with any of the reagents. Therefore, salt X can only be aluminum chloride.
A common mistake among those taking the Unified State Exam in chemistry is not understanding that in an aqueous solution ammonia forms a weak base - ammonium hydroxide due to the reaction occurring:
NH 3 + H 2 O<=>NH4OH
In this regard, an aqueous solution of ammonia gives a precipitate when mixed with solutions of metal salts that form insoluble hydroxides:
3NH 3 + 3H 2 O + AlCl 3 = Al(OH) 3 + 3NH 4 Cl
In a given transformation scheme
Cu X > CuCl 2 Y > CuI
substances X and Y are:
Answer: 35
Explanation:
Copper is a metal located in the activity series 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 (ll) chloride is possible in our case only by reaction with chlorine:
Cu + Cl 2 = CuCl 2
Iodide ions (I -) cannot coexist in the same solution with divalent copper ions, because are oxidized by them:
Cu 2+ + 3I - = CuI + I 2
Establish a correspondence between the reaction equation and the oxidizing substance in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.
REACTION EQUATION A) H 2 + 2Li = 2LiH B) N 2 H 4 + H 2 = 2NH 3 B) N 2 O + H 2 = N 2 + H 2 O D) N 2 H 4 + 2N 2 O = 3N 2 + 2H 2 O |
OXIDANT |
Write down the selected numbers in the table under the corresponding letters.
Answer: 1433
Explanation:
An oxidizing agent in a reaction is a substance that contains an element that lowers its oxidation state
Establish a correspondence between the formula of a substance and the reagents with each of which this substance can interact: for each position indicated by a letter, select the corresponding position indicated by a number.
FORMULA OF THE SUBSTANCE | REAGENTS |
A) Cu(NO 3) 2 | 1) NaOH, Mg, Ba(OH) 2 2) HCl, LiOH, H 2 SO 4 (solution) 3) BaCl 2, Pb(NO 3) 2, S 4) CH 3 COOH, KOH, FeS 5) O 2, Br 2, HNO 3 |
Write down the selected numbers in the table under the corresponding letters.
Answer: 1215
Explanation:
A) Cu(NO 3) 2 + NaOH and Cu(NO 3) 2 + Ba(OH) 2 – similar interactions. A salt reacts with a metal hydroxide if the starting substances are soluble, and the products contain a precipitate, gas or slightly dissociating substance. For both the first and second reactions, both requirements are met:
Cu(NO 3) 2 + 2NaOH = 2NaNO 3 + Cu(OH) 2 ↓
Cu(NO 3) 2 + Ba(OH) 2 = Na(NO 3) 2 + Cu(OH) 2 ↓
Cu(NO 3) 2 + Mg - a salt reacts with a metal if the free metal is more active than what is included in the salt. Magnesium in the activity series is located to the left of copper, which indicates its greater activity, therefore, the reaction proceeds:
Cu(NO 3) 2 + Mg = Mg(NO 3) 2 + Cu
B) Al(OH) 3 – metal hydroxide in the oxidation state +3. Metal hydroxides in the oxidation state +3, +4, as well as the hydroxides Be(OH) 2 and Zn(OH) 2 as exceptions, are classified as amphoteric.
By definition, amphoteric hydroxides are those that react with alkalis and almost all soluble acids. For this reason, we can immediately conclude that answer option 2 is appropriate:
Al(OH) 3 + 3HCl = AlCl 3 + 3H 2 O
Al(OH) 3 + LiOH (solution) = Li or Al(OH) 3 + LiOH(sol.) =to=> LiAlO 2 + 2H 2 O
2Al(OH) 3 + 3H 2 SO 4 = Al 2 (SO 4) 3 + 6H 2 O
C) ZnCl 2 + NaOH and ZnCl 2 + Ba(OH) 2 – interaction of the “salt + metal hydroxide” type. The explanation is given in paragraph A.
ZnCl 2 + 2NaOH = Zn(OH) 2 + 2NaCl
ZnCl 2 + Ba(OH) 2 = Zn(OH) 2 + BaCl 2
It should be noted that with an excess of NaOH and Ba(OH) 2:
ZnCl 2 + 4NaOH = Na 2 + 2NaCl
ZnCl 2 + 2Ba(OH) 2 = Ba + BaCl 2
D) Br 2, O 2 are strong oxidizing agents. The only metals that do not react are silver, platinum, and gold:
Cu + Br 2 t° > CuBr 2
2Cu + O2 t° >2CuO
HNO 3 is an acid with strong oxidizing properties, because oxidizes not with hydrogen cations, but with an acid-forming element - nitrogen N +5. Reacts with all metals except platinum and gold:
4HNO 3(conc.) + Cu = Cu(NO 3)2 + 2NO 2 + 2H 2 O
8HNO 3(dil.) + 3Cu = 3Cu(NO 3) 2 + 2NO + 4H 2 O
Establish a correspondence between the general formula of a homologous series and the name of a substance belonging to this series: 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.
Answer: 231
Explanation:
From the proposed list of substances, select two substances that are isomers of cyclopentane.
1) 2-methylbutane
2) 1,2-dimethylcyclopropane
3) penten-2
4) hexene-2
5) cyclopentene
Write down the numbers of the selected substances in the answer field.
Answer: 23
Explanation:
Cyclopentane has the molecular formula C5H10. Let's write the structural and molecular formulas of the substances listed in the condition
Substance name | Structural formula | Molecular formula |
cyclopentane | C5H10 | |
2-methylbutane | C5H12 | |
1,2-dimethylcyclopropane | C5H10 | |
penten-2 | C5H10 | |
hexene-2 | C6H12 | |
cyclopentene | 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) methylpropane
Write down the numbers of the selected substances in the answer field.
Answer: 15
Explanation:
Of the hydrocarbons that react with an aqueous solution of potassium permanganate, those that contain structural formula C=C or C≡C bonds, as well as homologues of benzene (except benzene itself).
Methylbenzene and styrene are suitable in this way.
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 the numbers of the selected substances in the answer field.
Answer: 24
Explanation:
Phenol has weak acidic properties, more pronounced than in alcohols. For this reason, phenols, unlike alcohols, react with alkalis:
C 6 H 5 OH + NaOH = 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 an orienting agent of the first kind, that is, it facilitates substitution reactions in the ortho and para positions:
From the proposed list of substances, select two substances that undergo hydrolysis.
1) glucose
2) sucrose
3) fructose
5) starch
Write down the numbers of the selected substances in the answer field.
Answer: 25
Explanation:
All of the substances listed are carbohydrates. Of carbohydrates, monosaccharides do not undergo hydrolysis. Glucose, fructose and ribose are monosaccharides, sucrose is a disaccharide, and starch is a polysaccharide. Therefore, sucrose and starch from the above list are subject to hydrolysis.
The following scheme of substance transformations is specified:
1,2-dibromoethane → X → bromoethane → Y → ethyl formate
Determine which of the indicated substances are substances X and Y.
2) ethanal
4) chloroethane
5) acetylene
Write down the numbers of the selected substances under the corresponding letters in the table.
Answer: 31
Explanation:
Establish a correspondence between the name of the starting substance and the product, which is mainly formed when this substance reacts with bromine: 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.
Answer: 2134
Explanation:
Substitution at the secondary carbon atom occurs to a greater extent than at the primary one. Thus, the main product of propane bromination is 2-bromopropane, not 1-bromopropane:
Cyclohexane is a cycloalkane with a ring size of more than 4 carbon atoms. Cycloalkanes with a ring size of more than 4 carbon atoms, when interacting with halogens, enter into a substitution reaction with preservation of the cycle:
Cyclopropane and cyclobutane - cycloalkanes with a minimum ring size preferentially undergo addition reactions accompanied by ring rupture:
The replacement of hydrogen atoms at the tertiary carbon atom occurs to a greater extent than at the secondary and primary ones. Thus, the bromination of isobutane proceeds mainly as follows:
Establish a correspondence between the reaction scheme and the organic substance that is the product of this reaction: 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.
Answer: 6134
Explanation:
Heating aldehydes with freshly precipitated copper hydroxide leads to the oxidation of the aldehyde group to a carboxyl group:
Aldehydes and ketones are reduced with hydrogen in the presence of nickel, platinum or palladium to alcohols:
Primary and secondary alcohols are oxidized by hot CuO to aldehydes and ketones, respectively:
When concentrated sulfuric acid reacts with ethanol upon heating, two different products may form. When heated to a temperature below 140 °C, intermolecular dehydration occurs predominantly with the formation of diethyl ether, and when heated above 140 °C, intramolecular dehydration occurs, resulting in the formation of ethylene:
From the proposed list of substances, select two substances whose thermal decomposition reaction is redox.
1) aluminum nitrate
2) potassium bicarbonate
3) aluminum hydroxide
4) ammonium carbonate
5) ammonium nitrate
Write down the numbers of the selected substances in the answer field.
Answer: 15
Explanation:
Redox reactions are those reactions in which one or more chemical elements change their oxidation state.
The decomposition reactions of absolutely all nitrates are redox reactions. Metal nitrates from Mg to Cu inclusive decompose to metal oxide, nitrogen dioxide and molecular oxygen:
All metal hydrocarbonates decompose even with slight heating (60 o C) to metal carbonate, carbon dioxide and water. In this case, no change in oxidation states occurs:
Insoluble oxides decompose when heated. The reaction is not redox because Not a single chemical element changes its oxidation state as a result:
Ammonium carbonate decomposes when heated into carbon dioxide, water and ammonia. The reaction is not redox:
Ammonium nitrate decomposes into nitric oxide (I) and water. The reaction relates to OVR:
From the proposed list, select two external influences that lead to an increase in the rate of reaction of nitrogen with hydrogen.
1) decrease in temperature
2) increase in pressure in the system
5) use of an inhibitor
Write down the numbers of the selected external influences in the answer field.
Answer: 24
Explanation:
1) temperature decrease:
The rate of any reaction decreases as the temperature decreases
2) increase in pressure in the system:
Increasing pressure increases the rate of any reaction in which at least one gaseous substance takes part.
3) decrease in hydrogen concentration
Decreasing the concentration always reduces the reaction rate
4) increase in nitrogen concentration
Increasing the concentration of reagents always increases the reaction rate
5) use of an inhibitor
Inhibitors are substances that slow down the rate of a reaction.
Establish a correspondence between the formula of a substance and the products of electrolysis of an aqueous solution of this substance on inert electrodes: 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.
Answer: 5251
Explanation:
A) NaBr → Na + + Br -
Na+ cations and water molecules compete with each other for the cathode.
2H 2 O + 2e — → H 2 + 2OH —
2Cl - -2e → Cl 2
B) Mg(NO 3) 2 → Mg 2+ + 2NO 3 —
Mg 2+ cations and water molecules compete with each other for the cathode.
Alkali metal cations, as well as magnesium and aluminum, are not able to be reduced in an aqueous solution due to their high activity. For this reason, water molecules are reduced instead according to the equation:
2H 2 O + 2e — → H 2 + 2OH —
NO3 anions and water molecules compete with each other for the anode.
2H 2 O - 4e - → O 2 + 4H +
So answer 2 (hydrogen and oxygen) is appropriate.
B) AlCl 3 → Al 3+ + 3Cl -
Alkali metal cations, as well as magnesium and aluminum, are not able to be reduced in an aqueous solution due to their high activity. For this reason, water molecules are reduced instead according to the equation:
2H 2 O + 2e — → H 2 + 2OH —
Cl anions and water molecules compete with each other for the anode.
Anions consisting of one chemical element(except F -) win competition from water molecules for oxidation at the anode:
2Cl - -2e → Cl 2
Therefore, answer option 5 (hydrogen and halogen) is appropriate.
D) CuSO 4 → Cu 2+ + SO 4 2-
Metal cations to the right of hydrogen in the activity series are easily reduced under aqueous solution conditions:
Cu 2+ + 2e → Cu 0
Acidic residues containing an acid-forming element in highest degree oxidation, lose competition to water molecules for oxidation at the anode:
2H 2 O - 4e - → O 2 + 4H +
Thus, answer option 1 (oxygen and metal) is appropriate.
Establish a correspondence between the name of the salt and the medium of the aqueous solution of this salt: 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.
Answer: 3312
Explanation:
A) iron(III) sulfate - Fe 2 (SO 4) 3
formed by a weak “base” Fe(OH) 3 and a strong acid H 2 SO 4. Conclusion - the environment is acidic
B) chromium(III) chloride - CrCl 3
formed by the weak “base” Cr(OH) 3 and the strong acid HCl. Conclusion - the environment is acidic
B) sodium sulfate - Na 2 SO 4
Formed by the strong base NaOH and the strong acid H 2 SO 4. Conclusion - the environment is neutral
D) sodium sulfide - Na 2 S
Formed by the strong base NaOH and the weak acid H2S. Conclusion - the environment is alkaline.
Establish a correspondence between the method of influencing the equilibrium system
CO (g) + Cl 2 (g) COCl 2 (g) + Q
and the direction of the shift in chemical equilibrium as a result of this effect: 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.
Answer: 3113
Explanation:
The equilibrium shift under external influence on the system occurs in such a way as to minimize the effect of this external influence(Le Chatelier's principle).
A) An increase in the concentration of CO causes the equilibrium to shift toward the forward reaction because it results in a decrease in the amount of CO.
B) An increase in temperature will shift the equilibrium towards an endothermic reaction. Since the forward reaction is exothermic (+Q), the equilibrium will shift towards the reverse reaction.
C) A decrease in pressure will shift the equilibrium towards the reaction that results in an increase in the amount of gases. As a result of the reverse reaction, more gases are formed than as a result of the direct reaction. Thus, the equilibrium will shift towards the opposite reaction.
D) An increase in the concentration of chlorine leads to a shift in the equilibrium towards the direct reaction, since as a result it reduces the amount of chlorine.
Establish a correspondence between two substances and a reagent that can be used to distinguish these substances: for each position indicated by a letter, select the corresponding position indicated by a number.
SUBSTANCES A) FeSO 4 and FeCl 2 B) Na 3 PO 4 and Na 2 SO 4 B) KOH and Ca(OH) 2 D) KOH and KCl |
REAGENT |
Write down the selected numbers in the table under the corresponding letters.
Answer: 3454
Explanation:
It is possible to distinguish two substances with the help of a third one only if these two substances interact with it differently, and, most importantly, these differences are externally distinguishable.
A) Solutions of FeSO 4 and FeCl 2 can be distinguished using a solution of barium nitrate. In the case of FeSO 4, a white precipitate of barium sulfate forms:
FeSO 4 + BaCl 2 = BaSO 4 ↓ + FeCl 2
In the case of FeCl 2 there are no visible signs of interaction, since the reaction does not occur.
B) Solutions of Na 3 PO 4 and Na 2 SO 4 can be distinguished using a solution of MgCl 2. The Na 2 SO 4 solution does not react, and in the case of Na 3 PO 4 a white precipitate of magnesium phosphate precipitates:
2Na 3 PO 4 + 3MgCl 2 = Mg 3 (PO 4) 2 ↓ + 6NaCl
C) Solutions of KOH and Ca(OH) 2 can be distinguished using a solution of Na 2 CO 3. KOH does not react with Na 2 CO 3, but Ca(OH) 2 gives a white precipitate of calcium carbonate with Na 2 CO 3:
Ca(OH) 2 + Na 2 CO 3 = CaCO 3 ↓ + 2NaOH
D) Solutions of KOH and KCl can be distinguished using a solution of MgCl 2. KCl does not react with MgCl 2, and mixing solutions of KOH and MgCl 2 leads to the formation of a white precipitate of magnesium hydroxide:
MgCl 2 + 2KOH = Mg(OH) 2 ↓ + 2KCl
Establish a correspondence between the substance and its area of application: 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.
Answer: 2331
Explanation:
Ammonia - used in the production of nitrogenous fertilizers. In particular, ammonia is a raw material for the production nitric acid, from which fertilizers, in turn, are obtained - 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 produce high molecular weight compounds (polymers), namely polyethylene.
The answer to tasks 27–29 is a number. Write this number in the answer field in the text of the work, while maintaining the specified degree of accuracy. Then transfer this number to ANSWER FORM No. 1 to the right of the number of the corresponding task, starting from the first cell. Write each character in a separate box in accordance with the samples given in the form. Units of measurement physical quantities no need to write. In a reaction whose thermochemical equation is MgO (tv.) + CO 2 (g) → MgCO 3 (tv.) + 102 kJ, 88 g of carbon dioxide entered. How much heat will be released in this case? (Write the number to the nearest whole number.) Answer: ___________________________ kJ. Answer: 204 Explanation: Let's calculate the amount of carbon dioxide: n(CO 2) = n(CO 2)/ M(CO 2) = 88/44 = 2 mol, According to the reaction equation, when 1 mole of CO 2 reacts with magnesium oxide, 102 kJ is released. In our case, the amount of carbon dioxide is 2 mol. Designating the amount of heat released as x kJ, we can write the following proportion: 1 mol CO 2 – 102 kJ 2 mol CO 2 – x kJ Therefore, the equation is valid: 1 ∙ x = 2 ∙ 102 Thus, the amount of heat that will be released when 88 g of carbon dioxide participates in the reaction with magnesium oxide is 204 kJ. Determine the mass of zinc that reacts with hydrochloric acid to produce 2.24 L (N.S.) of hydrogen. (Write the number to the nearest tenth.) Answer: ___________________________ g. Answer: 6.5 Explanation: Let's write the reaction equation: Zn + 2HCl = ZnCl 2 + H 2 Let's calculate the amount of hydrogen substance: n(H 2) = V(H 2)/V m = 2.24/22.4 = 0.1 mol. Since in the reaction equation there are equal coefficients in front of zinc and hydrogen, this means that the amounts of zinc substances that entered into the reaction and the hydrogen formed as a result of it are also equal, i.e. n(Zn) = n(H 2) = 0.1 mol, therefore: m(Zn) = n(Zn) ∙ M(Zn) = 0.1 ∙ 65 = 6.5 g.
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Preparation for the Unified State Exam in chemistry is covered by our specialists in this section - analysis of problems, reference data and theoretical material. You can now prepare for the Unified State Exam easily and for free with our sections on each subject! We are confident that you will pass the Unified State Exam in 2019 with the maximum score!
General information about the exam
The Unified State Examination in Chemistry consists of two parts and 34 tasks .
First part contains 29 short answer tasks, including 20 tasks basic level difficulties: No. 1–9, 12–17, 20–21, 27–29. Nine tasks higher level difficulties: No. 9–11, 17–19, 22–26.
Second part contains 5 tasks high level Difficulties with a detailed answer: No. 30–34
Tasks of a basic level of complexity with a short answer test the mastery of the content of the most important sections school course chemistry: theoretical foundations chemistry, Not organic chemistry, organic chemistry, methods of knowledge in chemistry, chemistry and life.
Quests increased level of difficulty with a short answer are focused on checking the mandatory elements of the content of the main educational programs in chemistry not only at a basic level, but also at an advanced level. In comparison with the tasks of the previous group, they involve performing a greater variety of actions to apply knowledge in a changed, non-standard situation (for example, to analyze the essence of the studied types of reactions), as well as the ability to systematize and generalize the acquired knowledge.
Tasks with a detailed answer , unlike tasks of the two previous types, provide for a comprehensive test of assimilation at an in-depth level of several content elements from various content blocks.
Unified State Exam 2017 Chemistry Typical test tasks Medvedev
M.: 2017. - 120 p.
Typical test tasks in chemistry contain 10 variant sets of tasks, compiled taking into account all the features and requirements of the Unified state exam in 2017. The purpose of the manual is to provide readers with information about the structure and content of the 2017 KIM in chemistry, the degree of difficulty of the tasks. The collection contains answers to all test options and provides solutions to all tasks of one of the options. In addition, samples of forms used in the Unified State Exam for recording answers and solutions are provided. The author of the assignments is a leading scientist, teacher and methodologist who is directly involved in the development of control measuring instruments Unified State Exam materials. The manual is intended for teachers to prepare students for the chemistry exam, as well as for high school students and graduates - for self-preparation and self-control.
Format: pdf
Size: 1.5 MB
Watch, download:drive.google
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
Solving problems of option 10 99
Part 1 99
Part 2 113
Present training manual is a collection of tasks to prepare for passing the Unified State Exam (USE) in chemistry, which is like the final exam for the course high school, and the entrance exam to the university. The benefit structure reflects modern requirements to the procedure passing the Unified State Exam in chemistry, which will allow you to better prepare for new forms of final certification and for admission to universities.
The manual consists of 10 variants of tasks, which in form and content are close to the demo version of the Unified State Exam 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 03/05/2004).
Content presentation level educational material in the assignments it is correlated with the requirements of the state standard for the preparation of secondary (full) school graduates in chemistry.
The control measurement materials of the Unified State Exam use tasks of three types:
- tasks of a basic level of difficulty with a short answer,
- tasks of an increased level of complexity with a short answer,
- tasks of a high level of complexity with a detailed answer.
Every option exam paper built according to a single plan. The work consists of two parts, including a total of 34 tasks. Part 1 contains 29 short answer questions, including 20 basic level tasks and 9 advanced level tasks. Part 2 contains 5 tasks of a high level of complexity, with detailed answers (tasks numbered 30-34).
In tasks of a high level of complexity, the text of the solution is written on a special form. Tasks of this type make up the bulk of written work in chemistry for university entrance exams.