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00:00:08

hello guys, my name is

00:00:10

Tatyana Gennadievna Belousova, I’m a chemistry teacher.

00:00:13

Today we continue to prepare

00:00:15

for the unified national test and are

00:00:18

solving the next option, so question

00:00:20

number one sounds like this: the

00:00:23

number of the period in the periodic table of

00:00:26

chemical elements is shown among the

00:00:29

options here is the charge of the nucleus the number of

00:00:31

neutrons the number of electrons layers, the number of

00:00:34

electrons and the number of protons, which

00:00:38

shows the period number, the period number

00:00:40

shows the number of

00:00:43

energy levels of

00:00:46

energy layers or levels, and the

00:00:50

period number reflects exactly this moment and reflects,

00:00:52

for example, for hydrogen, period number 1

00:00:54

means it has one energy

00:00:57

level, then question number two, the

00:01:01

salt undergoes complete hydrolysis in

00:01:03

order to answer this question, and here

00:01:06

among the given substances there is

00:01:08

magnesium sulfate, potassium carbonate, sodium chloride,

00:01:11

silver bromide and aluminum sulfide, I

00:01:15

strongly advise you to open and look at the

00:01:17

periodic solubility table, you

00:01:20

will see for one of the substances, and this is

00:01:23

aluminum sulfide, what is in its

00:01:27

solubility table there is such a sign, this

00:01:30

sign means that complete hydrolysis is taking place

00:01:32

and such a salt

00:01:35

will not exist in solution, but what will

00:01:37

happen to it is just the equation,

00:01:38

let’s show it happens very

00:01:42

quickly, since the salt is formed by a weak

00:01:45

base, it will be obtained here

00:01:48

aluminum already three times precipitates

00:01:50

and hydrogen sulfide gas is released since this is

00:01:54

also a weak acid, it is released

00:01:57

during this irreversible hydrolysis, here we

00:02:00

will put 2 here 3 and accordingly here

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six I draw your attention to the fact that such

00:02:06

tasks also occur when you

00:02:08

might be asked exactly this

00:02:11

equation or for example what level

00:02:12

corresponds to this hydrolysis in

00:02:15

all other cases,

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hydrolysis will occur except for

00:02:20

sodium chloride since this salt

00:02:22

is formed by a strong base and a strong

00:02:25

acid, but hydrolysis will occur to the end

00:02:28

if only both weak

00:02:31

electrolytes, then the next third

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question is the question of deduction, starch turns blue when

00:02:39

present in solution, and so on among the

00:02:42

answer options there is bromine 2 as that as that

00:02:48

2 there is fluorine in the cake 2 chlorine 2 and j2

00:02:55

but I really hope that you know about such a

00:02:58

special feature of iodine that iodine

00:03:01

has a qualitative reaction to starch,

00:03:04

this is a mutual reaction, that is, I can

00:03:07

say iodine can be detected with the help of

00:03:09

starch or starch is detected with the

00:03:12

help of iodine indeed in this case a

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blue color appears,

00:03:17

this is a qualitative reaction specifically to

00:03:21

starch and j fourth question why do you

00:03:25

have semiconductor properties

00:03:27

among the given answers there is corbin

00:03:31

graphite carbon black fullerene diamond I draw your

00:03:35

attention to that that the substances I have listed

00:03:38

are all allotropic modifications of

00:03:41

carbon, every single one, and so the

00:03:47

substance graphite has semiconducting properties

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graphite is a substance that conducts current,

00:03:53

unlike all the others, diamonds do not

00:03:55

conduct, fullerene does not conduct and,

00:03:58

accordingly, the remaining substances are also in

00:04:01

this group in the fourth group of semi-conductors

00:04:05

Silicon also has semiconductor properties; it is a

00:04:09

semiconductor, so the next fifth

00:04:12

question is glycerin,

00:04:16

which type of substance is an integral part, but first,

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let’s recall that the formula of glycerin in a

00:04:21

folded form is c3 h8 o 3, this substance

00:04:27

is a 3-atomic alcohol, but

00:04:32

triatomic alcohol, glycerin, is an

00:04:34

integral part of substances which we

00:04:37

call fat and fats are esters

00:04:42

that are formed precisely by glycerol,

00:04:47

glycerol and,

00:04:50

accordingly, fatty carboxylic

00:04:53

acids,

00:04:54

so all the

00:04:57

resulting fats are sometimes called the word

00:05:00

glycerides, strained, palmitic

00:05:03

acid reigns, and the fat and only

00:05:07

glycerin contain no other alcohol and this does not

00:05:10

include the next question, sixth question

00:05:13

is so common for starch and cellulose

00:05:15

is, but first let’s remember

00:05:19

that these substances are polymers, the

00:05:23

formula of these substances obeys the general

00:05:27

formula c 6 h 10 again, such a formula in

00:05:31

collapsed form

00:05:33

really corresponds to both starch and

00:05:36

cellulose, but these substances are different and

00:05:40

that’s what we need the question is asked what unites them,

00:05:44

so the first answer sounds like passion the

00:05:47

structure of the simplest link it will not be the

00:05:50

same because at least starch

00:05:53

consists of alpha glucose residues for

00:05:58

cellulose the component would be

00:06:00

glucose, which means the structural link

00:06:04

they have the same in general cannot

00:06:06

be further structure of this macromolecule,

00:06:10

the structure of macromolecules and the consequences of this

00:06:12

also cannot be the same, so

00:06:14

we will also reject this answer, the degree

00:06:18

of polymerization no, the degree of polymerization

00:06:20

of starch is significantly less than that of

00:06:24

cellulose, then the molecular weight is no

00:06:28

and due to the fact that the degree of polymerization

00:06:30

they still have different molecular

00:06:33

weight There can’t be one such polymer, ok,

00:06:35

but the composition of the simplest unit,

00:06:38

the composition is exactly the amount of carbon,

00:06:41

hydrogen and oxygen, so it will be

00:06:44

exactly the same for them, and so the common

00:06:47

property is the composition of their simplest

00:06:52

unit,

00:06:54

but these are different substances, these are polymers, but

00:06:58

polymers have different unit structures so the

00:07:02

seventh question the question sounds like this:

00:07:05

they ask in 0 2 mole of the substance

00:07:10

that we are named here hexa,

00:07:14

how many molecules does it contain and

00:07:19

so asks the number of molecules the number of

00:07:21

molecules is found by the formula by whose words of

00:07:25

the frame multiplied by the number mole therefore

00:07:28

whether it is hexane or any other

00:07:30

thing, but here it’s not fundamentally

00:07:33

Avogadro’s number, we have 602 over 10 to the 23rd

00:07:38

power of mole to the minus first power, and

00:07:41

for the amount of substance 0 2 mole of mine, let’s

00:07:46

substitute the moles moles will obviously cancel

00:07:49

and it will be one integer

00:07:52

204 over 10 23 this will be the answer and here

00:07:59

here I would like to draw your attention

00:08:00

to the fact that instead of hexane there could be

00:08:03

heptane and ubthane and methane here in general this

00:08:07

in this case does not matter, it is important

00:08:10

that the problem is solved precisely according to this

00:08:12

formula eighth task eighth task

00:08:15

this task we are told that when acting

00:08:19

at

00:08:21

90 grams of a 70 percent solution of

00:08:24

acetic acid, this acid acts

00:08:27

on 84 grams of sodium bicarbonate,

00:08:31

they ask what volume of carbon dioxide

00:08:34

will be released here, and so the conditions of the

00:08:37

problem are that a

00:08:38

solution of acetic acid is taken and

00:08:43

this solution is 90 grams and there is a badolho maz in

00:08:48

it of acid 70 percent for this solution

00:08:53

this solution they process

00:08:55

sodium bicarbonate

00:08:57

sodium bicarbonate is sodium, as many as

00:09:00

three baking soda is called the trivial

00:09:03

name

00:09:04

84 grams and asks us about the volume of

00:09:07

carbon dioxide released in this case,

00:09:10

the conditions are normal and so let's start with the

00:09:13

reaction equation acetic acid

00:09:17

will react with sodium bicarbonate,

00:09:20

each of you has probably done this experiment

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or I saw because quenching

00:09:26

soda with vinegar is a popular reaction in everyday life;

00:09:29

in this case,

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sodium acetate is obtained,

00:09:34

but the carbonic acid released in the process

00:09:38

breaks down into co2 and h2o and

00:09:42

you see gas bubbles in this case,

00:09:44

acetic acid releases carbonic

00:09:48

acid, if only because carbonic

00:09:50

acid is weaker and so we

00:09:53

know acetic acid 10 we

00:09:57

know

00:09:58

sodium bicarbonate

00:10:00

its pulse stop in the opinion of the reaction 10 and we

00:10:03

are asked about co2 here also 10 but the

00:10:06

substance is not given explicitly and

00:10:09

therefore we will do the calculation and so

00:10:12

first we will calculate how much us

00:10:15

the mass of the acid itself

00:10:17

for this is 90 grams this is one hundred percent x

00:10:22

gram this is 70 means 90 multiplied by 0 7 and

00:10:27

in this case we get the mass of the

00:10:31

acid solution in the solution is 63 grams and

00:10:35

therefore a mule of acetic acid

00:10:39

will be 63 grams we divide by the molar

00:10:43

mass of acetic acid I highly recommend

00:10:46

you often find this acid in calculation

00:10:49

problems, remember, and massana 60 is 12

00:10:54

plus one knife for 3 + 12 two oxygens

00:10:58

is another 32 and plus hydrogen one for 60

00:11:02

grams per mole,

00:11:03

therefore the acid is taken as one whole

00:11:08

050, let’s substitute this given above the acid

00:11:12

1050 and

00:11:15

sodium bicarbonate given by mass

00:11:18

let's find molly of sodium bicarbonate the

00:11:22

same formula works so mass

00:11:26

per molar is 84 grams if we calculate the

00:11:29

molar mass of this substance 23 + 1 + 12

00:11:34

plus 16 multiplied by 3 it will be 84 grams

00:11:39

per mole which means we have taken 10 from our court

00:11:43

at this stage we can see that

00:11:47

to find

00:11:50

the moles of co2 we need to determine which

00:11:53

substance is taken in excess since their

00:11:55

moles according to the reaction equation are equal, we

00:11:59

can compare these numbers and it is clearly

00:12:01

seen that sodium bicarbonate is a

00:12:04

lack of nitrogen, acetic acid is taken

00:12:07

in excess and and a small amount

00:12:09

is excess but and x which we

00:12:13

see from the equation will obviously

00:12:16

coincide with moles

00:12:18

of bicarbonate, which means it will be 10 co2 and

00:12:23

since we are talking about gas co2 video

00:12:28

volume, then all that remains is to calculate it

00:12:31

will be in m multiplied by mole the molar

00:12:34

volume of any gas 20 24

00:12:36

leader at sea multiplied by 10

00:12:39

therefore, this volume will be equal to 20

00:12:43

2 and 4 liters, this will be the correct answer,

00:12:48

so you have solved up to 8 tasks in the

00:12:51

next lesson, we will continue to solve this

00:12:54

option, so I invite you to the

00:12:56

next lesson, and for today,

00:12:58

goodbye guys

00:13:01

[music]

00:13:10

[music]

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