Molar volume of gas. Molar volume of gas. The relationship between the volume of a substance and its quantity

One of the basic units in the International System of Units (SI) is the unit of quantity of a substance is the mole.

mole – this is such an amount of a substance that contains as many structural units of a given substance (molecules, atoms, ions, etc.) as there are carbon atoms in 0.012 kg (12 g) of a carbon isotope 12 WITH .

Given that the value of the absolute atomic mass for carbon is m(C) \u003d 1.99 10  26 kg, you can calculate the number of carbon atoms N A contained in 0.012 kg of carbon.

A mole of any substance contains the same number of particles of this substance (structural units). The number of structural units contained in a substance with an amount of one mole is 6.02 10 23 and called Avogadro's number (N A ).

For example, one mole of copper contains 6.02 10 23 copper atoms (Cu), and one mole of hydrogen (H 2) contains 6.02 10 23 hydrogen molecules.

molar mass(M) is the mass of a substance taken in an amount of 1 mol.

The molar mass is denoted by the letter M and has the unit [g/mol]. In physics, the dimension [kg/kmol] is used.

In the general case, the numerical value of the molar mass of a substance numerically coincides with the value of its relative molecular (relative atomic) mass.

For example, the relative molecular weight of water is:

Mr (H 2 O) \u003d 2Ar (H) + Ar (O) \u003d 2 ∙ 1 + 16 \u003d 18 a.m.u.

The molar mass of water has the same value, but is expressed in g/mol:

M (H 2 O) = 18 g/mol.

Thus, a mole of water containing 6.02 10 23 water molecules (respectively 2 6.02 10 23 hydrogen atoms and 6.02 10 23 oxygen atoms) has a mass of 18 grams. 1 mole of water contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.

1.3.4. The relationship between the mass of a substance and its quantity

Knowing the mass of a substance and its chemical formula, and hence the value of its molar mass, one can determine the amount of a substance and, conversely, knowing the amount of a substance, one can determine its mass. For such calculations, you should use the formulas:

where ν is the amount of substance, [mol]; m is the mass of the substance, [g] or [kg]; M is the molar mass of the substance, [g/mol] or [kg/kmol].

For example, to find the mass of sodium sulfate (Na 2 SO 4) in the amount of 5 mol, we find:

1) the value of the relative molecular weight of Na 2 SO 4, which is the sum of the rounded values of the relative atomic masses:

Mr (Na 2 SO 4) \u003d 2Ar (Na) + Ar (S) + 4Ar (O) \u003d 142,

2) the value of the molar mass of the substance numerically equal to it:

M (Na 2 SO 4) = 142 g/mol,

3) and, finally, a mass of 5 mol of sodium sulfate:

m = ν M = 5 mol 142 g/mol = 710 g

Answer: 710.

1.3.5. The relationship between the volume of a substance and its quantity

Under normal conditions (n.o.), i.e. at pressure R , equal to 101325 Pa (760 mm Hg), and temperature T, equal to 273.15 K (0 С), one mole of various gases and vapors occupies the same volume, equal to 22.4 l.

The volume occupied by 1 mole of gas or vapor at n.o. is called molar volumegas and has the dimension of a liter per mole.

V mol \u003d 22.4 l / mol.

Knowing the amount of gaseous substance (ν ) and molar volume value (V mol) you can calculate its volume (V) under normal conditions:

V = ν V mol,

where ν is the amount of substance [mol]; V is the volume of the gaseous substance [l]; V mol \u003d 22.4 l / mol.

Conversely, knowing the volume ( V) of a gaseous substance under normal conditions, you can calculate its amount (ν) :

Theoretical material, see the page "Molar volume of gas".

Basic formulas and concepts:

From Avogadro's law, for example, it follows that under the same conditions, 1 liter of hydrogen and 1 liter of oxygen contain the same number of molecules, although their sizes vary greatly.

The first corollary of Avogadro's law:

The volume that occupies 1 mole of any gas under normal conditions (n.s.) is 22.4 liters and is called molar volume of gas(Vm).

V m \u003d V / ν (m 3 / mol)

What is called normal conditions (n.o.):

normal temperature = 0°C or 273 K;
normal pressure = 1 atm or 760 mmHg or 101.3 kPa

From the first consequence of Avogadro's law it follows that, for example, 1 mole of hydrogen (2 g) and 1 mole of oxygen (32 g) occupy the same volume, equal to 22.4 liters at n.o.

Knowing V m, you can find the volume of any quantity (ν) and any mass (m) of gas:

V=V m ν V=V m (m/M)

Typical task 1: What is the volume at n.o.s. occupies 10 moles of gas?

V=V m ν=22.4 10=224 (l/mol)

Typical task 2: What is the volume at n.o.s. takes 16 g of oxygen?

V(O 2)=V m ·(m/M) M r (O 2)=32; M(O 2) \u003d 32 g / mol V (O 2) \u003d 22.4 (16/32) \u003d 11.2 l

The second corollary of Avogadro's law:

Knowing the density of the gas (ρ=m/V) at n.o., we can calculate the molar mass of this gas: M=22.4 ρ

The density (D) of one gas is otherwise called the ratio of the mass of a certain volume of the first gas to the mass of a similar volume of the second gas, taken under the same conditions.

Sample problem 3: Determine the relative density of carbon dioxide from hydrogen and air.

D hydrogen (CO 2) \u003d M r (CO 2) / M r (H 2) \u003d 44/2 \u003d 22 D air \u003d 44/29 \u003d 1.5

one volume of hydrogen and one volume of chlorine give two volumes of hydrogen chloride: H 2 + Cl 2 \u003d 2HCl
two volumes of hydrogen and one volume of oxygen give two volumes of water vapor: 2H 2 + O 2 \u003d 2H 2 O

Task 1 . How many moles and molecules are contained in 44 g of carbon dioxide.

Solution:

M(CO 2) \u003d 12 + 16 2 \u003d 44 g / mol ν \u003d m / M \u003d 44/44 \u003d 1 mol N (CO 2) \u003d ν NA \u003d 1 6.02 10 23 \u003d 6.02 10 23

Task 2 . Calculate the mass of one ozone molecule and an argon atom.

Solution:

M(O 3) \u003d 16 3 \u003d 48 g m (O 3) \u003d M (O 3) / NA \u003d 48 / (6.02 10 23) \u003d 7.97 10 -23 g M (Ar) \u003d 40 g m (Ar) \u003d M (Ar) / NA \u003d 40 / (6.02 10 23) \u003d 6.65 10 -23 g

Task 3 . What is the volume at n.o. occupies 2 moles of methane.

Solution:

ν \u003d V / 22.4 V (CH 4) \u003d ν 22.4 \u003d 2 22.4 \u003d 44.8 l

Task 4 . Determine the density and relative density of carbon monoxide (IV) for hydrogen, methane and air.

Solution:

M r (CO 2 )=12+16·2=44; M(CO 2)=44 g/mol M r (CH 4)=12+1 4=16; M(CH 4)=16 g/mol M r (H 2)=1 2=2; M(H 2)=2 g/mol M r (air)=29; M (air) \u003d 29 g / mol ρ \u003d m / V ρ (CO 2) \u003d 44 / 22.4 \u003d 1.96 g / mol D (CH 4) \u003d M (CO 2) / M (CH 4) = 44/16=2.75 D(H 2)=M(CO 2)/M(H 2)=44/2=22 D(air)=M(CO 2)/M(air)=44/24= 1.52

Task 5 . Determine the mass of the gas mixture, which includes 2.8 cubic meters of methane and 1.12 cubic meters of carbon monoxide.

Solution:

M r (CO 2 )=12+16·2=44; M(CO 2)=44 g/mol M r (CH 4)=12+1 4=16; M(CH 4) \u003d 16 g / mol 22.4 cubic meters CH 4 \u003d 16 kg 2.8 cubic meters CH 4 \u003d xm (CH 4) \u003d x \u003d 2.8 16 / 22.4 \u003d 2 kg 22.4 cubic meters CO 2 \u003d 28 kg 1.12 cubic meters CO 2 \u003d xm (CO 2) \u003d x \u003d 1.12 28 / 22.4 \u003d 1.4 kg m (CH 4) + m (CO 2) \u003d 2 + 1, 4=3.4 kg

Task 6 . Determine the volumes of oxygen and air required for the combustion of 112 cubic meters of divalent carbon monoxide with the content of non-combustible impurities in it in volume fractions of 0.50.

Solution:

determine the volume of pure CO in the mixture: V (CO) \u003d 112 0.5 \u003d 66 cubic meters
determine the volume of oxygen required to burn 66 cubic meters of CO: 2CO + O 2 \u003d 2CO 2 2mol + 1mol 66m 3 + X m 3 V (CO) \u003d 2 22.4 \u003d 44.8 m 3 V (O 2) \u003d 22 .4 m 3 66 / 44.8 \u003d X / 22.4 X \u003d 66 22.4 / 44.8 \u003d 33 m 3 or 2V (CO) / V (O 2) \u003d V 0 (CO) / V 0 (O 2) V - molar volumes V 0 - calculated volumes V 0 (O 2) \u003d V (O 2) (V 0 (CO) / 2V (CO))

Task 7 . How will the pressure change in a vessel filled with hydrogen and chlorine gases after they react? Likewise for hydrogen and oxygen?

Solution:

H 2 + Cl 2 \u003d 2HCl - as a result of the interaction of 1 mol of hydrogen and 1 mol of chlorine, 2 mol of hydrogen chloride is obtained: 1 (mol) + 1 (mol) \u003d 2 (mol), therefore, the pressure will not change, since the resulting volume of the gas mixture is the sum of the volumes of the components involved in the reaction.
2H 2 + O 2 \u003d 2H 2 O - 2 (mol) + 1 (mol) \u003d 2 (mol) - the pressure in the vessel will decrease by one and a half times, since 2 volumes of the gas mixture were obtained from 3 volumes of the components that entered into the reaction.

Task 8 . 12 liters of a gas mixture of ammonia and tetravalent carbon monoxide at n.o.s. have a mass of 18 g. How much is in the mixture of each of the gases?

Solution:

V(NH 3)=x l V(CO 2)=y l M(NH 3)=14+1 3=17 g/mol M(CO 2)=12+16 2=44 g/mol m( NH 3) \u003d x / (22.4 17) g m (CO 2) \u003d y / (22.4 44) g System of equations mixture volume: x + y \u003d 12 mixture mass: x / (22.4 ) 17)+y/(22.4 44)=18 After solving we get: x=4.62 l y=7.38 l

Task 9 . How much water will be obtained as a result of the reaction of 2 g of hydrogen and 24 g of oxygen.

Solution:

2H 2 + O 2 \u003d 2H 2 O

It can be seen from the reaction equation that the number of reactants does not correspond to the ratio of stoichiometric coefficients in the equation. In such cases, calculations are carried out on the substance, which is less, i.e., this substance will finish first in the course of the reaction. To determine which of the components is in short supply, you need to pay attention to the coefficient in the reaction equation.

Amounts of starting components ν(H 2)=4/2=2 (mol) ν(O 2)=48/32=1.5 (mol)

However, there is no need to rush. In our case, for the reaction with 1.5 moles of oxygen, 3 moles of hydrogen (1.5 2) are needed, and we have only 2 moles of it, i.e., 1 mole of hydrogen is not enough for all one and a half moles of oxygen to react. Therefore, we will calculate the amount of water by hydrogen:

ν (H 2 O) \u003d ν (H 2) \u003d 2 mol m (H 2 O) \u003d 2 18 \u003d 36 g

Task 10 . At a temperature of 400 K and a pressure of 3 atmospheres, the gas occupies a volume of 1 liter. What volume will this gas occupy at n.o.s.?

Solution:

From the Clapeyron equation:

P V/T = P n V n /T n V n = (PVT n)/(P n T) V n = (3 1 273)/(1 400) = 2.05 l

In the study of chemicals, important concepts are such quantities as molar mass, substance density, molar volume. So, what is the molar volume, and how is it different for substances in different states of aggregation?

Molar volume: general information

To calculate the molar volume of a chemical, the molar mass of that substance is divided by its density. Thus, the molar volume is calculated by the formula:

where Vm is the molar volume of the substance, M is the molar mass, p is the density. In the International SI system, this value is measured in cubic meters per mol (m 3 / mol).

Rice. 1. Molar volume formula.

The molar volume of gaseous substances differs from substances in liquid and solid states in that a gaseous element with an amount of 1 mol always occupies the same volume (if the same parameters are observed).

The volume of gas depends on temperature and pressure, so the calculation should take the volume of gas under normal conditions. Normal conditions are considered to be a temperature of 0 degrees and a pressure of 101.325 kPa.

The molar volume of 1 mol of gas under normal conditions is always the same and equal to 22.41 dm 3 /mol. This volume is called the molar volume of an ideal gas. That is, in 1 mole of any gas (oxygen, hydrogen, air), the volume is 22.41 dm 3 / m.

The molar volume under normal conditions can be derived using the equation of state for an ideal gas, which is called the Claiperon-Mendeleev equation:

where R is the universal gas constant, R=8.314 J/mol*K=0.0821 l*atm/mol K

Volume of one mole of gas V=RT/P=8.314*273.15/101.325=22.413 l/mol, where T and P are temperature (K) and pressure values under normal conditions.

Rice. 2. Table of molar volumes.

Avogadro's law

In 1811, A. Avogadro put forward the hypothesis that equal volumes of different gases under the same conditions (temperature and pressure) contain the same number of molecules. Later, the hypothesis was confirmed and became a law bearing the name of the great Italian scientist.

Rice. 3. Amedeo Avogadro.

The law becomes clear if we remember that in a gaseous form the distance between the particles is incomparably greater than the size of the particles themselves.

Thus, the following conclusions can be drawn from Avogadro's law:

Equal volumes of any gases taken at the same temperature and at the same pressure contain the same number of molecules.
1 mole of completely different gases under the same conditions occupies the same volume.
One mole of any gas under normal conditions occupies a volume of 22.41 liters.

The consequence of Avogadro's law and the concept of molar volume are based on the fact that a mole of any substance contains the same number of particles (for gases - molecules), equal to the Avogadro constant.

To find out the number of moles of a solute contained in one liter of a solution, it is necessary to determine the molar concentration of the substance by the formula c \u003d n / V, where n is the amount of the solute, expressed in moles, V is the volume of the solution, expressed in liters C - molarity.

What have we learned?

In the school curriculum in chemistry of the 8th grade, the topic "Molar volume" is studied. One mole of gas always contains the same volume, equal to 22.41 cubic meters / mol. This volume is called the molar volume of the gas.

Topic quiz

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Target:
To acquaint students with the concepts of "amount of substance", "molar mass" to give an idea of the Avogadro constant. Show the relationship between the amount of a substance, the number of particles and the Avogadro constant, as well as the relationship between the molar mass, mass and amount of a substance. Learn to do calculations.

1) What is the amount of substance?
2) What is a mole?
3) How many structural units are contained in 1 mole?
4) Through what quantities can the amount of a substance be determined?
5) What is the molar mass, what does it numerically coincide with?
6) What is molar volume?

The amount of a substance is a physical quantity that means a certain number of structural elements (molecules, atoms, ions) Denoted n (en) measured in the international system of units (Ci) mol
Avogadro's number - shows the number of particles in 1 mol of a substance Denoted by NA measured in mol-1 has a numerical value of 6.02*10^23
The molar mass of a substance is numerically equal to its relative molecular mass. Molar mass - a physical quantity that shows the mass in 1 mole of a substance. It is denoted by M measured in g / mol M \u003d m / n
Molar volume - a physical quantity that shows the volume that any gas occupies with the amount of substance 1 mol. It is denoted by Vm measured in l / mol Vm \u003d V / n Vm=22.4l/mol
A MOLE is a QUANTITY of SUBSTANCE equal to 6.02. 10 23 structural units of a given substance - molecules (if the substance consists of molecules), atoms (if it is an atomic substance), ions (if the substance is an ionic compound).
1 mole (1 M) water = 6 . 10 23 H 2 O molecules,

1 mole (1 M) iron = 6 . 10 23 Fe atoms,

1 mole (1 M) chlorine = 6 . 10 23 Cl 2 molecules,

1 mol (1 M) chloride ion Cl - = 6 . 10 23 ions Cl - .

1 mol (1 M) electrons e - = 6 . 10 23 electrons e - .

Tasks:
1) How many moles of oxygen are contained in 128 g of oxygen?

2) During lightning discharges in the atmosphere, the following reaction occurs: N 2 + O 2 ® NO 2. Equalize the response. How many moles of oxygen will be required to completely convert 1 mole of nitrogen into NO 2? How many grams of oxygen will that be? How many grams of NO 2 is formed?

3) 180 g of water is poured into a glass. How many water molecules are in a glass? How many moles of H 2 O is this?

4) Mixed 4 g of hydrogen and 64 g of oxygen. The mixture was blown up. How many grams of water did you get? How many grams of oxygen are left unused?

Homework: paragraph 15, ex. 1-3.5

Molar volume of gaseous substances.
Target: educational - to systematize students' knowledge about the concepts of the amount of a substance, Avogadro's number, molar mass, on their basis to form an idea of the molar volume of gaseous substances; reveal the essence of Avogadro's law and its practical application;

developing - to form the ability for adequate self-control and self-esteem; develop the ability to think logically, put forward hypotheses, draw reasoned conclusions.

During the classes:
1. Organizational moment.
2. Announcement of the topic and objectives of the lesson.

3.Updating basic knowledge
4. Problem solving

Avogadro's Law- this is one of the most important laws of chemistry (formulated by Amadeo Avogadro in 1811), stating that "in equal volumes of different gases, which are taken at the same pressure and temperature, the same number of molecules is contained."

Molar volume of gases is the volume of gas containing 1 mol of particles of this gas.

Normal conditions– temperature 0 С (273 K) and pressure 1 atm (760 mm Hg or 101 325 Pa).

Answer the questions:

1. What is called an atom? (Atom is the smallest chemically indivisible part of a chemical element, which is the carrier of its properties).

2. What is a mole? (A mole is the amount of a substance, which is equal to 6.02.10 ^ 23 structural units of this substance - molecules, atoms, ions. This is the amount of a substance containing as many particles as there are atoms in 12 g of carbon).

3. How is the amount of a substance measured? (In moles).

4. How is the mass of a substance measured? (The mass of a substance is measured in grams).

5. What is molar mass and how is it measured? (Molar mass is the mass of 1 mol of a substance. It is measured in g/mol).

Consequences of Avogadro's law.

Two consequences follow from Avogadro's law:

1. One mole of any gas occupies the same volume under the same conditions. In particular, under normal conditions, i.e. at 0 ° C (273 K) and 101.3 kPa, the volume of 1 mole of gas is 22.4 liters. This volume is called the molar volume of the gas Vm. This value can be recalculated to other temperatures and pressures using the Mendeleev-Clapeyron equation (Figure 3).

The molar volume of a gas under normal conditions is a fundamental physical constant widely used in chemical calculations. It allows you to use the volume of gas instead of its mass. The value of the molar volume of gas at n.o. is the coefficient of proportionality between the Avogadro and Loschmidt constants

2. The molar mass of the first gas is equal to the product of the molar mass of the second gas and the relative density of the second of the first gas. This position was of great importance for the development of chemistry, because. it made it possible to determine the partial weight of bodies that are capable of passing into a vapor or gaseous state. Therefore, the ratio of the mass of a certain volume of one gas to the mass of the same volume of another gas, taken under the same conditions, is called the density of the first gas according to the second

1. Fill in the blanks:

Molar volume is a physical quantity that shows ..............., denoted by .............. .., measured in ............... .

2. Write down the formula by the rule.

The volume of a gaseous substance (V) is equal to the product of the molar volume

(Vm) by the amount of substance (n) ............................. .

3. Using the material of task 3, derive formulas for calculation:

a) the volume of a gaseous substance.

b) molar volume.

Homework: paragraph 16, ex. 1-5

Solving problems for calculating the amount of matter, mass and volume.

Generalization and systematization of knowledge on the topic "Simple substances"
Target: generalize and systematize students' knowledge about the main classes of compounds
Progress:

1) Organizational moment

2) Generalization of the studied material:

a) Oral survey on the topic of the lesson

b) Completion of task 1 (finding oxides, bases, acids, salts among the given substances)

c) Completion of task 2 (compilation of formulas for oxides, bases, acids, salts)

3. Consolidation (independent work)

5. Homework

2)
a)
What two groups can substances be divided into?

What substances are called simple?

What two groups are simple substances divided into?

What substances are called complex?

What complex substances are known?

What substances are called oxides?

What substances are called bases?

What substances are called acids?

What substances are called salts?

b)
Write out oxides, bases, acids, salts separately:

KOH, SO 2, HCI, BaCI 2, P 2 O 5,

NaOH, CaCO 3 , H 2 SO 4 , HNO 3 ,

MgO, Ca (OH) 2, Li 3 PO 4

Name them.

v)
Write formulas for oxides corresponding to bases and acids:

Potassium Hydroxide-Potassium Oxide

Iron(III) hydroxide-iron(III) oxide

Phosphoric acid-phosphorus(V) oxide

Sulfuric acid-sulfur(VI) oxide

Write the formula for barium nitrate salt; by ion charges, oxidation states of elements write down

formulas of the corresponding hydroxides, oxides, simple substances.

1. The oxidation state of sulfur is +4 in the compound:

2. Oxides include a substance:

3. Sulfurous acid formula:

4. The basis is the substance:

5. Salt K 2 CO 3 is called:

1- potassium silicate

2-potassium carbonate

3-potassium carbide

4- calcium carbonate

6. In a solution of what substance will litmus change color to red:

2- in alkali

3- in acid

Homework: repeat paragraphs 13-16

Examination No. 2
"Simple Substances"

Oxidation state: binary compounds

Purpose: to teach how to make molecular formulas of substances consisting of two elements according to the degree of oxidation. continue to consolidate the skill of determining the degree of oxidation of an element by the formula.
1. The oxidation state (s. o.) is conditional charge of the atoms of a chemical element in a complex substance, calculated on the basis of the assumption that it consists of simple ions.

Should know!

1) In connections with. O. hydrogen = +1, except for hydrides.
2) In compounds with. O. oxygen = -2, except for peroxides and fluorides
3) The oxidation state of metals is always positive.

For metals of the main subgroups of the first three groups With. O. constant:
Group IA metals - p. O. = +1,
Group IIA metals - p. O. = +2,
Group IIIA metals - p. O. = +3.
4) For free atoms and simple substances p. O. = 0.
5) Total s. O. all elements in the compound = 0.

2. Method of formation of names two-element (binary) compounds.

3.

Tasks:
Make formulas of substances by name.

How many molecules are contained in 48 g of sulfur oxide (IV)?

The oxidation state of manganese in the K2MnO4 compound is:

Chlorine exhibits the maximum oxidation state in a compound whose formula is:

Homework: paragraph 17, ex. 2,5,6

Oxides. Volatile hydrogen compounds.
Target: the formation of students' knowledge about the most important classes of binary compounds - oxides and volatile hydrogen compounds.

Questions:
What substances are called binary?
What is the degree of oxidation?
What oxidation state will the elements have if they donate electrons?
What oxidation state will the elements have if they accept electrons?
– How to determine how many electrons will give or receive elements?
– What oxidation state will single atoms or molecules have?
- What will the compounds be called if sulfur is in second place in the formula?
- What will the compounds be called if chlorine is in second place in the formula?
- What will the compounds be called if hydrogen is in second place in the formula?
- What will the compounds be called if nitrogen is in second place in the formula?
- What will the compounds be called if oxygen is in second place in the formula?
Exploring a new topic:
What do these formulas have in common?
– What will be the name of such substances?

SiO 2, H 2 O, CO 2, AI 2 O 3, Fe 2 O 3, Fe 3 O 4, CO.
oxides- a class of substances of inorganic compounds widespread in nature. Oxides include such well-known compounds as:

Sand (silicon dioxide SiO2 with a small amount of impurities);

Water (hydrogen oxide H2O);

Carbon dioxide (carbon dioxide CO2 IV);

Carbon monoxide (CO II carbon monoxide);

Clay (aluminum oxide AI2O3 with a small amount of other compounds);

Most ferrous ores contain oxides, such as red iron ore - Fe2O3 and magnetic iron ore - Fe3O4.

Volatile hydrogen compounds- the most practically important group of compounds with hydrogen. These include substances commonly found in nature or used in industry, such as water, methane and other hydrocarbons, ammonia, hydrogen sulfide, hydrogen halides. Many of the volatile hydrogen compounds are in the form of solutions in soil waters, in the composition of living organisms, as well as in gases formed during biochemical and geochemical processes, therefore their biochemical and geochemical role is very large.
Depending on the chemical properties, there are:

Salt-forming oxides:

o basic oxides (for example, sodium oxide Na2O, copper (II) oxide CuO): metal oxides, the oxidation state of which is I-II;

o acidic oxides (for example, sulfur oxide (VI) SO3, nitric oxide (IV) NO2): metal oxides with an oxidation state of V-VII and oxides of non-metals;

o amphoteric oxides (for example, zinc oxide ZnO, aluminum oxide Al2O3): metal oxides with oxidation states III-IV and exceptions (ZnO, BeO, SnO, PbO);

Non-salt-forming oxides: carbon monoxide (II) CO, nitric oxide (I) N2O, nitric oxide (II) NO, silicon oxide (II) SiO.

Homework: paragraph 18, exercise 1,4,5

Foundations.
Target:
to introduce students to the composition, classification and representatives of the base class

continue the formation of knowledge about ions on the example of complex hydroxide ions

continue the formation of knowledge about the degree of oxidation of elements, chemical bonds in substances;

give the concept of qualitative reactions and indicators;

to form skills in handling chemical glassware and reagents;

develop a caring attitude towards your health.

In addition to binary compounds, there are complex substances, such as bases, which consist of three elements: metal, oxygen, and hydrogen.
Hydrogen and oxygen are included in them in the form of a hydroxo group OH -. Therefore, the hydroxo group OH- is an ion, but not simple, like Na + or Cl-, but complex - OH- - hydroxide ion.

Foundations - These are complex substances consisting of metal ions and one or more hydroxide ions associated with them.
If the charge of the metal ion is 1+, then, of course, one hydroxo group OH- is associated with the metal ion, if 2+, then two, etc. Therefore, the composition of the base can be written by the general formula: M (OH) n, where M is the metal , m - the number of OH groups and at the same time the charge of the ion (oxidation state) of the metal.

The names of the bases consist of the word hydroxide and the name of the metal. For example, Na0H is sodium hydroxide. Ca(OH)2 - calcium hydroxide.
If the metal exhibits a variable degree of oxidation, then its value, just as for binary compounds, is indicated by a Roman numeral in brackets and pronounced at the end of the name of the base, for example: CuOH - copper (I) hydroxide, read "copper hydroxide one"; Cr (OH), - copper (II) hydroxide, reads "copper hydroxide two."

In relation to water, the bases are divided into two groups: soluble NaOH, Ca (OH) 2, K0H, Ba (OH)? and insoluble Cr(OH)7, Re(OH)2. Soluble bases are also called alkalis. You can find out whether a base is soluble or insoluble in water using the table "Solubility of bases, acids and salts in water."

Sodium hydroxide NaOH- solid white substance, hygroscopic and therefore deliquescent in air; dissolves well in water, and heat is released. A solution of sodium hydroxide in water is soapy to the touch and very caustic. It corrodes leather, textiles, paper and other materials. For this property, sodium hydroxide is called caustic soda. Sodium hydroxide and its solutions must be handled with care, being careful not to get them on clothes, shoes, and even more so on hands and face. On the skin from this substance, wounds that do not heal for a long time are formed. NaOH is used in soap making, leather and pharmaceutical industries.

Potassium hydroxide KOH- also a solid white substance, highly soluble in water, with the release of a large amount of heat. A solution of potassium hydroxide, like a solution of caustic soda, is soapy to the touch and very caustic. Therefore, potassium hydroxide is otherwise called caustic potash. It is used as an additive in the production of soap, refractory glass.

Calcium hydroxide Ca (OH) 2 or slaked lime is a loose white powder, slightly soluble in water (in the solubility table against the formula Ca (OH) a there is the letter M, which means a poorly soluble substance). It is obtained by the interaction of quicklime CaO with water. This process is called quenching. Calcium hydroxide is used in construction during masonry and plastering of walls, for whitewashing trees, to obtain bleach, which is a disinfectant.

A clear solution of calcium hydroxide is called lime water. When CO2 is passed through lime water, it becomes cloudy. This experience serves to recognize carbon dioxide.

Reactions by which certain chemicals are recognized are called qualitative reactions.

For alkalis, there are also qualitative reactions, with the help of which solutions of alkalis can be recognized among solutions of other substances. These are reactions of alkalis with special substances - indicators (lat. "pointers"). If a few drops of an indicator solution are added to an alkali solution, it will change its color.

Homework: paragraph 19, exercises 2-6, table 4

Before solving problems, you should learn the formulas and rules for how to find the volume of gas. Remember Avogadro's law. And the volume of gas itself can be calculated using several formulas, choosing the appropriate one from them. When selecting the necessary formula, environmental conditions, in particular temperature and pressure, are of great importance.

Avogadro's law

It says that at the same pressure and the same temperature, in the same volumes of different gases, the same number of molecules will be contained. The number of gas molecules contained in one mole is Avogadro's number. It follows from this law that: 1 Kmol (kilomole) of an ideal gas, and any one, at the same pressure and temperature (760 mm Hg and t \u003d 0 * C) always occupies one volume = 22.4136 m3.

How to determine the volume of gas

The formula V=n*Vm is most often found in problems. Here, the volume of gas in liters is V, Vm is the molar volume of gas (l / mol), which under normal conditions = 22.4 l / mol, and n is the amount of substance in moles. When there is no amount of matter in the conditions, but at the same time there is a mass of matter, then we proceed as follows: n=m/M. Here M is g / mol (molar mass of the substance), and the mass of the substance in grams is m. In the periodic table, it is written under each element, like its atomic mass. Add up all the masses and get the desired.
So, how to calculate the volume of gas. Here is the task: dissolve 10 g of aluminum in hydrochloric acid. Question: how much hydrogen can be released at n. at.? The reaction equation looks like this: 2Al + 6HCl (ex.) = 2AlCl3 + 3H2. At the very beginning, we find aluminum (amount) that reacted according to the formula: n(Al)=m(Al)/M(Al). We take the mass of aluminum (molar) from the periodic table M (Al) \u003d 27 g / mol. Substitute: n(Al)=10/27=0.37mol. It can be seen from the chemical equation that 3 moles of hydrogen were formed by dissolving 2 moles of aluminum. It should be calculated how much hydrogen will be released from 0.4 moles of aluminum: n(H2)=3*0.37/2=0.56mol. Substitute the data in the formula and find the volume of this gas. V=n*Vm=0.56*22.4=12.54l.