moscow city exam

7/29/2019 Moscow City Exam

1/13

8th Grade

1.8 Using the periodic table can predict the properties of unknown elements, by analogy with the

known properties. Complete the equation of the following reactions (they are).

Sr (OH) 2 + HBr = SrCl 2 + Rb 2 SO 4 = Sr + Cl 2 =Sr + O 2 Sr + HBr = Sr + H 2 O =

SrCO 3 + HBr = Sr + Fe 2 O 3 SrO + HBr =

Sr (ClO 4 ) 2 + C SrO + H 2 O = SrCl 2 + AgNO 3 =

8.2 Solution substance A in liquid B contains only hydrogen, oxygen and sodium. Solution of the

substance in the liquid B contains only hydrogen, oxygen and chlorine. When pumping these solutions

generates heat. When added to a solution ofa solution of the substance D precipitate containing only

the silver and oxygen.

What may be the substance A , B , In , and D ? Write the equation of all these reactions.

8.3 Shapoklyak old woman decided to cook guncotton (C 6 H 7 O 2 (NO 3 ) 3 ) n by the reaction

(C 6 H 10 O 5 ) n 3 + n HNO 3 = (C 6 H 7 O 2 (NO 3 ) 3 ) n 3 + n H 2 O, and bought a bottle containing 20 kg of

acid.

1. What you can do a lot of pyroxylin of this nitric acid?

2. What other reagents should purchase Shapoklyak old woman?

8.4 The young chemist decided to inflate hydrogen balloon volume of 1 m 3 (based on the

condition). Hydrogen for this, he decided to get out of zinc and hydrochloric acid. Zinc is 240 $ / kg, 36%

hydrochloric acid technical costs 5 $ / kg.

How much will this experiment the parents of the young chemist?

5.8 The figure shows the dependence of the density of acetic acid concentration.

Someone mixed 1.000 liters of water and 1,000 liters of acetic acid (CH 3 COOH).

1. What will be the mass fraction of acetic acid in the resulting solution?

2. What will be the volume of the resulting solution?

3. How many moles of acid will be present in 1 liter of this solution?

4. How many water molecules have for one molecule of acetic acid, with a maximum density?

SOLUTION

8/1. Sr (OH) 2 + 2HBr = 2H 2 O + SrBr 2

SrCl 2 + Rb 2 SO 4 = SrSO 4 + 2RbCl

Sr + Cl 2 = SrCl 2

2Sr + O 2 2SrO

Sr + 2HBr SrBr = 2 + H 2

Sr + 2H 2 O = Sr (OH) 2 + H 2


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SrCO 3 + 2HBr = SrBr 2 + H 2 O + CO 2

3Sr + Fe 2 O 3t 3SrO 2Fe +

SrO + 2HBr SrBr = 2 + H 2 O

Sr ( ClO 4 ) 2 + 4C SrCl 2 + 4CO 2

SrO + H 2 O = Sr (OH) 2

SrCl 2 + 2AgNO 3 = 2AgCl + Sr (NO 3 ) 2

8.2. A = NaOH, B = H 2 O, in = HCl, or any of the oxygen-chlorine acid, D - any soluble silver salt.

NaOH + HCl = NaCl + H 2 O

+ 2NaOH 2AgNO 3 = 2NaNO 3 + Ag 2 O .

8.3. A 1 mol of HNO 3 receives 1/3 mol (C 6 H 7 O 2 (NO 3 ) 3 .)

Number of nitric acid 20000/63 = 317 mol.

So count pyroxylin (per monomer) 106 mol .

This count has a mass of 106 297 = 31 400 g = 4.31 kg.

B. Wool and sulfuric acid

8.4. chemist to 1000/22, 4 = 44.6 moles of hydrogen

Zn + 2HCl = ZnCl 2 + H 2

This requires 44.6 mol of zinc and 89.2 moles of hydrochloric acid.

weight Zn 44.6 ? 65 = 2900 It will cost 240 2.9 = 696 p.

weight HCl solution will be 89.2 36.5:0.36 = 9044 g

It will cost 9.044 5 = 45.22 rubles. Total - 742 rubles.

8.5. figure 1 liter of water weighs 1 kg, 1 acetic acid - 1,050 kg.

A. The total weight will be 2.050 kg, so the mass fraction of acetic acid 1.050/2.050 = 0.512.

B. The density of the resulting p-pa (from the graph) 1.058 kg / l.

Hence, its volume will be 2.050 / 1.058 = 1.938 liters.

B. This volume contains the 1050/60 = 17.5 mol.

So, in 1 liter of 17.5 / 1.938 = 9.03 mol.

G. In this area, 23 g of water for 77 grams of vinegar.

ratio is (23/18): (77/60) = 1:1.


3/13

9th Grade

1.9 orange color pigment ink - red lead - a composition Pb 3 O 4 .

1. What is the possible oxidation of lead in this compound?

2. What class of chemical compounds can be attributed this stuff?

3. In the interaction of red lead and coal can be metallic lead. Write the reaction.

4. How many grams of lead can be obtained from 6.85 g of red lead, if the output of the theoretically

possible reaction is 97%?

9.2 The substance weighing 14.9 g, formed by the two types of chemical species with identical electronic

structure 1s 2 2s 2 2p 6 3s 2 3p 6 , fully reacted with 20.0 g of 98% sulfuric acid.

1. Write the reaction.

2. Determine the mass of the resulting salt.

3. Write the reaction of an aqueous solution of the resulting salt with magnesium metal.

3.9 The reaction of aqueous solutions containing 36.5 g of hydrochloric acid and 40 g of sodium

hydroxide, to allocate 57 kJ of heat.

1. Write long and short ionic equation for the reaction.2. How much heat is released during the interaction 150 g of a 10% solution of sulfuric acid and 50 g of

11.2% aqueous solution of potassium hydroxide?

4.9 If the solid sodium nitrate added concentrated phosphoric acid, when heated, it can be a mixture of

nitric acid to drive away.

1. Write the reaction (phosphoric acid in excess).

2. Does this experiment, that phosphoric acid is stronger nitric?

3. Can I get the reaction of nitric acid 30% solution of sodium nitrate and phosphoric acid?

9.5 The reaction of 69.8 g of a mixture of carbonate and bicarbonate of the same alkali metal with

hydrochloric acid released 30.8 g of carbon dioxide (IV).Identify the alkali metal salts, and the mass of the original mixture.

9.6 The vessel at 300 K (degrees of absolute temperature) and 2 bar contains 1 mole of hydrogen and 1

mole of chlorine. After the mixture with ultraviolet light and the reaction vessel temperature of 600 K.

1. Write the reaction.

2. Why this reaction may start with light?

3. What process is the first stage of the reaction when illuminated?

4. What pressure (atm) in the vessel was measured temperature after the reaction?

5. How many grams of hydrogen chloride contained in the vessel after the reaction?

SOLUTION

9-1. 1) Pb 2 PbO 4 - oxidation states +2 and +4 or (not quite true, but estimated) Pb (PbO 2 ) two oxidation

states of +2 and +3 lead

2) This is a lead salt lead (leaded) acid

3) Pb 3 O 4 + 2 C = 3 Pb + 2 CO 2

4) 6.85 g of 0.01 mol minium, he will make 0.01 moles or 6.21 g of lead. Based out 6.21 g of lead x 0.97 =

6.02 g


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9-2. Electronic structure correspond to KCl, CaS, ScP and combinations of these cations and anions.

20 g of sulfuric acid is 20/98 = 0.204 mol

Then 14.9 g corresponds to 0.2 or 0.4 moles of salt.

Suitable KCl (74 and 5)

correspond to 14.9 g CaS 0.207 mol salt from it with sulfuric acid is formed only CaSO 4 , almost insoluble

in water and does not react in solution with magnesium. 14.9 g ScP correspond 0.196 mole of salt, with

sulfuric acid in the negative. Formal approaches K 2 S - completed secondary and hydrogen sulfate.

0.2 mol KCl and H 2 SO 4

1) KCl + H 2 SO 4 = HCl + KHSO 4

2) Get 0.2M or 27.2 g acid salt

3) 2 KHSO 4 + Mg = K 2 SO 4 + MgSO 4 + H 2

9-3. 1) HCl + NaOH = NaCl + H 2 O

H + + OH - = H 2 O

refers to the thermal effect of the neutralization reaction: H + + OH - = H 2 O + Q

2) and 36.5 g of hydrogen chloride 40 g of sodium hydroxide up to 1 mol, ie heat of neutralization Q = 57

kJ / mol. In 150 g of a 10% solution of 15 g of sulfuric acid, or 0,15-eaten

in 50 g of 11.2% aqueous solution of 5.6 g of potassium hydroxide and 0.1 mol. Sulfuric acid in excess,calculated from KOH

0.1 mol of water formed during the neutralization highlighted 57h0, 1 = 5.7 kJ

9-4. 1) H 3 PO 4 + NaNO 3 = NaH 2 PO 4 + HNO 3 (when heated),

2) the strength of acid is determined by its ability to dissociate in aqueous solution, as described in the

reaction equilibrium is shifted to the right due to the difference of volatility acids but not their

strength. IePhosphoric acid can be less nitrogen (as it really is.)

3) in aqueous solution at equilibrium will be a few ions. Since the reaction is not allocated gas does not

form sludge or water , nitric acid, in this case, can not be obtained.

9.5. 1) The reactions in general: MHCO 3 + HCl = MCl + CO 2 + H 2 OM 2 CO 3 + 2 HCl = 2 MCl + CO 2 + H 2 O

2) can be seen that for 1 mol CO 2 requires more carbonate by weight than hydrogen. Amount of CO 2 is

30.8 / 44 = 0.7 mol. If 69.8 g would be a pure hydrogen (0.7 mol MHCO 3 ), then the conditional atomic

weight of an alkali metal would be: (69.8 / 0.7) - 61 = 99.7 - 61 = 38.7

Similarly for pure carbonate obtain: (99.7 - 60) / 2 = 19.9

Therefore, we have a mixture of carbonate and bicarbonate, an alkali metal with an atomic mass of

between 19.9 and 38.7 - suitable only sodium.

3) Now we get a system of two equations with two unknowns:

For X mol Na 2 CO 3 (106) and V mol NaHCO 3 (84), we write

106H + 84U = 69.8

X + Y = 0.7

Solving the system, we get: X = 0.5, Y = 0.2

A: 0.5 M Na 2 CO 3 (53.0 g), 0.2 mol of NaHCO 3 (16.8 g) (double: the author's solution (A.E.Leontev):

The total stoichiometry. scheme for carbon:xMe 2 CO 3 + yMencia 3 = (x + y) CO 2

M (Me 2 CO 3 ) = 2 andM + 60 (Mensah 3 ) = a + 61


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n (CO 2 ) = 0.70 mol =

We get a system of two equations with three unknowns.

Since the total amount of material on both sides of the stoichiometric scheme are (x + y), we can

determine the numerical value of the average molar mass of the salt mixture.

= M = = 100 g / mol

Assume that the system consists of only one substance: or Me 2 CO 3 , or MENS 3 . In this case, should be

performed accordingly one of the equations.

For Me 2 CO 3 : 2a + 60 = 100, hence a = 20

For Mens 3 : a + 61 = 100 = 39 and hence the

assumption allows us to find the left and right edge of the interval that contains the desired value of

a. Obtain the double inequality: 20 < a


6/13

10th Grade

Job grade 10

1.10 Qualitative analysis revealed the presence of a mineral leonita it of potassium, magnesium and

sulfate ions. Ignition mineral specimen weighing 7.32 g, its mass is reduced to 1.44, the same linkage ofthe mineral when dissolved in water and the subsequent addition of an excess of barium chloride

solution forms a precipitate 9.32 g.

Determine the formula leonita.

10.02 A mixture of ethane, ethylene and propene has a density of hydrogen 15.9. To 1 liter of this

mixture was added 1 liter of hydrogen, and the mixture is passed over a platinum catalyst. The total

volume of gas at the outlet of the reactor was 1.5 liters.

Calculate the composition of the mixture of hydrocarbons in percent by volume, except that the

reaction was quantified (all volumes are measured at standard conditions).

3.10 The interaction of a mixture of two monobromalkanov with metallic sodium gave a mixture of

hydrocarbons A , B , and in (in order of molecular weight). Bromination And you can get two isomeric

monobromproizvodnyh with density in the air 4.72. Bromination in the results in three products with

the air density 6.66.

1. Determine the structural formulas of all these substances.

2. Give an equation for the reaction

3. How do you think which of the three brominated hydrocarbon products in produced in large

quantities, and why?

4.10 Gas A is obtained by heating pieces of wax with a simple substance X . Gas B formed by the

reaction of some metals with acid in the presence of water vapor. Gases A and B react with each other

in a volume ratio of 2:1, this gives substance X .

Substances Which is it? Write the reactions mentioned.

5.10 Here is an excerpt from the book by Astrid Lindgren's "The Adventures of Calle Blyumkvist" which

omitted certain words or phrases.

- This is a common bulb into which I pour _____ and I put a few pieces of ___. Here hydrogen is released,

right? If we now enter here arsenic in any desired form, you get gas _____. Hence the gas comes to dry

up in the dry ______, and then here in this narrow tube. Here we heat the gas in the spirit lamp, and it is

divided into _____ and _____. _____ Being deposited on the walls of the tube in the form of shiny gray-

black coating. The so-called arsenic _____ - I hope you've heard of him, my young friend. His young

friend about anything not heard, but it with unflagging interest the all the preparations. - But do not

forget - I do not say that chocolate does have arsenic - said Kalle, when he finally lit the spirit lamp. - I'mjust about to do a little experience and I sincerely hope that my suspicions were unfounded.

1. Insert the missing words

2. Write the reactions

3. Perhaps you know the name of the specified method of detecting arsenic. Which materials determine

arsenic (qualitative and quantitative), and for what?


7/13

6.10 Simple substance A interacts with a liquid consisting of the elements ofB , and In . This releases

the gas B 2 , and from the solution can be identified easily volatilized substance ofAB 3 , containing

33.33% of the element in the . AB 3 reacts with the gas B 2 , forming a substance A and the other

connection elements B and In .

Determine of any substances in question. Write the reactions.

SOLUTION

1.10 . calcining sample crystal water is removed.

define its mass: 7.32 - 1.44 = 5.88 g

By adding barium chloride barium sulfate is obtained:

M = 233, n = 9.32 / 233 = 0, 04 mol.

Thus, the original linkage contains 0.04 moles of sulfate ions. Mass of sulfate ions in the rigging - 3.84 g,

the mass amounts of metals - 2.04, the

formula to calculate the mineral form a system of equations:

2 n (Mg) + n (K) = 2 n (SO 42 - ) (condition electroneutrality.)

24 n (Mg) + 39 n (K) = 2.04 (mass of metal rigging mineral).

Solving the system of equations, we obtain n (Mg) = 0.02, n (K) = .04.Value K: Mg: SO 4 = 2: 1: 2. Thus, the formula - K 2 Mg (SO 4 ) 2 . Now you have to find the amount of

water of crystallization. 1.44 g of water is 0.08 mol - two times more than the number of moles of

sulfate.

Consequently, formula leonita - K 2 Mg (SO 4 ) 2 4 H2 O

10-2. By passing the mixture over a platinum catalyst is the hydrogenation of ethylene and propylene,

ethane remains unchanged.

Let the mixture contains ethane X l, Y l ethylene and propylene Zl.

C 2 H 4 + H 2 = C 2 H 6

C 3 H 6 + H 2 = C 3 H 8

Since one volume of the hydrocarbon reacts with 1 volume of hydrogen, and the result is a one volume,

a decrease in volume of the mixture (0.5 L) corresponds to the amount of reacted hydrogen, and the

total volume of ethylene and propylene in the mixture.

Thus, Y + Z = 0 , 5. Since the total amount of a mixture of 1 liter, then X = 1 - 0.5 = 0.5.

mixture density of hydrogen is described as follows:

15 + X 14 21 Y + Z = 15,9, (X, Y and Z a respective volume fractions of hydrocarbons in the mixture, as

the volume of a mixture of 1 l, and 15, 14 and 21 - is the density of hydrogen of ethane, ethylene and

propylene, respectively). Since X = 0,5, we get

0,5 14 15 + Y + Z = 15,9 21

14 21 Y + Z = 8,4. In addition, it is known that

Y + Z = 0,5

The system of equations gives Y = 0,3, Z = 0,2.

Thus, the initial mixture contains 50% ethane, 30% ethylene and 20% propylene.

10-3. 1) Suppose that the starting materials of the formula R 1 R Br and 2 Br

hydrocarbons Then A , B , and to have the formula R 1 - R 1 , R 1 - R 2 , and R 2 - R 2

The molar mass of the hydrocarbon products bromination A is 137, less bromine (80) plus hydrogen (1)

the molar mass of A is 58, A = C 4 H 10


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Molecular weight brominated products in equal to 193 minus 80 plus 1 gives 114, it corresponds to =

C 8 H 18 . Obviously, B = C 6 H 14 .

starting bromides - C 2 H 5 Br, and C 4 H 9 Br.

Identify structural formula can be based on the fact that the A forms two monobromproizvodnyh

and in - three. In addition, the molecules A and to be symmetrical.

Substance A - n-butane (isobutane molecule is asymmetric)

Symmetrical octane isomers:

n-octane, CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 - gives four different products in bromination.

These isomers can form respectively one and four monobromproizvodnyh.

Thus, the structural formula of the substance in :

Substance B :

Initial bromides: RBr: CH 3 CH 2 Br, and (CH 3 ) 2 CHCH 2 Br.

2) the equation:

Wurtz reaction C 2 H 5 Br + C 4 H 9 Br Na + C 4 H 10 + C 6 H 14 + C 8 H 18 + NaBr

Bromination of hydrocarbons RH + Br 2 = RBr + HBr

3) The relative amounts of products substituted by two factors. First fakor - a high mobility of the

hydrogen atom at the tertiary atom with (CH) compared with secondary (group CH 2 ) and primary

(group CH 3 ).On the other hand, the number of H atoms with atoms in the primary - 12, in the

secondary - 4, and with tertiary - only 2. Clearly, these factors act in opposite directions, so the finalresult is difficult to predict, not knowing how many times the rate constant for the tertiary atom is

greater than the primary or secondary.

10-4. A = hydrogen sulphide

sulfur X =

B = SO 2

V conc. sulfuric acid


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reactions: C 40 H 82 + 41 S = 41 H 2 S + 40 C

Cu + 2 H 2 SO 4 = SO 2 + CuSO 4 + 2 H 2 O

SO 2 + 2 H 2 S = 3 S + 2 H 2 O

10-5.

1. (1) Hydrochloric acid (or the other, which is suitable for hydrogen production - meaning no nitrogen)

(2) - zinc (or other metal, which reacts with the acid, with evolution of hydrogen)

(3) AsH 3 ( arsine)

(4) calcium chloride or any other dryer

(5), (6), arsenic and hydrogen or hydrogen and arsenic

(7) Arsenic

(8) mirror

2. could write several versions of reactions.

example, Zn + 2HCl = ZnCl 2 + H 2

AsO 33 - + 9 H (atomic hydrogen) = AsH 3 + 3 H 2 O

is another option: K 3 AsO 3 + 3 Zn + 9 HCl = 3 ZnCl 2 + AsH 3 + 3 H 2 O + 3 KCl

2 AsH 3 = 2 As + 3 H 2 .

3. determine Arsenic in the environment , because it is toxic and its content developed standards (MPC)

for various objects. Examples of the environment can be water, soil, and air. Arsenic is also determined

in food. We are talking about the quantitative determination. In the field of criminology arsenic

determined in connection with cases of poisoning - for example, in the above example assumes that the

arsenic was added to chocolate. If a person has been poisoned, the presence of arsenic in the tissues

(usually hair) shows that there is arsenic poisoning in these cases rather qualitative method.

10-6.

1) Presumably, AB 3 is the oxide. (B = oxygen)

define the molar mass of the element A (x)48 / (x + 48) = 0.3333. Hence x = 96. Element - molybdenum.

Thus, A = Mo,

B, and C = hydrogen and oxygen. Liquid consisting of them - H 2 O 2 .

AB 3 = MoO 3

Another compound B and C elements - water.

2) The reaction

Mo + 2 H 2 O 2 MoO 3 + H 2 + H 2 O 2 or Mo + 3 H 2 O 2 2 MoO 3 + 3 H 2

MoO 3 + 3 H 2 = Mo + 3H 2 O.


10/13

11st Grade

1.11 Sample alloy of zinc and aluminum weighing 10.0 g was dissolved in an excess of hydrochloric

acid. In this case stood out 7.03 liters of gas (reduced to standard conditions).

Determine the mass fractions of metals in the alloy. What gas will escape the dissolution of the same

alloy in hot concentrated alkali solution?

2.11 In bromination of some aromatic compounds in the presence of a catalyst AlBr 3 in the dark

produced only one monobromproizvodnoe, and 1.00 g of the starting material can be obtained 1.57 g of

monobromproizvodnogo.

Determine structure described organic matter, knowing that the yield of the reaction product is 90%.

3.11 Weigh 3.7 g ofD dissolved in excess aqueous NaOH. The resulting solution was neutralized with

nitric acid to pH = 7 and added an excess of calcium nitrate. The precipitated white solid was filtered,

dried and weighed - his weight was 3.1, the solution was added to the remaining excess of silver nitrate,

which caused a loss of another 8.61 g white solid.

Identify substance D and write the equations of all the reactions mentioned in the problem. What

substances can be formed by the reaction ofD with ammonia?

11.4 Substance K is a volatile liquid with a vapor density of hydrogen 42. When processing to a dilute

solution of NaOH and the subsequent acidification of the acid form unstable N , which is then readily

decarboxylated giving acetone.

Identify substance K and L . How can I get stuffto industrial and laboratory? Give the reaction of a

substance to a: a) ammonia (molar mass of the product is 101 g / mol)

b) benzene in the presence of aluminum chloride (molar mass of the product is 162 g / mol).

11.5 Analyzing the old shelves in the laboratory chemists discovered a white crystalline substance. The

substance is highly soluble in water, forming an acid solution. When added to a solution of 1.00 g of the

substance, a solution of AgNO 3 was formed 2.73 g of white curd cake. With further addition ofAgNO 3 observed loss of black sludge and release of gas with a density of hydrogen 14.

Determine what substance found chemists and write the equation of its reaction with AgNO 3 .

6.11 Organic Matter And with gentle heating in concentrated phosphoric acid decomposes to form a

gas mixture with a density of 25 to hydrogen, and the solution is not organic matter. If the resulting gas

mixture passed through bromine water, its volume is reduced by half, and the subsequent passage

through the remainder of its alkaline solution completely absorbed. Identify the substance And if we

know that the hydrogen atoms in it are indistinguishable. Write the reactions mentioned. A process for

producing the substance A . The purposes for which a substance A is widely used in organic

synthesis? Give an example.

ANSWERS

"... I didn't say it would be easy. I just said it would be the truth ..."

Morpheus, The Matrix


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1. W (Zn) = 60% W (Al) = 40% (range within 2% adopted for the correct answer) When dissolved inalkali highlighted 14.06 liters (0.626 mol) of gas.

2. Aromatic -p -xylene (molecular mass = 106), theproduct of bromination - 1-bromo-2 ,5-dimethylbenzene (molar mass = 185)

3. Substance D - POCl 3 (molar mass = 153.5)The reaction products D c ammonia may be the following:

4.

5. Detection of substances - Hydrazine dihydrochloride N 2 H 6 Cl 2 (molar mass = 17 cation * n,where n - the number of chlorine atoms).

Reactions:

N 2 H 6 Cl 2 + 2AgNO 3 = N 2 H 6 (NO 3 ) 2 + 2AgCl N 2 H 6 (NO 3 ) 2 + 4AgNO 3 = 4Ag + 6HNO 3 + N 2

6. Reaction Getting A (a common abbreviation - Boc 2 O) is used as a defense alcohol groups inorganic synthesis.


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SOLUTION

1. We write the equation of the reaction of dissolution of metals in acid:Zn + 2HCl = ZnCl 2 + H 2

2Al + 6HCl = 2AlCl 3 + 3H 2

Let x - mass fraction of zinc alloy, and y - mass fraction of aluminum. Then we can make the

system of equations:

x + y = 1

10 * x/65 + (10 * y/27) * 3/2 = 7.03 / 22.4

Solving it we get the answer: W (Zn) = 60%; W (Al) = 40%

To determine the amount of hydrogen released during dissolution of 20.0 g of this alloy in

alkaline solution, not necessarily carry out the calculation. Since the dissolution of both the acid

and the alkali is essentially a hydrogen reduction, the amount of released gas to the mass of the

metal does not change. Therefore, when dissolved 20.0 g stand out twice as many hydrogen -

14.06 liters.

2. The molar mass of an aromatic compound in the reaction of bromination is amended asfollows:

M (monobromproizvodnogo) = M (connection) + 80 (Br) - 1 (the hydrogen atoms)denote the molar mass of the aromatic compound for x and form the equation:

1.57 * 0.9 / (x + 80 - 1) = 1,00 / x

Where x = 106.

Among the benzene derivatives of the molar mass of the corresponding ethyl benzene, xylenes,

and benzaldehyde. Of these substances only one monobromproizvodnoe can form vapor-xylene

(due to the equivalence of all positions) and benzaldehyde (due to preferential substitution

in the meta -position). However, much of benzaldehyde oxidizes when bromination under these

conditions, and formed (in lesser amounts) ortho - and steam -bromzameschennye

products. Therefore, the maximum score in this task set for the response steam -xylene.

3. Sludge precipitated by adding silver nitrate probably - AgCl. Imagine a substance D Aclformula x . Then the molar mass of A is equal to:M (A) = 3.07 / (8.61 / x * 143.5) - 35.5 * x = x * (3.07 / (8.61 / 143.4) - 35.5)

where 143.5 - molar mass of AgCl. Hence, we find A = 47 and x = 3 (for other x to be not an

integer value A).

precipitate precipitated by adding calcium nitrate can be fluoride, carbonate or phosphate

(calcium sulfate falls only by heating the solution.) And so should contain C, F or P. Reasonable

response is obtained only if the last version - A = PO. Thus D = POCl 3 .

Reaction ofD ammonia most likely replacement of the chlorine atoms (oxygen more tightly

bound to the phosphorus).

4. Acid, which forms the decarboxylation acetone - CH 3 C (O) CH 2 COOH (acetoacetic acid). It has amolar mass of 102 g / mol, which is different from the molar mass of the substance to (42 * 2 =

84) to 18 g / mol, that is, the water molecule. Cleavage of the water molecule atsetouksunoy

acid can obtain the following structural formula: Formula 1 and 2 may in fact be the

tautomers. Formula 3 is less preferred because it contains unstable allene moiety. True to the

structure of matter to , determine that can be based mainly on the knowledge, the formula 1 -

diketene. Get diketene ketene dimerization CH 2 = C = O, and it in turn - HCl elimination of acetyl

chloride CH 3 COCl under the tertiary amines (in the laboratory) or by pyrolysis of acetic acid

itself (in the industry). main focus is the nucleophilic reactions of diketene attack at the carbonyl

group. Reaction with ammonia and benzene proceed exactly according to this mechanism,


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which explains the structure of the products (see FAQ).

5. White curdy precipitate formed after the addition of silver nitrate - again AgCl. By the formulasimilar to that given in the task 3 the molar mass of the substance of the cation A:

M (A) = x * (1.00 / (2.73 / 143.4) - 35.5), M (A) = 17 * x

The addition of an excess of AgNO 3 to the solution obviously is oxidized (a precipitate of silver)

has a molar mass of

14 * 2 = 28. This can be a CO, N 2 and C 2 H 4 . Comparing with the molar mass of the cation, we

find the right solution for the test substance - N 2 H 6 Cl 2 hydrazine dihydrochloride. Cation has a

molar mass of 34 = 17 * 2, and the oxidation of nitrogen forms.

6. Organic substance A at low heat in concentrated phosphoric acid decomposes to form a gasmixture with a density of 25 to hydrogen, and the solution is not organic matter. If the resulting

gas mixture passed through bromine water, its volume is reduced by half, and the subsequentpassage through the remainder of its alkaline solution completely absorbed. Identify the

substance And if we know that the hydrogen atoms in it are indistinguishable. Write the

reactions mentioned. A process for producing the substance A . The purposes for which a

substance A is widely used in organic synthesis? Give an example.

mixture of gases formed during the decomposition ofA has a molar mass of 25 * 2 = 50 g / mol,

and probably one of the gases is an alkene or alkyne (absorbed bromine water), and the second

- CO 2(absorbed by alkali). Knowing that the ratio of gases in a mixture of 1:1 (volume is reduced

by half when passed through bromine water), we find the molar mass of the first gas:

M (gas) = 50 * 2 - 44 = 56, which corresponds to the molar mass of butene.

Given that in the parent compound all hydrogen atoms should be equivalent, it is reasonable to

assume that it contains the tert-butyl substituents, and the gas, respectively, is isobutylene. He,

in turn, can be formed in the dehydration oftert-butyl alcohol. Combination of fragments ofCO 2 and (CH 3 ) 3 CO-we obtain the substance A - Boc 2 O:

moscow city exam

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