Quite often schoolchildren and students have to compose the so-called. ionic reaction equations. In particular, Problem 31, proposed at the Unified State Exam in Chemistry, is devoted to this topic. In this article, we will discuss in detail the algorithm for writing short and complete ionic equations, we will analyze many examples of different levels of complexity.

Why do we need ionic equations

Let me remind you that when many substances are dissolved in water (and not only in water!), A process of dissociation occurs - the substances decompose into ions. For example, HCl molecules in an aqueous medium dissociate into hydrogen cations (H +, more precisely, H 3 O +) and chlorine anions (Cl -). Sodium bromide (NaBr) is in an aqueous solution not in the form of molecules, but in the form of hydrated ions Na + and Br - (by the way, ions are also present in solid sodium bromide).

Writing down "ordinary" (molecular) equations, we do not take into account that it is not molecules that enter into the reaction, but ions. For example, here's what the equation for the reaction between hydrochloric acid and sodium hydroxide looks like:

HCl + NaOH = NaCl + H 2 O. (1)

Of course, this diagram does not quite correctly describe the process. As we have already said, there are practically no HCl molecules in an aqueous solution, but there are H + and Cl - ions. The same is the case with NaOH. It would be more correct to write the following:

H + + Cl - + Na + + OH - = Na + + Cl - + H 2 O. (2)

That's what it is complete ionic equation... Instead of "virtual" molecules, we see particles that are actually present in the solution (cations and anions). For now, we will not dwell on the question of why we wrote down H 2 O in molecular form. This will be explained later. As you can see, there is nothing complicated: we have replaced the molecules with ions that are formed during their dissociation.

However, even the complete ionic equation is not perfect. Indeed, take a closer look: both on the left and on the right sides of equation (2) there are identical particles - Na + cations and Cl - anions. During the reaction, these ions do not change. Why, then, are they needed at all? Let's take them away and get short ionic equation:

H + + OH - = H 2 O. (3)

As you can see, it all comes down to the interaction of H + and OH - ions with the formation of water (neutralization reaction).

All, complete and concise ionic equations are written down. If we solved problem 31 on the exam in chemistry, we would receive the maximum mark for it - 2 points.

So, once again about the terminology:

• HCl + NaOH = NaCl + H 2 O - molecular equation ("ordinary" equations, schematically reflecting the essence of the reaction);
• H + + Cl - + Na + + OH - = Na + + Cl - + H 2 O - complete ionic equation (real particles in solution are visible);
• H + + OH - = H 2 O - a short ionic equation (we removed all the "garbage" - particles that are not involved in the process).

Algorithm for writing ionic equations

1. We compose the molecular equation of the reaction.
2. All particles dissociating in a solution to an appreciable degree are written in the form of ions; we leave substances that are not prone to dissociation "in the form of molecules".
3. We remove from the two parts of the equation the so-called. observer ions, i.e. particles that do not participate in the process.
4. We check the coefficients and get the final answer - a short ionic equation.

Example 1... Write a complete and concise ionic equation describing the interaction of aqueous solutions of barium chloride and sodium sulfate.

Solution... We will act in accordance with the proposed algorithm. Let's first compose the molecular equation. Barium chloride and sodium sulfate are two salts. Let's look at the section of the reference book "Properties of inorganic compounds". We see that salts can interact with each other if a precipitate forms during the reaction. Let's check:

Exercise 2... Complete the equations for the following reactions:

1. KOH + H 2 SO 4 =
2. H 3 PO 4 + Na 2 O =
3. Ba (OH) 2 + CO 2 =
4. NaOH + CuBr 2 =
5. K 2 S + Hg (NO 3) 2 =
6. Zn + FeCl 2 =

Exercise # 3... Write the molecular equations of the reactions (in aqueous solution) between: a) sodium carbonate and nitric acid, b) nickel (II) chloride and sodium hydroxide, c) phosphoric acid and calcium hydroxide, d) silver nitrate and potassium chloride, e) phosphorus oxide (V) and potassium hydroxide.

I sincerely hope you have no problem completing these three assignments. If this is not the case, it is necessary to return to the topic "Chemical properties of the main classes of inorganic compounds".

How to turn a molecular equation into a complete ionic equation

The fun begins. We must understand which substances should be recorded as ions and which should be left in "molecular form". We'll have to remember the following.

In the form of ions, write down:

• soluble salts (I emphasize, only salts that are readily soluble in water);
• alkalis (let me remind you that alkalis are water-soluble bases, but not NH 4 OH);
• strong acids (H 2 SO 4, HNO 3, HCl, HBr, HI, HClO 4, HClO 3, H 2 SeO 4, ...).

As you can see, this list is not difficult to remember: it includes strong acids and bases and all soluble salts. By the way, to especially vigilant young chemists, who may be outraged by the fact that strong electrolytes (insoluble salts) were not included in this list, I can tell you the following: NOT including insoluble salts in this list does not at all deny that they are strong electrolytes.

All other substances must be present in the ionic equations in the form of molecules. For those demanding readers who are not satisfied with the vague term "all other substances", and who, following the example of the hero of a famous film, demand to "read out the full list" I give the following information.

In the form of molecules, write down:

• all insoluble salts;
• all weak bases (including insoluble hydroxides, NH 4 OH and similar substances);
• all weak acids (H 2 CO 3, HNO 2, H 2 S, H 2 SiO 3, HCN, HClO, almost all organic acids ...);
• in general, all weak electrolytes (including water !!!);
• oxides (all types);
• all gaseous compounds (in particular, H 2, CO 2, SO 2, H 2 S, CO);
• simple substances (metals and non-metals);
• almost all organic compounds (with the exception of water-soluble salts of organic acids).

Phew, I think I haven't forgotten anything! Although easier, in my opinion, it is still to remember list No. 1. Of the fundamentally important in list No. 2, I will once again note water.

Let's train!

Example 2... Write a complete ionic equation describing the interaction of copper (II) hydroxide and hydrochloric acid.

Solution... Let's start, naturally, with the molecular equation. Copper (II) hydroxide is an insoluble base. All insoluble bases react with strong acids to form salt and water:

Cu (OH) 2 + 2HCl = CuCl 2 + 2H 2 O.

And now we find out which substances to write in the form of ions, and which ones - in the form of molecules. The lists above will help us. Copper (II) hydroxide is an insoluble base (see table of solubility), weak electrolyte. Insoluble bases are recorded in molecular form. HCl is a strong acid; in solution it almost completely dissociates into ions. CuCl 2 is a soluble salt. We write in ionic form. Water - only in the form of molecules! We get the complete ionic equation:

Cu (OH) 2 + 2H + + 2Cl - = Cu 2+ + 2Cl - + 2H 2 O.

Example 3... Write the complete ionic equation for the reaction of carbon dioxide with an aqueous solution of NaOH.

Solution... Carbon dioxide is a typical acidic oxide, NaOH is an alkali. When acidic oxides react with aqueous solutions of alkalis, salt and water are formed. We compose the molecular equation of the reaction (do not forget, by the way, about the coefficients):

CO 2 + 2NaOH = Na 2 CO 3 + H 2 O.

CO 2 - oxide, gaseous compound; we keep the molecular shape. NaOH - strong base (alkali); we write in the form of ions. Na 2 CO 3 - soluble salt; we write in the form of ions. Water is a weak electrolyte, practically does not dissociate; leave in molecular form. We get the following:

CO 2 + 2Na + + 2OH - = Na 2+ + CO 3 2- + H 2 O.

Example 4... Sodium sulfide in aqueous solution reacts with zinc chloride to form a precipitate. Write the complete ionic equation for this reaction.

Solution... Sodium sulfide and zinc chloride are salts. When these salts interact, zinc sulfide precipitates:

Na 2 S + ZnCl 2 = ZnS ↓ + 2NaCl.

I will immediately write down the complete ionic equation, and you will analyze it yourself:

2Na + + S 2- + Zn 2+ + 2Cl - = ZnS ↓ + 2Na + + 2Cl -.

I offer you several tasks for independent work and a small test.

Exercise 4... Write molecular and complete ionic equations for the following reactions:

1. NaOH + HNO 3 =
2. H 2 SO 4 + MgO =
3. Ca (NO 3) 2 + Na 3 PO 4 =
4. CoBr 2 + Ca (OH) 2 =

Exercise # 5... Write the complete ionic equations describing the interaction of: a) nitric oxide (V) with an aqueous solution of barium hydroxide, b) a solution of cesium hydroxide with hydroiodic acid, c) aqueous solutions of copper sulfate and potassium sulfide, d) calcium hydroxide and an aqueous solution of iron nitrate ( III).

Reactions between different kinds of chemicals and elements are one of the main subjects of study in chemistry. To understand how to draw up a reaction equation and use them for your own purposes, you need a fairly deep understanding of all the laws governing the interaction of substances, as well as processes with chemical reactions.

Drawing up equations

One way of expressing a chemical reaction is a chemical equation. It records the formula of the initial substance and the product, coefficients that show how many molecules each substance has. All known chemical reactions are divided into four types: substitution, combination, exchange and decomposition. Among them are: redox, exogenous, ionic, reversible, irreversible, etc.

Learn more about how to write chemical equations:

1. It is necessary to determine the name of the substances interacting with each other in the reaction. We write them on the left side of our equation. As an example, consider the chemical reaction that formed between sulfuric acid and aluminum. Place the reagents on the left: H2SO4 + Al. Next, we write the "equal" sign. In chemistry, you can see the sign "arrow", which points to the right, or two oppositely directed arrows, they mean "reversibility." The result of the interaction of a metal and an acid is salt and hydrogen. Write down the products obtained after the reaction after the "equal" sign, that is, on the right. H2SO4 + Al = H2 + Al2 (SO4) 3. So, we see the scheme of the reaction.
2. To draw up a chemical equation, it is imperative to find the coefficients. Let's go back to the previous scheme. Let's look at the left side of it. The composition of sulfuric acid contains atoms of hydrogen, oxygen and sulfur, in an approximate ratio of 2: 4: 1. On the right side there are 3 sulfur atoms and 12 oxygen atoms in the salt. Two hydrogen atoms are contained in a gas molecule. On the left, the ratio of these elements is 2: 3: 12
3. To equalize the number of oxygen and sulfur atoms, which are in the composition of aluminum (III) sulfate, it is necessary to put a coefficient 3 in front of the acid in the left side of the equation. Now we have 6 hydrogen atoms on the left side. In order to equalize the number of elements of hydrogen, you need to put 3 in front of hydrogen on the right side of the equation.
4. Now all that remains is to equalize the amount of aluminum. Since the salt contains two metal atoms, then on the left side in front of aluminum we set the coefficient 2. As a result, we get the reaction equation of this scheme: 2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2

Having understood the basic principles of how to compose the equation for the reaction of chemicals, in the future it will not be difficult to write down any, even the most exotic, from the point of view of chemistry, reaction.

The main subject of comprehension in chemistry is the reactions between different chemical elements and substances. A great awareness of the validity of the interaction of substances and processes in chemical reactions makes it possible to guide them and use them for their own purposes. A chemical equation is a method of expressing a chemical reaction, in which formulas of initial substances and products are written, indicators showing the number of molecules of any substance. Chemical reactions are divided into compound, substitution, decomposition and exchange reactions. Also among them it is allowed to distinguish redox, ionic, reversible and irreversible, exogenous, etc.

Instructions

1. Determine which substances interact with each other in your reaction. Write them down on the left side of the equation. For example, consider the chemical reaction between aluminum and sulfuric acid. Place the reagents on the left: Al + H2SO4 Next, put the equal sign, as in the mathematical equation. In chemistry, you can see an arrow pointing to the right, or two oppositely directed arrows, a “reversibility sign.” As a result of the interaction of a metal with an acid, salt and hydrogen are formed. Write the reaction products after the equal sign, on the right. Al + H2SO4 = Al2 (SO4) 3 + H2 This is the reaction scheme.

2. In order to create a chemical equation, you need to find indicators. On the left side of the previously obtained scheme, the sulfuric acid contains atoms of hydrogen, sulfur and oxygen in a ratio of 2: 1: 4, on the right side there are 3 sulfur atoms and 12 oxygen atoms in the salt composition and 2 hydrogen atoms in the H2 gas molecule. On the left side, the ratio of these 3 elements is 2: 3: 12.

3. In order to equalize the number of sulfur and oxygen atoms in the composition of aluminum (III) sulfate, put the indicator 3 in front of the acid on the left side of the equation. Now there are six hydrogen atoms on the left side. To equalize the number of hydrogen elements, put an indicator 3 in front of it on the right side. Now the ratio of atoms in both parts is 2: 1: 6.

4. It remains to equalize the number of aluminum. Because the salt contains two metal atoms, put 2 in front of aluminum on the left side of the diagram, as a result, you will get the reaction equation for this diagram. 2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2

A reaction is the transformation of some chemicals into others. And the formula for writing them with the help of special symbols is the equation of this reaction. There are different types of chemical interactions, but the rule for writing their formulas is identical.

You will need

• periodic table of chemical elements D.I. Mendeleev

Instructions

1. On the left side of the equation, the initial substances are written, which react. They are called reagents. The recording is made with the help of special symbols that denote any substance. A plus sign is placed between the reagent substances.

2. On the right side of the equation, the formula of the obtained one or several substances is written, which are called reaction products. Instead of an equal sign, an arrow is placed between the left and right sides of the equation, which indicates the direction of the reaction.

3. Later, writing down the formulas of reagents and reaction products, it is necessary to arrange the indicators of the reaction equation. This is done so that, according to the law of conservation of mass of matter, the number of atoms of the same element in the left and right sides of the equation remains identical.

4. In order to correctly arrange the indicators, you need to make out any of the substances that react. For this, one of the elements is taken and the number of its atoms on the left and right is compared. If it is different, then it is necessary to find a number that is a multiple of numbers denoting the number of atoms of a given substance in the left and right sides. After that, this number is divided by the number of atoms of the substance in the corresponding part of the equation, and an exponent is obtained for each of its parts.

5. From the fact that the indicator is placed in front of the formula and refers to each substance included in it, the next step will be to compare the data obtained with the number of another substance that is included in the formula. This is carried out in the same way as with the first element and taking into account the more closely available indicator for each formula.

6. Later, after all the elements of the formula have been disassembled, a final check of the correspondence of the left and right sides is carried out. Then the reaction equation can be considered complete.

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Note!
In the equations of chemical reactions, it is impossible to rearrange the left and right sides. Otherwise, you get a diagram of a completely different process.

Helpful advice
The number of atoms of both individual reagent substances and substances that make up the reaction products is determined using the periodic system of chemical elements of D.I. Mendeleev

How unsurprising nature is for humans: in winter it envelops the earth with a snowy duvet, in spring it reveals that popcorn flakes, all living things, in the summer it riots with a riot of colors, in the fall it sets fire to the plants with red fire ... And only if you think about it and take a closer look, you can see what is behind all these familiar changes are difficult physical processes and CHEMICAL REACTIONS. And in order to study all living things, you need to be able to solve chemical equations. The main requirement for equalizing chemical equations is the knowledge of the law of conservation of the number of matter: 1) the number of matter before the reaction is equal to the number of matter after the reaction; 2) the total number of the substance before the reaction is equal to the total number of the substance after the reaction.

Instructions

1. In order to equalize the chemical "example" you need to perform several steps. the equation reactions in general. To do this, designate unknown indicators in front of the formulas of substances with letters of the Latin alphabet (x, y, z, t, etc.). Let it be required to equalize the reaction of combining hydrogen and oxygen, which will result in water. Before the molecules of hydrogen, oxygen and water, put the Latin letters (x, y, z) - indicators.

2. For any element, based on physical equilibrium, compose mathematical equations and obtain a system of equations. In the above example, for hydrogen on the left, take 2x, because it has an index “2”, on the right - 2z, tea he also has an index “2”., It turns out 2x = 2z, hence, x = z. For oxygen on the left, take 2y, because there is an index “2”, on the right - z, there is no index for tea, which means it is equal to one, which it is customary not to write. It turns out, 2y = z, and z = 0.5y.

Note!
If a larger number of chemical elements participate in the equation, then the task does not become more complicated, but increases in volume, which should not be scared.

Helpful advice
It is allowed to equalize reactions using the theory of probability, using the valences of chemical elements.

Tip 4: How to compose a redox reaction

Redox reactions are reactions with a change in oxidation states. It often happens that the initial substances are given and it is necessary to write the products of their interaction. Occasionally, the same substance can give different final products in different environments.

Instructions

1. Depending not only on the reaction medium, but also on the oxidation state, the substance behaves differently. A substance in its highest oxidation state is invariably an oxidizing agent, in the lowest - a reducing agent. In order to make an acidic environment, sulfuric acid (H2SO4) is traditionally used, less often nitric (HNO3) and hydrochloric (HCl). If necessary, we use sodium hydroxide (NaOH) and potassium hydroxide (KOH) to create an alkaline medium. Below we will consider some examples of substances.

2. Ion MnO4 (-1). In an acidic environment, it turns into Mn (+2), a colorless solution. If the medium is neutral, then MnO2 is formed, and a brown precipitate is formed. In an alkaline environment, we get MnO4 (+2), a green solution.

3. Hydrogen peroxide (H2O2). If it is an oxidizing agent, i.e. accepts electrons, then in neutral and alkaline media it transforms according to the scheme: H2O2 + 2e = 2OH (-1). In an acidic medium, we get: H2O2 + 2H (+1) + 2e = 2H2O. Provided that hydrogen peroxide is a reducing agent, i.e. gives up electrons, in an acidic environment O2 is formed, in an alkaline one - O2 + H2O. If H2O2 enters an environment with a strong oxidizing agent, it will itself be a reducing agent.

4. The Cr2O7 ion is an oxidizing agent; in an acidic environment, it turns into 2Cr (+3), which are green. From the Cr (+3) ion in the presence of hydroxide ions, i.e. in an alkaline environment, yellow CrO4 (-2) is formed.

5. Here is an example of how the reaction is composed: KI + KMnO4 + H2SO4 - In this reaction, Mn is in its highest oxidation state, i.e. it is an oxidizing agent, accepting electrons. The medium is acidic, as shown by sulfuric acid (H2SO4). The reducing agent here is I (-1), it gives up electrons, while increasing its oxidation state. We write down the reaction products: KI + KMnO4 + H2SO4 - MnSO4 + I2 + K2SO4 + H2O. We arrange the indicators by the electronic equilibrium method or by the half-reaction method, we get: 10KI + 2KMnO4 + 8H2SO4 = 2MnSO4 + 5I2 + 6K2SO4 + 8H2O.

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Note!
Do not forget to place indicators in reactions!

Chemical reactions are the interaction of substances accompanied by a change in their composition. In other words, the substances that enter into the reaction do not correspond to the substances that result from the reaction. A person encounters similar interactions hourly, every minute. Tea processes occurring in his body (respiration, protein synthesis, digestion, etc.) are also chemical reactions.

Instructions

1. Any chemical reaction must be recorded correctly. One of the main requirements is that the number of atoms of the entire element of the substances on the left side of the reaction (they are called "initial substances") correspond to the number of atoms of the same element in the substances on the right side (they are called "reaction products"). In other words, the reaction record must be equalized.

2. Let's take a look at a specific example. What happens when the gas burner is lit in the kitchen? Natural gas reacts with oxygen in the air. This oxidation reaction is so exothermic, that is, accompanied by the release of heat, that a flame appears. With the help of which you either cook food or reheat more cooked food.

3. To make it easier, assume that natural gas consists of only one of its components - methane, which has the formula CH4. Because what is it to compose and equalize this reaction?

4. When carbon-containing fuel is burned, that is, when carbon is oxidized with oxygen, carbon dioxide is formed. You are familiar with its formula: CO2. And what is formed during the oxidation of hydrogen contained in methane with oxygen? Of course, water is in the form of steam. Even the person farthest from chemistry knows her formula by heart: H2O.

5. It turns out, write down the initial substances on the left side of the reaction: СН4 + О2. On the right side, respectively, there will be the reaction products: СО2 + Н2О.

6. An advance record of this chemical reaction will be further: CH4 + O2 = CO2 + H2O.

7. Equalize the above reaction, that is, achieve the basic rule: the number of atoms of the entire element in the left and right sides of the chemical reaction must be identical.

8. You can see that the number of carbon atoms is the same, but the number of oxygen and hydrogen atoms is different. On the left side there are 4 hydrogen atoms, and on the right - only 2. Therefore, put the indicator 2 in front of the water formula. Get: CH4 + O2 = CO2 + 2H2O.

9. The atoms of carbon and hydrogen are equalized, now it remains to do the same with oxygen. On the left side of the oxygen atoms there are 2, and on the right - 4. Putting the indicator 2 in front of the oxygen molecule, you get the final record of the methane oxidation reaction: CH4 + 2O2 = CO2 + 2H2O.

A reaction equation is a conditional notation of a chemical process in which some substances are converted into others with a change in properties. To record chemical reactions, formulas of substances and skills about the chemical properties of compounds are used.

Instructions

1. Write the formulas correctly according to their names. For example, aluminum oxide Al? O?, Index 3 from aluminum (corresponds to its oxidation state in this compound) put near oxygen, and the index 2 (oxidation state of oxygen) near aluminum. If the oxidation state is +1 or -1, then the index is not set. For example, you need to write down the formula for ammonium nitrate. Nitrate is an acidic residue of nitric acid (-NO ?, s.o. -1), ammonium (-NH ?, s.o. +1). So the formula for ammonium nitrate is NH? NO ?. Occasionally, the oxidation state is indicated in the name of the compound. Sulfur oxide (VI) - SO ?, silicon oxide (II) SiO. Some primitive substances (gases) are written with the index 2: Cl ?, J ?, F ?, O ?, H? etc.

2. You need to know what substances react. Visible signs of reaction: gas evolution, color metamorphosis and precipitation. Hefty often reactions pass without visible changes. Example 1: reaction of neutralization H? SO? + 2 NaOH? Na? SO? + 2 H? O Sodium hydroxide reacts with sulfuric acid to form a soluble salt of sodium sulfate and water. The sodium ion is cleaved off and combines with the acidic residue, replacing hydrogen. The reaction takes place without external signs. Example 2: iodoform test С? H? OH + 4 J? + 6 NaOH? CHJ ?? + 5 NaJ + HCOONa + 5 H? O The reaction proceeds in several stages. The final result is the precipitation of yellow iodoform crystals (good reaction to alcohols). Example 3: Zn + K? SO? ? The reaction is unthinkable, because in the series of metal voltages, zinc is later than potassium and cannot displace it from compounds.

3. The law of conservation of mass states: the mass of substances that have entered into a reaction is equal to the mass of the substances formed. Competent recording of a chemical reaction is half the furore. You need to arrange the indicators. Start equalizing with those compounds whose formulas contain large indices. K? Cr? O? + 14 HCl? 2 CrCl? + 2 KCl + 3 Cl ?? + 7 H? O Start placing the indicators with potassium dichromate, because its formula contains the largest index (7). Such accuracy in recording reactions is needed to calculate mass, volume, concentration, released energy and other quantities. Be careful. Remember the most common formulas for acids and bases, as well as acid residues.

Tip 7: How to Determine Redox Equations

A chemical reaction is the process of reincarnation of substances that occurs with a change in their composition. Those substances that enter into a reaction are called initial, and those that are formed as a result of this process are called products. It so happens that in the course of a chemical reaction, the elements that make up the initial substances change their oxidation state. That is, they can accept other people's electrons and give up their own. And in fact, and in another case, their charge changes. Such reactions are called redox reactions.

Instructions

1. Write down the exact equation for the chemical reaction you are considering. See what elements are included in the initial substances, and what are the oxidation states of these elements. Later, compare these indicators with the oxidation states of the same elements on the right side of the reaction.

2. If the oxidation state has changed, this reaction is redox. If the oxidation states of all the elements remain the same, no.

3. Here is, say, the well-known good-quality reaction for the detection of the sulfate ion SO4 ^ 2-. Its essence is that barium sulfate salt, which has the formula BaSO4, is actually insoluble in water. When formed, it instantly precipitates as a dense, heavy white precipitate. Write down some equation for a similar reaction, say, BaCl2 + Na2SO4 = BaSO4 + 2NaCl.

4. It turns out that from the reaction you see that in addition to the precipitate of barium sulfate, sodium chloride was formed. Is this reaction a redox reaction? No, it is not, because not a single element that is part of the initial substances has changed its oxidation state. On both the left and right sides of the chemical equation, barium has an oxidation state of +2, chlorine -1, sodium +1, sulfur +6, oxygen -2.

5. But the reaction is Zn + 2HCl = ZnCl2 + H2. Is it redox? Elements of the initial substances: zinc (Zn), hydrogen (H) and chlorine (Cl). See what are their oxidation states? For zinc it is equal to 0 as in any simple substance, for hydrogen +1, for chlorine -1. And what are the oxidation states of the same elements on the right side of the reaction? For chlorine, it remained unwavering, that is, equal to -1. But for zinc it became equal to +2, and for hydrogen - 0 (from the fact that hydrogen was released in the form of a simple substance - gas). Consequently, this reaction is redox.

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The canonical equation of the ellipse is composed of those considerations that the sum of the distances from any point of the ellipse to its 2 foci is invariably continuous. By fixing this value and moving the point along the ellipse, it is possible to determine the equation of the ellipse.

You will need

• A sheet of paper, ballpoint pen.

Instructions

1. Specify two fixed points F1 and F2 on the plane. Let the distance between the points be equal to some fixed value F1F2 = 2s.

2. Draw a straight line on a piece of paper, which is the coordinate line of the abscissa axis, and draw the points F2 and F1. These points represent the foci of the ellipse. The distance from the entire focal point to the origin must be equal to the same value, equal to c.

3. Draw the y-axis, thus forming a Cartesian coordinate system, and write the basic equation that defines the ellipse: F1M + F2M = 2a. Point M represents the current point of the ellipse.

4. Determine the size of the segments F1M and F2M with the help of the Pythagorean theorem. Keep in mind that point M has the current coordinates (x, y) relative to the origin, and relative to, say, point F1, point M has coordinates (x + c, y), that is, the "x" coordinate is shifted. Thus, in the expression of the Pythagorean theorem, one of the terms must be equal to the square of the value (x + c), or the value (x-c).

5. Substitute the expressions for the moduli of the vectors F1M and F2M into the main relation of the ellipse and square both sides of the equation by moving one of the square roots to the right side of the equation in advance and opening the brackets. After canceling identical terms, divide the resulting ratio by 4a and raise it again to the second power.

6. Give similar terms and collect terms with the same factor of the square of the "x" variable. Pull out the square of the "x" variable outside the parenthesis.

7. Designate the square of some quantity (say, b) the difference between the squares of the quantities a and c, and divide the resulting expression by the square of this new quantity. Thus, you got the canonical equation of an ellipse, on the left side of which is the sum of the squares of coordinates divided by the values ​​of the axes, and on the left side is one.

Helpful advice
In order to check the fulfillment of the task, you can use the law of conservation of mass.