Tests to prepare for the exam in chemistry. How to solve chemistry problems, ready-made solutions

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Hundreds of Unified State Exam tasks. Word problems and probability theory. Simple and easy to remember algorithms for solving problems. Geometry. Theory, reference material, analysis of all types of Unified State Examination tasks. Stereometry. Tricky solutions, useful cheat sheets, development of spatial imagination. Trigonometry from scratch to problem 13. Understanding instead of cramming. Clear explanations of complex concepts. Algebra. Roots, powers and logarithms, function and derivative. Basis for solution complex tasks 2 parts of the Unified State Exam.

Methods for solving problems in chemistry

When solving problems, you must be guided by a few simple rules:

Read the task conditions carefully;
Write down what is given;
Convert, if necessary, units of physical quantities into SI units (some non-system units are allowed, for example liters);
Write down, if necessary, the reaction equation and arrange the coefficients;
Solve a problem using the concept of the amount of a substance, and not the method of drawing up proportions;
Write down the answer.

In order to successfully prepare for chemistry, you should carefully consider the solutions to the problems given in the text, and also solve a sufficient number of them yourself. It is in the process of solving problems that the basic theoretical principles of the chemistry course will be reinforced. It is necessary to solve problems throughout the entire time of studying chemistry and preparing for the exam.

You can use the problems on this page, or you can download a good collection of problems and exercises with solutions to standard and complicated problems (M. I. Lebedeva, I. A. Ankudimova): download.

Mole, molar mass

Molar mass– is the ratio of the mass of a substance to the amount of substance, i.e.

M(x) = m(x)/ν(x), (1)

where M(x) is the molar mass of substance X, m(x) is the mass of substance X, ν(x) is the amount of substance X. The SI unit of molar mass is kg/mol, but the unit g/mol is usually used. Unit of mass – g, kg. The SI unit for quantity of a substance is the mole.

Any chemistry problem solved through the amount of substance. You need to remember the basic formula:

ν(x) = m(x)/ M(x) = V(x)/V m = N/N A , (2)

where V(x) is the volume of the substance X(l), V m is the molar volume of the gas (l/mol), N is the number of particles, N A is Avogadro’s constant.

1. Determine mass sodium iodide NaI amount of substance 0.6 mol.

Given: ν(NaI)= 0.6 mol.

Find: m(NaI) =?

Solution. The molar mass of sodium iodide is:

M(NaI) = M(Na) + M(I) = 23 + 127 = 150 g/mol

Determine the mass of NaI:

m(NaI) = ν(NaI) M(NaI) = 0.6 150 = 90 g.

2. Determine the amount of substance atomic boron contained in sodium tetraborate Na 2 B 4 O 7 weighing 40.4 g.

Given: m(Na 2 B 4 O 7) = 40.4 g.

Find: ν(B)=?

Solution. The molar mass of sodium tetraborate is 202 g/mol. Determine the amount of substance Na 2 B 4 O 7:

ν(Na 2 B 4 O 7) = m(Na 2 B 4 O 7)/ M(Na 2 B 4 O 7) = 40.4/202 = 0.2 mol.

Recall that 1 mole of sodium tetraborate molecule contains 2 moles of sodium atoms, 4 moles of boron atoms and 7 moles of oxygen atoms (see sodium tetraborate formula). Then the amount of atomic boron substance is equal to: ν(B) = 4 ν (Na 2 B 4 O 7) = 4 0.2 = 0.8 mol.

Calculations according to chemical formulas. Mass fraction.

Mass fraction of a substance is the ratio of the mass of a given substance in a system to the mass of the entire system, i.e. ω(X) =m(X)/m, where ω(X) is the mass fraction of substance X, m(X) is the mass of substance X, m is the mass of the entire system. Mass fraction is a dimensionless quantity. It is expressed as a fraction of a unit or as a percentage. For example, the mass fraction of atomic oxygen is 0.42, or 42%, i.e. ω(O)=0.42. The mass fraction of atomic chlorine in sodium chloride is 0.607, or 60.7%, i.e. ω(Cl)=0.607.

3. Determine the mass fraction water of crystallization in barium chloride dihydrate BaCl 2 2H 2 O.

Solution: The molar mass of BaCl 2 2H 2 O is:

M(BaCl 2 2H 2 O) = 137+ 2 35.5 + 2 18 = 244 g/mol

From the formula BaCl 2 2H 2 O it follows that 1 mol of barium chloride dihydrate contains 2 mol of H 2 O. From this we can determine the mass of water contained in BaCl 2 2H 2 O:

m(H 2 O) = 2 18 = 36 g.

We find the mass fraction of water of crystallization in barium chloride dihydrate BaCl 2 2H 2 O.

ω(H 2 O) = m(H 2 O)/ m(BaCl 2 2H 2 O) = 36/244 = 0.1475 = 14.75%.

4. Silver weighing 5.4 g was isolated from a rock sample weighing 25 g containing the mineral argentite Ag 2 S. Determine the mass fraction argentite in the sample.

Given: m(Ag)=5.4 g; m = 25 g.

Find: ω(Ag 2 S) =?

Solution: we determine the amount of silver substance found in argentite: ν(Ag) =m(Ag)/M(Ag) = 5.4/108 = 0.05 mol.

From the formula Ag 2 S it follows that the amount of argentite substance is half as much as the amount of silver substance. Determine the amount of argentite substance:

ν(Ag 2 S)= 0.5 ν(Ag) = 0.5 0.05 = 0.025 mol

We calculate the mass of argentite:

m(Ag 2 S) = ν(Ag 2 S) M(Ag 2 S) = 0.025 248 = 6.2 g.

Now we determine the mass fraction of argentite in a rock sample weighing 25 g.

ω(Ag 2 S) = m(Ag 2 S)/ m = 6.2/25 = 0.248 = 24.8%.

Deriving compound formulas

5. Determine the simplest formula of the compound potassium with manganese and oxygen, if the mass fractions of elements in this substance are 24.7, 34.8 and 40.5%, respectively.

Given: ω(K) =24.7%; ω(Mn) =34.8%; ω(O) =40.5%.

Find: formula of the compound.

Solution: for calculations we select the mass of the compound equal to 100 g, i.e. m=100 g. The masses of potassium, manganese and oxygen will be:

m (K) = m ω(K); m (K) = 100 0.247 = 24.7 g;

m (Mn) = m ω(Mn); m (Mn) =100 0.348=34.8 g;

m (O) = m ω(O); m(O) = 100 0.405 = 40.5 g.

We determine the amounts of atomic substances potassium, manganese and oxygen:

ν(K)= m(K)/ M(K) = 24.7/39= 0.63 mol

ν(Mn)= m(Mn)/ М(Mn) = 34.8/ 55 = 0.63 mol

ν(O)= m(O)/ M(O) = 40.5/16 = 2.5 mol

We find the ratio of the quantities of substances:

ν(K) : ν(Mn) : ν(O) = 0.63: 0.63: 2.5.

By dividing right side equality to a smaller number (0.63) we get:

ν(K) : ν(Mn) : ν(O) = 1: 1: 4.

Therefore, the simplest formula for the compound is KMnO 4.

6. The combustion of 1.3 g of a substance produced 4.4 g of carbon monoxide (IV) and 0.9 g of water. Find the molecular formula substance if its hydrogen density is 39.

Given: m(in-va) =1.3 g; m(CO 2)=4.4 g; m(H 2 O) = 0.9 g; D H2 =39.

Find: formula of a substance.

Solution: Let's assume that the substance we are looking for contains carbon, hydrogen and oxygen, because during its combustion, CO 2 and H 2 O were formed. Then it is necessary to find the amounts of CO 2 and H 2 O substances in order to determine the amounts of atomic carbon, hydrogen and oxygen substances.

ν(CO 2) = m(CO 2)/ M(CO 2) = 4.4/44 = 0.1 mol;

ν(H 2 O) = m(H 2 O)/ M(H 2 O) = 0.9/18 = 0.05 mol.

We determine the amounts of atomic carbon and hydrogen substances:

ν(C)= ν(CO 2); ν(C)=0.1 mol;

ν(H)= 2 ν(H 2 O); ν(H) = 2 0.05 = 0.1 mol.

Therefore, the masses of carbon and hydrogen will be equal:

m(C) = ν(C) M(C) = 0.1 12 = 1.2 g;

m(N) = ν(N) M(N) = 0.1 1 =0.1 g.

We determine the qualitative composition of the substance:

m(in-va) = m(C) + m(H) = 1.2 + 0.1 = 1.3 g.

Consequently, the substance consists only of carbon and hydrogen (see the problem statement). Let us now determine its molecular weight based on the given condition tasks hydrogen density of a substance.

M(v-va) = 2 D H2 = 2 39 = 78 g/mol.

ν(С) : ν(Н) = 0.1: 0.1

Dividing the right side of the equality by the number 0.1, we get:

ν(С) : ν(Н) = 1: 1

Let us take the number of carbon (or hydrogen) atoms as “x”, then multiplying “x” by atomic masses carbon and hydrogen and equating this sum to the molecular mass of the substance, we solve the equation:

12x + x = 78. Hence x = 6. Therefore, the formula of the substance is C 6 H 6 - benzene.

Molar volume of gases. Laws of ideal gases. Volume fraction.

The molar volume of a gas is equal to the ratio of the volume of the gas to the amount of substance of this gas, i.e.

V m = V(X)/ ν(x),

where V m is the molar volume of gas - a constant value for any gas under given conditions; V(X) – volume of gas X; ν(x) – amount of gas substance X. Molar volume of gases under normal conditions ( normal pressure pH = 101,325 Pa ≈ 101.3 kPa and temperature Tn = 273.15 K ≈ 273 K) is V m = 22.4 l/mol.

In calculations involving gases, it is often necessary to switch from these conditions to normal ones or vice versa. In this case, it is convenient to use the formula following from the combined gas law of Boyle-Mariotte and Gay-Lussac:

──── = ─── (3)

Where p is pressure; V – volume; T - temperature in Kelvin scale; the index “n” indicates normal conditions.

The composition of gas mixtures is often expressed using the volume fraction - the ratio of the volume of a given component to the total volume of the system, i.e.

where φ(X) is the volume fraction of component X; V(X) – volume of component X; V is the volume of the system. Volume fraction is a dimensionless quantity; it is expressed in fractions of a unit or as a percentage.

7. Which volume will take at a temperature of 20 o C and a pressure of 250 kPa ammonia weighing 51 g?

Given: m(NH 3)=51 g; p=250 kPa; t=20 o C.

Find: V(NH 3) =?

Solution: determine the amount of ammonia substance:

ν(NH 3) = m(NH 3)/ M(NH 3) = 51/17 = 3 mol.

The volume of ammonia under normal conditions is:

V(NH 3) = V m ν(NH 3) = 22.4 3 = 67.2 l.

Using formula (3), we reduce the volume of ammonia to these conditions [temperature T = (273 +20) K = 293 K]:

p n TV n (NH 3) 101.3 293 67.2

V(NH 3) =──────── = ───────── = 29.2 l.

8. Define volume, which will be occupied under normal conditions by a gas mixture containing hydrogen, weighing 1.4 g, and nitrogen, weighing 5.6 g.

Given: m(N 2)=5.6 g; m(H 2)=1.4; Well.

Find: V(mixtures)=?

Solution: find the amounts of hydrogen and nitrogen substances:

ν(N 2) = m(N 2)/ M(N 2) = 5.6/28 = 0.2 mol

ν(H 2) = m(H 2)/ M(H 2) = 1.4/ 2 = 0.7 mol

Since under normal conditions these gases do not interact with each other, the volume of the gas mixture will be equal to the sum volumes of gases, i.e.

V(mixtures)=V(N 2) + V(H 2)=V m ν(N 2) + V m ν(H 2) = 22.4 0.2 + 22.4 0.7 = 20.16 l.

Calculations using chemical equations

Calculations according to chemical equations(stoichiometric calculations) are based on the law of conservation of mass of substances. However, in real chemical processes Due to the incomplete course of the reaction and various losses of substances, the mass of the resulting products is often less than that which should be formed in accordance with the law of conservation of mass of substances. The yield of the reaction product (or mass fraction of yield) is the ratio, expressed as a percentage, of the mass of the actually obtained product to its mass, which should be formed in accordance with the theoretical calculation, i.e.

η = /m(X) (4)

Where η is the product yield, %; m p (X) is the mass of product X obtained in the real process; m(X) – calculated mass of substance X.

In those tasks where the product yield is not specified, it is assumed that it is quantitative (theoretical), i.e. η=100%.

9. How much phosphorus needs to be burned? for getting phosphorus (V) oxide weighing 7.1 g?

Given: m(P 2 O 5) = 7.1 g.

Find: m(P) =?

Solution: we write down the equation for the combustion reaction of phosphorus and arrange the stoichiometric coefficients.

4P+ 5O 2 = 2P 2 O 5

Determine the amount of substance P 2 O 5 resulting in the reaction.

ν(P 2 O 5) = m(P 2 O 5)/ M(P 2 O 5) = 7.1/142 = 0.05 mol.

From the reaction equation it follows that ν(P 2 O 5) = 2 ν(P), therefore, the amount of phosphorus required in the reaction is equal to:

ν(P 2 O 5)= 2 ν(P) = 2 0.05= 0.1 mol.

From here we find the mass of phosphorus:

m(P) = ν(P) M(P) = 0.1 31 = 3.1 g.

10. Magnesium weighing 6 g and zinc weighing 6.5 g were dissolved in excess hydrochloric acid. What volume hydrogen, measured under standard conditions, will stand out wherein?

Given: m(Mg)=6 g; m(Zn)=6.5 g; Well.

Find: V(H 2) =?

Solution: we write down the reaction equations for the interaction of magnesium and zinc with hydrochloric acid and arrange the stoichiometric coefficients.

Zn + 2 HCl = ZnCl 2 + H 2

Mg + 2 HCl = MgCl 2 + H 2

We determine the amounts of magnesium and zinc substances that reacted with hydrochloric acid.

ν(Mg) = m(Mg)/ М(Mg) = 6/24 = 0.25 mol

ν(Zn) = m(Zn)/ M(Zn) = 6.5/65 = 0.1 mol.

From the reaction equations it follows that the amounts of metal and hydrogen substances are equal, i.e. ν(Mg) = ν(H 2); ν(Zn) = ν(H 2), we determine the amount of hydrogen resulting from two reactions:

ν(H 2) = ν(Mg) + ν(Zn) = 0.25 + 0.1 = 0.35 mol.

We calculate the volume of hydrogen released as a result of the reaction:

V(H 2) = V m ν(H 2) = 22.4 0.35 = 7.84 l.

11. When a volume of 2.8 liters of hydrogen sulfide (normal conditions) was passed through an excess solution of copper (II) sulfate, a precipitate weighing 11.4 g was formed. Determine the exit reaction product.

Given: V(H 2 S)=2.8 l; m(sediment)= 11.4 g; Well.

Find: η =?

Solution: we write down the equation for the reaction between hydrogen sulfide and copper (II) sulfate.

H 2 S + CuSO 4 = CuS ↓+ H 2 SO 4

We determine the amount of hydrogen sulfide involved in the reaction.

ν(H 2 S) = V(H 2 S) / V m = 2.8/22.4 = 0.125 mol.

From the reaction equation it follows that ν(H 2 S) = ν(СuS) = 0.125 mol. This means we can find the theoretical mass of CuS.

m(СuS) = ν(СuS) М(СuS) = 0.125 96 = 12 g.

Now we determine the product yield using formula (4):

η = /m(X)= 11.4 100/ 12 = 95%.

12. Which one weight ammonium chloride is formed by the interaction of hydrogen chloride weighing 7.3 g with ammonia weighing 5.1 g? Which gas will remain in excess? Determine the mass of the excess.

Given: m(HCl)=7.3 g; m(NH 3)=5.1 g.

Find: m(NH 4 Cl) =? m(excess) =?

Solution: write down the reaction equation.

HCl + NH 3 = NH 4 Cl

This task is about “excess” and “deficiency”. We calculate the amounts of hydrogen chloride and ammonia and determine which gas is in excess.

ν(HCl) = m(HCl)/ M(HCl) = 7.3/36.5 = 0.2 mol;

ν(NH 3) = m(NH 3)/ M(NH 3) = 5.1/ 17 = 0.3 mol.

Ammonia is in excess, so we calculate based on the deficiency, i.e. for hydrogen chloride. From the reaction equation it follows that ν(HCl) = ν(NH 4 Cl) = 0.2 mol. Determine the mass of ammonium chloride.

m(NH 4 Cl) = ν(NH 4 Cl) М(NH 4 Cl) = 0.2 53.5 = 10.7 g.

We have determined that ammonia is in excess (in terms of the amount of substance, the excess is 0.1 mol). Let's calculate the mass of excess ammonia.

m(NH 3) = ν(NH 3) M(NH 3) = 0.1 17 = 1.7 g.

13. Technical calcium carbide weighing 20 g was treated with excess water, obtaining acetylene, which, when passed through excess bromine water, formed 1,1,2,2-tetrabromoethane weighing 86.5 g. Determine mass fraction CaC 2 in technical carbide.

Given: m = 20 g; m(C 2 H 2 Br 4) = 86.5 g.

Find: ω(CaC 2) =?

Solution: we write down the equations for the interaction of calcium carbide with water and acetylene with bromine water and arrange the stoichiometric coefficients.

CaC 2 +2 H 2 O = Ca(OH) 2 + C 2 H 2

C 2 H 2 +2 Br 2 = C 2 H 2 Br 4

Find the amount of tetrabromoethane.

ν(C 2 H 2 Br 4) = m(C 2 H 2 Br 4)/ M(C 2 H 2 Br 4) = 86.5/ 346 = 0.25 mol.

From the reaction equations it follows that ν(C 2 H 2 Br 4) = ν(C 2 H 2) = ν(CaC 2) = 0.25 mol. From here we can find the mass of pure calcium carbide (without impurities).

m(CaC 2) = ν(CaC 2) M(CaC 2) = 0.25 64 = 16 g.

We determine the mass fraction of CaC 2 in technical carbide.

ω(CaC 2) =m(CaC 2)/m = 16/20 = 0.8 = 80%.

Solutions. Mass fraction of solution component

14. Sulfur weighing 1.8 g was dissolved in benzene with a volume of 170 ml. The density of benzene is 0.88 g/ml. Define mass fraction sulfur in solution.

Given: V(C 6 H 6) = 170 ml; m(S) = 1.8 g; ρ(C 6 C 6) = 0.88 g/ml.

Find: ω(S) =?

Solution: to find the mass fraction of sulfur in a solution, it is necessary to calculate the mass of the solution. Determine the mass of benzene.

m(C 6 C 6) = ρ(C 6 C 6) V(C 6 H 6) = 0.88 170 = 149.6 g.

Find the total mass of the solution.

m(solution) = m(C 6 C 6) + m(S) = 149.6 + 1.8 = 151.4 g.

Let's calculate the mass fraction of sulfur.

ω(S) =m(S)/m=1.8 /151.4 = 0.0119 = 1.19%.

15. Iron sulfate FeSO 4 7H 2 O weighing 3.5 g was dissolved in water weighing 40 g. Determine mass fraction of iron (II) sulfate in the resulting solution.

Given: m(H 2 O)=40 g; m(FeSO 4 7H 2 O) = 3.5 g.

Find: ω(FeSO 4) =?

Solution: find the mass of FeSO 4 contained in FeSO 4 7H 2 O. To do this, calculate the amount of the substance FeSO 4 7H 2 O.

ν(FeSO 4 7H 2 O)=m(FeSO 4 7H 2 O)/M(FeSO 4 7H 2 O)=3.5/278=0.0125 mol

From the formula of iron sulfate it follows that ν(FeSO 4) = ν(FeSO 4 7H 2 O) = 0.0125 mol. Let's calculate the mass of FeSO 4:

m(FeSO 4) = ν(FeSO 4) M(FeSO 4) = 0.0125 152 = 1.91 g.

Considering that the mass of the solution consists of the mass of iron sulfate (3.5 g) and the mass of water (40 g), we calculate the mass fraction of ferrous sulfate in the solution.

ω(FeSO 4) =m(FeSO 4)/m=1.91 /43.5 = 0.044 =4.4%.

Problems to solve independently

50 g of methyl iodide in hexane were exposed to metallic sodium, and 1.12 liters of gas were released, measured under normal conditions. Determine the mass fraction of methyl iodide in the solution. Answer: 28,4%.
Some alcohol was oxidized to form a monocarboxylic acid. When 13.2 g of this acid was burned, carbon dioxide was obtained, the complete neutralization of which required 192 ml of KOH solution with a mass fraction of 28%. The density of the KOH solution is 1.25 g/ml. Determine the formula of alcohol. Answer: butanol.
The gas obtained by reacting 9.52 g of copper with 50 ml of an 81% nitric acid solution with a density of 1.45 g/ml was passed through 150 ml of a 20% NaOH solution with a density of 1.22 g/ml. Determine the mass fractions of dissolved substances. Answer: 12.5% NaOH; 6.48% NaNO 3 ; 5.26% NaNO2.
Determine the volume of gases released during the explosion of 10 g of nitroglycerin. Answer: 7.15 l.
A sample of organic matter weighing 4.3 g was burned in oxygen. The reaction products are carbon monoxide (IV) with a volume of 6.72 l (normal conditions) and water with a mass of 6.3 g. The vapor density of the starting substance with respect to hydrogen is 43. Determine the formula of the substance. Answer: C 6 H 14.

Chemistry is not the most suitable subject for testing knowledge in a test format. The test involves answer options, and the correct answer becomes obvious, or doubts arise due to close answer options. This greatly interferes with the student’s ability to concentrate and answer questions. Of course, it is much easier for poor students to pass chemistry in the Unified State Exam format than in the classic version. But for the rest of the students, the Unified State Exam in chemistry became a big problem.

How to pass the Unified State Exam in Chemistry?

As with any exam, the Unified State Exam in Chemistry requires careful preparation. Answering test questions requires precise knowledge, and not approximate numbers, which are enough for the classic answer. If in writing a reaction by hand the conditions can be written in a range, then the Unified State Exam requires an exact answer to the question posed. Therefore, preparing for the Unified State Exam in chemistry is somewhat different from preparing for other exams. First of all, the role of practice and preparedness for such issues increases. They can best teach you how to pass the Unified State Exam in preparatory courses for college. Professors who may have participated in the preparation of assignments take part in the training. Therefore, they know better than anyone the subtleties of the questions, and the prepared traps that tend to bring down the student. But not everyone has the opportunity to attend expensive courses. In addition, some people do not necessarily need a high score in chemistry, but they still need to pass the Unified State Exam.

Online Unified State Exam tests - a type of self-preparation for the exam

In such cases, cooking itself comes to the fore. Even a school cannot provide a student with sufficient preparation for such a difficult exam. All responsibility falls on the student himself. One of the best ways self-training are considered online tests Unified State Exam. You can access the site online on the educational portal Unified State Exam test in chemistry, to independently prepare for the upcoming exam. The online tests on our website are different in that you do not need to register or enter any personal data to take them. The online Unified State Exam is available to everyone an unlimited number of times. Another advantage is unlimited time. If you are faced with a difficult question, you can open a textbook or search the Internet for the answer to the question. In this way, knowledge gaps can be identified and addressed. Constant training also allows you to get used to the Unified State Exam format and learn to extract the exact knowledge from textbooks that is necessary to answer exam questions.

State final certification 2019 in chemistry for 9th grade graduates educational institutions is carried out to assess the level of general education training of graduates in this discipline. The tasks test knowledge of the following sections of chemistry:

The structure of the atom.
Periodic law and Periodic table chemical elements D.I. Mendeleev.
The structure of molecules. Chemical bond: covalent (polar and non-polar), ionic, metallic.
Valency of chemical elements. The degree of oxidation of chemical elements.
Simple and complex substances.
Chemical reaction. Conditions and signs of chemical reactions. Chemical equations.
Electrolytes and non-electrolytes. Cations and anions. Electrolytic dissociation of acids, alkalis and salts (average).
Ion exchange reactions and conditions for their implementation.
Chemical properties simple substances: metals and non-metals.
Chemical properties of oxides: basic, amphoteric, acidic.
Chemical properties of bases. Chemical properties of acids.
Chemical properties of salts (average).
Pure substances and mixtures. Rules for safe work in a school laboratory. Chemical pollution environment and its consequences.
The degree of oxidation of chemical elements. Oxidizing agent and reducing agent. Redox reactions.
Calculation of mass fraction chemical element in matter.
Periodic law D.I. Mendeleev.
Initial information about organic matter. Biologically important substances: proteins, fats, carbohydrates.
Determination of the nature of the solution environment of acids and alkalis using indicators. Qualitative reactions to ions in solution (chloride, sulfate, carbonation, ammonium ion). Qualitative reactions to gaseous substances (oxygen, hydrogen, carbon dioxide, ammonia).
Chemical properties of simple substances. Chemical properties of complex substances.

Date of passing the OGE in chemistry 2019:
June 4 (Tuesday).

There are no changes in the structure and content of the 2019 examination paper compared to 2018.

In this section you will find online tests that will help you prepare for taking the OGE (GIA) in chemistry. We wish you success!

The standard OGE test (GIA-9) of the 2019 format in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. In this regard, in this test Only the first part is presented (i.e. the first 19 tasks). According to the current exam structure, among these tasks, answer options are offered only in 15. However, for the convenience of passing tests, the site administration decided to offer answer options in all tasks. But for tasks in which the compilers of real test and measurement materials (CMMs) do not provide answer options, the number of answer options has been significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

The standard OGE test (GIA-9) of the 2019 format in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. In this regard, only the first part (i.e., the first 19 tasks) is presented in this test. According to the current exam structure, among these tasks, answer options are offered only in 15. However, for the convenience of passing tests, the site administration decided to offer answer options in all tasks. But for tasks in which the compilers of real test and measurement materials (CMMs) do not provide answer options, the number of answer options has been significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

The standard OGE test (GIA-9) of the 2018 format in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. In this regard, only the first part (i.e., the first 19 tasks) is presented in this test. According to the current exam structure, among these tasks, answer options are offered only in 15. However, for the convenience of passing tests, the site administration decided to offer answer options in all tasks. But for tasks in which the compilers of real test and measurement materials (CMMs) do not provide answer options, the number of answer options has been significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

The standard OGE test (GIA-9) of the 2017 format in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. In this regard, only the first part (i.e., the first 19 tasks) is presented in this test. According to the current exam structure, among these tasks, answer options are offered only in 15. However, for the convenience of passing tests, the site administration decided to offer answer options in all tasks. But for tasks in which the compilers of real test and measurement materials (CMMs) do not provide answer options, the number of answer options has been significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

The standard OGE test (GIA-9) of the 2016 format in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. In this regard, only the first part (i.e., the first 19 tasks) is presented in this test. According to the current exam structure, among these tasks, answer options are offered only in 15. However, for the convenience of passing tests, the site administration decided to offer answer options in all tasks. But for tasks in which the compilers of real test and measurement materials (CMMs) do not provide answer options, the number of answer options has been significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

The standard OGE test (GIA-9) of the 2015 format in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. In this regard, only the first part (i.e., the first 19 tasks) is presented in this test. According to the current exam structure, among these tasks, answer options are offered only in 15. However, for the convenience of passing tests, the site administration decided to offer answer options in all tasks. But for tasks in which the compilers of real test and measurement materials (CMMs) do not provide answer options, the number of answer options has been significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

When completing tasks A1-A19, select only one correct option .
When completing tasks B1-B3, select two correct options.

When completing tasks A1-A15, select only one correct option.

When completing tasks A1-A15, choose only one correct option.

■ Is there a guarantee that after classes with you we will pass the Unified State Exam in Chemistry with the required score?

More than 95% graduates who completed a full year's course of study with me and regularly completed their homework, entered the chosen university. Students who took the test Unified State Exam in September with 20-30 points scored above 80 in May! Your achievements will depend on you: if you are willing to work hard, success will come!

■ We are moving to 11th grade, our knowledge of chemistry is zero. Is it too late or is there still a chance to enroll?

There's definitely a chance! I’ll tell you a secret: 80% of the applicants whom I will begin preparing for the Unified State Exam in Chemistry in September will study in the group for beginners. These are the statistics: 80% of eleventh graders learned virtually nothing from school chemistry lessons. But the same statistics say that most of them will successfully pass the Unified State Exam and enter the university of their dreams. The main thing is to study seriously!

■ Is preparing for the Unified State Exam in Chemistry very difficult?

First of all, it's very interesting! My main task is to change the school idea of chemistry as boring, confusing, of little use in real life science. Yes, the student will have to work during class. Yes, he will have to do extensive homework. But if you can get him interested in chemistry, this work will be a joy!

■ According to what textbooks you are working?

Mainly on our own. I have been polishing my own system of preparing for the Unified State Exam for more than 10 years, and over the years it has proven its effectiveness. You don't need to worry about purchasing educational literature- I will provide you with everything you need. For free!

■ How (technically) can I sign up for your classes?

Very simple!

Call me on: 8-903-280-81-91 . You can call any day until 23.00.
We will arrange a first meeting for preliminary testing and to determine the level of the group.
You choose the lesson time and group size that is convenient for you (individual lessons, pair lessons, mini-groups).
That's it, work begins at the appointed time.

Good luck!

Or you can simply use it on this site.

■ How effective is group learning? Isn't it better to choose the format of individual lessons?

Classes in groups are the most acceptable in terms of price-quality ratio. The question of their effectiveness is a question of: 1) the qualifications of the tutor, 2) the number of students in the group, 3) correct selection composition of the group.

The fears of parents are understandable: the phrase “group classes” brings to mind school classes in which 30 - 35 children with different levels training and, to put it mildly, different levels of intelligence.

A qualified tutor will not allow anything like this. First of all, I follow the sacred rule: “No more than 5 people in a group!” In my opinion this is maximum amount people, which can be taken into account Individual characteristics every student. A more numerous composition is “in-line production”.

Secondly, everyone who begins preparing for the Unified State Exam undergoes mandatory testing. Groups are formed from students with approximately the same level of knowledge. The situation in which one person in the group perceives the material, and the rest are simply bored, is excluded! All participants will receive equal attention and we will ensure that ALL students fully understand each topic!

■ But are individual lessons still possible?

Of course they are possible! Call me (8-903-280-81-91) - we will discuss which option will be best for you.

■ Do you go to students’ homes?

Yes, I'm leaving. To any district of Moscow (including areas beyond the Moscow Ring Road) and to the near Moscow region. Moreover, not only individual but also group classes can be conducted at students’ homes.

■ And we live far from Moscow. What to do?

Study remotely. Skype is ours best helper. Distance learning is no different from face-to-face learning: the same methodology, the same educational materials. My login: repetitor2000. Contact us! Let's do a trial lesson and see how simple it is!

■ Is it possible to start preparing for the Unified State Exam in 10th grade?

Of course you can! And not only is it possible, but it is also recommended. Imagine that at the end of 10th grade a student is almost ready for the Unified State Exam. If there are any problems left, there will be time in 11th grade to correct them. If everything goes well, 11th grade can be devoted to preparing for the chemistry Olympiads (and a decent performance at the Lomonosov Olympiad, for example, practically guarantees admission to leading universities, including Moscow State University). The sooner you start practicing, the greater your chances of success.

■ We are interested not only in preparing for the Unified State Exam in chemistry, but also in biology. Can you help?

I don't teach biology, but I can recommend you a qualified tutor in this subject. The Unified State Exam in Biology is much easier than the Unified State Exam in Chemistry, but, of course, you also need to prepare seriously for this exam.

■ We will not be able to start classes in September. Is it possible to join the group a little later?

Such issues are resolved individually. If there is free space, if the rest of the group does not object, and if testing shows that your level of knowledge corresponds to the level of the group, I will gladly accept you. Call me (8-903-280-81-91), we will discuss your situation.

■ How much will the Unified State Examination 2019 in chemistry differ from the Unified State Examination 2018?

Changes are planned, but they are not structural, but rather cosmetic. If in 10th grade you have already studied in one of my groups and completed the full course of preparation for the Unified State Exam, there is not the slightest need to take it again: you have all the necessary knowledge. If you are planning to expand your horizons, I invite you to join the group for those preparing for Chemistry Olympiads.