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 the amount of a substance that contains as many structural units of a given substance (molecules, atoms, ions, etc.) as there are carbon atoms contained in 0.012 kg (12 g) of a carbon isotope 12 WITH .
Considering that the value of the absolute atomic mass for carbon is equal to m(C) = 1.99 10 26 kg, the number of carbon atoms can be calculated 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 is 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 mole.
Molar mass is designated by the letter M and has the dimension [g/mol]. In physics they use the unit [kg/kmol].
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:
Мr(Н 2 О) = 2Аr (Н) + Аr (O) = 2∙1 + 16 = 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. Water, with an amount of substance of 1 mole, contains 2 moles of hydrogen atoms and one 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 therefore the value of its molar mass, you can determine the amount of the substance and, conversely, knowing the amount of the substance, you can determine its mass. For such calculations you should use the formulas:
where ν is the amount of substance, [mol]; m– mass of the substance, [g] or [kg]; M – molar mass of the substance, [g/mol] or [kg/kmol].
For example, to find the mass of sodium sulfate (Na 2 SO 4) in an amount of 5 moles, we find:
1) the value of the relative molecular mass of Na 2 SO 4, which is the sum of the rounded values of the relative atomic masses:
Мr(Na 2 SO 4) = 2Аr(Na) + Аr(S) + 4Аr(O) = 142,
2) a numerically equal value of the molar mass of the substance:
M(Na 2 SO 4) = 142 g/mol,
3) and, finally, the 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
At normal conditions(n.s.), 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 different gases and vapors occupies the same volume equal to 22.4 l.
The volume occupied by 1 mole of gas or vapor at ground level is called molar volumegas and has the dimension liter per mole.
V mol = 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 – volume of gaseous substance [l]; V mol = 22.4 l/mol.
And, conversely, knowing the volume ( V) of a gaseous substance under normal conditions, its quantity (ν) can be calculated :
For 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.
First corollary of Avogadro's law:
The volume occupied by 1 mole of any gas under normal conditions (n.s.) is 22.4 liters and is called molar volume gas(Vm).
V m =V/ν (m 3 /mol)
What are called normal conditions (n.s.):
- normal temperature = 0°C or 273 K;
- normal pressure = 1 atm or 760 mm Hg. or 101.3 kPa
From the first corollary 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 ground level.
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 problem 1: What is the volume at no. occupies 10 moles of gas?
V=V m ·ν=22.4·10=224 (l/mol)
Typical problem 2: What is the volume at no. takes up 16 g of oxygen?
V(O 2)=V m ·(m/M) M r (O 2)=32; M(O 2)=32 g/mol V(O 2)=22.4·(16/32)=11.2 l
Second corollary of Avogadro's law:
Knowing the gas density (ρ=m/V) at normal conditions, we can calculate the molar mass of this gas: M=22.4·ρ
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.
Typical task 3: Determine the relative density of carbon dioxide compared to hydrogen and air.
D hydrogen (CO 2) = M r (CO 2)/M r (H 2) = 44/2 = 22 D air = 44/29 = 1.5
- one volume of hydrogen and one volume of chlorine give two volumes of hydrogen chloride: H 2 +Cl 2 =2HCl
- two volumes of hydrogen and one volume of oxygen give two volumes of water vapor: 2H 2 + O 2 = 2H 2 O
Task 1. How many moles and molecules are contained in 44 g of carbon dioxide?
Solution:
M(CO 2) = 12+16 2 = 44 g/mol ν = m/M = 44/44 = 1 mol N(CO 2) = ν N A = 1 6.02 10 23 = 6.02 ·10 23
Task 2. Calculate the mass of one molecule of ozone and an argon atom.
Solution:
M(O 3) = 16 3 = 48 g m(O 3) = M(O 3)/N A = 48/(6.02 10 23) = 7.97 10 -23 g M(Ar) = 40 g m(Ar) = M(Ar)/N A = 40/(6.02 10 23) = 6.65 10 -23 g
Task 3. What is the volume at standard conditions? occupies 2 moles of methane.
Solution:
ν = V/22.4 V(CH 4) = ν 22.4 = 2 22.4 = 44.8 l
Task 4. Determine the density and relative density of carbon monoxide (IV) from 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)=29 g/mol ρ=m/V ρ(CO 2)=44/22.4=1.96 g/mol D(CH 4)=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) = 16 g/mol 22.4 cubic meters CH 4 = 16 kg 2.8 cubic meters CH 4 = x m(CH 4) = x = 2.8 16/22.4 = 2 kg 22.4 cubic meters CO 2 = 28 kg 1.12 cubic meters CO 2 = x m(CO 2)=x=1.12·28/22.4=1.4 kg m(CH 4)+m(CO 2)=2+1, 4=3.4 kg
Task 6. Determine the volumes of oxygen and air required to burn 112 cubic meters of divalent carbon monoxide when it contains non-combustible impurities in a volume fraction of 0.50.
Solution:
- determine the volume of pure CO in the mixture: V(CO)=112·0.5=66 cubic meters
- determine the volume of oxygen required to burn 66 cubic meters of CO: 2CO+O 2 =2CO 2 2mol+1mol 66m 3 +X m 3 V(CO)=2·22.4 = 44.8 m 3 V(O 2)=22 .4 m 3 66/44.8 = X/22.4 X = 66 22.4/44.8 = 33 m 3 or 2V(CO)/V(O 2) = V 0 (CO)/V 0 (O 2) V - molar volumes V 0 - calculated volumes V 0 (O 2) = 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? Is it the same for hydrogen and oxygen?
Solution:
- H 2 +Cl 2 =2HCl - as a result of the interaction of 1 mole of hydrogen and 1 mole of chlorine, 2 moles of hydrogen chloride are obtained: 1 (mol) + 1 (mol) = 2 (mol), therefore, the pressure will not change, since the resulting volume of the gas mixture is equal to the sum of the volumes of the components that reacted.
- 2H 2 + O 2 = 2H 2 O - 2 (mol) + 1 (mol) = 2 (mol) - the pressure in the vessel will decrease by one and a half times, since from 3 volumes of the components that reacted, 2 volumes of the gas mixture are obtained.
Task 8. 12 liters of a gas mixture of ammonia and tetravalent carbon monoxide at no. have a mass of 18 g. How much of each gas is in the mixture?
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)=x/(22.4·17) g m(CO 2)=y/(22.4·44) g System of equations volume of mixture: x+y=12 mass of mixture: x/(22.4· 17)+y/(22.4·44)=18 After solving we get: x=4.62 l y=7.38 l
Task 9. What amount of 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 =2H 2 O
From the reaction equation it is clear that the number of reactants does not correspond to the ratio of the stoichiometric coefficients in the equation. In such cases, calculations are carried out using a substance that is less abundant, i.e., this substance will end up first during the reaction. To determine which of the components is deficient, 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, to react with 1.5 moles of oxygen, 3 moles of hydrogen (1.5 2) are needed, but we only have 2 moles, i.e., 1 mole of hydrogen is missing for all one and a half moles of oxygen to react. Therefore, we will calculate the amount of water using hydrogen:
ν(H 2 O)=ν(H 2)=2 mol m(H 2 O) = 2 18=36 g
Problem 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 zero level?
Solution:
From the Clapeyron equation:
P·V/T = Pn ·Vn/Tn Vn = (PVT n)/(Pn T) Vn = (3·1·273)/(1·400) = 2.05 l
When studying chemical substances, important concepts are such quantities as molar mass, density of a substance, and molar volume. So, what is molar volume, and how does it differ for substances in different states of aggregation?
Molar volume: general information
To calculate the molar volume of a chemical substance, it is necessary to divide the molar mass of this substance 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 meter per mole (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 mole always occupies the same volume (if the same parameters are met).
The volume of gas depends on temperature and pressure, so when calculating, you 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 mole 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 at normal conditions can be derived using the equation of state for an ideal gas, called the Clayperon-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 the value of temperature (K) and pressure 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 gaseous form the distance between 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 corollary to Avogadro's law and the concept of molar volume are based on the fact that a mole of any substance contains same number particles (for gases - molecules), equal to Avogadro's constant.
To find out the number of moles of solute contained in one liter of solution, it is necessary to determine the molar concentration of the substance using the formula c = n/V, where n is the amount of solute, expressed in moles, V is the volume of the solution, expressed in liters C is molarity.
What have we learned?
IN school curriculum in 8th grade chemistry 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.
Test on the topic
Evaluation of the report
Average rating: 4.2. Total ratings received: 64.
Target:
Introduce students to the concepts of “amount of substance”, “molar mass” and give an idea of Avogadro’s constant. Show the relationship between the amount of substance, the number of particles and Avogadro's constant, as well as the relationship between molar mass, mass and amount of substance. Learn to make 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 molar mass, and 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 structural elements(molecules, atoms, ions) Denoted by n (en) measured in the international system of units (Si) mole
Avogadro's number - shows the number of particles in 1 mole 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 is a physical quantity that shows the mass of 1 mole of a substance. Denoted by M, measured in g/mol M = m/n
Molar volume is a physical quantity that shows the volume occupied by any gas with an amount of substance of 1 mol. Designated by Vm, measured in l/mol Vm = V/n At zero. Vm=22.4l/mol
A MOL is an AMOUNT 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 mol (1 M) water = 6 .
10 23 molecules H 2 O,
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) chlorine ions Cl - = 6 . 10 23 Cl - ions.
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 reaction. How many moles of oxygen are required to completely convert 1 mole of nitrogen into NO 2? How many grams of oxygen will this be? How many grams of NO 2 are produced?
3) 180 g of water was poured into a glass. How many water molecules are in a glass? How many moles of H2O 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 remain unused?
Homework: paragraph 15, ex. 1-3.5
Molar volume of gaseous substances.
Target: educational – to systematize students’ knowledge about the concepts of quantity 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;
developmental – to form the ability for adequate self-control and self-esteem; develop the ability to think logically, put forward hypotheses, and draw reasoned conclusions.
During the classes:
1. Organizational moment.
2. Announcement of the topic and goals of the lesson.
3.Updating basic knowledge
4.Problem solving
Avogadro's law is one of the most important laws of chemistry (formulated by Amadeo Avogadro in 1811), which states that “equal volumes of different gases, taken at the same pressure and temperature, contain the same number of molecules.”
Molar volume of gases– volume of gas containing 1 mole of particles of this gas.
Normal conditions– temperature 0 C (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 chemical element, which is the bearer of its properties).
2. What is a mole? (A mole is an amount of a substance that is equal to 6.02.10^23 structural units of this substance - molecules, atoms, ions. This is an amount of a substance containing the same number of 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 mole 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 at normal conditions is a fundamental physical constant widely used in chemical calculations. It allows you to use the volume of a gas instead of its mass. The value of the molar volume of gas at no. is the proportionality coefficient 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 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. Consequently, 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.................. .., measured in...................... .
2. Write down the formula according to the rule.
The volume of a gaseous substance (V) is equal to the product of the molar volume
(Vm) per amount of substance (n) ..................................
3. Using the material from task 3, derive formulas for calculation:
a) volume of a gaseous substance.
b) molar volume.
Homework: paragraph 16, ex. 1-5
Solving problems on 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) Completing task 1 (finding oxides, bases, acids, salts among given substances)
c) Completing task 2 (drawing up formulas of oxides, bases, acids, salts)
3. Fastening ( 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 down 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)
Draw up formulas of 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
Create a formula for barium nitrate salt; write down the ion charges and oxidation states of elements
formulas of the corresponding hydroxides, oxides, simple substances.
1. The oxidation state of sulfur is +4 in the compound:
2. The following substances belong to oxides:
3. Formula of sulfurous acid:
4. The base 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 which substance will litmus change color to red:
2- in alkali
3- in acid
Homework: repeat paragraphs 13-16
Test No. 2
"Simple substances"
Oxidation state: binary compounds
Goal: to teach how to compose molecular formulas of substances consisting of two elements according to their oxidation state. continue to consolidate the skill of determining the oxidation state of an element using the formula.
1. Oxidation state (s.o.) is the conventional 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.
You should know!
1) In connections with. O. hydrogen = +1, except hydrides.
2) In connections with. O. oxygen = -2, except 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) In free atoms and simple substances p. O. = 0.
5) Total s. O. all elements in the connection = 0.
2. Method of formation of names two-element (binary) compounds.
3. 
Tasks:
Make up formulas for substances by name.
How many molecules are there in 48 g of sulfur(IV) oxide?
The oxidation state of manganese in the K2MnO4 compound is equal to:
Chlorine exhibits its maximum oxidation state in a compound whose formula is:
Homework: paragraph 17, ex. 2,5,6
Oxides. Volatile hydrogen compounds.
Target: developing students' knowledge about the most important classes of binary compounds - oxides and volatile hydrogen compounds.
Questions:
– What substances are called binary?
– What is the oxidation state called?
– 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 elements will give or accept?
– 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?
Learning a new topic:
– What do these formulas have in common?
– What will such substances be called?
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 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 metal 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, and hydrogen halides. Many of the volatile hydrogen compounds are found in the form of solutions in soil waters, in living organisms, as well as in gases formed during biochemical and geochemical processes, so their biochemical and geochemical role is very large.
Depending on the chemical properties distinguish:
Salt-forming oxides:
o basic oxides (for example, sodium oxide Na2O, copper(II) oxide CuO): metal oxides whose oxidation state is I-II;
o acidic oxides (for example, sulfur oxide(VI) SO3, nitrogen oxide(IV) NO2): metal oxides with oxidation state V-VII and non-metal oxides;
o amphoteric oxides (for example, zinc oxide ZnO, aluminum oxide Al2O3): metal oxides with oxidation state III-IV and exclusion (ZnO, BeO, SnO, PbO);
Non-salt-forming oxides: carbon oxide (II) CO, nitrogen oxide (I) N2O, nitrogen oxide (II) NO, silicon oxide (II) SiO.
Homework: paragraph 18, exercises 1,4,5
Grounds.
Target:
introduce students to the composition, classification and representatives of the class of bases
continue to develop knowledge about ions using the example of complex hydroxide ions
continue to develop knowledge about the degree of oxidation of elements, chemical bonds in substances;
give an idea of qualitative reactions and indicators;
develop skills in handling chemical utensils and reagents;
develop a careful attitude towards your health.
In addition to binary compounds, there are complex substances, for example bases, which consist of three elements: metal, oxygen and hydrogen.
Hydrogen and oxygen are included in them in the form of the hydroxo group OH -. Consequently, the hydroxo group OH- is an ion, not a simple one like Na+ or Cl-, but a complex one - OH- - hydroxide ion.
Grounds
- 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. Consequently, the composition of the base can be written by the general formula: M(OH)n, where M is metal , m is the number of OH groups and at the same time the charge of the metal ion (oxidation state).
The names of the bases consist of the word hydroxide and the name of the metal. For example, Na0H is sodium hydroxide. Ca(0H)2 - calcium hydroxide.
If the metal exhibits a variable oxidation state, then its value, as for binary compounds, is indicated with 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, read “copper hydroxide two”.
In relation to water, bases are divided into two groups: soluble NaOH, Ca(OH)2, K0H, Ba(OH)? and insoluble Cr(OH)7, Ke(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- a solid white substance, hygroscopic and therefore deliquescent in air; It dissolves well in water and releases heat. A solution of sodium hydroxide in water is soapy to the touch and very caustic. It corrodes leather, fabrics, paper and other materials. For this property, sodium hydroxide is called caustic soda. Sodium hydroxide and its solutions must be handled carefully, being careful not to get them on clothes, shoes, and even more so on your hands and face. This substance causes wounds on the skin that take a long time to heal. NaOH is used in soap making, leather and pharmaceutical industries.
Potassium hydroxide KOH- also a solid white substance, highly soluble in water, releasing a large amount of heat. A solution of potassium hydroxide, like a solution of sodium hydroxide, is soapy to the touch and very caustic. Therefore, potassium hydroxide is also called potassium hydroxide. It is used as an additive in the production of soap and refractory glass.
Calcium hydroxide Ca(OH)2 or slaked lime is a loose white powder, slightly soluble in water (in the solubility table, the formula Ca(OH)a has the letter M, which means a slightly soluble substance). It is obtained by reacting quicklime CaO with water. This process is called quenching. Calcium hydroxide is used in construction for masonry and plastering of walls, for whitewashing trees, and for producing 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 chemical substances are recognized are called qualitative reactions.
There are also ones for alkalis 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 (Latin for “pointers”). If you add a few drops of an indicator solution to an alkali solution, it will change its color
Homework: paragraph 19, exercises 2-6, table 4
Before solving problems, you should know the formulas and rules of how to find the volume of gas. We should 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 required formula, great importance have environmental conditions, in particular temperature and pressure.
Avogadro's law
It says that at the same pressure and the same temperature, the same volumes of different gases will contain the same number of molecules. The number of gas molecules contained in one mole is Avogadro's number. From this law it follows that: 1 Kmol (kilomol) of an ideal gas, any gas, at the same pressure and temperature (760 mm Hg and t = 0*C) always occupies one volume = 22.4136 m3.
How to determine gas volume
- The formula V=n*Vm can most often be 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 the conditions do not have the amount of a substance, but there is a mass of the substance, then we proceed this way: 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, as it is atomic mass. Let's add up all the masses and get what we are looking for.
- So, how to calculate the volume of gas. Here's the task: hydrochloric acid dissolve 10 g of aluminum. Question: how much hydrogen can be released at u.? The reaction equation looks like this: 2Al+6HCl(g)=2AlCl3+3H2. At the very beginning, we find the aluminum (quantity) 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) = 27 g/mol. Let's substitute: n(Al)=10/27=0.37 mol. From the chemical equation it can be seen that 3 moles of hydrogen are formed when 2 moles of aluminum are dissolved. It is necessary to calculate how much hydrogen will be released from 0.4 moles of aluminum: n(H2)=3*0.37/2=0.56mol. Let's substitute the data into the formula and find the volume of this gas. V=n*Vm=0.56*22.4=12.54l.
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