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For students all specialities



INDIVIDUAL WORK-2

For students all specialities

MODUL 2

 

In the maximal possible exited state the total spin of an atom makes 3. Find the number of the group, in which located this atom:

a

2

b

 

c

 

d

 

f

 

 

 

Among the given electronic configurations the smallest value of energy of ionization corresponds to:

a

1 s 22 s 22 p 3

b

1 s 22 s 22 p 63 s 2

c

1 s 22 s 22 p 6

d

1 s 22 s 22 p 63 s 1

 

 

The reason of periodic change of properties of elements is:

a

Increase of a charge of a nucleus of atom

b

Increase of radius of atoms in groups

c

Periodic repetition of similar electronic structures

d

Increase of number of electrons in atoms

 

At interaction of 1,5 g of an oxide of alkaline metal with water was formed 1,65 g of an alkaline. This metal is:

a

Lithium

b

Potassium

c

Sodium

d

Cesium

f

Rubidium

 

 

Mass of 1l of an hydrogen compound of an element, which is in the seventh group of periodic system, under normal conditions is 1,63 l. This element is:

a

Bromine

b

Astatium

c

Fluorine

d

Iodine

f

Chlorine

 

 

 

At the decomposition of 25 g of a carbonate of twovalent metal was liberated 5,6 l carbon dioxide under normal conditions. The formula of the salt looks like:

a

SrCO3

b

CaCO3

c

MgCO3

d

BeCO3

f

BaCO3

 

 

Find covalence and oxidation state of sulphur in a molecule Н2S:

a

3, +2

b

6, +2

c

4, –2

d

6, –2

f

2, –2

 

 

Find charge of the complex ione, charge and coordination number of the central atom in complex compound K4[Fe(CN)6]:

a

4–, 2+, 4

b

2–, 6+, 4

c

3–, 3+, 3

d

4–, 2+, 6

f

2–, 3+, 6

 

 

Find concentration of complex ion [Co(NH3)5]3+, that forms according the first step of dissociation (first instability constant is К1 = 4×10–5) in 0,01 M solution of hexaamminecobalt (ІІІ) chloride, if concentration of complex ion [Co(NH3)6]3+ at its dissociation practically does not change:

a

mol/

b

4×10–5×0,01 mol/

c

0,01 mol/

d

mol/

f

mol/

 

 

Thermochemical equations of formation of nitrogen and carbon oxides look like:

1/2N2(g)+ 1/2O2(g)= NO(g), DНо = 90,25 kJ;

С(crist)+ 1/2O2(g)= СО(g), DНо = –110,5 kJ.

It means, that in this reactions:

a

200,75 kJ of heat is liberated

b

90,25 kJ is absorbed and 110,5 kJ of heat is absorbed

c

90,25 kJ is absorbed and 110,5 kJ of heat is liberated

d

200,75 kJ of heat is absorbed

f

90,25 kJ is liberated and 110,5 kJ of heat is absorbed

 

 

The rate constant of chemical reaction

2SO2+ O2= 2SO3

is 1,6. Initial concentrations of oxygen and oxide of sulfur are 0,16 mol/l and 1,2 mol/l accordingly. After some time the concentration of oxygen has decreased down to 0,06 mol/l. Find thus the concentration of sulfur (IV) oxide and the reaction rate:

a

С(SО2) = 1,0 mol/l; 9,6×10–2mol/l×s

b

С(SО2) = 0,12 mol/l; 1,76×10–5mol/l×s

c

С(SО2) = 1,1 mol/l; 3,46×10–3mol/l×s

d

С(SО2) = 1,14 mol/l; 2,56×10–3mol/l×s

f

С(SО2) = 0,08 mol/l; 9,84×10–4mol/l×s

 

 

Equilibrium of reaction

H2+ Cl2= 2НСl(g)

was established at the following concentrations of substances: [Н2] = 0,25 mol/l; [Cl2] = 0,05 mol/l; [НСl] = 0,9 mol/l. Find the initial concentration of chlorine:

a

0,7 mol/l

b

0,5 mol/l

c

0,8 mol/l

d

0,2 mol/l

f

0,6 mol/l

 


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