أهم 100سؤال فى مراجعة مادة الكيمياء لغات الجزء الثامن ثالث ثنوي

من طرف **حنين الصمت** الجمعة 17 مايو 2013, 2:40 pm

الدرس في شكل نص مقروء

أهم 100سؤال فى مادة الكيمياء لغات - الجزء الثامن

16) Caculate the equilibrium constant of this reaction

H2 + I2 2HI

Given that the concentrations are H2=0.1 , H2=0.1 , HI = 1 mole/liter

Solution

H2 + I2 2HI

Kc = = = 100

17) Caculate the equilibrium constant of this reaction

N2 + 3H2 2NH3

Given that the pressure are N2= 0.3 , H2=0.1 , NH3 = 0.09 atmospheric pressure

Solution

N2 + 3H2 2NH3

Kp = = = 27

18) Caculate the degree of dissociation of 0.08 molar of HCN if the Ka = 2 x 10-8

Solution

Ka = α2 x C

α= = 5 x 10-4

19) Caculate the concentration of hydrogen ion of 0.2 molar of acetic acid then find its

pH given that Ka = 5 x 10-8

Solution

[H+] = = 1 x 10-4

pH = -log (H+) = -log( 1 x 10-4 ) = 4

20) Caculate the pH of ammonia solution its concentration 0.1 molar given that Kb=1x10-5

Solution

[OH-] = = 1 x 10-3

pOH = -log (OH-) = -log( 1 x 10-3 ) = 3

pH + pOH = 14 pH = 14 – 3 = 11

21) Find ionization constant (Ka) of a wak acid its ionization degree 1% its conc.= o.5M

Solution

Ka = α2 x C

Ka = ( )2 x 0.5 = 5 x 10-5

أهم 100سؤال فى مراجعة مادة الكيمياء لغات الجزء الثامن ثالث ثنوي 1

أهم 100سؤال فى مراجعة مادة الكيمياء لغات الجزء الثامن ثالث ثنوي 1

22) If the degree of solubility of silver bromide AgBr = 0.2 x 10-3 mole/ liter

calculate the solubility product of silver bromide.

Solution

AgBr Ag+ + Br -

Ksp = [Ag+] [Br -]

Ksp = [0.2 x 10-3] [0.2 x 10-3] = 4 x 10-8

23) Calculate the degree of
solubility of CaCl2 in gram/liter given that its solubility product in
water = 8 x10-9( Ca=40 , Cl=35.5)

Solution

CaCl2 Ca2+ + 2Cl -

Ksp = [Ca2+] [2Cl-]2 say degree of solubility = X

8 x 10-9 = [ X ] [ 2 X]2

8 x 10-9 = 4 (X)3

X = 2 x 10-3 one mole of CaCl2= 40 + 71 = 111

Degree of solubility (g/l) = 2 x 10-3 x 111 = 0.222 g/l

24) Whate the effect of pressure and temp. on the concentration of hydrogen in this reaction

H2 + CO2 H2O + CO , ΔH = +41 kj

Solution

• increase the temperature conc. of hydrogen decreases

• increase the prerssure has no effect

25) This reaction has two equilibrium constant Kc=50 at 500oC
and Kc = 80 at 800 oC deduce this reaction is exothermic or indothermic

N2 + O2 2NO

Solution

Increase the temp. Kc increases means thr reaction proceed forward = endothermic reaction

Draw a cell diagram of:

1- Galvanic cell formed from tin and silver if the reduction potential of Sn=-0.14 and that of Ag=0.8 [ Sn2+ , Ag+]

أهم 100سؤال فى مراجعة مادة الكيمياء لغات الجزء الثامن ثالث ثنوي 2

أهم 100سؤال فى مراجعة مادة الكيمياء لغات الجزء الثامن ثالث ثنوي 2

Answer

Reduction of silver > reduction of tin

So silver acts as cathode

Cell diagram Sn/Sn2+ // 2Ag+/2Ag

e.m.f = difference between two reduction

e.m.f = 0.8 – (-0.14) = 0.94 volts

2- galvanic cell in which this reaction occurs

Ni2+ + Fe Ni + Fe2+

show a) cathode and anode

b) direction of the current

Answer

Cell diagram Fe/Fe2+ // Ni2+/Ni

cathode is Ni and anode is Fe

direction of the current from iron to nickel

3- in this equation CuO + H2 Cu + H2O

If the oxidation potential of copper = -0.34 v

Is this represent a galvanic cell

a) if yes draw the expression of the galvanic cell

b) if no say why?

Answer

Copper is reduced and hydrogen is oxidized

e.m.f.= oxidation of anode – oxidation of cathode

e.m.f.= 0 – (-0.34) = + 0.34 volt

e.m.f. = +ve value so it is galvanic cell

its diagram H2/2H+ // Cu+2/Cu

4) In this equation CuCl2 Cu2+ + 2Cl-

If the reduction potential of copper = 0.34 v and that of chlorine =1.36 v

a) Write the equations occurs at anode and at cathode

b) Is this reaction spontaneous or nonspontameous

Answer

At anode 2Cl- Cl2 + 2e- ( oxidation)

At cathode Cu+2 +2e- Cu (reduction)

e.m.f.= redction of cathode – reduction of anode

e.m.f.= 0.34 – 1.36 = - 1.02 volt

e.m.f. = -ve value so it is non spontaneous reaction

Calculate

1- The no. of faraday necessary to deposit gram-atom of aluminum during electrolysis of Al2O3

Answer

Amount of electricity needed to deposit gram/atom = ZF ( valency X faraday)

Amount of electricity = 3 faraday

2- the no. of faraday necessary to deposit 21.6 gram of silver during electroplating process if the reaction at cathode

Ag+ + e- Ag [Ag=108]

Answer

Eq.wt. = 108/1=108

Mass = faraday X eq.wt

21.6 = faraday X 108

Electricity = 21.6/108= 0.2 faraday

أهم 100سؤال فى مراجعة مادة الكيمياء لغات الجزء الثامن ثالث ثنوي 3

أهم 100سؤال فى مراجعة مادة الكيمياء لغات الجزء الثامن ثالث ثنوي 3

3- minutes needed to produce 10500 coulomb from current of 25 ampere

Answer

Coulomb = ampere X time (in second) 10500 = 25 X time

Time = 10500/25 = 420 second

= 7 minutes

4- the mass of silver ppt. when electric current its intensity = 2 ampere passes for an hour in silver nitrate solution [Ag=108]

reaction at cathode Ag+ + e- Ag

Answer

eq.wt.= 108/1 = 108

Mass = = = 8.058 gm.

5- the volume of chlorine gas evolved during electrolysis of chloride solution using 2 amperes for half hour [Cl=35.5]

Answer

eq.wt.= 35.5/1 = 35.5

Mass = = = 1.32 gm.

Volume of Cl2 = = = 0.418 liter