Unified State Exam test in chemistry. Tests by topic
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How to competently pass the Unified State Exam (and Unified State Exam) in chemistry? If you only have 2 months and you’re not ready yet? And don’t be friends with chemistry...
It offers tests with answers for each topic and task, by passing which you can study the basic principles, patterns and theory found in the Unified State Exam in chemistry. Our tests allow you to find answers to most questions encountered in the Unified State Examination in chemistry, and our tests allow you to consolidate the material, find weak points, and work on the material.
All you need is the Internet, stationery, time and a website. It is best to have a separate notebook for formulas/solutions/notes and a dictionary of trivial names of compounds.
- From the very beginning, you need to assess your current level and the number of points you need, for this it is worth going through. If everything is very bad and you need excellent performance, congratulations, even now all is not lost. You can train yourself to pass successfully without the help of a tutor.
Decide on the minimum number of points you want to score, this will allow you to understand how many tasks you must solve accurately to get the score you need.
Naturally, take into account that everything may not go so smoothly and solve as many problems as possible, or better yet, all of them. The minimum that you have determined for yourself - you must decide ideally. - Let's move on to the practical part - training for the solution.
The most effective way is the following. Select only the exam you are interested in and solve the corresponding test. About 20 solved tasks guarantee that you will meet all types of problems. As soon as you begin to feel that you know how to solve every task you see from beginning to end, proceed to the next task. If you don’t know how to solve a task, use the search on our website. There is almost always a solution on our website, otherwise just write to the tutor by clicking on the icon in the lower left corner - it’s free. - At the same time, we repeat the third point for everyone on our website, starting with.
- When the first part is given to you at least at an average level, you begin to decide. If one of the tasks is difficult, and you made a mistake in completing it, then return to the tests on this task or the corresponding topic with tests.
- Part 2. If you have a tutor, focus on studying this part with him. (provided that you are able to solve the rest at least 70%). If you started part 2, then you should score a passing grade without any problems 100% of the time. If this does not happen, it is better to stay on the first part for now. When you are ready for part 2, we recommend that you get a separate notebook where you will write down only the solutions to part 2. The key to success is solving as many tasks as possible, just like in part 1.
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 at all with insoluble hydroxides, 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 a failure to understand 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 for 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 |
OXIDIZER |
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 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 are those that contain C=C or C≡C bonds in their structural formula, 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 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 bicarbonates 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 in accordance with 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 in accordance with 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 with 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 the highest oxidation state 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 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 of nitric acid, from which, in turn, fertilizers 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. There is no need to write units of measurement of physical quantities. 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.
|
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 and 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 materials for the Unified State Exam. 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
This textbook is a collection of tasks for preparing for the Unified State Exam (USE) in chemistry, which is both a final exam for a high school course and an entrance exam to a university. The structure of the manual reflects modern requirements for the procedure for 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 defined by the Federal component of the state standard of general education. Chemistry (Order of the Ministry of Education No. 1089 of 03/05/2004).
The level of presentation of the content of educational material in tasks 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.
Each version of the examination paper is 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 the written work in chemistry in university entrance exams.
