Calculator for cheesy priests. Rozrakhunks for cheeky ryvnyanny. Rozrakhunks for cheeky rivnyanny
In case of viral rozrakhunkovy chemical tasks, it is necessary to keep the calculation for the equal chemical reactions. The lesson of assignment to the algorithm of rozrakhunkiv masi (obsyagu, kіlkostі) one of the participants of the reaction for the given masoi (volume, number) of the first participant of the reaction.
Topic: Records and Revisions
lesson:Rozrakhunki for the level of chemistry reactions
A clear-cut reactionary response to the adoption of the lead from simple speeches:
2H 2 + O 2 = 2H 2 O
It can be said that two molecules of water and one molecule of water are made up of two molecules of water. From the side, write down to talk about those who, for the approval of the skin two moles of water, it is necessary to take two blessed water one mole sour.
The younger the participation of the participants in the reactions of additional assistance to the virality is important for the chemical synthesis of rosary. Put such rozrakhunіv clearly.
ENTITY 1. Apparently masu vodi, scho established itself as a result of burning a water of 3.2 g of sour.
If you want to know what you want to know, it is necessary to write down the data on the problem and write down the data above it.
Yakby we knew a few words of speech entered into a reaction of sourness, then they could only have a few words of words. And then, we’d fuck with the voices, I know the number of words. In order to know the number of words, it is necessary to know how many words are used on the first molar basis.
The molar mass is numerically expensive. For souring, the value should be set to 32. Adds to the formula: the number of words souring for delivery is 3.2 g to 32 g / mol. It was 0.1 mol.
For the knowledge of the number of speeches, the proportion is too much;
for 0.1 mole of sourness, it is impossible to bring the amount of speech of water, and for 1 mole of sourness, 2 good water is brought.
The rate of speech is 0.2 mol.
However, you need to know the amount of water, you need to know the value of the amount of water and multiply by її molar weight, so that 0.2 mol is multiplied by 18 g / mol, and 3.6 g of water can be recognized.
Small. 1. Registration of the record short wash i solution Zavdannya 1
Krym masi, you can rozrakhovuvati obsyag gas-like participant in the reaction (under n.u.), vikoristovuyu, as you can see the formula, apparently to what kind of gas under n.o. add a number of words to gas for molar volume. The butt of the solution of tasks is clearly visible.
ENDORSEMENT 2. Rozrakhumo obsyag kisnyu (under n.u.), which was seen when 27g of water was distributed.
We can write down the equal reaction and the given of the problems. In order to know a lot of words, you need to know a few words of talk through the masu, because of the equal reaction, there is a lot of talk of money, if you can get rid of it at present.
The number of words in the drive and the transportation of the weight of the drive to the її molar weight. A value of 1.5 mol is accepted.
In a way, in proportion: from 1.5 to good water and to pretend not to be sour, from 2 moles of water to pretend to be 1 mole of sour. Zvidsi kіlkist kisnyu one 0.75 blessed. Rozrahuєmo obsyag kisnu under n.u. Win dorivnyu dobutku kilkosti sisnyu on molar volume. The molar volume of a gas-like speech at n.u. doorway 22.4 l / mol. Having entered the numerical values into the formula, we can take a note of sourness, which is 16.8 liters.
Small. 2. Registration of the record of a short speech and solution Zavdannya 2
Knowing the algorithm for verifying other employees, it is possible to develop a masu, either a number of speeches from one of the participants in the reaction after a maso, about a number of speeches from one of the participants in the reaction.
1. Zbirnik zabdan і right from chemistry: 8th grade: to Navchan. P.A. Orzhekovsky i іn. "Khimiya. 8 class "/ P.A. Orzhekovskiy, N.A. Titov, F.F. Hegel. - M.: AST: Astrel, 2006. (p. 40-48)
2. Ushakova O.V. A working man sows from chemistry: 8th grade: before the assistant P.A. Orzhekovsky i іn. "Khimiya. 8 class "/ O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovskiy; pid. ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006. (p. 73-75)
3. Chemistry. 8 class. Textbook. for zagalnoosvіtnіkh. ustanovka / P.A. Orzhekovskiy, L.M. Meshcheryakova, M.M. Shalashova. - M .: Astrel, 2013. (§23)
4. Khimiya: 8th class: navch. for zagalnoosvіtnіkh. ustanovka / P.A. Orzhekovskiy, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005. (§29)
5. Chemistry: non-organ. chemistry: navch. for 8kl. zagalnoosvіtnіkh. mouth. / G.E. Rudzitis, F.G. Feldman. - M.: Education, VAT "Moskovski pidruchniki", 2009. (p.45-47)
6. Encyclopedia for children. Volume 17. Хімія / Chapter. ed. by V.A. Volodin, led. scientific. ed. І. Leenson. - M .: Avanta +, 2003.
Dodatkovi web resources
2.United collection of digital educational resources ().
Home zavdannya
1) p. 73-75 No. 2, 3, 5 s robotized from chemistry: 8th grade: before the assistant P.A. Orzhekovsky i іn. "Khimiya. 8 class "/ O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovskiy; pid. ed. prof. P.A. Orzhekovsky - M .: AST: Astrel: Profizdat, 2006.
2) p.135 No. 3,4 from the assistant P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova "Khimiya: 8kl.", 2013 p
The lesson of the assignments of the continuation of the vivchennya by those "Equivalent chemistry reactions". In the process of looking at the simplest rosary for the equal chemical reactions, related to the number of speeches, to take part in the reaction.
Topic: Pochatkovy Chemistry Announcement
Lesson: Chemistry of Chemistry
1. Spіvvіdnoshennya number of speeches, to take part in the reaction
The performance in the reaction does not show only the number of molecules of the cutaneous speech, but the ratio of the number of words, which will take part in the reaction. So, according to the reaction: 2H2 + O2 = 2H2O - it is possible to use it, but for the establishment of a singing water (for example, 2 mol), you need a bottle of water (2 mol) and two times less than a simple acid ). Put some rosary on top of it.
2. Zavdannya 1
FUNCTIONAL 1. It is significant that the number of words is sour, for the result of the distribution of 4 moles of water.
ALGORITHM for solving problems:
1. The areas of equal reaction
2. The proportions, meaning the number of words for the equal reaction and for the brain problem (denote the number of words for x mol).
3. Rivnyannya flanks (proportions).
4. Virishiti rivnyannya, know x.
Small. 1. Designing a short mind and solution of tasks 1
3. Zavdannya 2ZAVDANNA 2. How much sour is required for the repeated burning of 3 mol mіdі?Scoring by the algorithm the revision of the building from the victories of the chemical reaction.
Small. 2. Designing a short mind and solution of tasks 2.
It is important to add algorithms and write it to the program
I... Vikoristovuchi algorithm, check the self-advancement:
1. Calculate the amount of speech in aluminum oxide, which is accepted as a result of the interaction of aluminum oxide, 0.27 mol with a sufficient amount of acid (4Al + 3O 2 = 2Al2 O3).
2. Calculate the amount of speech to sodium oxide, after confirming the result of the interaction of sodium to the amount of speech 2.3 mol with a sufficient amount of sour(4Na + O2 = 2Na2 O).
algorithm # 1
The number of speeches is enumerated according to the number of speeches, so that we will take part in the reaction.
Butt. Count the number of speech words, which was seen as a result of the distribution of water and a number of words of 6 mol.
II. Vikoristovuchi algorithm, check the self-advancement:
1. Calculate the mass of gray, necessary for removing the oxide of gray (IV), a quantity of 4 mol (S + O 2
= SO2).
2. Calculate the weight of lithium required for removing lithium chloride with a lithium concentration of 0.6 mol (2Li + Cl2 = 2LiCl).
algorithm # 2
Enumeration of the mass of speech for the view of many of the speech, so that we will take part in the reaction.
Butt: Calculate the mass of aluminum required for removing aluminum oxide with a quantity of 8 mol.
III. Vikoristovuchi algorithm, check the self-advancement:
1. Calculate the amount of speech to sodium sulfide, when the reaction with sodium is added with a weight of 12.8 g (2Na + S = Na2S).
2. Calculate the number of words accepted as medium, if the reaction with water enters the oxide medium (II) weighing 64 g (CuO + H2 = Cu + H2 O).
It is important to implement the algorithm and write it into the file
algorithm # 3
Enumeration of a number of speeches for all kinds of speeches, so that you can take part in the reaction.
Butt. Calculate the amount of speech to oxide midi (I), as in the reaction with acidic entry into the mass of 19.2g.
It is important to implement the algorithm and write it into the file
IV. Vikoristovuchi algorithm, check the self-advancement:
1. Calculate the mass of sour, necessary for the reaction with a mass of 112 g
(3Fe + 4O2 = Fe3 O4).
algorithm # 4
The enumeration of the mass of speech for the view of the mass of the speech, that will take care of the fate of the reaction
Butt. Calculate the mass of sour, necessary for the combustion of phosphorus, in the mass of 0.31 g.
ZAVDANNYA FOR SELF-ROCKER
1. Calculate the amount of aluminium oxide, having accepted the result of the interaction of aluminum oxide with 0.27 mol with a sufficient amount of acid (4Al + 3O2 = 2Al2 O3).
2. Calculate the amount of speech for sodium oxide, after confirming the result of the interaction of sodium with a amount of 2.3 mol with a sufficient amount of sour acid (4Na + O2 = 2Na2 O).
3. Calculate the mass of serum required for removing serum oxide (IV) from the amount of 4 mol (S + O2 = SO2).
4. Calculate the lithium mass required to remove lithium chloride with a lithium chloride of 0.6 mol (2Li + Cl2 = 2LiCl).
5. Calculate the amount of speech sodium sulfide, when sodium sulfide enters into the reaction with a mass of 12.8 g (2Na + S = Na2 S).
6. Count the number of words accepted as medium, if the reaction with water enters the oxide medium (II) with a mass of 64 g (CuO + H2 = Cu + H2 O).
EXERCISE
Simulator No. 1 - Analysis of the chemical reaction
Exercise machine No. 6 - Technological dimensions
For rozrakhuniv, it is even more important to vibrate one to one single unit of mass, the volume of the same number of words. You can use the table 7 for quick reference.
Table 7
Spіvvіdnoshennya deyakim units of physical and chemical values
For this purpose, it is possible to speed up the problem with the help of an offensive algorithm.
- The area of the same chemistry reaction.
- Convert dan zavdannya (masu abo obsyag) to number of words (mol, kmol, mmol).
Just as for the intellectual task, speech enters into the reaction, and to revenge the houses, then there is a need to rely on a lot of pure speech, and then to develop this number; If in the task of moving the word about the discourse, then the collection of the demand must be counted the mass of the disclosed speech, which can then be translated into the number of the discourse.
- Above the general formula, write down the number of words above the formula, and denote the number of words over the formulas of the words used through x and y.
- Know a few words of talk, I know, that there are a few of those words between words of talk about the factors before these formulas in the simple reactions.
- Change the knowledge of the number of speeches in the masu abo obsyag.
- Submit a view.
Supervisor 1. Root up the water supply (n. U.), which will be required for the exchange of 480 kg of oxide of zaliza (III). Count the number of speeches in the drive, as if you were at it.
2. a) In order to know the number of words for the oxide of zaliza (III), the range of growth of this molar mass (in this type of decay it is polymolar, as the mass of oxide zaliza (III) is given in kilograms (div. Table 7)):
Mr (Fe 2 O 3) = 56 2 + 16 3 = 160;
M (Fe 2 O 3) = 160 kg / kmol.
b) We know the number of words for the oxide of zaliza (III) in kilomols, as it is given in the factory proponated in kilograms:
3. Above the formula of the speeches - the oxide of zaliz (III) - in the usual reaction, the number of speeches is 3 kmol, and the number of speeches is 3 kmol, and the number of speeches is water and water, and meaningfully over these formulas, as follows:
4. a) According to the reaction, 1 kmol of zaliza (III) oxide together with 3 kmol of water. Otzhe, 3 kmol oxide of zaliza (III) is reduced by three times more number speech vodnyu, i.e. x = 9 kmol.
b) Rozrakhumo obsyag vodnyu according to the known number of words:
5. As a matter of fact, prior to the reaction, 3 kmol of salt (III) oxide is set at 3 kmol of water, and 3 kmol of oxide of salt (III) is three times as much as the amount of water, i.e. Y = 9 mol.
Type: V (H 2) = 201.6 m 3; n (H 2 O) = 9 kmol.
Zabdannya 2. Yakiy obsyag povitrya (n. U.) Will it be necessary for the exchange of 270 g of aluminum, why should we get 20% of the houses? What is the amount of speech for aluminum oxide with a whole species?
3. Above the formula of aluminum in an equal reaction, the written down quantity is 8 mol, and the amount of acid and aluminum oxide is signifi- cantly indicated through x and y:
4. According to the previous reaction, 4 mol of aluminum is combined with 3 mol of acid, as a result of which 2 mol of aluminum oxide is added. From the same, 8 mol of aluminum is converted into 6 mol of acid and 4 mol of aluminum oxide.
n (O 2) = 6 mol; n (A1 2 O 3) = 4 mol.
5. Rosrakhumo obsyag sisnu according to the known number of words:
V (O 2) = n (O 2) Vm;
V (O 2) = 6 mol 22.4 L / mol = 134.4 L.
6. However, in the tasks, it is necessary to know that the obsyag is not sour, but povtrya. In every case, 21% of sourness for volume is avenged. Reconstructing the formula φ = V (O 2) / V (pov), we know the volume of the wind:
V (pov) = V (O 2): φ (O 2) = 134.4: 0.21 = 640 (l).
Type: V (ref) = 640 l; n (Al 2 O 3) = 4 mol.
Zabdannya 3. Yakiy obsyag vodnyu (n. U.) To see if 730 g of 30% hydrochloric acid and zinc, necessary for the reaction, are needed? How will you feel a bit of speech?
3. Above the chlorine-water formula in an equal reaction, the number of words is 6 mol, and the number of words for zinc and water is signifi- cantly indicated through x and y:
4. According to the reaction, 2 mol of chlorine water is combined with 1 mol of zinc, as a result of which 1 mol of water is consumed. From the same, 6 moles of chlorine water give 3 moles of zinc and 3 moles of water.
5. Rozrakhumo obsyag vodnyu according to the known number:
V (H 2) = n (H 2) V m;
V (H 2) = 3 mol 22.4 L / mol = 67.2 L.
Type: V (H 2) = 67.2 l; n (Zn) = 3 mol.
Key words and words
- One of the most important values.
- Algorithm for calculating for equal reactions.
A robot with a computer
- Become up to electronic dodatkom. Vivchit material lesson and vikonayte proponated zavdannya.
- Know in the Internet the electronic addresses, which can serve as additional dzherels, so that you can open the change of key words and the word of the paragraph. Give the reader your help in preparing a new lesson - to learn more about the key words and words of the offensive paragraph.
Power supply and power supply
- Yakiy obsyag vodnyu (n. U.) The first part of the speech of salt is established with the interaction of hydrochloric acid from 540 mg of aluminum, why should 4% of the houses be removed?
- Yaka masa to calcium oxide weyde with 250 kg of calcium carbonate, why should 20% of the houses be covered? Yakiy obsyag (n. U.) Oxide in carbon (IV) when you see it?
- Molecules of sour molecules and what kind of water (n.U.)
- Think up the mind of the tasks, in which it is necessary to vikoristovuvati aim at the bottom of the line, and vyrishite:
H 3 PO 5 + 3NaOH = Na 3 PO 4 + 3H 2 O.
- Think up and solve the problem, in the minds of what kind of speech will be given the mass of the line of speech with the singing mass fraction of the speech, and you will need to know the number of speech of one of the speeches and the generals of the person. When you have created a manager, you can use the same reaction:
Zn + H 2 SO 4 = ZnSO 4 + H 2.
Why don't you come, you
vchishsya for yourself.
Petroniy
Meta lesson:
- Learn about the main ways of finding a factory for chemistry:
- I know the number, the amount and the amount of reaction products for the number,
- to promote the form of a navichok robot with the text. primary tasks, Vmіnnya warehousing of chemical reactions.
- develop vmіnnya analizuvati, razvnyuvati, vidіlyati smut, put together a plan for diy, robiti visnovka.
- vikhovuvati tolerance to others, self-reliance in accepting decisions, while objectively evaluating the results of their work.
Forms of robots: frontal, individual, pair, group.
Lesson type: combinations from ІKT stasis
I Organizational moment.
Privit lads. Today, we will be able to see you with the help of your own chemistry reactions. Slide 1 (div. Presentation).
Meta lesson Slide 2.
II.Actualization of knowledge, mind, navichok.
Chemistry is even more tricky, and science is foldable at the same hour. For the sake of the nobility and the intelligence of chemistry, it is not only necessary to acquire the material, but to fix the knowledge. If there were any signs to indicate an overload of chemical reactions, there were some indications of chemical reactions. I am grateful that, for good reason, I have acquired those and without the help of my power supply
I don’t know the cheeky re-creation:
a) setting up a siege; c) zmіna obsyagu;
b) seeing gas; d) scent. slide 3
Add in numbers:
a) equal reactions
b) the equal reaction of the deputy
c) equal reactions to the distribution slide 4
In order for you to see the result, you need to use an algorithm for doing it, so that the end result is important.
Algorithm for rozrakhunku for cheeky ryvnyanny (at the skin specialist on the table)
5. Write down the message.
Start up until the date of the project, start the algorithm
The enumeration of the mass of speech for the view of the mass of the speech, that will take care of the fate of the reaction
Calculate the amount of money that was seen in the results of the distribution
water port with a weight of 9 m
We know the molar mass of water and sour:
M (H 2 O) = 18 g / mol
M (O 2) = 32 g / mol slide 6
Let's write down the chemistry reaction:
2H 2 O = 2H 2 + O 2
Above the formula in the equal reaction, we can write down
the value of the number of words, and under the formulas of the words -
stoichiometric relationship, as displayed
cheeky people
0.5 good x mol
2H 2 O = 2H 2 + O 2
2mol 1mol
The number of speeches is numbered, which is necessary to know.
For the whole warehouse in a proportion
0.5 good = hmol
2mol 1mol
stars x = 0.25 mol slide 7
Already, n (O 2) = 0.25 mol
We know Masu rechovini, yak needs to be counted
m (O 2) = n (O 2) * M (O 2)
m (O 2) = 0.25 mol 32 g / mol = 8 g
write down
Type: m (О 2) = 8 g slide 8
The enumeration of the ossuary of speech for the vіdomoyu masoyu іnshoї speeches, so take care of the fate in the reaction
Calculate the volume of water (n.U.)
V (0 2) =? L (n.o.)
M (H 2 O) = 18 g / mol
Vm = 22.4 l / mol slide 9
Recordable reaction. configurable functionality
2H 2 O = 2H 2 + O 2
Above the formula in the equal reaction, we can write down the meaning of the number of speech, and under the formulas of the speech, it is stoichiometric, as it appears to the cheeky people
0.5 good - x mol
2H 2 O = 2H 2 + O 2 Slide10
2mol - 1mol
The number of speeches is numbered, which is necessary to know. For the whole foldable proportion
stars x = 0.25 mol
We know that there is a lot of speech, which needs to be counted
V (0 2) = n (0 2) Vm
V (O 2) = 0.25 mol 22.4 L / mol = 5.6 L (n.U.)
Type: 5.6 l slide 11
III. Securing the implanted material.
Head for an independent solution:
1.When the oxides of Fe 2 O 3 and SnO 2 were renewed, 20 g of Fe and Sn were removed each. How many grams of dermal oxide are taken?
2.In which case, pretend to be more water:
a) with the addition of 10 g of midi (I) oxide (Cu 2 O) abo
b) with 10 g of medium (II) oxide (CuO)? slide 12
Revise the solution of tasks 1
M (Fe 2 O 3) = 160g / mol
M (Fe) = 56g / mol, 
m (Fe 2 O 3) =, m (Fe 2 O 3) = 0.18 * 160 = 28.6 g
Type: 28.6g
slide 13
Revise the solution of tasks 2
M (CuO) = 80 g / mol
4. 
x mol = 0.07 mol,
n (H 2 O) = 0.07 mol
m (H 2 O) = 0.07 mol * 18g / mol = 1.26g
slide 14
CuO + H 2 = Cu + H 2 O
n (CuO) = m / M (CuO)
n (CuO) = 10g / 80g / mol = 0.125 mol
0.125 mol hmol
CuO + H 2 = Cu + H 2 O
1 mol 1 mol
x mol = 0.125 mol, n (H 2 O) = 0.125 mol
m (H 2 O) = n * M (H 2 O);
m (H 2 O) = 0.125 mol * 18 g / mol = 2.25 g
Type: 2.25g slide 15
Home office: vivchiti material of the handler with. 45-47, Virishity task
Yaku masu calcium oxide and yaky ob'm in carbon dioxide (n.u.)
can be trimmed with 250g calcium carbonate?
CaCO 3 = CaO + CO Slide 16.
literature
1. Gabrielian O.S. Program for the course of chemistry for 8-11 grades in English. M. Bustard 2006r.
2. Gabrielian O.S. Chemia. 8 class. Pidruchnik for outbound installation. Bustard. M. 2005r.
3. Gorbuntsova S.V. Tests of the main distributions of the school course хіііі. 8 - 9 classrooms, VAKO, Moscow, 2006
4. Gorkovenko M.Yu. Before the assistants of O.S. Gabrielian, L. S. Guzei, V. V. Sorokin, R. P. Surovtseva and G. E. Rudzitis, F. G. Feldman. Grade 8, VAKO, Moscow, 2004
5. Gabrielian O.S. Chemia. 8th class: Controlling conversion robots. - M .: Bustard, 2003.
6.Radetskiy A.M., Gorshkova V.P. Didactic material from chemistry for 8-9 grades: Book for the reader. - M .: Education, 2000.
Dodatok.
Rozrakhunks for cheeky rivnyanny
Algorithm diy.
In order to complete a rozrahunkov task in chemistry, you can quickly use the next algorithm - to create five crocs:
1. Areas of the same chemistry reaction.
2. Above the formulas of words, write down types and sizes with specific units of vimir (only for pure words, without houses). As soon as for the intellectual task, speech enters into the reaction, and to revenge the houses, then a handful of things is needed instead of pure speech.
3. According to the formulas of the words with the given and unavailable records of the given values of the values, known for the equal reactions.
4. The folds and proportions.
5. Write down the message.
Spіvіdnіchnya deyakіkh fіsiko-khіmіchnіy values і їх odinitsa
Maca (m): g; kg; mg
Number of substances (n): mol; kmol; mmol
Molar mass (M): g / mol; kg / kmol; mg / mmol
Obsyag (V): l; m 3 / kmol; ml
Molar volume (Vm): l / mol; m 3 / kmol; ml / mmol
Number of particles (N) 6 1023 (Avagadro number - N A); 6 1026; 6 1020
Zavdannya. Skіlki lіtrіv sisnyu (n. U.) Enter into the reaction when 4.8 g of magnesium is burned?
Periodic law (PS) and periodic system (PS)
D. I. Elementiv Mendeleva
The announcement of the PS and the Pobudova PS was the pinnacle of the development of chemistry in the 19th century (1869). D.I. Mendeleev roztashuv in all vіdomі at that hour of the element (63) in the order of growth of the atomic masses and in tsomu having appeared the ring of authorities chemical elements with the atomic masses, as she fell in that, through the singing intervals of the power of the elements were repeated. D. I. Mendeleev having formulated the periodic law as follows: The power of simple words, as well as the form and power of a single element, are found in periodic depletion due to the value of atomic oils.
Unimportant for all the magnitude of the significance of such a visnovka, PZ and PS represented a deprivation of the genially empirical (experimental) knowledge of facts, and physical zmist For the last hour, I was unreasonable. The reason for this is that in the 19th century there was absolutely no information about the folding of the atom.
Most often there are three options for PS:
1. short-period;
2.withdrawn (all elements of the 4th and 5th periods are in one line of 18 elements;
3. long-term (in one line, all s, p, d and f elements.
The short form of PS is stored for 7 periods and 8 groups.
Period - a single horizontal row, which can be repaired with metal(For the first period) and end with an inert element(Except this period).
The first, the other and the third period are stored in one row and are called malimi. Quarters, ideas and the first time are stacked in two rows and are called great. all in periodic systems 10 rows The top row is guy, the bottom row is unpaired. Paired rows take revenge only threw and the power of the elements, there is little change to the right. The guy's row of the great period will end in three ways, follow the power of the elements: the trinity. Unpaired rows were revenged by metal and non-metal, in them there is evil to the right and there are steps going from metal powers to non-metal ones.
In the case of the post-lanthanum period La (No. 57), 14 elements with similar powers (No. 58 - 71) are roasted: lanthanum. All the stinks are reactive, react with water, they have a strong horizontal analogy.
In the same period of time the actinia Ac (No. 89) has a similar arrangement of 14 elements (No. 90 - 103), similar acts: actinous. The nuclei of їх atoms in the edges of the nestіyki, so that they are radioactive.
The skin group is stored in two groups: head and side.
Pidgruppies, which include elements of small and great periods, are called heads (A). Pidgroups, which include elements of only the great periods, are called side-by-side (B). Pidgroup combine most of the same items.
For elements of one group, the following laws are characteristic:
1. All the elements, except for noble gases, set up sour spheres.
2. Food valency and food is positive step of oxidation depending on the number of the group. Vinyats: 1) in the 8th group only ruthenia Ru and Os valency VIII; Cu +1, Cu +2; O -2; F-1.
3. The elements of the head group from IV to VIII of the group set up the holes in the water. The valence of the number in the number of 8 and the number of the group. For example, N is in the V-th group and the valence of the road is 8 - 5 = 3 in NH 3.
budova atom
At the XIX century. respected that the atom is not a part of it, as it does not change when chemical reactions... At the end of the XIX - the ear of the XX century. bully vіdkrіtі roentgenіvske vipromіnuvannya (nіmetsky K. Roentgenom, 1895), radioactivity (French student A. Becquerel, 1896), elektron (English clergy J. Thomson, 1897). Massa m (e) = 9.109 × 10 -28 g і negative charge q (e) = 1.602 × 10 -19 C. The magnitude of the electric charge is taken as a unit of the elementary electric charge.
In 1903, J. Thomson proponated the model of the atom’s budget, then there was a positive charge equal to the distribution of the atom and neutralization by impregnated electrons. Razvivayuchi cі uyavlennya, E. Rutherford in 1911r. proponuvav planetary model of the budovi atom. For the whole theory in the center of the atom there is a positively charged nucleus, near which the electronics collapse. The number of electrons in the atom is called yogo electronic shell. In 1913, D. Mozli's English doctrine showed that the value of the positive charge of the nucleus of an atom is important to the ordinal number of an element in the periodic system of elements D.I. Mendeleva. Atom is electrically neutral, since the number of electrons in the electronic shell of the atom for the nuclear charge Z or the ordinal number of the element in the periodic system.
In 1932, D. D. Ivanenko's and E. N. Gapon's radianskі vchenі, right from them, V. Heisenberg's note was taken proton-neutron theory of budov and nuclei. Proton p - tse part with a mass, scho dorіvnyuє 1 а. eat.
(1.66 × 10 -24 g), i charge + 1. neutron n is an entire electroneutral particle with a mass close to the mass of a proton. Protons and neutrons are called nucleons.
The charge of the nucleus of an atom is determined by the number of protons. already, the number of protons in the nucleus of an atom is also relevant to the ordinal number of an element in a periodic system... The initial number of protons and neutrons is called the mass number (A). Vono to the rounded up to an integer number of values of the atomic mass.
Zavdannya. What is the charge of the nucleus and the amount of electrons, protons, neutrons in the zinc atom?
Z = + 30, p = 30, e = 30, n = 65-30 = 35.
isotopi
The types of atoms of one element, which have the same charges of nuclei, and different mass numbers (the same number of protons and the same number of neutrons), are called isotopes. Cheerful authorities all isotopes of one element are the same.
Leather isotopes are characterized by two values: mass number (put down at the top of the evil sign from the chemical sign) і serial number (put down at the bottom of the evil sign from the chemical sign) і signified by the symbol of the specific element. For example, an element of water has three isotopes. H - counter (1 p); D (H) - deuterium (1p, 1 n); T (H) - tritium (1 p, 2 n).