The decision about the need to maintain such a notebook did not come immediately, but gradually, with the accumulation of work experience.
In the beginning, this was a space at the end of the workbook - a few pages for writing down the most important definitions. Then the most important tables were placed there. Then came the realization that most students, in order to learn to solve problems, need strict algorithmic instructions, which they, first of all, must understand and remember.
That’s when the decision came to keep, in addition to the workbook, another mandatory notebook in chemistry - a chemical dictionary. Unlike workbooks, of which there may even be two during one academic year, a dictionary is a single notebook for the entire chemistry course. It is best if this notebook has 48 sheets and a durable cover.
We arrange the material in this notebook as follows: at the beginning - the most important definitions, which the children copy from the textbook or write down under the dictation of the teacher. For example, in the first lesson in 8th grade, this is the definition of the subject “chemistry”, the concept of “chemical reactions”. During the school year in the 8th grade, more than thirty of them accumulate. I conduct surveys on these definitions in some lessons. For example, an oral question in a chain, when one student asks a question to another, if he answered correctly, then he already asks the next question; or, when one student is asked questions by other students, if he cannot answer, then they answer themselves. In organic chemistry, these are mainly definitions of classes of organic substances and main concepts, for example, “homologues”, “isomers”, etc.
At the end of our reference book, material is presented in the form of tables and diagrams. On the last page is the very first table “Chemical elements. Chemical signs." Then the tables “Valence”, “Acids”, “Indicators”, “Electrochemical series of metal voltages”, “Electronegativity series”.
I especially want to dwell on the contents of the table “Correspondence of acids to acid oxides”:
| Correspondence of acids to acid oxides | ||||
| Acid oxide | Acid | |||
| Name | Formula | Name | Formula | Acid residue, valence |
| carbon(II) monoxide | CO2 | coal | H2CO3 | CO3(II) |
| sulfur(IV) oxide | SO 2 | sulfurous | H2SO3 | SO3(II) |
| sulfur(VI) oxide | SO 3 | sulfuric | H2SO4 | SO 4 (II) |
| silicon(IV) oxide | SiO2 | silicon | H2SiO3 | SiO3(II) |
| nitric oxide (V) | N2O5 | nitrogen | HNO3 | NO 3 (I) |
| phosphorus(V) oxide | P2O5 | phosphorus | H3PO4 | PO 4 (III) |
Without understanding and memorizing this table, it is difficult for 8th grade students to compile equations for the reactions of acid oxides with alkalis.
When studying the theory of electrolytic dissociation, we write down diagrams and rules at the end of the notebook.
Rules for composing ionic equations:
1. The formulas of strong electrolytes soluble in water are written in the form of ions.
2. The formulas of simple substances, oxides, weak electrolytes and all insoluble substances are written in molecular form.
3. The formulas of poorly soluble substances on the left side of the equation are written in ionic form, on the right - in molecular form.
When studying organic chemistry, we write into the dictionary general tables on hydrocarbons, classes of oxygen- and nitrogen-containing substances, and diagrams on genetic connections.
| Physical quantities | |||
| Designation | Name | Units | Formulas |
| amount of substance | mole | = N / N A ; = m / M; V / V m (for gases) |
|
| N A | Avogadro's constant | molecules, atoms and other particles | N A = 6.02 10 23 |
| N | number of particles | molecules, atoms and other particles |
N = N A |
| M | molar mass | g/mol, kg/kmol | M = m / ; /M/ = M r |
| m | weight | g, kg | m = M ; m = V |
| Vm | molar volume of gas | l/mol, m 3/kmol | Vm = 22.4 l / mol = 22.4 m 3 / kmol |
| V | volume | l, m 3 | V = V m (for gases); |
| density | g/ml; | =m/V; M / V m (for gases) |
|
Over the 25-year period of teaching chemistry at school, I had to work using different programs and textbooks. At the same time, it was always surprising that practically no textbook teaches how to solve problems. At the beginning of studying chemistry, to systematize and consolidate knowledge in the dictionary, my students and I compile a table “Physical quantities” with new quantities:
When teaching students how to solve calculation problems, I attach great importance to algorithms. I believe that strict instructions for the sequence of actions allow a weak student to understand the solution of problems of a certain type. For strong students, this is an opportunity to reach a creative level in their further chemical education and self-education, since first you need to confidently master a relatively small number of standard techniques. On the basis of this, the ability to correctly apply them at different stages of solving more complex problems will develop. Therefore, I have compiled algorithms for solving calculation problems for all types of school course problems and for elective classes.
I will give examples of some of them.
Algorithm for solving problems using chemical equations.
1. Briefly write down the conditions of the problem and compose a chemical equation.
2. Write the problem data above the formulas in the chemical equation, and write the number of moles under the formulas (determined by the coefficient).
3. Find the amount of substance, the mass or volume of which is given in the problem statement, using the formulas:
M/M; = V / V m (for gases V m = 22.4 l / mol).
Write the resulting number above the formula in the equation.
4. Find the amount of a substance whose mass or volume is unknown. To do this, reason according to the equation: compare the number of moles according to the condition with the number of moles according to the equation. If necessary, make a proportion.
5. Find the mass or volume using the formulas: m = M; V = Vm.
This algorithm is the basis that the student must master so that in the future he will be able to solve problems using equations with various complications.
Problems with excess and deficiency.
If in the problem conditions the quantities, masses or volumes of two reacting substances are known at once, then this is a problem with excess and deficiency.
When solving it:
1. You need to find the quantities of two reacting substances using the formulas:
M/M; = V/V m .
2. Write the resulting mole numbers above the equation. Comparing them with the number of moles according to the equation, draw a conclusion about which substance is given in deficiency.
3. Based on the deficiency, make further calculations.
Problems on the fraction of the yield of the reaction product practically obtained from the theoretically possible.
Using the reaction equations, theoretical calculations are carried out and theoretical data for the reaction product are found: theor. , m theor. or V theory. . When carrying out reactions in the laboratory or in industry, losses occur, so the practical data obtained are practical. ,
m pract. or V practical. always less than theoretically calculated data. The yield share is designated by the letter (eta) and is calculated using the formulas:
(this) = practical. / theory = m pract. / m theor. = V practical / V theor.
It is expressed as a fraction of a unit or as a percentage. Three types of tasks can be distinguished:
If in the problem statement the data for the starting substance and the fraction of the yield of the reaction product are known, then you need to find a practical solution. , m practical or V practical. reaction product.
Solution procedure:
1. Carry out a calculation using the equation based on the data for the starting substance, find the theory. , m theor. or V theory. reaction product;
2. Find the mass or volume of the reaction product practically obtained using the formulas:
m pract. = m theoretical ; V practical = V theor. ; pract. = theoretical .
If in the problem statement the data for the starting substance and practice are known. , m practical or V practical. the resulting product, and you need to find the yield fraction of the reaction product.
Solution procedure:
1. Calculate using the equation based on the data for the starting substance, find
Theor. , m theor. or V theory. reaction product.
2. Find the yield fraction of the reaction product using the formulas:
Pract. / theory = m pract. / m theor. = V practical /V theor.
If the practical conditions are known in the problem conditions. , m practical or V practical. the resulting reaction product and its yield fraction, while you need to find data for the starting substance.
Solution procedure:
1. Find theory, m theory. or V theory. reaction product according to the formulas:
Theor. = practical / ; m theor. = m pract. / ; V theor. = V practical / .
2. Perform calculations using the equation based on the theory. , m theor. or V theory. product of the reaction and find the data for the starting substance.
Of course, we consider these three types of problems gradually, practicing the skills of solving each of them using the example of a number of problems.
Problems on mixtures and impurities.
A pure substance is the one that is more abundant in the mixture, the rest are impurities. Designations: mass of mixture – m cm, mass of pure substance – m p.h., mass of impurities – m approx. , mass fraction of pure substance - p.h.
The mass fraction of a pure substance is found using the formula: p.h. = m h.v. / m cm, it is expressed in fractions of one or as a percentage. Let's distinguish 2 types of tasks.
If the problem statement gives the mass fraction of a pure substance or the mass fraction of impurities, then the mass of the mixture is given. The word “technical” also means the presence of a mixture.
Solution procedure:
1. Find the mass of a pure substance using the formula: m h.v. = h.v. m cm
If the mass fraction of impurities is given, then you first need to find the mass fraction of the pure substance: p.h. = 1 - approx.
2. Based on the mass of the pure substance, make further calculations using the equation.
If the problem statement gives the mass of the initial mixture and n, m or V of the reaction product, then you need to find the mass fraction of the pure substance in the initial mixture or the mass fraction of impurities in it.
Solution procedure:
1. Calculate using the equation based on the data for the reaction product and find n p.v. and m h.v.
2. Find the mass fraction of the pure substance in the mixture using the formula: p.h. = m h.v. / m see and mass fraction of impurities: approx. = 1 - h.v
Law of volumetric relations of gases.
The volumes of gases are related in the same way as their quantities of substances:
V 1 / V 2 = 1 / 2
This law is used when solving problems using equations in which the volume of a gas is given and you need to find the volume of another gas.
Volume fraction of gas in the mixture.
Vg / Vcm, where (phi) is the volume fraction of gas.
Vg – gas volume, Vcm – volume of gas mixture.
If the volume fraction of the gas and the volume of the mixture are given in the problem statement, then, first of all, you need to find the volume of the gas: Vg = Vcm.
The volume of the gas mixture is found using the formula: Vcm = Vg /.
The volume of air spent on combustion of a substance is found through the volume of oxygen found by the equation:
Vair = V(O 2) / 0.21
Derivation of formulas of organic substances using general formulas.
Organic substances form homologous series that have common formulas. This allows:
1. Express the relative molecular weight in terms of the number n.
M r (C n H 2n + 2) = 12 n + 1 (2n + 2) = 14n + 2.
2. Equate M r, expressed through n, to the true M r and find n.
3. Draw up reaction equations in general form and make calculations based on them.
Deriving formulas of substances based on combustion products.
1. Analyze the composition of combustion products and draw a conclusion about the qualitative composition of the burned substance: H 2 O -> H, CO 2 -> C, SO 2 -> S, P 2 O 5 -> P, Na 2 CO 3 -> Na, C.
The presence of oxygen in the substance requires verification. Denote the indices in the formula by x, y, z. For example, CxHyOz (?).
2. Find the amount of substances in combustion products using the formulas:
n = m / M and n = V / Vm.
3. Find the amounts of elements contained in the burned substance. For example:
n (C) = n (CO 2), n (H) = 2 ћ n (H 2 O), n (Na) = 2 ћ n (Na 2 CO 3), n (C) = n (Na 2 CO 3) etc.
4. If a substance of unknown composition has burned, then it is imperative to check whether it contained oxygen. For example, CxНyОz (?), m (O) = m in–va – (m (C) + m(H)).
b) if the relative density is known: M 1 = D 2 M 2, M = D H2 2, M = D O2 32,
M = D air 29, M = D N2 28, etc.
Method 1: find the simplest formula of the substance (see previous algorithm) and the simplest molar mass. Then compare the true molar mass with the simplest one and increase the indices in the formula by the required number of times.
Method 2: find the indices using the formula n = (e) Mr / Ar(e).
If the mass fraction of one of the elements is unknown, then it needs to be found. To do this, subtract the mass fraction of the other element from 100% or from unity.
Gradually, in the course of studying chemistry in the chemical dictionary, algorithms for solving problems of various types occur. And the student always knows where to find the right formula or the necessary information to solve a problem.
Many students like keeping such a notebook; they themselves supplement it with various reference materials.
As for extracurricular activities, my students and I also keep a separate notebook for writing down algorithms for solving problems that go beyond the scope of the school curriculum. In the same notebook, for each type of problem we write down 1-2 examples; they solve the rest of the problems in another notebook. And, if you think about it, among the thousands of different problems that appear on the chemistry exam in all universities, you can identify 25 - 30 different types of problems. Of course, there are many variations among them.
In developing algorithms for solving problems in elective classes, A.A.’s manual helped me a lot. Kushnareva. (Learning to solve problems in chemistry, - M., School - press, 1996).
The ability to solve problems in chemistry is the main criterion for creative mastery of the subject. It is through solving problems of various levels of complexity that a chemistry course can be effectively mastered.
If a student has a clear understanding of all possible types of problems and has solved a large number of problems of each type, then he will be able to cope with the chemistry exam in the form of the Unified State Exam and when entering universities.
Modern symbols for chemical elements were introduced into science in 1813 by Berzelius. According to his proposal, elements are designated by the initial letters of their Latin names. For example, oxygen (Oxygenium) is designated by the letter O, sulfur by the letter S, hydrogen (Hydrogenium) by the letter H. In cases where the names of several elements begin with the same letter, one of the subsequent ones is added to the first letter. Thus, carbon (Carboneum) has the symbol C, calcium, copper, etc.
Chemical symbols are not only abbreviated names of elements: they also express certain quantities (or masses), i.e., each symbol represents either one atom of an element, or one mole of its atoms, or the mass of an element equal to (or proportional to) the molar mass of this element. For example, C means either one carbon atom, or one mole of carbon atoms, or 12 mass units (usually ) of carbon.
Formulas of substances also indicate not only the composition of the substance, but also its quantity and mass. Each formula represents either one molecule of a substance, or one mole of a substance, or a mass of a substance equal to (or proportional to) its molar mass. For example, it means either one molecule of water, or one mole of water, or 18 units of mass (usually) of water.
Simple substances are also indicated by formulas that show how many atoms a molecule of a simple substance consists of: for example, the formula for hydrogen. If the atomic composition of a molecule of a simple substance is not precisely known or the substance consists of molecules containing a different number of atoms, and also if it has an atomic or metallic structure rather than a molecular one, the simple substance is designated by the symbol of the element.
For example, the simple substance phosphorus is denoted by the formula P, since, depending on conditions, phosphorus can consist of molecules with a different number of atoms or have a polymer structure.
The formula of a substance is determined based on the results of its analysis. For example, according to analysis, glucose contains (wt.) carbon, (wt.) hydrogen and (wt.) oxygen. Therefore, the masses of carbon, hydrogen and oxygen are related to each other as . Let us denote the required formula for glucose, where are the numbers of carbon, hydrogen and oxygen atoms in the molecule. The masses of the atoms of these elements are respectively equal. Therefore, the glucose molecule contains carbon, hydrogen and oxygen. The ratio of these masses is equal to . But we have already found this relationship based on glucose analysis data. Hence:
According to the properties of proportion:
Therefore, in a glucose molecule there are two hydrogen atoms and one oxygen atom per carbon atom. This condition is satisfied by formulas, etc. The first of these formulas is called the simplest or empirical formula; it has a molecular weight of 30.02. In order to find out the true or molecular formula, you need to know the molecular weight of a given substance. When heated, glucose is destroyed without turning into gas. But its molecular weight can be determined by the methods described in Chapter VII: it is equal to 180. From a comparison of this molecular weight with the molecular weight corresponding to the simplest formula, it is clear that the formula corresponds to glucose.
Having become familiar with the derivation of chemical formulas, it is easy to understand how the exact values of molecular masses are determined. As already mentioned, existing methods for determining molecular weights in most cases do not give entirely accurate results. But, knowing at least approximately the molecular weight and percentage composition of a substance, it is possible to establish its formula, which expresses the atomic composition of the molecule. Since the molecular mass is equal to the sum of the atomic masses of the atoms that form it, then by adding the atomic masses of the atoms that make up the molecule, we determine the molecular mass of the substance. The accuracy of the molecular mass found will correspond to the accuracy with which the substance was analyzed.
Key words of the abstract: Chemical elements, signs of chemical elements.
In chemistry a very important concept is "chemical element"(the word "element" in Greek means "component"). To understand its essence, remember how mixtures and chemical compounds differ.
For example, iron and sulfur retain their properties in the mixture. Therefore, it can be argued that a mixture of iron powder and sulfur powder consists of two simple substances - iron and sulfur. Since the chemical compound iron sulfide is formed from simple substances - iron and sulfur, I would like to argue that iron sulfide also consists of iron and sulfur. But having become acquainted with the properties of iron sulfide, we understand that this cannot be said. This, formed as a result of chemical interaction, has completely different properties than the original substances. Because the composition of complex substances does not include simple substances, but atoms of a certain type.
A CHEMICAL ELEMENT is a specific type of atom.
So, for example, all oxygen atoms, regardless of whether they are part of oxygen molecules or water molecules, are the chemical element oxygen. All atoms of hydrogen, iron, sulfur are, respectively, the chemical elements hydrogen, iron, sulfur, etc.
There are currently 118 different types of atoms known, i.e. 118 chemical elements. From the atoms of this relatively small number of elements a huge variety of substances is formed. (The concept of “chemical element” will be clarified and expanded in further notes).
Using the concept of “chemical element”, we can clarify the definitions: SIMPLE substances are substances that consist of atoms of one chemical element. COMPLEX substances are substances that consist of atoms of different chemical elements.
It is necessary to distinguish between concepts "simple matter" And "chemical element" , although their names are in most cases the same. Therefore, every time we come across the words “oxygen”, “hydrogen”, “iron”, “sulfur”, etc., we need to understand what we are talking about - a simple substance or a chemical element. If, for example, they say: “Fish breathe oxygen dissolved in water,” “Iron is a metal that is attracted by a magnet,” this means that we are talking about simple substances - oxygen and iron. If they say that oxygen or iron is part of a substance, then they mean oxygen and iron as chemical elements.
Chemical elements and the simple substances they form can be divided into two large groups: metals and non-metals. Examples of metals are iron, aluminum, copper, gold, silver, etc. Metals are ductile, have a metallic luster, and conduct electricity well. Examples of non-metals are sulfur, phosphorus, hydrogen, oxygen, nitrogen, etc. The properties of non-metals are varied.
Chemical element signs
Each chemical element has its own name. For simplified designation of chemical elements, use chemical symbolism. A chemical element is designated by the initial or initial and one of the subsequent letters of the Latin name of this element. Thus, hydrogen (lat. hydrogenium - hydrogenium) is designated by the letter N, mercury (lat. hydrargyrum - hydrargyrum) - letters Hg etc. Modern chemical symbolism was proposed by the Swedish chemist J. J. Berzelius in 1814
Abbreviated letter designations for chemical elements are signs(or symbols) chemical elements. Chemical symbol (chemical sign) means one atom of a given chemical element .
You already know the symbols of some chemical elements.
What does a chemical symbol show?
1) Designates a chemical element (give a name);
2) one atom of this element;
3) by the symbol you can determine the place of the element in the periodic table D.I. Mendeleev;
4) using the periodic table, you can determine the relative atomic mass of an element.
Let's look at an example.
Chemical element symbol - Cu
1) Chemical element - copper.
2) one copper atom;
3) Copper is in the periodic table of elements in period 4, group 1, serial number - 29.
4) Ar(Cu)=64
Let us summarize the information known to us that the chemical formula contains.
Table. Information contained in a chemical formula.
Example: HNO3 - nitric acid
| 1. High-quality composition | 1. The molecule consists of atoms of three chemical elements: H, N, O |
| 2. Quantitative composition | 2. the molecule contains five atoms: one hydrogen atom, one nitrogen atom, three oxygen atoms |
| 3. Relative molecular weight | 3.Mr(HNO3)= 1 1+14 1+16 3=63 |
| 4. Mass of the molecule | 4. mm(HNO3)= 1a.u.m. ·1+ 14 amu ·1+ 16 amu ·3=63 a.m.u. |
| 5. Mass fractions of elements | 5.ω(H) = Ar(H) 1 / Mr(HNO3)= 1 1/63=0.016 or 1.6% ω(N)= Ar (N) 1 /Mr(HNO3)= ω(O)= Ar (O) 3 /Mr(HNO3)= |
Complete the task in your workbook by analogy
Summarizing
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Modern symbols for chemical elements consist of the first letter or the first and one of the following letters of the Latin name of the elements. In this case, only the first letter is capitalized. For example, H - hydrogen (lat. Hydrogenium), N - nitrogen (lat. Nitrogenium), Ca - calcium (lat. Calcium), Pt - platinum (lat. Platinum) and so on.
Metals discovered in the 15th-18th centuries - bismuth, zinc, cobalt - began to be designated by the first letters of their names. At the same time, symbols for complex substances appeared, associated with their names. For example, the sign for wine spirit is made up of the letters S and V (lat. spiritus vini). Signs of strong vodka (lat. aqua fortis) - nitric acid, and aqua regia (lat. aqua regis), mixtures of hydrochloric and nitric acids, are made up of the sign for water and the capital letters F and R, respectively. Glass sign (lat. vitrum) is formed from two letters V - straight and inverted. 
A.-L. Lavoisier, working on a new classification and nomenclature, proposed a very cumbersome system of chemical symbols for elements and compounds. Attempts to streamline ancient chemical signs continued until the end of the 18th century. A more appropriate sign system was proposed in 1787 by J.-A. Gassenfratz and P.-O. Ade; their chemical signs are already adapted to Lavoisier’s antiphlogistic theory and have some features that were subsequently preserved. They proposed introducing symbols in the form of simple geometric figures and letter designations as common to each class of substances, as well as straight lines drawn in various directions to designate the “true elements” - light and caloric, as well as elementary gases - oxygen, nitrogen and hydrogen Thus, all metals had to be designated by circles with the initial letter (sometimes two letters, the second lowercase) of the French name for the metal in the middle; all alkalis and alkaline earths (also classified by Lavoisier as elements) - in variously arranged triangles with Latin letters in the middle, etc.
In 1814, Berzelius detailed a system of chemical symbolism based on the designation of elements by one or two letters of the element's Latin name; the number of atoms of an element was proposed to be indicated by superscript digital indices (the currently accepted indication of the number of atoms by subscripts was proposed in 1834 by Justus Liebig). The Berzelius system received universal recognition and has been preserved to this day. In Russia, the first printed message about Berzelius's chemical signs was made by the Moscow doctor I. Ya. Zatsepin.
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An excerpt characterizing the Symbols of chemical elements
The friends were silent. Neither one nor the other began to speak. Pierre glanced at Prince Andrei, Prince Andrei rubbed his forehead with his small hand.“Let’s go have dinner,” he said with a sigh, getting up and heading to the door.
They entered the elegantly, newly, richly decorated dining room. Everything, from napkins to silver, earthenware and crystal, bore that special imprint of novelty that happens in the household of young spouses. In the middle of dinner, Prince Andrei leaned on his elbow and, like a man who has had something on his heart for a long time and suddenly decides to speak out, with an expression of nervous irritation in which Pierre had never seen his friend before, he began to say:
– Never, never get married, my friend; Here's my advice to you: don't get married until you tell yourself that you did everything you could, and until you stop loving the woman you chose, until you see her clearly; otherwise you will make a cruel and irreparable mistake. Marry an old man, good for nothing... Otherwise, everything that is good and lofty in you will be lost. Everything will be spent on little things. Yes Yes Yes! Don't look at me with such surprise. If you expect something from yourself in the future, then at every step you will feel that everything is over for you, everything is closed except for the living room, where you will stand on the same level as a court lackey and an idiot... So what!...
He waved his hand energetically.
Pierre took off his glasses, causing his face to change, showing even more kindness, and looked at his friend in surprise.
“My wife,” continued Prince Andrei, “is a wonderful woman.” This is one of those rare women with whom you can be at peace with your honor; but, my God, what I wouldn’t give now not to be married! I’m telling you this alone and first, because I love you.
Prince Andrei, saying this, looked even less like than before that Bolkonsky, who was lounging in Anna Pavlovna’s chair and, squinting through his teeth, spoke French phrases. His dry face was still trembling with the nervous animation of every muscle; the eyes, in which the fire of life had previously seemed extinguished, now shone with a radiant, bright shine. It was clear that the more lifeless he seemed in ordinary times, the more energetic he was in these moments of almost painful irritation.
“You don’t understand why I’m saying this,” he continued. – After all, this is a whole life story. You say Bonaparte and his career,” he said, although Pierre did not talk about Bonaparte. – You say Bonaparte; but Bonaparte, when he worked, walked step by step towards his goal, he was free, he had nothing but his goal - and he achieved it. But tie yourself to a woman, and like a shackled convict, you lose all freedom. And everything that you have in you of hope and strength, everything only weighs you down and torments you with remorse. Living rooms, gossip, balls, vanity, insignificance - this is a vicious circle from which I cannot escape. I am now going to war, to the greatest war that has ever happened, but I know nothing and am no good for anything. “Je suis tres aimable et tres caustique, [I am very sweet and very eater,” continued Prince Andrei, “and Anna Pavlovna listens to me.” And this stupid society, without which my wife and these women cannot live... If only you could know what it is toutes les femmes distinguees [all these women of good society] and women in general! My father is right. Selfishness, vanity, stupidity, insignificance in everything - these are women when they show everything as they are. If you look at them in the light, it seems that there is something, but nothing, nothing, nothing! Yes, don’t get married, my soul, don’t get married,” Prince Andrei finished.
“It’s funny to me,” said Pierre, “that you consider yourself incapable, that your life is a spoiled life.” You have everything, everything is ahead. And you…
He didn’t say you, but his tone already showed how highly he valued his friend and how much he expected from him in the future.
“How can he say that!” thought Pierre. Pierre considered Prince Andrei to be a model of all perfections precisely because Prince Andrei united to the highest degree all those qualities that Pierre did not have and which can be most closely expressed by the concept of willpower. Pierre was always amazed at Prince Andrei's ability to calmly deal with all kinds of people, his extraordinary memory, erudition (he read everything, knew everything, had an idea about everything) and most of all his ability to work and study. If Pierre was often struck by Andrei’s lack of ability for dreamy philosophizing (to which Pierre was especially prone), then in this he saw not a disadvantage, but a strength.
In the best, most friendly and simple relationships, flattery or praise is necessary, just as greasing is necessary for the wheels to keep them moving.
“Je suis un homme fini, [I am a finished man,” said Prince Andrei. - What can you say about me? Let’s talk about you,” he said, after a pause and smiling at his comforting thoughts.
This smile was reflected on Pierre’s face at the same instant.
– What can we say about me? - said Pierre, spreading his mouth into a carefree, cheerful smile. -What am I? Je suis un batard [I am an illegitimate son!] - And he suddenly blushed crimson. It was clear that he made a great effort to say this. – Sans nom, sans fortune... [No name, no fortune...] And well, that’s right... - But he didn’t say that’s right. – I’m free for now, and I feel good. I just don’t know what to start. I wanted to seriously consult with you.
Prince Andrei looked at him with kind eyes. But his glance, friendly and affectionate, still expressed the consciousness of his superiority.
– You are dear to me, especially because you are the only living person among our entire world. You feel good. Choose what you want; it does not matter. You will be good everywhere, but one thing: stop going to these Kuragins and leading this life. So it doesn’t suit you: all these carousings, and hussarism, and everything...
“Que voulez vous, mon cher,” said Pierre, shrugging his shoulders, “les femmes, mon cher, les femmes!” [What do you want, my dear, women, my dear, women!]
“I don’t understand,” Andrey answered. – Les femmes comme il faut, [Decent women] is another matter; but les femmes Kuragin, les femmes et le vin, [Kuragin’s women, women and wine,] I don’t understand!
Pierre lived with Prince Vasily Kuragin and took part in the wild life of his son Anatole, the same one who was going to be married to Prince Andrei’s sister for correction.
“You know what,” said Pierre, as if an unexpectedly happy thought had come to him, “seriously, I’ve been thinking about this for a long time.” With this life I can neither decide nor think about anything. My head hurts, I have no money. Today he called me, I won’t go.
- Give me your word of honor that you won’t travel?
- Honestly!
It was already two o'clock in the morning when Pierre left his friend. It was a June night, a St. Petersburg night, a gloomless night. Pierre got into the cab with the intention of going home. But the closer he got, the more he felt it was impossible to fall asleep that night, which seemed more like evening or morning. It was visible in the distance through the empty streets. Dear Pierre remembered that that evening the usual gambling society was supposed to gather at Anatole Kuragin's place, after which there would usually be a drinking party, ending with one of Pierre's favorite amusements.
“It would be nice to go to Kuragin,” he thought.
But he immediately remembered his word of honor given to Prince Andrei not to visit Kuragin. But immediately, as happens with people called spineless, he so passionately wanted to once again experience this dissolute life so familiar to him that he decided to go. And immediately the thought occurred to him that this word meant nothing, because even before Prince Andrei, he also gave Prince Anatoly the word to be with him; Finally, he thought that all these honest words were such conventional things that had no definite meaning, especially if you realized that maybe tomorrow he would either die or something so extraordinary would happen to him that there would no longer be any honest , nor dishonest. This kind of reasoning, destroying all his decisions and assumptions, often came to Pierre. He went to Kuragin.
Having arrived at the porch of a large house near the horse guards barracks in which Anatole lived, he climbed onto the illuminated porch, onto the stairs, and entered the open door. There was no one in the hall; there were empty bottles, raincoats, and galoshes lying around; there was a smell of wine, and distant talking and shouting could be heard.
The game and dinner were already over, but the guests had not yet left. Pierre took off his cloak and entered the first room, where the remains of dinner were standing and one footman, thinking that no one was seeing him, was secretly finishing off unfinished glasses. From the third room you could hear fuss, laughter, screams of familiar voices and the roar of a bear.
About eight young people crowded anxiously around the open window. The three were busy with a young bear, which one was dragging on a chain, frightening the other with it.
- I'll give Stevens a hundred! - one shouted.
- Be careful not to support! - shouted another.
- I am for Dolokhov! - shouted the third. - Take them apart, Kuragin.
- Well, leave Mishka, there’s a bet here.
“One spirit, otherwise it’s lost,” shouted the fourth.
- Yakov, give me a bottle, Yakov! - shouted the owner himself, a tall handsome man standing in the middle of the crowd wearing only a thin shirt open at the middle of his chest. - Stop, gentlemen. Here he is Petrusha, dear friend,” he turned to Pierre.
