The textbook contains a summary of the main theoretical provisions, tasks for all sections of the general physics course, answers and suggestions that can be used when solving problems. BBC 22.3

And 1% argon, find mass 1 mole air. Find volume of a mixture of nitrogen weight 2.8 kg and oxygen weight 3.2 kg ... and neutrons are contained in the nuclei of isotopes carbon: 1) 16°С; 2) "^C; 3) "|C; 4) 5) !64С; 6) 1|С? 183 What time...

1. mol. The amount of substance equivalent (equivalent) and the molar mass of the equivalent (equivalent mass) of simple and complex substances. Law of Equivalents

kg/ mole. Relative molecular weight Mr is the molar weight substance referred to molar masses atom carbon... Cn and molar mass equivalent (me) of a solute, the titer is easily find according to the formula...

1. In the hydrated manganese (II) sulfate, the mass fraction of manganese is 0.268. Determine the amount of water substance per 1 mole of crystalline hydrate. Write the formula for salt. 2

Water per 1 mole crystalline hydrate. Write the formula of salt... g organic matter(M= 93 g/ mole) formed 8.8 g of oxide carbon(IV) , 2.1 g of water and... Find mass acetic acid in the original mixture. 7. There is a mixture of iron, carbon ...

Task 3.1.1 How many moles of iron (II) sulfide are contained in 8.8 g of FeS?

Solution: We define molar mass(M) iron(II) sulfide.

M (FeS) \u003d 56 + 32 \u003d 88 g / mol

Calculate how many moles are contained in 8.8 g of FeS

Task 3.1.2 How many molecules are there in 54 g of water? What is the mass of one molecule of water?

Solution: Determine the molar mass of water.

M (H 2 O) \u003d 18 g / mol

Therefore, 54 g of water contains 54/18 = 3 mol H 2 O. One mole of any substance contains 6.02  10 23 molecules. Then 3 moles (54g H 2 O) contain 6.02  10 23  3 = 18.06  10 23 molecules.

Determine the mass of one molecule of water.

Task 3.1.3 How many moles and molecules are contained in 1 m 3 of any gas at normal conditions?

Solution: 1 mole of any gas under normal conditions occupies a volume of 22.4 liters. Therefore, 1m 3 (= 1000l) will contain 44.6 moles of gas:

1 mole of any gas contains 6.02  10 23 molecules. It follows from this that 1 m 3 of any gas under normal conditions contains

6.02  10 23  44.6 \u003d 2.68  10 25 molecules.

Task 3.1.4 Express in moles: a) 6.02  10 23 C 2 H 2 molecules;

b) 1.80  10 24 nitrogen atoms; c) 3.01  10 23 NH 3 molecules. What is the molar mass of these substances?

Solution: A mole is the amount of a substance that contains the number of particles of any particular kind, equal to the Avogadro constant. From here:

a)

b)

in)

The mass of a mole of a substance is expressed in g/mol . The molar mass of a substance in grams is numerically equal to its relative molecular (atomic) mass, expressed in amu.

Therefore, the molar masses of these substances are equal:

a) M (C 2 H 2) \u003d 26 g / mol; b) М(N) = 14 g/mol;

c) M (NH 3) \u003d 17 g / mol;

Task 3.1.5 Determine the molar mass of a gas if, under normal conditions, 0.824 g it is occupied by a volume of 0.260 l.

Solution: Under normal conditions, 1 mole of any gas occupies a volume of 22.4 liters. By calculating the mass of 22.4 liters of a given gas, we find out its molar mass.

0.824 g of gas occupy a volume of 0.260 liters

X g of gas occupy a volume of 22.4 liters

Therefore, the molar mass of the gas is 71 g/mol.

3.2 Equivalent. Equivalence factor. Molar mass equivalents.

Task 3.2.1 Calculate the equivalent, equivalence factor, and molar mass of H equivalents 3 RO 4 in exchange reactions, as a result of which acidic and normal salts are formed.

Solution Let us write down the reaction equations for the interaction of phosphoric acid with alkali:

H 2 PO 4 + NaOH = NaH 2 PO 4 + H 2 O; (one)

H 2 RO 4 H 2 RO 4 + 2NaOH \u003d Na 2 HPO 4 + 2H 2 O; (2)

H 2 PO 4 + 3NaOH \u003d Na 3 PO 4 + 3H 2 O; (3)

Because phosphoric acid is a tribasic acid, it forms two acid salts: NaH 2 PO 4 - sodium dihydrogen phosphate and Na 2 HPO 4 - sodium hydrogen phosphate and one average salt Na 3 PO 4 - sodium phosphate.

In reaction (1), phosphoric acid exchanges one hydrogen atom for a metal, i.e. behaves like a monobasic acid, so

f e (H 3 RO 4) in reaction (1) is 1;

E (H 3 RO 4) \u003d H 3 RO 4; M (1 H 3 RO 4) \u003d 98 g / mol.

In reaction (2), phosphoric acid exchanges three f e (H 3 PO 4) hydrogen atoms for a metal, i.e. behaves like a dibasic acid, therefore f e (H 3 PO 4) in reaction (2) is equal to 1/2;

E (H 3 RO 4) \u003d 1/2H 3 RO 4; M (1/2 H 3 RO 4) \u003d 49 g / mol.

In reaction (3), phosphoric acid behaves like a tribasic acid, so f e (H 3 PO 4) in this reaction is 1/3;

E (H 3 RO 4) \u003d 1/3H 3 RO 4; M (1/3 H 3 RO 4) \u003d 32.67 g / mol.

Task 3.2.2 An excess of potassium hydroxide acted on solutions of: a) potassium dihydrogen phosphate; b) dihydroxovismuth (III) nitrate. Write the reaction equations for these substances with KOH and determine their equivalent and molar mass equivalents.

Solution: Let's write the equations of the occurring reactions:

KN 2 RO 4 + 2KOH \u003d K 3 RO 4 + 2 H 2 O

Bi(OH) 2 NO 3 + KOH = Bi(OH) 3 + KNO 3

Potassium dihydrogen phosphate, when interacting with KOH, exchanges two hydrogen atoms for a metal, i.e. f e salt KH 2 RO 4 = 1/2;

E (KN 2 RO 4) \u003d 1/2 KN 2 RO 4;

M (1/2 KH 2 RO 4) \u003d 1/2 (39 + 2 + 31 + 64) \u003d 68 g / mol

Nitrate of dihydroxovismuth (III) during the reaction with KOH exchanges one group of NO 3 - for an ion OH -, i.e. one bond is rearranged in the basic salt molecule, therefore:

E (Bi (OH) 2 NO 3) \u003d Bi (OH) 2 NO 3;

M (1 Bi (OH) 2 NO 3) \u003d 305 g / mol

Class: 8

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.

Lesson type: lesson of studying and primary consolidation of new knowledge.

During the classes

I. Organizational moment

II. Checking d / z on the topic: "Types of chemical reactions"

III. Learning new material

1. Amount of substance - mole

Substances react in strictly defined proportions. For example, to obtain the substance water, you need to take so much hydrogen and oxygen that for every two molecules of hydrogen there is one molecule of oxygen:

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

To obtain the substance iron sulfide, you need to take so much iron and sulfur that for each atom of iron there is one atom of sulfur.

To obtain the substance phosphorus oxide, you need to take so many molecules of phosphorus and oxygen that for four molecules of phosphorus there are five molecules of oxygen.

It is impossible to determine the number of atoms, molecules and other particles in practice - they are too small and not visible to the naked eye. To determine the number of structural units (atoms, molecules) in chemistry, a special value is used - amount of matter ( v - nude). The unit of quantity of a substance is mole.

• A mole is the amount of a substance that contains as many structural particles (atoms, molecules) as there are atoms in 12 g of carbon.

It has been experimentally established that 12 g of carbon contains 6·10 23 atoms. This means that one mole of any substance, regardless of its state of aggregation, contains the same number of particles - 6 10 23.

• 1 mole of oxygen (O 2) contains 6 10 23 molecules.
• 1 mol of hydrogen (H 2) contains 6 10 23 molecules.
• 1 mol of water (H 2 O) contains 6 10 23 molecules.
• 1 mole of iron (Fe) contains 6 10 23 molecules.

Exercise: Using the information you received, answer the following questions:

a) how many oxygen atoms are there in 1 mole of oxygen?

– 6 10 23 2 = 12 10 23 atoms.

b) how many hydrogen and oxygen atoms are there in 1 mole of water (H 2 O)?

– 6 10 23 2 = 12 10 23 hydrogen atoms and 6 10 23 oxygen atoms.

Number 6 10 23 is called Avogadro's constant in honor of the Italian scientist of the 19th century and is designated NA. Units of measurement are atoms/mol or molecules/mol.

2. Solving problems for finding the amount of substance

Often you need to know how many particles of a substance are contained in a certain amount of a substance. Or to find the amount of substance by a known number of molecules. These calculations can be done using the formula:

where N is the number of molecules, NA is the Avogadro constant, v- amount of substance. From this formula, you can express the amount of substance.

Task 1. How many atoms are there in 2 moles of sulfur?

N = 2 6 10 23 = 12 10 23 atoms.

Task 2. How many atoms are there in 0.5 moles of iron?

N = 0.5 6 10 23 = 3 10 23 atoms.

Task 3. How many molecules are there in 5 moles of carbon dioxide?

N = 5 6 10 23 = 30 10 23 molecules.

Task 4. How much of a substance is 12 10 23 molecules of this substance?

v= 12 10 23 / 6 10 23 \u003d 2 mol.

Task 5. What amount of a substance is 0.6 10 23 molecules of this substance?

v= 0.6 10 23 / 6 10 23 \u003d 0.1 mol.

Task 6. How much of a substance is 3 10 23 molecules of this substance?

v= 3 10 23 / 6 10 23 \u003d 0.5 mol.

3. Molar mass

For chemical reactions, you need to take into account the amount of substance in moles.

Q: But how in practice to measure 2, or 2.5 moles of a substance? What is the best unit to measure the mass of substances?

For convenience in chemistry, molar mass is used.

Molar mass is the mass of one mole of a substance.

It is designated - M. It is measured in g / mol.

The molar mass is equal to the ratio of the mass of a substance to the corresponding amount of the substance.

Molar mass is a constant value. The numerical value of the molar mass corresponds to the value of the relative atomic or relative molecular weight.

Q: How can I find relative atomic or relative molecular weights?

Mr(S) = 32; M (S) \u003d 32 g / mol - which corresponds to 1 mole of sulfur

Mr (H 2 O) = 18; M (H 2 O) \u003d 18 g / mol - which corresponds to 1 mole of water.

4. Solving problems on finding the mass of matter

Task 7. Determine the mass of 0.5 mol of iron.

Task 8. Determine the mass of 0.25 mol of copper

Task 9. Determine the mass of 2 moles of carbon dioxide (CO 2)

Task 10. How many moles of copper oxide - CuO make up 160 g of copper oxide?

v= 160 / 80 = 8 mol

Task 11. How many moles of water correspond to 30 g of water

v= 30/18 = 1.66 mol

Task 12. How many moles of magnesium corresponds to its 40 grams?

v= 40/24 = 1.66 mol

IV. Anchoring

Front poll:

1. What is the amount of substance?
2. What is 1 mole of any substance equal to?
3. What is molar mass?
4. Is there a difference between the terms "mole of molecules" and "mole of atoms"?
5. Explain using the example of the ammonia molecule NH3.
6. Why is it important to know formulas when solving problems?

Tasks:

1. How many molecules are there in 180 grams of water?
2. How many molecules make up 80 g of carbon dioxide?

V. Homework

Study the text of the paragraph, make two tasks: to find the amount of substance; to find the mass of a substance.

Literature:

1. Gara N.N. Chemistry. Lessons in Grade 8: A Teacher's Guide. _ M.: Enlightenment, 2009.
2. Rudzites G.E., Feldman F.G. Chemistry. Grade 8: Textbook for general educational institutions - M .: Education, 2009.

16. Moth. Molar mass. Molar mass of the equivalent. Equivalent.

17. Avogadro's law: Consequences from Avogadro's law. Density of one gas over another

18. Clapeyron-Mendeleev equation for ideal gases.

19. The law of conservation of mass and energy.

20. The law of constancy of composition: daltonides, bertolides.

21. Law of equivalents (three forms).

Practical tasks/tasks

6. How many molecules are contained in 62 g of oxygen at n.o.?

7. Determine the molar mass of the equivalent of a metal if, during the combustion of 432 grams of it, 464 grams of oxide is formed.

8. Manganese oxide contains 22.56% oxygen. Calculate the molar mass equivalent of manganese in oxide and write the formula.

9. What volume is occupied by 16 g of oxygen at 20 ◦ C, pressure 250 mm. rt. Art. (760 mm Hg = 1 atm = 101.3 kPa)?

10. At n.s., 300 ml of gas have a mass of 0.857 g. Determine the molar mass of the gas, its density in air.

Part 4. Theoretical inorganic chemistry. The structure of the atom, chemical bonds

22. Dimensions, charges and masses of atoms and nucleons. Atomic Orbital The number of electrons, protons, and neutrons in an electrically neutral atom. The mass of an atom (its mass number).

23. Modern views about the structure of the atom in accordance with the principle of least energy, Klechkovsky's rule, Pauli's principle and Hund's rule. Klechkovsky's rule. Pauli principle.

24. Gund's rule. Louis De Broglie's equation, h is Planck's constant. Planck's equation Periodic law D.I. Mendeleev Properties of elements and their compounds in periodic dependence on charge atomic nuclei elements. Isobars. Isotons. Isotopes.

25. Electron affinity Ecp. Ionization energy of atoms Eion Chemical bond. The reason for the formation of chem. Connections. Bond energy and bond length. Covalent (or combined) chemical bond. Saturability. Orientation of communication. Spatial configuration of molecules with different types of hybridization, bond angle.

26. Ionic bond. Her properties. Metal connection. Intermolecular interaction: ion-dipole; dipole-dipole (orientation); induction; dispersion. Van der Waals forces. hydrogen bond

27. Hydrophilic-hydrophobic interaction. State of aggregation Solids are crystalline and amorphous. Anisotropy. Isotropy. liquid state. Liquid-crystalline state. Gas, steam Plasma.

Practical tasks/tasks

11. Specify type chemical bond, type of hybridization of the central atom, valence angle and shape of molecules and ions: BCl 3, CH 4, PO 4 3-, PF 5, [PCl 4] +, NO 2 -, AsO 4, NH 5, SH 6.

12. Compare the values ​​of the bond angles of the molecules and explain the nature of their change in the series CH 4 - 109.5 0; H 3 N - I07.3 0; H 2 O - 104.5 0.

13. Which atom is being completed electronic formula 4s 2 4p 4 ?

How many electrons, protons, neutrons are in the most common isotope, an electrically neutral atom with such an electronic formula?

Part 5. Methods for expressing the concentrations of solutions. Determination of the pH of solutions

28. Signs chemical reaction when a substance is dissolved.

29. The main ways of expressing the concentrations of solutions: mass fraction W, molar concentration C, molar concentration of Se equivalent, molar concentration b, titer T.

30. Hydrogen index, pH. Strong acids, strong bases.

31. The degree of dissociation for strong and weak electrolytes.

32. Calculation of pH for acids. Calculation of pH for bases (hydroxides). buffer solutions.

33. Hydrolysis of salts. pH of the medium in various cases of hydrolysis.

Practical tasks/tasks

14. Write the ionic and molecular equations for the hydrolysis of the MnCl 2 salt. Specify the reaction of the environment.

15. Calculate the pH of a 0.02 M solution of H 2 CO 3 (take the dissociation constant in the reference book).

16. How many ml of a 30% solution of MgCO 3 (take the density in the reference book) is needed to prepare 500 ml of a solution with Ce = 0.1 mol / l?

17. The product of the solubility of a solution of MgCO 3 take in the reference book. Calculate the concentration of Mg ions in a saturated salt solution.

18. What is the mass fraction of NaCl in solution with Ce = 0.1 mol / l (density in the reference book)?