To compose an equation means to express in mathematical form the relationship between the data (known) of the problem and the sought (unknown) values of it. Sometimes this connection is so clearly contained in the formulation of the problem that the drawing up of the equation is just a literal retelling of the problem, in the language of mathematical signs.
Example 1. Petrov received 160 rubles for his work. more than half of the amount that Ivanov received. Together they received 1120 rubles. How much did Petrov and Ivanov get for their work? Let us denote Ivanov's earnings by x. Half of his earnings are 0.5x; Petrov's monthly salary is 0.5x + 160 together they earn 1120 rubles; the mathematical notation of the last phrase will be
(0.5x + 160) + x = 1120.
The equation is drawn up. Solving it according to the established rules, we find Ivanov's earnings x = 640 rubles; Petrov's earnings are 0.5x + 160 = 480 (rubles).
More often, however, it happens that the relationship between the data and the desired values is not directly indicated in the problem; it must be established based on the conditions of the problem. In practical tasks, this is almost always the case. The example just given is contrived; in life, almost never such tasks are encountered.
Therefore, it is impossible to give completely exhaustive instructions for the compilation of the equation. However, at first, it is useful to be guided by the following. Let us take as the value of the desired quantity (or several quantities) some randomly taken number (or several numbers) and set ourselves the task of checking whether we have guessed the correct solution to the problem or not. If we were able to carry out this check and find either that our guess is correct, or that it is incorrect (most likely, the latter will happen, of course), then we can immediately compose the required equation (or several equations). Namely, we will write down the very actions that we performed for verification, only instead of a randomly taken number we will introduce an alphabetic sign of an unknown quantity. We will get the required equation.
Example 2. A piece of an alloy of copper and zinc with a volume of 1 dm3 weighs 8.14 kg. How much copper is in the alloy? (specific weight of copper 8.9 kg / dm3; zinc - 7.0 kg / dm3).
Let's take at random a number that expresses the required volume of copper, for example, 0.3 dm3. Let's check if we have taken this number successfully. Since 1 kg / dm3 of copper weighs 8.9 kg, then 0.3 dm3 weighs 8.9 * 0.3 = 2.67 (kg). The volume of zinc in the alloy is 1 - 0.3 = 0.7 (dm3). Its weight is 7.0 0.7 = 4.9 (kg). The total weight of zinc and copper is 2.67 + 4.9 = 7.57 (kg). Meanwhile, the weight of our piece, according to the problem, is 8.14 kg. Our guess is untenable. But then we will immediately get an equation whose solution will give the correct answer. Instead of a randomly taken number of 0.3 dm3, we denote the volume of copper (in dm3) by x. Instead of the product 8.9 0.3 = 2.67, we take the product 8.9 x. This is the weight of the copper in the alloy. Instead of 1 - 0.3 = 0.7 we take 1 - x; this is the volume of zinc. Instead of 7.0 0.7 = 4.9 we take 7.0 (1 - x); this is the weight of zinc. Instead of 2.67 + 4.9, we take 8.9x + 7.0 (1 - x); this is the total weight of zinc and copper. By condition, it is equal to 8.14 kg; so 8.9x +7.0 (1 - x) = 8.14.
Solving this equation gives x = 0.6. The guesswork of a decision taken can be done in a variety of ways; accordingly, different types of equations can be obtained for the same problem; all of them, however, will give the same solution for the desired quantity; such equations are called equivalent to each other.
Of course, after gaining the skills in drawing up equations, there is no need to check the number taken at random: you can take not a number for the value of the desired quantity, but some letter (x, y, etc.) and act as if this letter ( unknown) was the number we're going to check.
Word problem solutions for writing equations will be useful primarily for schoolchildren. The curriculum for grades 9, 10 covers a wide class of problems in which it is required to determine unknowns, draw up an equation and solve. Below is only a small part of the possible problems and the methodology for their calculations.
Example 1. The first cyclist travels 50 meters less than the second every minute, so he spends 2 hours more on the 120 km path than the second. Find the speed of the second cyclist (in km per hour).
Solution: The task is difficult for many, but in reality everything is simple.
Under the phrase "Drives 50 meters less every minute" is hidden a speed of 50 m / min. Since the rest of the data is in km and hours, then 50 m / min is reduced to km / h.
50/1000 * 60 = 3000/1000 = 3 (km / h).
Let's denote the speed of the second cyclist by V, and the time of movement by t.
By multiplying the speed by the time of movement, we get the path
V * t = 120.
The first cyclist rides slower and therefore takes longer. We compose the corresponding equation of motion
(V-3) (t + 2) = 120.
We have a system of two equations with two unknowns.
From the first equation we express the time of motion and substitute it into the second
t = 120 / V; (V-3) (120 / V + 2) = 120.
After multiplying by V / 2 and grouping similar terms, you can get such a quadratic equation
V ^ 2-3V-180 = 0.
Calculate the discriminant of the equation
D = 9 + 4 * 180 = 729 = 27 * 27
and roots
V = (3 + 27) / 2 = 15;
V = (3-27) / 2 = -12.
We reject the second, it has no physical meaning. The found value V = 15 km / h is the speed of the second cyclist.
Answer: 15 km / h.
Example 2. Sea water contains 5% salt by weight. How much fresh water should be added to 30 kg of sea water to reduce the salt concentration by 70%?
Solution: Find how much salt is in 30 kg of sea water
30 * 5/100 = 1.5 (kg).
In a new solution, this will be
(100% -70%) = 30% of 5%, make up the proportions
5% – 100%
X - 30%.
Performing calculations
X = 5 * 30/100 = 150/100 = 1.5%.
Thus, 1.5 kg of salt corresponds to 1.5% in the new solution. Add the proportions again
1.5 - 1.5% Y - 100%.
Find the mass of the sea water solution
Y = 1.5 * 100 / 1.5 = 100 (kg).
Subtract the masses of salt water to find the amount of fresh
100-30 = 70 (kg).
Answer: 70 kg of fresh water.
Example 3. The motorcyclist stayed at the barrier for 24 minutes. After that, increasing his speed by 10 kilometers per hour, he atoned for being late on the stretch of 80 km. Determine the speed of the motorcyclist before decelerating (in kilometers per hour).
Solution: The task of drawing up an equation for speed. Let's denote the initial speed of the motorcyclist through V, and the time during which he had to drive through t. There are two unknowns, therefore there should be 2 equations too. According to the condition, during this time he had to travel 80 km.
V * t = 80 (km).
Delayed means that the time has decreased by 24 minutes. It is also worth noting that in such tasks, the time must be converted into hours or minutes (depending on the condition) and then solved. We draw up the equation of motion taking into account less time and higher speed
(V + 10) (t-24/60) = 80.
There are two equations for time and speed. Since the problems ask for speed, we express the time from the first equation and substitute it into the second
t = 80 / V;
(V + 10) (80 / V-24/60) = 80.
Our goal is to teach you how to compose equations for problems, from which you can determine the required quantities.
Therefore, without going into details, the resulting equation by multiplying by 60 * V and dividing by 24 can be reduced to the following quadratic equation
V ^ 2 + 10 * V-2000 = 0.
Find the discriminant and the roots of the equation yourself. You should get the value
V = -50;
V = 40.
We discard the first meaning, it has no physical meaning. The second V = 40 km / h is the desired speed of the motorcyclist.
Answer: 40 km / h.
Example 4. The freight train was delayed on the way for 12 minutes, and then, at a distance of 112 kilometers, it made up for lost time, increasing its speed by 10 km / h. Find the initial train speed (in km / h).
Solution: We have a problem in which the unknowns are the speed of the train V and the time of movement t.
Since the problem according to the scheme of equations corresponds to the previous one, we write two equations for the unknowns
V * t = 112;
(V + 10) * (t-12/60) = 112.
Equations should be drawn up in just such notation. This makes it possible to express in a simple form from the first equation the time
t = 112 / V
and, substituting into the second, obtain an equation only for the velocity
(V + 10) * (112 / V -12/60) = 112.
If it is unsuccessful to choose the notation, then you can get an equation for the unknowns of such a plan
V * (t + 12) = 112;
(V + 10) * t = 112.
Here t corresponds to the time after the increase in speed by 10 km / h, but this is not the point. The above equations are also correct, but not convenient from the point of view of calculations.
Try to solve the first two equations and the last and you will understand that the second scheme should be avoided when writing equations. Therefore, think carefully about what notation to introduce in order to minimize the amount of computation.
The resulting equation
(V + 10) * (112 / V -12/60) = 112.
reduce to a quadratic equation (multiply by 60 * V / 12)
V ^ 2 + 10 * V-5600 = 0.
Without going into intermediate calculations, the roots will be
V = -80;
V = 70.
In problems of this type, we always get a negative root (V = -80) that needs to be discarded. The train speed is 70 km / h.
Example 5. Leaving the bus station 10 minutes later, the bus drove to the first stop at a speed of 16 km / h more than scheduled and arrived on time. What speed (in km / h) should the bus have on schedule if the distance from the bus station to the first stop is 16 kilometers?
Solution: Bus speed V and time t are unknown.
We draw up an equation, taking into account that the time of delay is set in minutes, not hours
V * t = 16 - this is how the bus had to go in normal mode;
(V + 16) (t-10/60) = 16 - equation of motion due to late departure of the bus.
There are two equations and two unknowns.
From the first equation, we express the time and substitute into the second
t = 16 / V;
(V + 16) (16 / V-1/6) = 16.
The resulting equation for speed is reduced to a square (* 6 * V)
V ^ 2 + 16 * V-1536 = 0.
The roots of the quadratic equation are
V = 32; V = -48.
The required bus speed is 32 km / h.
Answer: 32 km / h.
Example 6. The driver of the car stopped for a wheel change for 12 minutes. After that, increasing the speed of movement by 15 km / h, he redeemed the time spent on 60 kilometers. At what speed (in km / h) did he move after stopping?
Solution: The algorithm for solving the problem was given several times in the previous examples. We denote the speed and time by V, t as standard.
Remember to convert minutes to hours when you come up with your equation. The system of equations will have the form
V * t = 60;
(V + 15) (t-12/60) = 60.
You should also know or memorize further manipulations.
t = 60 / V;
(V + 15) (60 / V -12/60) = 60.
This equation can be reduced to a quadratic equation
V ^ 2 + 15 * V-4500 = 0.
Having solved the quadratic equation, we obtain the following values of the speeds
V = 60; V = -75.
The speed is never negative, so the only correct answer is V = 60 km / h.
Example 7. Some two-digit number is 4 times the sum and 3 times the product of its digits. Find this number.
Solution: Problem with numbers occupy an important place among problems of drawing up equations and are no less interesting in building solutions than problems with speed. All that is needed is a good understanding of the condition of the problem. Let's denote the number through ab, that is, the number is equal to 10 * a + b. By condition, we compose the system of equations
10 * a + b = 4 * (a + b);
10 * a + b = 3 * a * b.
Since the unknowns enter the first equation linearly, we write it down and express one of the unknowns through the other
10 * a + b-4 * a-4 * b = 0;
6 * a-3 * b = 0; b = 2 * a.
Substitute b = 2 * a in the second equation
10 * a + 2 * a = 3 * a * 2 * a;
6 * a2-12 * a = 0; a (a-2) = 0.
Hence a = 0; a = 2. It makes no sense to consider the first value, for a = 2 the second digit is b = 2 * a = 2 * 2 = 4, and the required number is 24.
Answer: the number is 24.
Chemistry is the science of substances, their properties and transformations.
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That is, if nothing happens to the substances around us, then this does not apply to chemistry. But what does it mean, "nothing happens"? If a thunderstorm suddenly found us in the field, and we all got wet, as they say "to the skin," is it not a transformation: after all, the clothes were dry, but became wet.
If, for example, you take an iron nail, file it, and then collect iron filings (Fe) , then is it also not a transformation: there was a nail - there was a powder. But if after that you assemble the device and hold obtaining oxygen (O 2): heat potassium permanganate(KMpO 4) and collect oxygen in a test tube, and then put these iron filings red-hot "to red" in it, then they will flare up with a bright flame and after combustion will turn into a brown powder. And this is the same transformation. So where is the chemistry? Despite the fact that in these examples the shape (iron nail) and the condition of the clothes (dry, wet) change - these are not transformations. The fact is that the nail itself, as a substance (iron), remained with it, despite its different shape, and as our clothes absorbed the water from the rain, then it evaporated into the atmosphere. The water itself hasn't changed. So what are transformations in terms of chemistry?
From the point of view of chemistry, transformations are those phenomena that are accompanied by a change in the composition of a substance. Let's take the same nail as an example. It does not matter what form he took after processing with a file, but after collected from him iron filings placed in an atmosphere of oxygen - it turned into iron oxide(Fe 2 O 3 ) ... So something has changed after all? Yes, it has changed. There was a substance called a nail, but under the influence of oxygen, a new substance was formed - element oxide gland. Molecular Equation this transformation can be represented by the following chemical symbols:
4Fe + 3O 2 = 2Fe 2 O 3 (1)
For the uninitiated person in chemistry, questions immediately arise. What is "molecular equation", what is Fe? Why are the numbers "4", "3", "2" put? What are the small numbers "2" and "3" in the Fe 2 O 3 formula? This means it's time to sort things out in order.
Signs of chemical elements.
Despite the fact that chemistry begins to be studied in the 8th grade, and some even earlier, the great Russian chemist D.I.Mendeleev is known to many. And, of course, his famous "Periodic Table of Chemical Elements". Otherwise, more simply, it is called the "Periodic Table".
In this table, the elements are arranged in the appropriate order. To date, about 120 of them are known. The names of many elements were known to us for a long time. These are: iron, aluminum, oxygen, carbon, gold, silicon. Previously, we did not hesitate to use these words, identifying them with objects: an iron bolt, aluminum wire, oxygen in the atmosphere, a gold ring, etc. etc. But in fact, all these substances (bolt, wire, ring) are composed of the corresponding elements. The whole paradox is that the element cannot be touched, taken in hand. How so? They are in the periodic table, but you cannot take them! Yes exactly. A chemical element is an abstract (that is, abstract) concept, and is used in chemistry, however, as in other sciences, for calculations, drawing up equations, and solving problems. Each element differs from the other in that it has its own electronic configuration of an atom. The number of protons in the nucleus of an atom is equal to the number of electrons in its orbitals. For example, hydrogen is element # 1. Its atom consists of 1 proton and 1 electron. Helium is element number 2. Its atom consists of 2 protons and 2 electrons. Lithium - cell number 3. Its atom consists of 3 protons and 3 electrons. Darmstadty - element number 110. Its atom consists of 110 protons and 110 electrons.
Each element is designated by a certain symbol, in Latin letters, and has a certain reading in translation from Latin. For example, hydrogen has the symbol "N", reads "hydrogenium" or "ash". Silicon has the symbol "Si" read as "silicium". Mercury has the symbol "Нg" and reads like "hydrargirum". Etc. All these designations can be found in any chemistry textbook for the 8th grade. The main thing for us now is to understand that when drawing up chemical equations, it is necessary to operate with the indicated symbols of the elements.
Simple and complex substances.
Designating single symbols of chemical elements for various substances (Hg Mercury, Fe iron, Cu copper, Zn zinc, Al aluminum) we essentially denote simple substances, that is, substances consisting of atoms of the same type (containing the same number of protons and neutrons in an atom). For example, if the substances iron and sulfur interact, then the equation will take the following form of writing:
Fe + S = FeS (2)
Simple substances include metals (Ba, K, Na, Mg, Ag), as well as non-metals (S, P, Si, Cl 2, N 2, O 2, H 2). Moreover, one should turn
special attention to the fact that all metals are designated by single symbols: K, Ba, Ca, Al, V, Mg, etc., and non-metals - either by simple symbols: C, S, P or may have different indices that indicate their molecular structure: H 2, Cl 2, O 2, J 2, P 4, S 8. In the future, this will be very important in the preparation of equations. It is not at all difficult to guess that complex substances are substances formed from atoms of different types, for example,
1). Oxides:
aluminium oxide Al 2 O 3, 
sodium oxide Na 2 O,
copper oxide CuO,
zinc oxide ZnO,
titanium oxide Ti 2 O 3,
carbon monoxide or carbon monoxide (+2) CO,
sulfur oxide (+6) SO 3
2). Reasons:
iron hydroxide(+3) Fe (OH) 3,
copper hydroxide Cu (OH) 2,
potassium hydroxide or potassium alkali KOH,
sodium hydroxide NaOH.
3). Acids:
hydrochloric acid HCl,
sulfurous acid H 2 SO 3,
Nitric acid HNO 3
4). Salts:
sodium thiosulfate Na 2 S 2 O 3,
sodium sulfate or Glauber's salt Na 2 SO 4,
calcium carbonate or limestone CaCO 3,
copper chloride CuCl 2
5). Organic matter:
sodium acetate CH 3 COONa,
methane CH 4,
acetylene C 2 H 2,
glucose S 6 N 12 O 6
Finally, after we have figured out the structure of various substances, we can begin to draw up chemical equations.
Chemical equation.
The word "equation" itself is derived from the word "equalize", i.e. divide something into equal parts. In mathematics, equations are almost the very essence of this science. For example, you can give such a simple equation in which the left and right sides are equal to "2":
40: (9 + 11) = (50 x 2): (80 - 30);
And in chemical equations the same principle: the left and right sides of the equation must correspond to the same number of atoms, the elements participating in them. Or, if the ionic equation is given, then it contains number of particles must also comply with this requirement. A chemical equation is a conditional notation of a chemical reaction using chemical formulas and mathematical signs. A chemical equation in its essence reflects one or another chemical reaction, that is, the process of interaction of substances, in the process of which new substances arise. For example, you need write a molecular equation reactions involved barium chloride BaCl 2 and sulphuric acid H 2 SO 4. As a result of this reaction, an insoluble precipitate is formed - barium sulfateВаSO 4 and hydrochloric acidНСl:
ВаСl 2 + H 2 SO 4 = BaSO 4 + 2HCl (3)
First of all, it is necessary to understand that the large number "2" in front of the HCl substance is called the coefficient, and the small numbers "2", "4" under the formulas BaCl 2, H 2 SO 4, BaSO 4 are called indices. And the coefficients and indices in chemical equations play the role of factors, not terms. To correctly write a chemical equation, you need arrange the coefficients in the reaction equation... Now let's start counting the atoms of the elements on the left and right sides of the equation. On the left side of the equation: the substance BaCl 2 contains 1 barium atom (Ba), 2 chlorine atoms (Cl). In the substance H 2 SO 4: 2 hydrogen atoms (H), 1 sulfur atom (S) and 4 oxygen atoms (O). On the right side of the equation: in the substance BaSO 4 there is 1 barium atom (Ba), 1 sulfur atom (S) and 4 oxygen atoms (O), in the HCl substance: 1 hydrogen atom (H) and 1 chlorine atom (Cl). Whence it follows that on the right side of the equation the number of hydrogen and chlorine atoms is half that on the left side. Therefore, in front of the HCl formula on the right side of the equation, you must put the coefficient "2". If we now add up the number of atoms of the elements participating in this reaction, both on the left and on the right, we get the following balance:
In both sides of the equation, the numbers of atoms of the elements participating in the reaction are equal, therefore, it is drawn up correctly.
Chemical equation and chemical reactions
As we have already found out, chemical equations are a reflection of chemical reactions. Chemical reactions are such phenomena in the process of which some substances are transformed into others. Among their diversity, two main types can be distinguished:
1). Compound reactions
2). Decomposition reactions.
In the overwhelming majority, chemical reactions belong to addition reactions, since changes in its composition can rarely occur with a single substance if it is not exposed to external influences (dissolution, heating, the action of light). Nothing characterizes a chemical phenomenon or reaction so much as the changes that occur when two or more substances interact. Such phenomena can occur spontaneously and be accompanied by an increase or decrease in temperature, light effects, a change in color, the formation of sediment, the release of gaseous products, and noise.
For clarity, we present several equations reflecting the reaction processes of the compound, in the course of which we obtain sodium chloride(NaCl), zinc chloride(ZnCl 2), silver chloride precipitate(AgCl), aluminum chloride(AlCl 3)
Cl 2 + 2Nа = 2NaCl (4)
СuCl 2 + Zn = ZnCl 2 + Сu (5)
AgNO 3 + КCl = AgCl + 2KNO 3 (6)
3HCl + Al (OH) 3 = AlCl 3 + 3H 2 O (7)
Among the reactions of the compound, the following should be specially noted : substitutions (5), exchange (6), and as a special case of the exchange reaction - the reaction neutralization (7).
Substitution reactions include those in which the atoms of a simple substance replace the atoms of one of the elements in a complex substance. In example (5), zinc atoms are replaced from the CuCl 2 solution for copper atoms, while zinc passes into the soluble ZnCl 2 salt, and copper is released from the solution in the metallic state.
Exchange reactions include those reactions in which two complex substances exchange their constituent parts. In the case of reaction (6), the soluble salts of AgNO 3 and KCl, when both solutions are merged, form an insoluble precipitate of the AgCl salt. At the same time, they exchange their constituent parts - cations and anions. Potassium cations K + are attached to NO 3 anions, and silver cations Ag + - to Cl - anions.
The neutralization reaction belongs to a special, particular case of exchange reactions. Neutralization reactions are reactions in which acids react with bases to form salt and water. In example (7), hydrochloric acid HCl, reacting with the base Al (OH) 3, forms the AlCl 3 salt and water. In this case, the aluminum cations Al 3+ from the base exchange with the Cl anions from the acid. As a result, it happens neutralization of hydrochloric acid.
Decomposition reactions include those in which two or more new simple or complex substances are formed from one complex, but of a simpler composition. As reactions can be cited those in the process of which decompose 1). Potassium nitrate(KNO 3) with the formation of potassium nitrite (KNO 2) and oxygen (O 2); 2). Potassium permanganate(KMnO 4): potassium manganate (K 2 MnO 4) is formed, manganese oxide(MnO 2) and oxygen (O 2); 3). Calcium carbonate or marble; in the process are formed carbonicgas(CO 2) and calcium oxide(CaO)
2КNO 3 = 2КNO 2 + O 2 (8)
2KMnO 4 = K 2 MnO 4 + MnO 2 + O 2 (9)
CaCO 3 = CaO + CO 2 (10)
In reaction (8), one complex substance and one simple substance are formed from a complex substance. In reaction (9) - two complex and one simple. In reaction (10) - two complex substances, but simpler in composition
All classes of complex substances undergo decomposition:
1). Oxides: silver oxide 2Ag 2 O = 4Ag + O 2 (11)
2). Hydroxides: iron hydroxide 2Fe (OH) 3 = Fe 2 O 3 + 3H 2 O (12)
3). Acids: sulphuric acid H 2 SO 4 = SO 3 + H 2 O (13)
4). Salts: calcium carbonate CaCO 3 = CaO + CO 2 (14)
5). Organic matter: alcoholic fermentation of glucose
C 6 H 12 O 6 = 2C 2 H 5 OH + 2CO 2 (15)
According to another classification, all chemical reactions can be divided into two types: reactions proceeding with the release of heat, they are called exothermic, and reactions proceeding with the absorption of heat - endothermic. The criterion for such processes is thermal effect of reaction. As a rule, exothermic reactions include oxidation reactions, i.e. interactions with oxygen, for example combustion of methane:
CH 4 + 2O 2 = CO 2 + 2H 2 O + Q (16)
and to endothermic reactions - decomposition reactions, already cited above (11) - (15). The Q sign at the end of the equation indicates whether heat is released during the reaction (+ Q) or absorbed (-Q):
CaCO 3 = CaO + CO 2 - Q (17)
You can also consider all chemical reactions by the type of change in the oxidation state of the elements involved in their transformations. For example, in reaction (17), the elements participating in it do not change their oxidation states:
Ca +2 C +4 O 3 -2 = Ca +2 O -2 + C +4 O 2 -2 (18)
And in reaction (16), the elements change their oxidation states:
2Mg 0 + O 2 0 = 2Mg +2 O -2
Reactions of this type are related to redox ... They will be considered separately. To draw up equations for reactions of this type, it is necessary to use half-reaction method and apply electronic balance equation.
After bringing various types of chemical reactions, you can proceed to the principle of drawing up chemical equations, otherwise, the selection of coefficients in the left and right sides.
Mechanisms for drawing up chemical equations.
Whatever type this or that chemical reaction belongs to, its record (chemical equation) must correspond to the condition of the equality of the number of atoms before and after the reaction.
There are equations (17) that do not require adjustment, i.e. placement of coefficients. But in most cases, as in examples (3), (7), (15), it is necessary to take actions aimed at equalizing the left and right sides of the equation. What principles should be followed in such cases? Is there any system in the selection of coefficients? There is, and not one. These systems include:
1). Selection of coefficients according to the given formulas.
2). Compilation of the valences of reactants.
3). Compilation of the oxidation states of the reactants.
In the first case, it is assumed that we know the formulas of the reacting substances both before and after the reaction. For example, given the following equation:
N 2 + О 2 → N 2 О 3 (19)
It is generally accepted that until equality is established between the atoms of the elements before and after the reaction, the equal sign (=) is not put in the equation, but is replaced by an arrow (→). Now let's get down to the actual adjustment. On the left side of the equation there are 2 nitrogen atoms (N 2) and two oxygen atoms (O 2), and on the right side there are two nitrogen atoms (N 2) and three oxygen atoms (O 3). It is not necessary to equalize it in terms of the number of nitrogen atoms, but in terms of oxygen it is necessary to achieve equality, since before the reaction two atoms took part, and after the reaction there were three atoms. Let's make the following scheme:
before reaction after reaction
О 2 О 3
Let's define the smallest multiple between the given numbers of atoms, it will be "6".
О 2 О 3
\ 6 /
Divide this number on the left side of the equation for oxygen by "2". We get the number "3", put it in the equation to be solved:
N 2 + 3O 2 → N 2 O 3
Let's also divide the number "6" for the right side of the equation by "3". We get the number "2", just put it in the equation to be solved:
N 2 + 3O 2 → 2N 2 O 3
The numbers of oxygen atoms in both the left and right sides of the equation became equal, respectively, 6 atoms each:
But the number of nitrogen atoms in both sides of the equation will not match:
On the left, there are two atoms, on the right, four atoms. Therefore, in order to achieve equality, it is necessary to double the amount of nitrogen on the left side of the equation by setting the coefficient "2":
Thus, the equality in nitrogen is observed and in general, the equation will take the form:
2N 2 + 3O 2 → 2N 2 O 3
Now in the equation you can put an equal sign instead of an arrow:
2N 2 + 3О 2 = 2N 2 О 3 (20)
Let's take another example. The following reaction equation is given:
P + Cl 2 → PCl 5
The left side of the equation has 1 phosphorus (P) and two chlorine (Cl 2) atoms, and the right side has one phosphorus (P) and five oxygen (Cl 5) atoms. It is not necessary to equalize it in terms of the number of phosphorus atoms, but in terms of chlorine it is necessary to achieve equality, since before the reaction two atoms took part, and after the reaction there were five atoms. Let's make the following scheme:
before reaction after reaction
Cl 2 Cl 5
Let's define the smallest multiple between the given number of atoms, it will be "10".
Cl 2 Cl 5
\ 10 /
Divide this number on the left side of the chlorine equation by "2". We get the number "5", put it in the equation to be solved:
Р + 5Cl 2 → РCl 5
Let's also divide the number "10" for the right side of the equation by "5". We get the number "2", just put it in the equation to be solved:
Р + 5Cl 2 → 2РCl 5
The number of chlorine atoms in both the left and right sides of the equation became equal, respectively, 10 atoms:
But the number of phosphorus atoms in both sides of the equation will not match:
Therefore, in order to achieve equality, it is necessary to double the amount of phosphorus on the left side of the equation by setting the coefficient "2":
Thus, the equality for phosphorus is met and, in general, the equation will take the form:
2Р + 5Cl 2 = 2РCl 5 (21)
When drawing up equations by valence must be given determination of valence and set values for the most famous elements. Valence is one of the previously used concepts that is currently not used in a number of school curricula. But with its help it is easier to explain the principles of drawing up the equations of chemical reactions. Valence is understood the number of chemical bonds that one or another atom can form with another, or other atoms ... Valence has no sign (+ or -) and is indicated by Roman numerals, usually above the symbols of chemical elements, for example:
Where do these values come from? How can they be applied to formulate chemical equations? The numerical values of the valencies of the elements coincide with their group number of the Periodic Table of Chemical Elements of D. I. Mendeleev (Table 1).
For other elements valency values may have different meanings, but never more than the number of the group in which they are located. Moreover, for even numbers of groups (IV and VI), the valencies of the elements take only even values, and for odd ones, they can have both even and odd values (Table 2).
Of course, there are exceptions in the values of valencies for some elements, but in each specific case, these points are usually specified. Now we will consider the general principle of drawing up chemical equations for given valences for certain elements. Most often, this method is acceptable in the case of drawing up the equations of chemical reactions of the compound of simple substances, for example, when interacting with oxygen ( oxidation reactions). Let's say you want to display the oxidation reaction aluminum... But let us recall that metals are designated by single atoms (Al), and non-metals in a gaseous state - with indices "2" - (O 2). First, let's write a general reaction scheme:
Al + О 2 → AlО
At this stage, it is not yet known what the correct spelling should be for aluminum oxide. And it is at this stage that knowledge of the valencies of the elements will come to our aid. For aluminum and oxygen, we put them above the assumed formula of this oxide:
III II
Al O
After that, we put the corresponding indices at the bottom of these symbols of the elements "cross" -on- "cross":
III II
Al 2 O 3
Composition of a chemical compound Al 2 O 3 is determined. The further scheme of the reaction equation will take the form:
Al + О 2 → Al 2 О 3
It remains only to equalize the left and right parts of it. We will proceed in the same way as in the case of drawing up equation (19). We will equalize the number of oxygen atoms, resorting to finding the smallest multiple:
before reaction after reaction
О 2 О 3
\ 6 /
Divide this number on the left side of the equation for oxygen by "2". We get the number "3", put it in the equation to be solved. Let's also divide the number "6" for the right side of the equation by "3". We get the number "2", just put it in the equation to be solved:
Al + 3О 2 → 2Al 2 О 3
To achieve equality for aluminum, it is necessary to adjust its amount on the left side of the equation by setting the coefficient "4":
4Al + 3О 2 → 2Al 2 О 3
Thus, the equality for aluminum and oxygen is observed and in general, the equation will take its final form:
4Al + 3О 2 = 2Al 2 О 3 (22)
Applying the method of valences, it is possible to predict what substance is formed in the course of a chemical reaction, how its formula will look. Suppose that nitrogen and hydrogen have entered into the reaction of the compound with the corresponding valences III and I. Let us write the general reaction scheme:
N 2 + H 2 → NH
For nitrogen and hydrogen, we put down the valences over the assumed formula of this compound:
As before, we put the corresponding indices at the bottom of these symbols of the elements:
III I
N N 3
The further scheme of the reaction equation will take the form:
N 2 + H 2 → NH 3
Equalizing by the already known method, through the smallest multiple for hydrogen, equal to "6", we obtain the desired coefficients, and the equation as a whole:
N 2 + 3H 2 = 2NH 3 (23)
When drawing up equations for oxidation states of reacting substances, it is necessary to recall that the number of electrons received or given up in the process of a chemical reaction is called the oxidation state of an element. Oxidation state in compounds basically, numerically coincides with the values of the element's valencies. But they differ in sign. For example, for hydrogen, the valency is I, and the oxidation state is (+1) or (-1). For oxygen, the valence is II and the oxidation state is (-2). For nitrogen, the valences are I, II, III, IV, V, and the oxidation states are (-3), (+1), (+2), (+3), (+4), (+5), etc. ... The oxidation states of the elements most commonly used in the equations are shown in Table 3.
In the case of compound reactions, the principle of drawing up equations by oxidation states is the same as when drawing up by valences. For example, let's give the equation for the oxidation of chlorine with oxygen, in which chlorine forms a compound with an oxidation state of +7. Let's write the proposed equation:
Cl 2 + О 2 → ClО
We put the oxidation states of the corresponding atoms over the putative ClO compound:
As in the previous cases, we will establish that the desired compound formula will take the form:
7 -2
Cl 2 O 7
The reaction equation will take the following form:
Cl 2 + О 2 → Cl 2 О 7
By equalizing for oxygen, finding the smallest multiple between two and seven, equal to "14", we will establish as a result equality:
2Cl 2 + 7О 2 = 2Cl 2 О 7 (24)
A slightly different method must be used with oxidation states in the preparation of exchange, neutralization, and substitution reactions. In some cases, it is difficult to find out: what compounds are formed during the interaction of complex substances?
How do you know: what happens during the reaction?
Indeed, how do you know: what reaction products can arise in the course of a particular reaction? For example, what is formed when barium nitrate and potassium sulfate react?
Ва (NO 3) 2 + К 2 SO 4 →?
Maybe BaK 2 (NO 3) 2 + SO 4? Or Ba + NO 3 SO 4 + K 2? Or something else? Of course, in the course of this reaction, compounds are formed: BaSO 4 and KNO 3. How is this known? And how to write formulas of substances correctly? Let's start with what is most often overlooked: the very concept of "exchange reaction". This means that during these reactions, substances change with each other in constituent parts. Since exchange reactions are mostly carried out between bases, acids or salts, the parts by which they will change are metal cations (Na +, Mg 2+, Al 3+, Ca 2+, Cr 3+), H + ions or OH -, anions - acid residues, (Cl -, NO 3 2-, SO 3 2-, SO 4 2-, CO 3 2-, PO 4 3-). In general terms, the exchange reaction can be summarized in the following entry:
Kt1An1 + Kt2An1 = Kt1An2 + Kt2An1 (25)
Where Kt1 and Kt2 are metal cations (1) and (2), and An1 and An2 are the corresponding anions (1) and (2). In this case, it must be taken into account that in the compounds before and after the reaction, cations are always established in the first place, and anions - in the second. Therefore, if the reaction enters potassium chloride and silver nitrate, both in a dissolved state
KCl + AgNO 3 →
then in the process of it the substances KNO 3 and AgCl are formed and the corresponding equation will take the form:
KCl + AgNO 3 = KNO 3 + AgCl (26)
In neutralization reactions, protons from acids (H +) will combine with hydroxyl anions (OH -) to form water (H 2 O):
НCl + KOH = КCl + Н 2 O (27)
The oxidation states of metal cations and the charges of the anions of acid residues are indicated in the table of solubility of substances (acids, salts and bases in water). The horizontal shows metal cations, and the vertical shows the anions of acid residues.
Proceeding from this, when drawing up the equation of the exchange reaction, it is first necessary to establish the oxidation state of the particles receiving in this chemical process in the left part of it. For example, if you want to write the equation of the interaction between calcium chloride and sodium carbonate, let's make the initial scheme of this reaction:
CaCl + NaCO 3 →
Ca 2+ Cl - + Na + CO 3 2- →
Having performed the already known action "cross" - on - "cross", we determine the real formulas of the initial substances:
CaCl 2 + Na 2 CO 3 →
Based on the principle of the exchange of cations and anions (25), we establish the preliminary formulas for the substances formed during the reaction:
CaCl 2 + Na 2 CO 3 → CaCO 3 + NaCl
We put the corresponding charges over their cations and anions:
Ca 2+ CO 3 2- + Na + Cl -
Formulas of substances written correctly, in accordance with the charges of cations and anions. Let's compose a complete equation by equating the left and right sides of it for sodium and chlorine:
CaCl 2 + Na 2 CO 3 = CaCO 3 + 2NаCl (28)
As another example, we give the equation of the neutralization reaction between barium hydroxide and phosphoric acid:
ВаОН + НРО 4 →
We put the corresponding charges over the cations and anions:
Ba 2+ OH - + H + PO 4 3- →
Let's define the real formulas of the starting materials:
Ва (ОН) 2 + Н 3 РО 4 →
Based on the principle of the exchange of cations and anions (25), we will establish the preliminary formulas of the substances formed during the reaction, taking into account that during the exchange reaction, one of the substances must necessarily be water:
Ва (ОН) 2 + Н 3 РО 4 → Ва 2+ РО 4 3- + Н 2 O
Let us determine the correct record of the salt formula formed during the reaction:
Ba (OH) 2 + H 3 PO 4 → Ba 3 (PO 4) 2 + H 2 O
Let's equalize the left side of the equation for barium:
3Ва (ОН) 2 + Н 3 РО 4 → Ва 3 (РО 4) 2 + Н 2 O
Since on the right side of the equation the remainder of phosphoric acid is taken twice, (PO 4) 2, then on the left it is also necessary to double its amount:
3Ва (ОН) 2 + 2Н 3 РО 4 → Ва 3 (РО 4) 2 + Н 2 O
It remains to match the number of hydrogen and oxygen atoms on the right side of the water. Since the total number of hydrogen atoms on the left is 12, then on the right it must also correspond to twelve, therefore, before the water formula, it is necessary put coefficient"6" (since there are already 2 hydrogen atoms in the water molecule). For oxygen, equality is also observed: on the left 14 and on the right 14. So, the equation has the correct form of writing:
3Ва (ОН) 2 + 2Н 3 РО 4 → Ва 3 (РО 4) 2 + 6Н 2 O (29)
Possibility of carrying out chemical reactions
The world consists of a great variety of substances. The number of variants of chemical reactions between them is also incalculable. But can we, having written this or that equation on paper, claim that a chemical reaction will correspond to it? There is a misconception that if correct place odds in the equation, then it will be feasible in practice. For example, if you take sulfuric acid solution and put into it zinc, then you can observe the process of hydrogen evolution:
Zn + H 2 SO 4 = ZnSO 4 + H 2 (30)
But if copper is lowered into the same solution, then the process of gas evolution will not be observed. The reaction is not feasible.
Cu + H 2 SO 4 ≠
If concentrated sulfuric acid is taken, it will react with copper:
Cu + 2H 2 SO 4 = CuSO 4 + SO 2 + 2H 2 O (31)
In reaction (23) between nitrogen and hydrogen gases, thermodynamic equilibrium, those. how many molecules ammonia NH 3 is formed per unit of time, the same amount of them and decompose back into nitrogen and hydrogen. Chemical equilibrium shift can be achieved by increasing the pressure and lowering the temperature
N 2 + 3H 2 = 2NH 3
If you take potassium hydroxide solution and add to it sodium sulfate solution, then no changes will be observed, the reaction will not be feasible:
KOH + Na 2 SO 4 ≠
Sodium chloride solution when interacting with bromine, it will not form bromine, despite the fact that this reaction can be attributed to a substitution reaction:
NaCl + Br 2 ≠
What are the reasons for such inconsistencies? The fact is that it turns out to be not enough just to correctly define compound formulas, it is necessary to know the specifics of the interaction of metals with acids, to skillfully use the table of solubility of substances, to know the rules of substitution in the row of activity of metals and halogens. This article outlines only the most basic principles of how arrange the coefficients in the reaction equations, how write molecular equations, how determine the composition of a chemical compound.
Chemistry, as a science, is extremely diverse and multifaceted. This article reflects only a small part of the processes occurring in the real world. Types, thermochemical equations, electrolysis, organic synthesis processes and much, much more. But more on that in the following articles.
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The solution to the problem usually comes down to finding the value of some quantity by logical reasoning and calculations. For example, find the speed, time, distance, mass of an object, or the amount of something.
This problem can be solved using an equation. To do this, the desired value is denoted through a variable, then, by logical reasoning, an equation is compiled and solved. Having solved the equation, a check is made to see whether the solution to the equation satisfies the conditions of the problem.
Lesson contentWriting expressions containing the unknown
Solving the problem is accompanied by drawing up an equation for this problem. At the initial stage of studying tasks, it is advisable to learn how to compose letter expressions describing a particular life situation. This stage is not difficult and can be studied in the process of solving the problem itself.
Let's consider several situations that can be written using a mathematical expression.
Problem 1... Father's age x years. Mom is two years younger. The son is 3 times younger than the father. Write down the age of each using expressions.
Solution:
Task 2... Father's age x years old, mother is 2 years younger than father. The son is 3 times younger than the father, the daughter is 3 times younger than the mother. Write down the age of each using expressions.
Solution:
Problem 3... Father's age x years old, mother is 3 years younger than father. The son is 3 times younger than the father, the daughter is 3 times younger than the mother. How old is each if the total age of father, mother, son and daughter is 92?
Solution:
In this task, in addition to writing expressions, it is necessary to calculate the age of each family member.
First, let's write down the ages of each family member using expressions. Per variable x Let's take the father's age, and then, using this variable, compose the rest of the expressions:
Now let's determine the age of each family member. To do this, we need to create and solve an equation. We have all the components of the equation ready. It remains only to collect them together.
The total age of 92 was obtained by adding the ages of dad, mom, son and daughter:
For each age, we put together a mathematical expression. These expressions will be the components of our equation. Let's put our equation together according to this diagram and the table that was given above. That is, we replace the words dad, mom, son, daughter with the expression corresponding to them in the table:
Mom's age expression x - 3, for clarity, it was bracketed.
Now let's solve the resulting equation. To begin with, you can expand the brackets where possible:
To free the equation from fractions, multiply both sides by 3
Let's solve the resulting equation using the well-known identity transformations:
We found the value of the variable x... This variable was responsible for the father's age. So the father's age is 36 years old.
Knowing the father's age, you can calculate the ages of the rest of the family. To do this, you need to substitute the value of the variable x in those expressions that are responsible for the age of a particular family member.
In the problem it was said that the mother is 3 years younger than the father. We designated her age through the expression x − 3. Variable value x now it is known, and to calculate the age of the mother, you need in the expression x - 3 instead of x substitute the found value 36
x - 3 = 36 - 3 = 33 years old mother.
The age of the rest of the family members is determined in a similar way:
Examination:
Problem 4... A kilogram of apples is worth x rubles. Write down an expression that calculates how many kilograms of apples you can buy for 300 rubles.
Solution
If a kilogram of apples is worth x rubles, then 300 rubles can buy a kilogram of apples.
Example... A kilogram of apples costs 50 rubles. Then you can buy 300 rubles, that is, 6 kilograms of apples.
Problem 5... On x rubles, 5 kg of apples were bought. Write down an expression that calculates how many rubles are one kilogram of apples.
Solution
If for 5 kg of apples was paid x rubles, then one kilogram will be worth rubles
Example... For 300 rubles, 5 kg of apples were bought. Then one kilogram of apples will be worth, that is, 60 rubles.
Problem 6... Tom, John and Leo went to the cafeteria during recess and bought a sandwich and a mug of coffee. The sandwich is worth x rubles, and a mug of coffee - 15 rubles. Determine the cost of a sandwich if you know that 120 rubles were paid for everything?
Solution
Of course, this problem is as simple as three kopecks and it can be solved without resorting to an equation. To do this, from 120 rubles, you need to subtract the cost of three cups of coffee (15 × 3), and divide the result by 3
But our goal is to form an equation for the problem and solve this equation. So the cost of a sandwich x rubles. Only three of them were bought. So, having increased the cost three times, we get an expression describing how many rubles were paid for three sandwiches
3x - cost of three sandwiches
And the cost of three cups of coffee can be written down as 15 × 3. 15 is the cost of one cup of coffee, and 3 is a multiplier (Tom, John and Leo) that triples that cost.
According to the condition of the problem, 120 rubles were paid for everything. We already have a rough diagram of what needs to be done:
We already have expressions describing the cost of three sandwiches and three cups of coffee. These are expressions 3 x and 15 × 3. Using the scheme, we compose an equation and solve it:
So, the cost of one sandwich is 25 rubles.
The problem is solved correctly only if the equation for it is composed correctly. Unlike ordinary equations, by which we learn to find roots, equations for solving problems have their own specific applications. Each component of such an equation can be described verbally. When composing an equation, it is imperative to understand why we include this or that component in its composition and why it is needed.
It is also necessary to remember that the equation is an equality, after the solution of which the left side will have to be equal to the right side. The equation you make should not contradict this idea.
Suppose the equation is a scale with two bowls and a screen showing the state of the scale.
The screen is currently showing an equal sign. It is clear why the left cup is equal to the right cup - there is nothing on the cups. We write down the state of the scales and the absence of anything on the bowls using the following equality:
0 = 0
Put a watermelon on the left side of the scale:
The left bowl outweighed the right bowl and the screen sounded the alarm, showing a not equal sign (≠). This sign indicates that the left cup is not equal to the right cup.
Now let's try to solve the problem. Suppose you want to know how much a watermelon weighs, which lies on the left bowl. But how do you know? After all, our scales are designed only to check whether the left pan is equal to the right one.
Equations come to the rescue. Recall that the equation by definition is equality containing the variable whose value you want to find. The balance in this case plays the role of this very equation, and the mass of the watermelon is a variable, the value of which must be found. Our goal is to correctly form this equation. Understand, align the scales so that you can calculate the mass of the watermelon.
You can place a heavy object on the right pan to level the scale. For example, let's put a 7 kg weight there.
Now, on the contrary, the right bowl outweighed the left one. The screen still shows that the bowls are not equal.
Let's try to put a weight of 4 kg on the left bowl
The scales are now leveled. The picture shows that the left bowl is at the same level as the right bowl. And the screen shows an equal sign. This sign indicates that the left cup is equal to the right cup.
Thus, we got an equation - an equality containing an unknown. The left bowl is the left side of the equation with 4 components and a variable x(the mass of a watermelon), and the right bowl is the right side of the equation, consisting of component 7.
Well, it's not hard to guess that the root of the equation 4 + x= 7 is equal to 3. So the mass of the watermelon is 3 kg.
The situation is similar with other tasks. To find some unknown value, various elements are added to the left or right side of the equation: terms, factors, expressions. In school problems, these elements are already given. It only remains to structure them correctly and build an equation. In this example, we were engaged in selection, trying weights of different masses in order to calculate the mass of a watermelon.
Naturally, the data that are given in the problem must first be reduced to a form in which they can be included in the equation. Therefore, as they say "Like it or not, but you have to think".
Consider the following problem. The age of the father is equal to the age of the son and daughter together. The son is twice his daughter's age and twenty years younger than his father. How old is everyone?
The daughter's age can be denoted by x... If the son is twice as old as the daughter, then his age will be designated as 2 x... In the condition of the problem it is said that together the age of the daughter and the son is equal to the age of the father. This means that the age of the father will be denoted by the sum x + 2x
The expression can contain similar terms. Then the father's age will be designated as 3 x
Now let's make the equation. We need to get the equality in which we can find the unknown x... Let's use weights. On the left side we put the father's age (3 x), and on the right side is the age of the son (2 x)
It is clear why the left bowl outweighed the right one and why the screen shows a sign (≠). After all, it is logical that the age of the father is greater than the age of the son.
But we need to equalize the scales so that we can calculate the unknown x... To do this, add some number to the right bowl. What number is indicated in the problem. The condition stated that the son was 20 years younger than his father. So 20 years is the very number that needs to be put on the scales.
The scales will even out if we add these 20 years to the right side of the scale. In other words, we will raise a son to the age of a father.
The scales are now leveled. The equation turned out 
, which can be easily solved:
x we marked the age of the daughter. Now we have found the value of this variable. Daughter is 20 years old.
And finally, let's calculate the age of the father. In the problem, it was said that it is equal to the sum of the ages of the son and daughter, that is, (20 + 40) years.
Let's go back to the middle of the problem and pay attention to one point. When we put the age of the father and the age of the son on the scales, the left pan outweighed the right
But we solved this problem by adding another 20 years to the right cup. As a result, the scales leveled off and we got the equality
But it was possible not to add these 20 years to the right cup, but to subtract them from the left. We would get equality and in that case
This time we get the equation 
... The root of the equation is still 20
That is, the equations and are equivalent. But we remember that the roots of equivalent equations coincide. If you look closely at these two equations, you can see that the second equation is obtained by transferring the number 20 from the right side to the left with the opposite sign. And this action, as indicated in the previous lesson, does not change the roots of the equation.
You also need to pay attention to the fact that at the beginning of solving the problem, the ages of each family member could be indicated through other expressions.
Let's say the age of the son is denoted by x and since he is two older than his daughter, then designate the daughter's age by (understand to make her twice younger than the son). And the age of the father, since it is the sum of the ages of the son and daughter, denote through the expression. And finally, to build a logically correct equation, add the number 20 to the son's age, because the father is twenty years older. The result is a completely different equation 
... Let's solve this equation
As you can see, the answers to the problem have not changed. The son is still 40 years old. The daughter is still years old, and the father is 40 + 20 years old.
In other words, the problem can be solved by different methods. Therefore, one should not despair that it is not possible to solve this or that problem. But you need to keep in mind that there are the simplest ways to solve the problem. You can get to the city center by different routes, but there is always the most convenient, fastest and safest route.
Examples of problem solving
Objective 1. There are only 30 notebooks in two packs. If 2 notebooks were transferred from the first pack to the second, then the first pack would have twice as many notebooks as the second. How many notebooks were there in each bundle?
Solution
Let us denote by x the number of notebooks that were in the first bundle. If there were 30 notebooks in total, and the variable x this is the number of notebooks from the first pack, then the number of notebooks in the second pack will be denoted by the expression 30 - x... That is, from the total number of notebooks we subtract the number of notebooks from the first pack and thus get the number of notebooks from the second pack.
and add these two notebooks to the second pack
Let's try to form an equation from the existing expressions. We put both bundles of notebooks on the scales
The left cup is heavier than the right. This is because the statement of the problem says that after two notebooks were taken from the first bundle and placed in the second, the number of notebooks in the first bundle became twice as large as in the second.
To align the scales and get the equation, double the right side. To do this, multiply it by 2
The equation turns out. Let's solve this equation:
We denoted the first pack through the variable x... Now we have found its meaning. Variable x equals 22. This means that there were 22 notebooks in the first pack.
And we denoted the second pack through the expression 30 - x and since the meaning of the change x now it is known, then you can calculate the number of notebooks in the second pack. It is equal to 30 - 22, that is, 8 pieces.
Task 2... Two people were peeling potatoes. One peeled two potatoes per minute, and the other three potatoes. Together they cleared 400 pieces. How long did each one work if the other worked 25 minutes more than the first?
Solution
Let us denote by x working time of the first person. Since the second person worked 25 minutes more than the first, then his time will be indicated through the expression
The first worker peeled 2 potatoes per minute, and since he worked x minutes, then in total he cleared 2 x potatoes.
The second person peeled three potatoes per minute, and since he worked for minutes, he peeled potatoes in total.
Together they peeled 400 potatoes
Let's compose and solve the equation from the available components. On the left side of the equation will be the potatoes peeled by each person, and on the right side will be their sum:
At the beginning of solving this problem through the variable x we have designated the working time of the first person. Now we have found the value of this variable. The first person worked for 65 minutes.
And the second person worked for minutes, and since the value of the variable x now it is known, then you can calculate the working time of the second person - it is equal to 65 + 25, that is, 90 minutes.
Problem from the Algebra Textbook by Andrey Petrovich Kiselev... A mixture of 32 kg is made from teas. A kilogram of the first grade costs 8 rubles, and the second grade costs 6 rubles. 50 kopecks How many kilograms are taken of both varieties, if a kilogram of the mixture costs (without profit and loss) 7 rubles. 10 kopecks?
Solution
Let us denote by x mass of first grade tea. Then the mass of the second grade tea will be denoted by the expression 32 - x
A kilogram of first grade tea costs 8 rubles. If these eight rubles are multiplied by the number of kilograms of first grade tea, then you can find out how many rubles cost x kg of first grade tea.
A kilogram of second-grade tea costs 6 rubles. 50 kopecks If these 6 rubles. 50 kopecks multiply by 32 - x, then you can find out how many rubles cost 32 - x kg of tea of the second grade.
The condition says that a kilogram of the mixture costs 7 rubles. 10 kopecks A total of 32 kg of the mixture was prepared. Multiply 7 rubles. 10 kopecks at 32 we can find out how much 32 kg of the mixture costs.
The expressions from which we will compose the equation now take the following form:
Let's try to form an equation from the existing expressions. Let's put on the left pan of the scales the cost of mixtures of first and second grade tea, and on the right pan we'll put the cost of 32 kg of the mixture, that is, the total cost of the mixture, which includes both varieties of tea:
At the beginning of solving this problem through the variable x we have designated the mass of the first grade tea. Now we have found the value of this variable. Variable x is equal to 12.8. This means that 12.8 kg of first grade tea was taken to prepare the mixture.
And through expression 32 - x we designated the mass of tea of the second grade and since the value of the change x now we know, we can calculate the mass of the second grade tea. It is equal to 32 - 12.8 that is 19.2. This means that 19.2 kg of second-grade tea was taken to prepare the mixture.
Problem 3... The cyclist covered some distance at a speed of 8 km / h. He had to return by another road, which was 3 km longer than the first, and, although returning, he drove at a speed of 9 km / h, he used up more minutes of time. How long were the roads?
Solution
Some tasks may address topics that the person may not have studied. This task belongs to such a range of tasks. It touches on the concepts of distance, speed and time. Accordingly, in order to solve such a problem, you need to have an idea of the things that are said in the problem. In our case, you need to know what distance, speed and time are.
In the problem, you need to find the distances of two roads. We have to write an equation that will calculate these distances.
Let's remember how distance, speed and time are interconnected. Each of these quantities can be described using a letter equation:
We will use the right side of one of these equations to compose our equation. To find out which one, you need to return to the text of the problem and look for what you can hook on
You can catch on to the moment where the cyclist on the way back took more minutes of time. This hint tells us that we can use the equation, namely its right side. This will allow us to write an equation that contains the variable S .
So, let us denote the length of the first road by S... The cyclist traveled this path at a speed of 8 km / h. The time during which he covered this path will be indicated by the expression, since time is the ratio of the distance traveled to speed
The way back for the cyclist was 3 km longer. Therefore, its distance will be denoted through the expression S+ 3. The cyclist traveled this road at a speed of 9 km / h. This means that the time during which he traveled this path will be indicated by the expression.
Now let's make an equation from the existing expressions
The right cup is heavier than the left. This is because the problem says that the cyclist spent more time on the way back.
To equalize the scales, add these same minutes to the left side. But first, let's convert minutes to hours, since in the problem the speed is measured in kilometers per hour, and not in meters per minute.
To convert minutes into hours, you need to divide them by 60 
Minutes are hours. We add these hours to the left side of the equation:
The equation turns out 
... Let's solve this equation. To get rid of fractions, both sides of the part can be multiplied by 72. Next, using the well-known identity transformations, we find the value of the variable S
Through variable S we marked the distance of the first road. Now we have found the value of this variable. Variable S equals 15. So the distance of the first road is 15 km.
And we denoted the distance of the second road through the expression S+ 3, and since the value of the variable S is now known, the distance of the second road can be calculated. This distance is equal to the sum of 15 + 3, that is, 18 km.
Problem 4... Two cars are going along the highway at the same speed. If the first one increases the speed by 10 km / h, and the second decreases the speed by 10 km / h, then the first one will pass as much in 2 hours as the second one in 3 hours. How fast do the cars go?
Solution
Let us denote by v the speed of each car. Further in the task, hints are given: increase the speed of the first car by 10 km / h, and the speed of the second, decrease by 10 km / h. Let's use this hint
It goes on to say that at such speeds (increased and decreased by 10 km / h), the first car will cover the same distance in 2 hours as the second in 3 hours. The phrase "as many" can be understood as "The distance covered by the first car will be equals distance traveled by the second car ".
The distance, as we remember, is determined by the formula. We are interested in the right side of this letter equation - it will allow us to compose an equation containing the variable v .
So at speed v + 10 km / h the first car will pass 2 (v + 10) km, and the second will pass 3 (v - 10) km... Under this condition, the machines will travel the same distance, therefore, to obtain an equation, it is enough to connect these two expressions with an equal sign. Then we get the equation. Let's solve it:
In the problem statement, it was said that the cars go at the same speed. We denoted this speed through the variable v... Now we have found the value of this variable. Variable v equals 50. So the speed of both cars was 50 km / h.
Problem 5... For 9 hours along the river, the ship goes the same way as for 11 hours against the current. Find your own speed of the ship if the river speed is 2 km / h.
Solution
Let us denote by v own speed of the ship. The river's speed is 2 km / h. Downstream of the river, the speed of the ship will be v + 2 km / h, and upstream - (v - 2) km / h.
In the condition of the problem it is said that in 9 hours along the river the motor ship passes the same path as in 11 hours upstream. The phrase "The same way" can be understood as “The distance covered by the motor ship along the river in 9 hours, equals the distance traveled by the motor ship against the river flow in 11 hours "... That is, the distances will be the same.
The distance is determined by the formula. Let's use the right side of this letter equation to compose our equation.
So, in 9 hours the motor ship will pass along the river 9 (v + 2) km, and 11 hours upstream - 11 (v - 2) km... Since both expressions describe the same distance, let's equate the first expression with the second. As a result, we get the equation. Let's solve it:
This means that the own speed of the ship is 20 km / h.
When solving problems, it is a useful habit to determine in advance on which one is looking for a solution for it.
Let us assume that the task required to find the time it takes for a pedestrian to cover the specified path. We denoted the time through the variable t, then they made an equation containing this variable and found its value.
From practice, we know that the time of movement of an object can take both integer values and fractional values, for example, 2 hours, 1.5 hours, 0.5 hours. Then we can say that the solution to this problem is sought on the set of rational numbers Q since each of the values 2 h, 1.5 h, 0.5 h can be represented as a fraction.
Therefore, after the unknown quantity has been designated as a variable, it is useful to indicate to which set this quantity belongs. In our example, the time t belongs to the set of rational numbers Q
t ∈ Q
You can also impose a constraint on the variable t, indicating that it can only take positive values. Indeed, if the object has spent a certain time on the path, then this time cannot be negative. Therefore, next to the expression t∈ Q we indicate that its value must be greater than zero:
t∈ R, t > 0
If we solve the equation, we get a negative value for the variable t, then it will be possible to conclude that the problem was solved incorrectly, since this solution will not satisfy the condition t∈ Q , t> 0 .
Another example. If we were solving a problem in which it was required to find the number of people to perform this or that work, then we would designate this number through the variable x... In such a problem, the solution would be sought on the set of natural numbers
x ∈ N
Indeed, the number of people is a whole number, for example 2 people, 3 people, 5 people. But not 1.5 (one whole person and half a person) or 2.3 (two whole people and another three-tenths of a person).
Here one could indicate that the number of people must be greater than zero, but the numbers included in the set of natural numbers N are themselves positive and greater than zero. In this set there are no negative numbers and the number 0. Therefore, the expression x> 0 need not be written.
Problem 6... To repair the school, a team arrived in which there were 2.5 times more painters than carpenters. Soon the foreman included four more painters in the brigade, and transferred two carpenters to another facility. As a result, there were 4 times more painters in the brigade than carpenters. How many painters and how many carpenters were in the brigade initially
Solution
Let us denote by x carpenters who originally arrived to renovate.
The number of carpenters is an integer greater than zero. Therefore, we point out that x belongs to the set of natural numbers
x∈N
There were 2.5 times more painters than carpenters. Therefore, the number of painters will be denoted as 2.5x.
And we will increase the number of painters by 4
Now the number of carpenters and painters will be indicated through the following expressions:
Let's try to compose an equation from the available expressions:
The right bowl is larger, because after the inclusion of four more painters in the brigade, and the movement of two carpenters to another object, the number of painters in the brigade turned out to be 4 times more than the carpenters. To equalize the scales, you need to increase the left pan by 4 times:
We got the equation. Let's solve it:
Through variable x the initial number of carpenters was indicated. Now we have found the value of this variable. Variable x equals 8. This means that 8 carpenters were in the brigade initially.
And the number of painters was indicated through the expression 2.5 x and since the value of the variable x now it is known, then you can calculate the number of painters - it is equal to 2.5 × 8, that is, 20.
We return to the beginning of the problem and make sure that the condition is met x∈ N. Variable x is equal to 8, and the elements of the set of natural numbers N these are all numbers starting with 1, 2, 3, and so on to infinity. This set also includes the number 8 that we found.
8 ∈N
The same can be said about the number of painters. The number 20 belongs to the set of natural numbers:
20 ∈N
To understand the essence of the problem and correctly draw up the equation, it is not at all necessary to use the model of scales with bowls. Other models can also be used: lines, tables, diagrams. You can come up with your own model that would describe the essence of the task well.
Problem 9... 30% of the milk was poured out of the can. As a result, 14 liters remained in it. How many liters of milk were in the can initially?
Solution
The value sought is the initial number of liters in the can. Let's represent the number of liters as a line and sign this line as X
It is said that 30% of the milk was cast from the can. Let's select in the figure about 30%
Percentage is, by definition, one hundredth of something. If 30% of the milk was discarded, then the remaining 70% remained in the can. These 70% account for the 14 liters indicated in the problem. Select the remaining 70% in the figure.
Now you can write the equation. Let's remember how to find the percentage of a number. To do this, the total amount of something is divided by 100 and the result is multiplied by the desired number of percent. Note that 14 liters, which is 70%, can be obtained in the same way: the initial number of liters X divide by 100 and multiply the result by 70. Equate all this to the number 14
Or get a simpler equation: write 70% as 0.70, then multiply by X and equate this expression to 14
This means that initially there were 20 liters of milk in the can.
Problem 9... We took two alloys of gold and silver. In one, the amount of these metals is in a ratio of 1: 9, and in the other 2: 3. How much of each alloy should be taken to get 15 kg of a new alloy, in which gold and silver would be in a ratio of 1: 4?
Solution
Let's first try to find out how much gold and silver will be contained in 15 kg of a new alloy. The problem says that the content of these metals should be in a ratio of 1: 4, that is, one part of the alloy should have gold, and four parts should have silver. Then the total parts in the alloy will be 1 + 4 = 5, and the mass of one part will be 15: 5 = 3 kg.
Let's determine how much gold will be contained in 15 kg of alloy. To do this, we multiply 3 kg by the number of parts of gold:
3 kg × 1 = 3 kg
Let's determine how much silver will be contained in 15 kg of alloy:
3 kg × 4 = 12 kg
This means that an alloy weighing 15 kg will contain 3 kg of gold and 12 kg of silver. Now let's go back to the original alloys. You need to use each of them. Let us denote by x the mass of the first alloy, and the mass of the second alloy can be denoted by 15 - x
Let us express as a percentage all the relations that are given in the problem and fill in the following table with them:
In the first alloy, gold and silver are in a ratio of 1: 9. Then the total parts will be 1 + 9 = 10. Of these, gold will be 
and silver 
.
Let's transfer this data to the table. 10% will be entered in the first line in the column "Percentage of gold in the alloy", 90% will also be entered in the first line of the column "Percentage of silver in the alloy", and in the last column "Alloy mass" let's add a variable x, since this is how we designated the mass of the first alloy:
We do the same with the second alloy. Gold and silver in it are in a ratio of 2: 3. Then there will be 2 + 3 = 5 parts in total. Of these, gold will be 
and silver 
.
Let's transfer this data to the table. We will enter 40% in the second line in the column "Percentage of gold in the alloy", 60% will also be entered in the second line of the column "Percentage of silver in the alloy", and in the last column "Alloy mass" add expression 15 - x, since this is how we designated the mass of the second alloy:
Let's fill in the last line. The resulting alloy weighing 15 kg will contain 3 kg of gold, which is 
alloy, and silver will 
alloy. In the last column we write down the mass of the resulting alloy 15
Now, from this table, you can make equations. We remember. If we add the gold of both alloys separately and equate this sum to the mass of gold of the resulting alloy, we can find out what the value is equal to x.
In the first alloy of gold it was 0.10 x, and in the second gold alloy it was 0.40 (15 - x). Then, in the resulting alloy, the mass of gold will be the sum of the gold masses of the first and second alloys, and this mass is 20% of the new alloy. And 20% of the new alloy is 3 kg of gold, which we calculated earlier. As a result, we obtain the equation 0,10x+ 0.40(15 − x) = 3 ... Let's solve this equation:
Initially through x we have designated the mass of the first alloy. Now we have found the value of this variable. Variable x is equal to 10. And we denoted the mass of the second alloy through 15 - x, and since the value of the variable x now it is known, then you can calculate the mass of the second alloy, it is equal to 15 - 10 = 5 kg.
This means that to obtain a new alloy weighing 15 kg in which gold and silver would be treated as 1: 4, you need to take 10 kg of the first and 5 kg of the second alloy.
The equation could be made using the second column of the resulting table. Then we would get the equation 0,90x+ 0.60(15 − x) = 12. The root of this equation is also 10
Problem 10... There is ore from two layers with copper grades of 6% and 11%. How much of a poor ore should be taken to get when mixed with a rich ore 20 tons with a copper content of 8%?
Solution
Let us denote by x a lot of poor ore. Since you need to get 20 tons of ore, then 20 rich ore will be taken - x... Since the copper content in the poor ore is 6%, then in x tons of ore will contain 0.06 x tons of copper. In rich ore, the copper content is 11%, and in 20 - x tonnes of high-grade ore will contain 0.11 (20 - x) tons of copper.
In the resulting 20 tons of ore, the copper content should be 8%. This means that 20 tons of copper ore will contain 20 × 0.08 = 1.6 tons.
Add expressions 0.06 x and 0.11 (20 - x) and equate this amount to 1.6. We get the equation 0,06x + 0,11(20 − x) = 1,6
Let's solve this equation:
This means that to obtain 20 tons of ore with a copper content of 8%, you need to take 12 tons of poor ore. The rich will take 20 - 12 = 8 tons.
Assignment 11... Having increased the average speed from 250 to 300 m / min, the athlete began to run the distance 1 minute faster. How long is the distance?
Solution
The length of the distance (or distance of the distance) can be described by the following letter equation:
Let's use the right side of this equation to compose our equation. Initially, the athlete ran the distance at a speed of 250 meters per minute. At this speed, the length of the distance will be described by the expression 250 t
Then the athlete increased her speed to 300 meters per minute. At this speed, the length of the distance will be described by the expression 300t
Note that the length of the distance is constant. From the fact that the athlete increases the speed or decreases it, the length of the distance will remain unchanged.
This allows us to equate the expression 250 t to expression 300 t since both expressions describe the length of the same distance
250t = 300t
But the problem says that at a speed of 300 meters per minute, the athlete began to run the distance 1 minute faster. In other words, at a speed of 300 meters per minute, the travel time will decrease by one. Therefore, in the equation 250 t= 300t on the right side, the time must be reduced by one:
At a speed of 250 meters per minute, the athlete runs the distance in 6 minutes. Knowing the speed and time, you can determine the length of the distance:
S= 250 × 6 = 1500 m
And at a speed of 300 meters per minute, the athlete runs the distance in t- 1, that is, in 5 minutes. As mentioned earlier, the length of the distance does not change:
S= 300 × 5 = 1500 m
Assignment 12... The rider overtakes the pedestrian 15 km ahead of him. In how many hours will the rider catch up with the pedestrian, if every hour the first one drives 10 km, and the second passes only 4 km?
Solution
This task is. It can be solved by determining the approach speed and dividing the initial distance between rider and pedestrian by this speed.
The approach speed is determined by subtracting the lower speed from the higher one:
10 km / h - 4 km / h = 6 km / h (approach speed)
Every hour the distance of 15 kilometers will be reduced by 6 kilometers. To find out when it will completely decrease (when the rider catches up with the pedestrian), you need to divide 15 by 6
15: 6 = 2.5 h
2,5 h that's two full hours and half an hour. And half an hour is 30 minutes. This means that the rider will catch up with the pedestrian in 2 hours and 30 minutes.
Let's solve this problem using the equation.
After that, a rider followed him at a speed of 10 km / h. And the pedestrian's speed is only 4 km / h. This means that the rider will overtake the pedestrian after a while. We need to find this time.
When the rider catches up with the pedestrian, it will mean that they have traveled the same distance together. The distance traveled by the rider and the pedestrian is described by the following equation:
Let's use the right side of this equation to compose our equation.
The distance traveled by the rider will be described by the expression 10 t... Since the pedestrian set out before the rider and managed to overcome 15 km, the distance traveled by him will be described by the expression 4 t + 15 .
At the moment when the rider catches up with the pedestrian, both of them will cover the same distance. This allows us to equate the distances traveled by the rider and the pedestrian:
It turned out to be the simplest equation. Let's solve it:
Tasks for independent solution
Problem 1. From one city to another, a passenger train arrives 45 minutes faster than a freight train. Calculate the distance between cities if the speed of a passenger train is 48 km / h, and a freight train is 36 km / h.
Solution
Train speeds in this problem are measured in kilometers per hour. Therefore, 45 minutes indicated in the problem will be converted into hours. 45 minutes is 0.75 hours
Let us denote the time it takes for a freight train to arrive in the city through the variable t... Since a passenger train arrives in this city 0.75 hours faster, the time of its movement will be denoted by the expression t - 0,75
The passenger train overcame 48 ( t - 0.75) km, and the commodity 36 t km. Since we are talking about the same distance, let us equate the first expression with the second. As a result, we obtain the equation 48(t - 0.75) = 36t ... Let's solve it:
Now let's calculate the distance between cities. For this, the speed of the freight train (36 km / h) is multiplied by the time of its movement t. Variable value t now it is known - it is equal to three hours
36 × 3 = 108 km
To calculate the distance, you can also use the speed of a passenger train. But in this case, the value of the variable
Variable value t is equal to 1.2. So the cars met in 1.2 hours. Answer: the cars met in 1.2 hours.
Problem 3. There are only 685 workers in the three workshops of the plant. In the second shop there are three times more workers than in the first, and in the third - 15 workers less than in the second shop. How many workers are there in each workshop?
Solution
Let be x workers were in the first shop. In the second shop there were three times more than in the first, so the number of workers in the second shop can be denoted through the expression 3 x... The third workshop had 15 fewer workers than the second. Therefore, the number of workers in the third workshop can be denoted through the expression 3 x - 15 .
The problem says that there were 685 workers in total.Therefore, you can add the expressions x, 3x, 3x - 15 and equate this sum to 685. As a result, we get the equation x + 3x + ( 3x - 15) = 685
Through variable x the number of workers in the first workshop was indicated. Now we have found the value of this variable, it is equal to 100. This means that there were 100 workers in the first workshop.
The second workshop had 3 x workers, that is, 3 × 100 = 300. And in the third workshop there were 3 x - 15, that is, 3 × 100 - 15 = 285
Answer: in the first shop there were 100 workers, in the second - 300, in the third - 285.
Task 4. Two repair shops within a week have to repair 18 motors according to the plan. The first workshop fulfilled the plan by 120%, and the second - by 125%, so 22 motors were repaired within a week. What plan did each workshop have to repair their motors for the week?
Solution
Let be x motors had to be repaired by the first workshop. Then the second workshop had to renovate 18 − x motors.
Since the first workshop fulfilled its plan by 120%, this means that it has repaired 1.2 x motors. And the second workshop fulfilled its plan by 125%, which means it repaired 1.25 (18 - x) motors.
The task says that 22 motors have been repaired. Therefore, you can add the expressions 1,2x and 1.25 (18 - x) , then equate this sum to the number 22. As a result, we obtain the equation 1,2x + 1,25(18- x) = 22
Through variable x the number of motors to be repaired by the first workshop was indicated. Now we have found the value of this variable, it is 10. So the first workshop had to repair 10 motors.
And through expression 18 - x the number of motors to be repaired by the second workshop was indicated. So the second workshop had to repair 18 - 10 = 8 motors.
Answer: the first workshop had to repair 10 motors, and the second - 8 motors.
Problem 5. The price of the goods has increased by 30% and is now 91 rubles. How much did the item cost before the price increase?
Solution
Let be x rubles cost the goods before the price increase. If the price has increased by 30% it means that it has increased by 0.30 x rubles. After the price increase, the goods began to cost 91 rubles. Add x to 0.30 x and equate this sum to 91. As a result, we obtain the equation When decreasing the number by 10%, it turned out 45. Find the initial value of the number. x - Answer: to obtain a 12% salt solution, add 0.25 kg of a 20% solution to 1 kg of a 10% solution.
Problem 12. Two solutions of salt in water are given, the concentrations of which are 20% and 30%. How many kilograms of each solution must be mixed in one vessel to obtain 25 kg of a 25.2% solution?
Solution
Let be x kg of the first solution must be taken. Since it is required to prepare 25 kg of solution, the mass of the second solution can be denoted through the expression 25 - x.
The first solution will contain 0.20x kg of salt and the second will contain 0.30 (25 - x) kg of salt. In the resulting solution, the salt content will be 25 × 0.252 = 6.3 kg. Add the expressions 0.20x and 0.30 (25 - x), then equate this sum to 6.3. As a result, we obtain the equation
So the first solution needs to be taken 12 kg, and the second 25 - 12 = 13 kg.
Answer: the first solution needs to be taken 12 kg, and the second 13 kg.
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Let's talk about how to write the equation of a chemical reaction. It is this question that mainly causes serious difficulties for schoolchildren. Some cannot understand the algorithm for drawing up product formulas, while others place the coefficients in the equation incorrectly. Given that all quantitative calculations are carried out precisely according to equations, it is important to understand the algorithm of actions. Let's try to figure out how to make up the equations of chemical reactions.
Drawing up formulas for valency
In order to correctly write down the processes occurring between different substances, you need to learn how to write down formulas. Binary compounds are composed taking into account the valencies of each element. For example, for metals of the main subgroups, it corresponds to the group number. When drawing up the final formula, the smallest multiple is determined between these indicators, then the indices are placed.
What is an equation
It is understood as a symbolic notation that displays the interacting chemical elements, their quantitative ratios, as well as those substances that are obtained as a result of the process. One of the tasks offered to ninth grade students at the final certification in chemistry has the following formulation: "Make up the equations of reactions characterizing the chemical properties of the proposed class of substances." In order to cope with the task at hand, students must master the algorithm of actions.
Algorithm of actions
For example, you need to write the process of burning calcium using symbols, coefficients, indices. Let's talk about how to create the equation of a chemical reaction using the procedure. On the left side of the equation, through "+", we write down the signs of the substances that participate in this interaction. Since combustion occurs with the participation of atmospheric oxygen, which belongs to diatomic molecules, we write its formula O2.
Behind the equal sign, we form the composition of the reaction product using the rules for arranging the valence:
2Ca + O2 = 2CaO.
Continuing the conversation about how to compose the equation of a chemical reaction, we note the need to use the law of constancy of composition, as well as preservation of the composition of substances. They allow you to carry out the adjustment process, to place the missing coefficients in the equation. This process is one of the simplest examples of interactions that occur in inorganic chemistry.
Important aspects
In order to understand how to compose the equation of a chemical reaction, let us note some theoretical questions related to this topic. The law of conservation of mass of substances, formulated by M.V. Lomonosov, explains the possibility of arranging the coefficients. Since the number of atoms of each element before and after the interaction remains unchanged, mathematical calculations can be performed.
When equalizing the left and right sides of the equation, the least common multiple is used, in the same way as the compound formula is drawn up, taking into account the valencies of each element.
Redox interactions
After the schoolchildren have worked out the algorithm of actions, they will be able to draw up an equation of reactions that characterize the chemical properties of simple substances. Now you can proceed to the analysis of more complex interactions, for example, those occurring with a change in the oxidation states of elements:
Fe + CuSO4 = FeSO4 + Cu.
There are certain rules according to which oxidation states are arranged in simple and complex substances. For example, in diatomic molecules this indicator is zero, in complex compounds the sum of all oxidation states should also be zero. When drawing up an electronic balance, atoms or ions are determined that donate electrons (reducing agent), accept them (oxidizing agent).
The smallest multiple is determined between these indicators, as well as the coefficients. The final stage in the analysis of the redox interaction is the arrangement of the coefficients in the scheme.
Ionic Equations
One of the important issues that is considered in the course of school chemistry is the interaction between solutions. For example, given the task of the following content: "Make the equation of the chemical reaction of ion exchange between barium chloride and sodium sulfate." It involves writing a molecular, complete, abbreviated ionic equation. To consider the interaction at the ionic level, it is necessary to indicate it according to the solubility table for each starting substance, reaction product. For example:
BaCl2 + Na2SO4 = 2NaCl + BaSO4
Substances that do not dissolve into ions are recorded in molecular form. The ion exchange reaction proceeds completely in three cases:
- sediment formation;
- gas evolution;
- obtaining a poorly dissociated substance, such as water.
If a substance has a stereochemical coefficient, it is taken into account when writing a complete ionic equation. After the complete ionic equation has been written, the reduction of those ions that were not bound in the solution is carried out. The end result of any task involving consideration of the process that takes place between solutions of complex substances will be a record of an abbreviated ionic reaction.
Conclusion
Chemical equations make it possible to explain with the help of symbols, indices, coefficients, the processes that are observed between substances. Depending on what kind of process is going on, there are certain subtleties of writing the equation. The general algorithm for composing reactions, considered above, is based on valence, the law of conservation of mass of substances, and the constancy of composition.
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