In chemistry, the values ​​​​of the absolute masses of molecules are not used, but the value of the relative molecular mass is used. It shows how many times the mass of a molecule is greater than 1/12 of the mass of a carbon atom. This value is denoted by M r .

The relative molecular weight is equal to the sum of the relative atomic masses of its constituent atoms. Calculate the relative molecular weight of water.

You know that a water molecule contains two hydrogen atoms and one oxygen atom. Then its relative molecular mass will be equal to the sum of the products of the relative atomic mass of each chemical element and the number of its atoms in a water molecule:

Knowing the relative molecular weights of gaseous substances, one can compare their densities, i.e., calculate the relative density of one gas from another - D (A / B). The relative density of gas A for gas B is equal to the ratio of their relative molecular masses:

Calculate the relative density of carbon dioxide for hydrogen:

Now we calculate the relative density of carbon dioxide for hydrogen:

D(co.g./hydrogen.) = M r (co. g.) : M r (hydrogen.) = 44:2 = 22.

Thus, carbon dioxide is 22 times heavier than hydrogen.

As you know, Avogadro's law applies only to gaseous substances. But chemists need to have an idea about the number of molecules and in portions of liquid or solid substances. Therefore, to compare the number of molecules in substances, chemists introduced the value - molar mass .

Molar mass is denoted M, it is numerically equal to the relative molecular weight.

The ratio of the mass of a substance to its molar mass is called amount of matter .

The amount of a substance is denoted n. This is a quantitative characteristic of a portion of a substance, along with mass and volume. The amount of a substance is measured in moles.

The word "mole" comes from the word "molecule". The number of molecules in equal amounts of a substance is the same.

It has been experimentally established that 1 mol of a substance contains particles (for example, molecules). This number is called Avogadro's number. And if you add a unit of measurement to it - 1 / mol, then it will be a physical quantity - the Avogadro constant, which is denoted N A.

Molar mass is measured in g/mol. The physical meaning of the molar mass is that this mass is 1 mole of a substance.

According to Avogadro's law, 1 mole of any gas will occupy the same volume. The volume of one mole of gas is called the molar volume and is denoted by V n .

Under normal conditions (and this is 0 ° C and normal pressure - 1 atm. Or 760 mm Hg or 101.3 kPa), the molar volume is 22.4 l / mol.

Then the amount of gas substance at n.o. can be calculated as the ratio of gas volume to molar volume.

TASK 1. What amount of substance corresponds to 180 g of water?

TASK 2. Let us calculate the volume at n.o., which will be occupied by carbon dioxide in the amount of 6 mol.

Bibliography

1. Collection of tasks and exercises in chemistry: 8th grade: to the textbook by P.A. Orzhekovsky and others. "Chemistry, Grade 8" / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. - M.: AST: Astrel, 2006. (p. 29-34)
2. Ushakova O.V. Chemistry workbook: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry. Grade 8” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; under. ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006. (p. 27-32)
3. Chemistry: 8th grade: textbook. for general institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005. (§§ 12, 13)
4. Chemistry: inorg. chemistry: textbook. for 8 cells. general institution / G.E. Rudzitis, F.G. Feldman. - M .: Education, JSC "Moscow textbooks", 2009. (§§ 10, 17)
5. Encyclopedia for children. Volume 17. Chemistry / Chapter. edited by V.A. Volodin, leading. scientific ed. I. Leenson. - M.: Avanta +, 2003.

1. A single collection of digital educational resources ().
2. Electronic version of the journal "Chemistry and Life" ().
3. Chemistry tests (online) ().

Homework

1.p.69 No. 3; p.73 Nos. 1, 2, 4 from the textbook "Chemistry: 8th grade" (P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M .: AST: Astrel, 2005).

2. №№ 65, 66, 71, 72 from the Collection of tasks and exercises in chemistry: 8th grade: to the textbook by P.A. Orzhekovsky and others. "Chemistry, Grade 8" / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. - M.: AST: Astrel, 2006.

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

V m = V(X) / n(X),

where V m - molar volume of gas - a constant value for any gas under given conditions;

V(X) is the volume of gas X;

n(X) is the amount of gas substance X.

The molar volume of gases under normal conditions (normal pressure p n \u003d 101 325 Pa ≈ 101.3 kPa and temperature T n \u003d 273.15 K ≈ 273 K) is V m \u003d 22.4 l / mol.

Laws of ideal gases

In calculations involving gases, it is often necessary to switch from these conditions to normal conditions 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:

pV / T = p n V n / T n

Where p is the pressure; V - volume; T is the temperature on the Kelvin scale; the index "n" indicates normal conditions.

Volume fraction

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

φ(X) = V(X) / V

where φ(X) - volume fraction of component X;

V(X) - volume of component X;

V is the volume of the system.

The volume fraction is a dimensionless quantity, it is expressed in fractions of a unit or as a percentage.

Example 1. What volume will take at a temperature of 20 ° C and a pressure of 250 kPa ammonia weighing 51 g?

1. Determine the amount of ammonia substance: n (NH 3) \u003d m (NH 3) / M (NH 3) \u003d 51/17 \u003d 3 mol. 2. The volume of ammonia under normal conditions is: V (NH 3) \u003d V m n (NH 3) \u003d 22.4 3 \u003d 67.2 l. 3. Using formula (3), we bring the volume of ammonia to these conditions (temperature T = (273 + 20) K = 293 K): V (NH 3) \u003d p n V n (NH 3) / pT n \u003d 101.3 293 67.2 / 250 273 \u003d 29.2 l. Answer: V (NH 3) \u003d 29.2 liters.

Example 2. Determine the volume that a gas mixture containing hydrogen, weighing 1.4 g and nitrogen, weighing 5.6 g, will take under normal conditions.

1. Find the amount of hydrogen and nitrogen matter: n (N 2) \u003d m (N 2) / M (N 2) \u003d 5.6 / 28 \u003d 0.2 mol n (H 2) \u003d m (H 2) / M (H 2) \u003d 1.4 / 2 \u003d 0.7 mol 2. Since under normal conditions these gases do not interact with each other, the volume of the gas mixture will be equal to the sum of the volumes of gases, i.e. V (mixtures) \u003d V (N 2) + V (H 2) \u003d V m n (N 2) + V m n (H2) \u003d 22.4 0.2 + 22.4 0.7 \u003d 20.16 l. Answer: V (mixture) \u003d 20.16 liters.

Law of Volumetric Relations

How to solve the problem using the "Law of Volumetric Relations"?

Law of volumetric ratios: The volumes of gases involved in a reaction are related to each other as small integers equal to the coefficients in the reaction equation.

The coefficients in the reaction equations show the number of volumes of reacting and formed gaseous substances.

Example. Calculate the volume of air required to burn 112 liters of acetylene.

1. We compose the reaction equation:

2. Based on the law of volumetric ratios, we calculate the volume of oxygen:

112/2 \u003d X / 5, whence X \u003d 112 5 / 2 \u003d 280l

3. Determine the volume of air:

V (air) \u003d V (O 2) / φ (O 2)

V (air) \u003d 280 / 0.2 \u003d 1400 l.

The purpose of the lesson: to form the concept of molar, millimolar and kilomolar volumes of gases and their units of measurement.

Lesson objectives:

• Educational- to consolidate the previously studied formulas and find the relationship between volume and mass, the amount of substance and the number of molecules, to consolidate and systematize the knowledge of students.
• Educational- to develop the skills and abilities to solve problems, the ability to think logically, to expand the horizons of students, their creative abilities, the ability to work with additional literature, long-term memory, interest in the subject.
• Educational- to educate individuals with a high level of culture, to form the need for cognitive activity.

Lesson type: Combined lesson.

Equipment and reagents: Table "Molar volume of gases", Avogadro's portrait, beaker, water, measuring cups with sulfur, calcium oxide, glucose in the amount of 1 mol.

Lesson plan:

1. Organizational moment (1 min.)
2. Knowledge testing in the form of a frontal survey (10 min.)
3. Completing the table (5 min.)
4. Explaining new material (10 min.)
5. Fixing (10 min.)
6. Summing up (3 min.)
7. Homework (1 min.)

During the classes

1. Organizational moment.

2. Frontal conversation on issues.

What is the mass of 1 mole of a substance called?

How to relate molar mass and amount of substance?

What is Avogadro's number?

What is the relationship between Avogadro's number and the amount of matter?

And how to relate the mass and number of molecules of a substance?

3. Now fill in the table by solving the problems - this is group work.

Formula, substances	Weight, g	Molar mass, g/mol	Amount of substance, mol	Number of molecules	Avogadro number, molecules/mol
ZnO	?	81 g/mol	? mole	18 10 23 molecules	6 10 23
MgS	5.6g	56 g/mol	? mole	?	6 10 23
BaCl2	?	? g/mol	0.5 mol	3 10 23 molecules	6 10 23

4. Learning new material.

“... We want not only to know how nature is organized (and how natural phenomena occur), but also, if possible, to achieve the goal, perhaps utopian and daring in appearance, to find out why nature is exactly this way and not another. In this, scientists find the highest satisfaction.
Albert Einstein

So, our goal is to find the highest satisfaction, like real scientists.

What is the volume of 1 mole of a substance called?

What does molar volume depend on?

What will be the molar volume of water if its M r = 18 and ρ = 1 g/ml?

(Of course 18 ml).

To determine the volume, you used the formula known from physics ρ = m / V (g / ml, g / cm 3, kg / m 3)

Let's measure this volume with measuring utensils. We measure the molar volumes of alcohol, sulfur, iron, sugar. They are different, because the density is different, (table of different densities).

How about gases? It turns out that 1 mole of any gas at n.o. (0 ° C and 760 mm Hg) occupies the same molar volume of 22.4 l / mol (shown in the table). What is the name of the volume of 1 kilomole? Kilomolar. It is equal to 22.4 m 3 / kmol. The millimolar volume is 22.4 ml/mol.

Where did this number come from?

It follows from Avogadro's law. Consequence from Avogadro's law: 1 mole of any gas at n.o. occupies a volume of 22.4 l/mol.

We will now hear a little about the life of the Italian scientist. (report on Avogadro's life)

And now let's see the dependence of values ​​on different indicators:

Substance formula	Aggregate state (at n.o.s.)	Weight, g	Density, g/ml	The volume of servings in 1 mol, l	Amount of substance, mol	Relationship between volume and amount of substance
NaCl	solid	58,5	2160	0,027	1	0,027
H2O	Liquid	18	1000	0,018	1	0,18
O2	Gas	32	1,43	22,4	1	22,4
H2	Gas	2	0,09	22,4	1	22,4
CO2	Gas	44	1,96	22,4	1	22,4
SO2	gas	64	2,86	22,4	1	22,4

From a comparison of the data obtained, draw a conclusion (the relationship between the volume and amount of a substance for all gaseous substances (at N.O.) is expressed by the same value, which is called the molar volume.)

It is denoted V m and measured in l / mol, etc. We derive a formula for finding the molar volume

Vm = V/v , from here you can find the amount of substance and volume of gas. Now let's recall the previously studied formulas, can they be combined? You can get universal formulas for calculations.

m/M = V/V m ;

V/Vm = N/Na

5. And now we will consolidate the acquired knowledge with the help of oral counting, so that knowledge through skills will be applied automatically, that is, they will turn into skills.

For the correct answer you will receive a point, by the number of points you will receive an assessment.

1. What is the formula for hydrogen?
2. What is its relative molecular weight?
3. What is its molar mass?
4. How many hydrogen molecules will be in each case?
5. What volume will be occupied at n.o.s. 3 g H2?
6. How much will 12 10 23 hydrogen molecules weigh?
7. What volume will these molecules occupy in each case?

Now let's solve problems in groups.

Task #1

Sample: What is the volume of 0.2 mol N 2 at n.o.?

1. What volume is occupied by 5 mol O 2 at n.o.?
2. What volume is occupied by 2.5 mol H 2 at n.o.?

Task #2

Sample: How much substance does 33.6 liters of hydrogen contain at n.o.?

Tasks for independent solution

Solve problems according to the given example:

1. What amount of a substance contains oxygen with a volume of 0.224 liters at n.o.?
2. What amount of substance contains carbon dioxide with a volume of 4.48 liters at n.o.?

Task #3

Sample: What volume will 56 g of CO gas at N.S. take up?

Tasks for independent solution

Solve problems according to the given example:

1. What volume will occupied by 8 g of gas O 2 at n.o.?
2. What volume will occupied by 64 g of SO 2 gas at N.O.?

Task #4

Sample: What volume contains 3 10 23 molecules of hydrogen H 2 at n.o.?

Tasks for independent solution

Solve problems according to the given example:

1. What volume contains 12.04 · 10 23 molecules of hydrogen CO 2 at n.o.?
2. What volume contains 3.01 10 23 molecules of hydrogen O 2 at n.o.?

The concept of the relative density of gases should be given on the basis of their knowledge of the density of the body: D = ρ 1 /ρ 2, where ρ 1 is the density of the first gas, ρ 2 is the density of the second gas. You know the formula ρ = m/V. Replacing m in this formula with M, and V with V m , we get ρ = M / V m . Then the relative density can be expressed using the right side of the last formula:

D \u003d ρ 1 / ρ 2 \u003d M 1 / M 2.

Conclusion: the relative density of gases is a number showing how many times the molar mass of one gas is greater than the molar mass of another gas.

For example, determine the relative density of oxygen by air, by hydrogen.

6. Summing up.

Solve problems for fixing:

Find the mass (n.o.): a) 6 l. About 3; b) 14 l. gas H 2 S?

What is the volume of hydrogen at n.o. formed by the interaction of 0.23 g of sodium with water?

What is the molar mass of the gas if 1 liter. its mass is 3.17 g? (Hint! m = ρ V)

Where m is mass, M is molar mass, V is volume.

4. Law of Avogadro. Established by the Italian physicist Avogadro in 1811. The same volumes of any gases, taken at the same temperature and the same pressure, contain the same number of molecules.

Thus, the concept of the amount of a substance can be formulated: 1 mole of a substance contains a number of particles equal to 6.02 * 10 23 (called the Avogadro constant)

The consequence of this law is that 1 mole of any gas occupies under normal conditions (P 0 \u003d 101.3 kPa and T 0 \u003d 298 K) a volume equal to 22.4 liters.

5. Boyle-Mariotte Law

At constant temperature, the volume of a given amount of gas is inversely proportional to the pressure under which it is:

6. Gay-Lussac's law

At constant pressure, the change in the volume of a gas is directly proportional to the temperature:

V/T = const.

7. The relationship between gas volume, pressure and temperature can be expressed the combined law of Boyle-Mariotte and Gay-Lussac, which is used to bring gas volumes from one condition to another:

P 0 , V 0 ,T 0 - volume pressure and temperature under normal conditions: P 0 =760 mm Hg. Art. or 101.3 kPa; T 0 \u003d 273 K (0 0 C)

8. Independent assessment of the value of molecular masses M can be done using the so-called equations of state for an ideal gas or the Clapeyron-Mendeleev equations :

pV=(m/M)*RT=vRT.(1.1)

Where R - gas pressure in a closed system, V- volume of the system, T - mass of gas T - absolute temperature, R- universal gas constant.

Note that the value of the constant R can be obtained by substituting the values ​​characterizing one mole of gas at N.C. into equation (1.1):

r = (p V) / (T) \u003d (101.325kPa 22.4 l) / (1 mol 273K) \u003d 8.31J / mol.K)

Examples of problem solving

Example 1 Bringing the volume of gas to normal conditions.

What volume (n.o.) will occupy 0.4×10 -3 m 3 of gas at 50 0 C and a pressure of 0.954×10 5 Pa?

Solution. To bring the volume of gas to normal conditions, use the general formula that combines the laws of Boyle-Mariotte and Gay-Lussac:

pV/T = p 0 V 0 /T 0 .

The volume of gas (n.o.) is, where T 0 \u003d 273 K; p 0 \u003d 1.013 × 10 5 Pa; T = 273 + 50 = 323 K;

M 3 \u003d 0.32 × 10 -3 m 3.

When (n.o.) gas occupies a volume equal to 0.32×10 -3 m 3 .

Example 2 Calculation of the relative density of a gas from its molecular weight.

Calculate the density of ethane C 2 H 6 from hydrogen and air.

Solution. It follows from Avogadro's law that the relative density of one gas over another is equal to the ratio of molecular masses ( M h) of these gases, i.e. D=M 1 /M 2. If M 1С2Н6 = 30, M 2 H2 = 2, the average molecular weight of air is 29, then the relative density of ethane with respect to hydrogen is D H2 = 30/2 =15.

Relative density of ethane in air: D air= 30/29 = 1.03, i.e. ethane is 15 times heavier than hydrogen and 1.03 times heavier than air.

Example 3 Determination of the average molecular weight of a mixture of gases by relative density.

Calculate the average molecular weight of a mixture of gases consisting of 80% methane and 20% oxygen (by volume) using the values ​​of the relative density of these gases with respect to hydrogen.

Solution. Often calculations are made according to the mixing rule, which is that the ratio of the volumes of gases in a two-component gas mixture is inversely proportional to the differences between the density of the mixture and the densities of the gases that make up this mixture. Let us denote the relative density of the gas mixture with respect to hydrogen through D H2. it will be greater than the density of methane, but less than the density of oxygen:

80D H2 - 640 = 320 - 20 D H2; D H2 = 9.6.

The hydrogen density of this mixture of gases is 9.6. average molecular weight of the gas mixture M H2 = 2 D H2 = 9.6×2 = 19.2.

Example 4 Calculation of the molar mass of a gas.

The mass of 0.327 × 10 -3 m 3 of gas at 13 0 C and a pressure of 1.040 × 10 5 Pa is 0.828 × 10 -3 kg. Calculate the molar mass of the gas.

Solution. You can calculate the molar mass of a gas using the Mendeleev-Clapeyron equation:

Where m is the mass of gas; M is the molar mass of the gas; R- molar (universal) gas constant, the value of which is determined by the accepted units of measurement.

If the pressure is measured in Pa, and the volume in m 3, then R\u003d 8.3144 × 10 3 J / (kmol × K).

3.1. When performing measurements of atmospheric air, air of the working area, as well as industrial emissions and hydrocarbons in gas pipelines, there is a problem of bringing the volumes of measured air to normal (standard) conditions. Often in practice, when conducting air quality measurements, the conversion of measured concentrations to normal conditions is not used, resulting in unreliable results.

Here is an excerpt from the Standard:

“Measurements are brought to standard conditions using the following formula:

C 0 \u003d C 1 * P 0 T 1 / R 1 T 0

where: C 0 - the result, expressed in units of mass per unit volume of air, kg / cu. m, or the amount of substance per unit volume of air, mol / cu. m, at standard temperature and pressure;

C 1 - the result, expressed in units of mass per unit volume of air, kg / cu. m, or the amount of substance per unit volume

air, mol/cu. m, at temperature T 1, K, and pressure P 1, kPa.

The formula for bringing to normal conditions in a simplified form has the form (2)

C 1 \u003d C 0 * f, where f \u003d P 1 T 0 / P 0 T 1

standard conversion factor for normalization. The parameters of air and impurities are measured at different temperatures, pressures and humidity. The results lead to standard conditions for comparing measured air quality parameters in different locations and different climates.

3.2 Industry normal conditions

Normal conditions are the standard physical conditions with which the properties of substances are usually correlated (Standard temperature and pressure, STP). Normal conditions are defined by IUPAC (International Union of Practical and Applied Chemistry) as follows: Atmospheric pressure 101325 Pa = 760 mm Hg. Air temperature 273.15 K = 0° C.

Standard conditions (Standard Ambient Temperature and Pressure, SATP) are normal ambient temperature and pressure: pressure 1 Bar = 10 5 Pa = 750.06 mm T. St.; temperature 298.15 K = 25 °C.

Other areas.

Air quality measurements.

The results of measurements of concentrations of harmful substances in the air of the working area lead to the following conditions: a temperature of 293 K (20°C) and a pressure of 101.3 kPa (760 mm Hg).

Aerodynamic parameters of pollutant emissions must be measured in accordance with current state standards. The volumes of exhaust gases obtained from the results of instrumental measurements must be brought to normal conditions (n.s.): 0 ° C, 101.3 kPa ..

Aviation.

The International Civil Aviation Organization (ICAO) defines the International Standard Atmosphere (ISA) at sea level with a temperature of 15°C, an atmospheric pressure of 101325 Pa, and a relative humidity of 0%. These parameters are used when calculating the movement of aircraft.

Gas economy.

The gas industry of the Russian Federation uses atmospheric conditions in accordance with GOST 2939-63 for settlements with consumers: temperature 20°C (293.15K); pressure 760 mm Hg. Art. (101325 N/m²); humidity is 0. Thus, the mass of a cubic meter of gas according to GOST 2939-63 is somewhat less than under “chemical” normal conditions.

Tests

For testing machines, instruments and other technical products, the following are taken as normal values ​​of climatic factors when testing products (normal climatic test conditions):

Temperature - plus 25°±10°С; Relative humidity - 45-80%

Atmospheric pressure 84-106 kPa (630-800 mmHg)

Verification of measuring instruments

The nominal values ​​of the most common normal influencing quantities are selected as follows: Temperature - 293 K (20°C), atmospheric pressure - 101.3 kPa (760 mmHg).

Rationing

The guidelines for setting air quality standards indicate that MPCs in ambient air are set under normal indoor conditions, i.e. 20 C and 760 mm. rt. Art.

The mass of 1 mole of a substance is called the molar mass. What is the volume of 1 mole of a substance called? Obviously, it is also called the molar volume.

What is the molar volume of water? When we measured 1 mol of water, we did not weigh 18 g of water on the scales - this is inconvenient. We used measuring utensils: a cylinder or a beaker, because we knew that the density of water is 1 g/ml. Therefore, the molar volume of water is 18 ml/mol. For liquids and solids, the molar volume depends on their density (Fig. 52, a). Another thing for gases (Fig. 52, b).

Rice. 52.
Molar volumes (n.a.):
a - liquids and solids; b - gaseous substances

If we take 1 mol of hydrogen H 2 (2 g), 1 mol of oxygen O 2 (32 g), 1 mol of ozone O 3 (48 g), 1 mol of carbon dioxide CO 2 (44 g) and even 1 mol of water vapor H 2 O (18 g) under the same conditions, for example, normal (in chemistry it is customary to call normal conditions (n.a.) a temperature of 0 ° C and a pressure of 760 mm Hg, or 101.3 kPa), it turns out that 1 mole of any of the gases will occupy the same volume, equal to 22.4 liters, and contain the same number of molecules - 6 × 10 23.

And if we take 44.8 liters of gas, then how much of its substance will be taken? Of course, 2 mol, since the given volume is twice the molar volume. Hence:

where V is the volume of gas. From here

Molar volume is a physical quantity equal to the ratio of the volume of a substance to the amount of a substance.

The molar volume of gaseous substances is expressed in l/mol. Vm - 22.4 l/mol. The volume of one kilomol is called kilomolar and is measured in m 3 / kmol (Vm = 22.4 m 3 / kmol). Accordingly, the millimolar volume is 22.4 ml/mmol.

Task 1. Find the mass of 33.6 m 3 of ammonia NH 3 (n.a.).

Task 2. Find the mass and volume (n.s.) that 18 × 10 20 molecules of hydrogen sulfide H 2 S have.

When solving the problem, let's pay attention to the number of molecules 18 × 10 20 . Since 10 20 is 1000 times smaller than 10 23 , obviously, calculations should be made using mmol, ml/mmol and mg/mmol.

Keywords and phrases

1. Molar, millimolar and kilomolar volumes of gases.
2. The molar volume of gases (under normal conditions) is 22.4 l / mol.
3. Normal conditions.

Work with computer

1. Refer to the electronic application. Study the material of the lesson and complete the suggested tasks.
2. Search the Internet for email addresses that can serve as additional sources that reveal the content of the keywords and phrases of the paragraph. Offer the teacher your help in preparing a new lesson - make a report on the key words and phrases of the next paragraph.

Questions and tasks

1. Find the mass and number of molecules at n. y. for: a) 11.2 liters of oxygen; b) 5.6 m 3 nitrogen; c) 22.4 ml of chlorine.
2. Find the volume which, at n. y. will take: a) 3 g of hydrogen; b) 96 kg of ozone; c) 12 × 10 20 nitrogen molecules.
3. Find the densities (mass of 1 liter) of argon, chlorine, oxygen and ozone at n. y. How many molecules of each substance will be contained in 1 liter under the same conditions?
4. Calculate the mass of 5 l (n.a.): a) oxygen; b) ozone; c) carbon dioxide CO 2.
5. Specify which is heavier: a) 5 liters of sulfur dioxide (SO 2) or 5 liters of carbon dioxide (CO 2); b) 2 liters of carbon dioxide (CO 2) or 3 liters of carbon monoxide (CO).