مراجعة ليلة الامتحان للثانوية العامة في الكيمياء لغات (الجزء 2)
Question ( no. 4 ) Problems:
1- Calculate the volume of 0.1 of CO2 gas at S.T.P.
Volume of CO2 gas at S.T.P = 0.1 × 22.4 = 2.24 liter
2- Calculate the density of oxygen (O2) at (STP) . [O=16]
Molecular mass = 2 × 16 = 32 grams
The molecular mass = The density × volume of one mole
32 = The density × 22.4
The density = 32 ÷ 22.4 = 1.43 grams / liter
3- Calculate the volume of chlorine gas evolved at (S.T.P) during the electrolysis of sodium chloride (NaCl) solution by passing an electric current, its intensity is 10 amperes for 20 minutes. ( Cl = 35.45 )
The equivalent mass of chlorine = = 35.45 gm
The mass of chlorine = = 4.4 gm
The number of chlorine moles = = 0.062 moles
The volume of chlorine at S.T.P = 0.062 × 22.4 = 1.3888 liters
4- Potassium super oxide KO2 is a compound used to purify atmospheric air from carbon dioxide in a closed atmosphere as following equation .
4KO2 + 2CO2 2K2CO3 + 3O2
If 14.2 gm of KO2 is used, calculate the volume of oxygen formed in liters
(K=39 , O=16)
4KO2 + 2CO2 2K2CO3 + 3O2
4( 39 + 2 × 16 ) 3 × 22.4 liters
284 grams 67.2 liters
14.2 grams (X) liters
Volume of Oxygen = = 3.36 liters
5- Calculate the mass of calcium oxide produced from the thermal decomposition of 50 gram calcium carbonate, then calculate the volume of carbon dioxide gas evolve. ( Ca = 40 , O = 16 , C = 12 )
CaCO3 CaO + CO2
40 + 12 + 3 × 16 40 + 16
100 grams 56 grams
50 grams (X) grams
(X) Mass of calcium oxide = = 28 grams
CaCO3 CaO + CO2
100 grams 22.4 liter
50 grams (X) liter
Volume of CO2 = = 11.2 liter
6- Calculate the number of ions produce by dissolving 7.1 grams of sodium sulphate (Na2SO4) in water.
Na2SO4 + H2O(I) 2Na+(aq) + SO42-(aq)
Molecular mass of sodium sulphate = (23 × 2) + (32 × 1) + (16 × 4) = 142 grams
Number of moles of sodium sulphate = = 0.05 mole
Each mole of sodium sulphate gives 3 moles of ions.
Total number of moles of the produced ions = 3 × 0.05 = 0.15 mole
Total number of the produced ions = 0.15 × 6.02 × 1023 = 0.9 × 1023 ions
7- A two gram of impure sodium chloride was dissolved in water, Excess of silver nitrate was added to precipitate 4.628 grams of silver chloride. Calculate the percentage of chlorine in the sample. Given that :
(0 = 16 , N = 14 , Na = 23 , Cl = 35.5 , Ag = 108 )
NaCl + AgNO3 AgCl + NaNO3
Mole of AgCl = (1× 108) + ( 1×35.5) = 143.5 gram
143.5 gram AgCl 35.5 gram chlorine
4.628 gram AgCl (X) gram chlorine
Therefore mass of chlorine in silver chloride = mass of chlorine in sodium chloride
= ـ = 1.145 = = 57.245 %
8- 25 ml of 0.5 molar hydrochloric acid was neutralized with 20 ml of calcium hydroxide solution. Calculate the concentration of calcium hydroxide (mole/liter)
2HCl+ Ca(OH)2 CaCl2 + H2O
= ® = Concentration of calcium hydroxide (M2) = = 0.3125 moles / liter
9- 10 ml of 0.1 molar sulphuric acid added to 0.2 grams of impure sample of calcium carbonate until complete reaction takes place. Calculate the percent of calcium carbonate in The equation for the reaction is :
CaCO3 + H2SO4 CaSO4 + CO2 + H2O
The number of moles of H2SO4 = = 0.001 mole
The balanced equation shows that :
The number of moles of CaCO3 = the number of moles of H2SO4
The number of moles of CaCO3 reacting with H2SO4 = 0.001 moles
One moles of CaCO3 = 40 + 12 + ( 3 × 16 ) = 100 gm.
Mass of one CaCO3 in the sample = 100 × 0.001 = 0.1 gm .
% of CaCO3 in the impure sample = = 50 %
10- 0.2 gm of a mixture of solid substance containing sodium hydroxide and sodium chloride was titrated wih 0.1 molar of hydrochloric. The complete reaction takes place by the consumption of 10 ml of the acid. Calaulate the percentage of sodium hydroxide in the mixture. (Na = 23. O = 16. H = 1 )
Number of moles of HCI = = 0.001 mole
Number of moles of NaOH = 0.001 mole
The mass of one mole of NaOH = 23 + 16 + 1 = 40 gm
The mass of NaOH in the mixture = 40 x 0.001 = 0.04 gm
The percentage of NaOH the mixture = = 20%
11- Four grams of impure Sodium chloride was dissolved in water and an excess of silver nitrate solution was added to precipitate 9.256 gram of silver choride. Calculate the percentage of sodium chloride in the sample.
(Na = 23,Cl = 35.5,Ag = 108)
NaCl + AgNo3 NaNo3 + AgCl
The molecular mass in grams of AgCl = 108 + 35.5 = 143.5 gm
The molecular mass in grams of NaCl = 23 + 35.5 = 58.5 gm .
Each 58.5gm. of NaCl 143.5gm.ofAgCl
(X) gm. of NaCl 9.256gm.ofAgCl
The mass of NaCl in the sample = = 3.7733gm.
% NaCl = x 100 = 94.3325%
12- A sample of 1.47grams of hydrated calcium chloride salt ( CaCl2. ×H2O). was heated several times till a constant mass of 1.11 grams., Find out the number of water molecules of crystallization in the molecule of hydrated calcium chloride.
(H = 1.O = 16.Ca = 40.Cl = 35.5)
The mass of water of crystallization in the sample = 1.47 – 1.11 = 0.36 gram.
Each 1.11gram of anhydrous (CaCl2) bind with 0.36 gram water of crystallization
The molecular mass of (CaCl2) = (35.5 x2) + 40 = 111gram
1.11 gram CaCl2 bind with 0.36 gram water of crystallization
111 gram CaCl2 bind with (X) gram water of crystallization
X (mass of water of crystallization) = ( 111 x 0.36)÷ 1.11 = 36 gram
The molecular mass of (H2O)= 16 + (2×1) = 18 gram
The number of molecules of water of crystallization in the molecule
of hydrate CaCl2 = 36 ÷ 18 = 2 molecules
13- Calculate the equilibrium constant ( Kp ) of the reaction :
N2(g) + 3H2(g) 2NH3(g) DH = -92 KJ
The pressures of gases are 2.3 atmosphere for N2 , 7.1 atmosphere for H2
and 0.6 atmosphere for NH3. Comment on the value of (Kp).
KP = = = 4.4 × 10-4
The small value of Kp indicating the small amount of resulting ammonia .
14- Calculate the equilibrium constant for the reaction : 12 + H2 2HI
Provided that concentrations of iodine , hydrogen and hydrogen iodide at equilibrium are 0.1105.0.1105 and 0.7815 mole / liter respectively .
Kc = = = 50.018
15- If the equilibrium constant for the following reaction is 15.75 Cl2 + PCl3 PCl5
And the concentrations of chlorine and phosphorus trichloride were 0.3 and 0.84 mole / liter respectively. Calculate the concentration of phosphorous pentachloride.
Equilibrium constant Kc = ® 15.75 = ® 15.75 =
The concentration of PCl5 = 15.75 × 0.252 = 3.969 mole / liter
16- Calculate the degree of dissociation in 0.1 molar hydrocyanic acid (HCN) solution at 25°C. Providing that the equilibrium constant of the acid Ka = 7.20 × 10-10
HCN + H2O H3O+ + CN-
Applying Ostwald law : Ka = a2 × C
7.2 × 10-10 = a2 × 0.1
a2 = = 72 × 10-10
a = = 8.5 × 10-5
The degree of dissociation of hydrogen ion = 8.5 × 10-5
17- Calculate the concentration of hydronium ions (H3O+) in a 0.2 molar solution of acetic acid at 25°C given that equilibrium constant of the acid is (1.8 × 10-5)
[H3O+] =
[H3O+] = = 1.897 × 10-3 molar
18- Calcium fluoride has Ksp of 3.9 × 10-11 . What is the fluoride ion concentration at equilibrium ?
CaF2 (s) Ca-2 (aq) + 2F-(aq)
(X) (X) (2x)
Ksp = [Ca-2] × [2f]2
3.9 × 10-11 = (X) × (2x)2
3.9 × 10-11 = 4 x3
x3 = C x = 2.1 × 10-4 mole / liter
19- Calculate the solubility product. (Ksp) of calcium phosphate Ca3(PO)4 which is sparingly soluble in water , given that the concentration of calcium ions is
( 1× 10-4 ) mole / litre.
Ca3(PO4)2 3Ca2+ + 2PO43-
Ksp = [Ca2+]3 [PO43-]2
= [1 × 10-8]3 [0.5 × 10-3]2 = 2.5 × 10-31
20- Two metals A and B their standard oxidation potentials are – 0.3 and 0.7 volts respectively, each of them is divalent :
1- Using a diagram how can you represent the cell which can be formed from these two metals?
2- Calculate the electromotive force of this cell.
3- Does this cell produce an electric current ? Why ?
1- B B+2 A+2 A
2- e.m.f = standard oxidation potential of anode – standard oxidation potential of cathode e.m.f = 0.7 – ( -0.3 ) = 1 volt
3- An electric current is produced because the e.m.f is positive value and the reaction inside the cell is spontaneous .
21- What is the number of faradays required to precipitate gram/atom of copper according to the following reaction :
Cu2- + 2e- Cu (at cathode)
2 Faradays
22- Calculate the electric current intensity required for passing 3.7 Faraday through an electrolyte for 40 minutes.
The quantity of electricity= current strength × time
3.7 × 96500 = I × 40 × 60
I = (3.7 × 96500) + (40× 60) = 148.7 ampere
23- What is the quantity of electricity (coulomb) necessary to separate 5.6 grams of iron from solution of Iron (III) chloride ? Where the cathode reaction is :
Fe3- + 3e- Fe0 (Fe = 55.86)
Equivalent mass of iron = 18.62 grams
Quantity of electricity (coulomb) =
= 29022.5 coulomb
24- Aluminum metal is produced from electrolysis of molten aluminum oxide.
Calculate the number of aluminum moles produced when an electric current of intensity 9.6 amperes is passed for 5 minutes . the cathode reaction equation is : Al3 + 3e- Al0 (Al = 27)
Quantity of electricity = electric current intensity intensity × time
= 9.65 × 5 × 60 = 2895 coulombs
Equivalent mass of Al = atomic mass / valency = = 27 / 3 = 9 gram
96500 coulombs 9 grams Al
285 coulombs (X) grams AI
X = = 0.27 gram
27 grams Al 1 mole Al
0.27 gram Al (X) mole Al
(X) = 0.01 mole Al
Question ( no. 5 )
Various questions
1- By using balanced symbolic chemical equations, how can you obtain :
Phenol from benzene .
Ethylene glycol from ethanol .
1- Obtaining phenol from benzene :
2- Ethylene glycol from ethanol
2- Show by balanced chemical equations : (Write the conditions of reaction ):
1- Conversion of carboxylic group of an organic compound into hydroxyl group
2- Conversion of an aromatic acid into an acid amide.
3- Show by balanced chemical equations, how to obtain :
A- Methane gas from ethanoic acid.
B- Picric acid from chlorobenzene .
4- Show by balanced chemical equations, how can you obtain each of the following:
1- aAcetone from 2- Bromopropane .
2- Ethyl benzoate ester from toluene.
5- Show by balanced chemical equations what happens in each of the following cases:
1- Addition of iron (Ill) chloride solution to ammonium thiocyanate solution.
2- Addition of yeast (zymase enzyme) to glucose solution.
3- Dry distillation of sodium benzoate.
4- The reaction of ethanol with concentrated sulphuric acid at 180°C.
6- Show by balanced chemical equations, how can you obtain each of the following :
1- Benzene sulphonic acid from sodium benzoate.
2- Ethyl chloride from acetic acid.
7- By using the balanced symbolic chemical equations, how can you obtain:
1- Nitrobenzene from sodium benzoate.
2- Dihydric alcohol from monohydric alcohol.
8- Illustrate by the balanced chemical equations how you can obtain the following:
1- Cycloalkane from normal alkane.
2- Pricric acid from chloro benzene.
2- propanol
1- propanol
Picric acid
2- methyl -1- propanol
2- methyl -2- propanol
Catecol
Choose from the above table, the compound (or compounds) that considered:
1- A Phenols .
2- A Secondary alcohol .
3- Alcohol which gives aldehyde on oxidation.
4- Alcohol which gives ketone on oxidation.
5- Product of nitration of phenol.
6- Disubstituted benzene.
7- Tetra substituted benzene .
1) Catecol. 2) 2-propanol.
3) 1-propanol. 4) 2-propanol.
5) Picric acid 6) Catecol.
7) Picric acid.
10-
Acetic acid
Formic acid
Oxalic acid
Ethyl formate
Methyl acetate
Ethyl acetate
From the above table find :
1- Two isomers.
2- Two compound that give acetamide on ammonolysis.
3- Compounds that give effervescence with NaHCO3 (one enough).
4- Compound that is named according to IUPAC ethyl methanoate.
1) Ethyl formate-methyl acetate (Two isomers) .
2) Methyl acetate – ethyl acetate (Two compounds from acetamide)
3) Acetic acid- formic acid – oxalic give effervescence with sodium bicarbonate (One acid is enough)
4) Ethyl formate.
11-
Methyl acetate
Sodium acetate
Ethanoic acid
Methyl formate
Potassium acetate
Ethyl formate
From the previous table illustrate the following :
1) The esters .
2) The salts of the carboxylic acids.
3) The compounds which take a name according to IUPAC system .
4) The isomeric compounds.
12-
Formic acid
Ethanoic acid
Oxalic acid
Salicylic acid
Butyric acid
Benzoic acid
From the previous table mention the following :
1) Aromatic monocarboxylic acid .
2) Dicarboxylic acid.
3) An acid in which the number of carboxylic groups equals the number of carbon atoms .
4) An acid contains two functional groups.
1) Benzoic acid. 2) Oxalic acid. 3) Oxalic acid. 4) Salicylic acid.
13- Give reason for each of the following :
Salt bridge must be used in Galvanic cells.
Cone. Sulphuric acid is added in ester formation and nitration.
The catalyst does not affect the position of equilibrium in the reversible reactions.
density of oxygen gas is more than of nitrogen. [N = 14 , O = 16 ]
1) To neutralize of excess ions in both half cells. Thus, it prevents the formation of electric potential between two solutions.
2) To absorb the water produced and prevent the reversible reaction.
3) Catalyst decreases the activation energy required to accelerate reversible and irreversible reaction at the same time.
4) Because molecular mass of O2 which is (32) is more than of N2 which is (28).
14- Give reasons for the following :
- Using a mixture of fluoride salts of aluminum, Sodium and calcium instead of cryolite containing a little amount of fluorspar in the extraction of aluminum from bauxite.
- Because this mixture gives with bauxite a melt which is characterized by a low melthing point and a lower density compared to the melt obtained from Cryolite.
15- Mention the structural formula of :
1- Naphthaline.
2- Diphyenyl
Do you consider these two compounds isomers? Why ?
1) Structural formula for naphthaline is
2) Structural formula for diphyenyl is
They are not isomers due to different molecular formula for each of them, where (C10H8) naphthaline – C12H10 diphenyl )
16- The molecular mass of a hydrocarbon is 58 grams, its mole contains 48 grams carbon ( H= 1 , C = 12)
1- Write the molecular formula for this compound.
2- This compound has two isomers (isomerism), write the structural formula for each isomer.
Mass of hydrogen = 58 – 48 = 10 grams
Number of carbon atoms = 48 ÷ 12 = 4 atoms
Number of hydrogen atoms = 10 atoms
The molecular formula of the hydrocarbon C4H10
The structural formulas for the two isomers are :
17- Read the following statement. Then answer the questions that follow it :
When glycerol reacts with nitric acid in the presence of concentrated sulphuric acid substance (x) is produced. Substance (x) is used to widen arteries in the treatment of heart crisis .
1- What is the name of the substance (x) ?
2- Mention another use for substance (x) .
3- Write the balanced chemical equation which describes the reaction of nitric acid with benzene in the presence of hot concentrated sulphuric acid (50°C).
1) Tri-nitroglycerine.
2) It is used in the prearation of explosive substance.
18- A and B are two organic compounds having the molecular formula [C2H6O]
1- Write the structural formula for each compound.
2- If compound (A) responds to oxidation reactions and compound (B) does not. How can you convert compound (A)) to a compound that has the function group of compound (B) ?
3- How can you obtain hydrogen from one of the two compounds?
19- When calcium reacts with carbon compound (A) is formed. On dropping water on compound (A) compound (B) is formed. On adding water to compound (B) in presence of catalyst and heating a liquid (C) is formed. On passing compound (B) through a red hot nickel tube a vapour of compound (D) is formed.
From the previous information :
1- Write the balanced chemical equations that illustrate the following :
a) How to obtain dichloroethane from compound (B) ?
b) The effect of concentrated sulphuric acid on compound (D).
2- Mention one use for compound (C).
2- Preparation of ethanoic acid (acetic)
20- An experiment was carried out on two organic compound (A) and (B). It was found that :
- Compound (A) reacts with sodium metal and not with sodium hydroxide.
- Compound (A) reacts with both sodium carbonate and sodium hydroxide.
1- To which group of compounds do (A) and (B) belong? Give an example for each group.
2- Write a symbolic equation for the reaction between (A) and (B) with reference to the conditions of the reaction.
Compound (A) is ethanol (C2H5OH)
Compound (B) is acetic acid (CH3COOH)
The reaction between (A) and (B) :
21- Explain the role of each of the following :
1- Analytical chemistry in agriculture.
2- Catalysts in industry.
1- Analytical chemistry in agriculture :
It is possible to know the constituents of a soil and rocks to decide whether the soil is suitable or not for cultivation.
2- Catalysts in industry :
It is possible to accelerate most of the slow reactions by using catalysts which increase the reaction rate without raising the temperature. This tends to save energy. Catalysts are used in more than 90% of the industrial processes such as foods, petrochemicals and fertilizer industries.
22- What is the role of each of the following :
1) Potassium hydroxide in the mercury cell.
2) Soda lime in preparation of methane .
3) Potassium permanganate in Baeyer’s reaction .
4) Concentrated sulphuric acid in the reaction of ester formation.
1- It is used as an electrolyte .
2- Soda lime is a mixture of sodium hydroxide and calcium oxide.
Sodium hydroxide reacts with sodium acetate forming methane and sodium carbonate calcium oxide helps in :
- Reducing the melting point of the mixture.
- Absorbing water vapour.
3- It acts as an oxidizing agent .
4- It absorbs the produced water and prevents the reversible reaction .
23- You are provided with three glasses containing : Ethanol – phenol – acetic acid. Explain how you can identify each of them practically.
Add acidified potassium dichromate dichromate to the same amount of the three solutions , heat the mixture in a water bath for 10 minutes. If the colour changes from orange to green then the solution is ethanol.
- Add iron III chloride to the three solutions. If violet colour appears then the solution is phenol.
- Acidic test: add sodium carbonate or bicarbonate, if effervescence takes place and carbon dioxide gas is evolved which turbids lime water then the solution is acetic acid.
24-How can you differentiate practically between ?
Dilute acetic acid and pure acetic acid.
Ethyl alcohol and phenol.
Methane and ethyne.
litmus solution and phenolphthalein solution.
1) By passing an electric current in each of them.
Electric conductivity of dilute acid is more than pure acid, In case of dilute acetic acid the lamp gives illumination – in case of pure acetic acid the lamp does not illuminate .
- BY adding FeCl3 solution
- If violet colour is formed, it is phenol .
- If no violet colour is formed , it is ethanol.
3- Experiment
Methane gas
Acetylene gas
By adding bromine dissolved in (CCl4) to each of them
No effect
The red colour of bromine disappears
4- By adding acid solution to both solutions,
– If the colour turns colourless, the solution is phenolphthalein
– If the colour turns red, the solution is litmus .
25- Show by practical experiment how you can detect the presence of carbon of carbon and hydrogen elements in an organic substance.
Put a small amount of an organic substance mixed with copper oxide in a glass tube and heat the test tube strongly then pass the resulting gases over anhydrous white copper sulphate , then through lime water.
Observation :
1- The white colour of anhydrous copper sulphate turns blue.
2- Lime water turns turbid.
Conclusion : The organic compound contains carbon and hydrogen elements.
The equations:
C + 2 CuO 2 Cu + CO2
2H + CuO Cu + H2O
26- How can you differentiate practically between the following :
1- Methane and ethylene.
2- Ethanol and 2- methyl -2- butanol.
1- Experiment
Methane gas
Ethene (ethylene) gas
By adding bromine dissolved in (CCl4) to each of them
No effect
The red colour of bromine disappears
2- Experiment
Ethanol
2- Methyl -2- butanol
By adding KMnO4 solution and conc. H2SO4 to each of them
The violet colour of potassium permanganate disappears
No effect
27- How can differentiate between :
( Illustrate your answer by observation and balanced chemical equations)
1- Ethanoic acid and carbolic acid (By acidity test ).
2- Ethylene gas and methane gas (By using potassium permanganate solution)
3-
1- Experiment
Ethanoic acid
Carbolic acid
By adding Na2CO3 solution to each of them
Effervescence takes places and CO2 evolves which turbids lime water
No effect
CH3COOH + NaHCO3 CH3COONa + Ca2 + H2O
C6H3OH + NaHCO3 no effect
2- Experiment
Methane gas
Ethene (ethylene) gas
b) By adding KMnO4 in an alkaline medium to each of them
No effect
The violet colour of KMnO4disappears
C2H4 + H2O + [O] C2H4 (OH) ethylene glycol
28- How can you differentiate practically between each of the following:
1-Litmus solution and phenolphthalein solution.
2- Ethanoic acid and carbolic acid.
1- Indicator
Colour in acidic medium
Colour in basic medium
Phenolphthalein
Colourless
Red
Litmus
Red
Blue
2- Experiment
Ethanoic acid
Carbolic acid
By adding Na2Co3 solution to each of them
Effervescence takes places and CO2evolves which turbids lime water
No effect
29- Draw the apparatus used for the preparation of ethyne in the laboratory then answer the following:
1- Show by practical experiment gow you can detect ethyne gas.
2- Write the balanced chemical equation which describes the reaction of ethyne with hydrogen bromide.
30- The apparatus used for preparation of ethane (ethylene) gas in laboratory:
1- Ethyne gas + bromine dissolved in CCl4 the red colour of bromine disappears.
2- The reaction of ethyne with hydrogen bromide.
31- Draw a labeled diagram for sector in dry cell.
32- Write the balanced chemical equation for preparation of methane in laboratory then draw the apparatus used in the preparation.
? The apparatus used :
Equation of reaction : CH3COONa + NaOH CH4 + Na2CO3
33- The degree of purity of metals which are prepared in industry is usually lower than the required degree of purity which is required for definite purposes.
Explain by drawing how to obtain copper of 99.95 % Purity degree.
The steps : Dip the pure copper and the impure copper in copper sulphate solution.
connect the pure copper to the negative pole of the battery.
( The pure copper acts as the cathode ) Connect the impure copper to the positive pole of the battery.
( The impure copper acts as the solution ) pass the electric current in the solution.
Observation :
Copper sulphate solution is ionized as follows :
CuSO4 Cu2+ + SO42-
At the anode : Cu Cu2- + 2e- (Oxidation)
At the cathode : Cu2+ + 2e- Cu (Reduction)
Therefore, copper will dissolve from the impure copper (anode) as copper ions (Cu2+), whereas copper ions in the solution will deposit (precipitate) as pure copper (Cu) at the pure copper (cathode).
In this process 99.95 % pure copper can be obtained.
34- If you have a copper spoon what are the different steps you must follow it to plate it by a layer of silver. Write down the equations (reactions) which occur at each of the cathode and anode.
The electroplanting of copper spoon by a layer of silver .
1) Clean the surface of the spoon completely.
2) Dip the spoon into electrolyte solution that contains silver ions.
3) Connect the spoon with the (-ve) electrode of the battery (cathode) and a rod of silver metal with the (+ve) electrode (anode).
4) Switch on the circuit.
When the electric current passes through the circuit :
- Oxidation process takes place at anode.
Ag Ag+ + e-
- Reduction process takes place at cathode.
Ag+ + e- Ag
So, silver atoms precipitate on the surface of the spoon.
35- Compare between each of the following :
1- Acidity of alcohol and acidity of phenol.
2- Dry cell and lead acid battery.
1- Acidity of alcohol
Acidity of phenol
1- Alcohols are less acidic than phenols because the alkyl group decreases the length of bond between oxygen and hydrogen atom in (O-H) group and this makes the separation of H+ ions is more difficult.
1- Phenols are more acidic than alcohols because the aryl group increases the length of bond between oxygen and hydrogen atom in (O-H) group and this makes the separation of H+ ions is easier.
2- Alcohols react with active metals such as sodium but do not react with NaOH.
2- Alcohols react with active metals such as sodium and react with NaOH.
P.O.C
Dry cell
Lead acid battery
1- Anode
Zinc electrode
Lead electrode
2- Cathode
Graphite electrode
Lead tetra oxide electrode
3- Electolite
Ammonium chloride
Dilute sulphuric acid
4- E.M.F
1.5 volt
2.0535 volt
5- Reactions
Zn0 + 2MnO2 + 2NH4
Zn2+ + 2Mn (OH) + 2NH3
At anode (Oxidation) :
Pb + SO42- PbSO4 + 2e-
At cathode (Reduction) :
PbO2 +4H+ + SO4 + 2e-
PbSO4 + 2H2O
Total reaction :
Pb + PbO2 + 4H+ + 2SO4
2PbSO4 + 2H2O
36- Compare between each of the following :
1- Addition polymerization and condensation polymerization .
2- Chemical equilibrium and ionic equilibrium.
1- Addition polymerization
Condensation polymerization
Combination of a large number of unsaturated simple molecules (monomers) to give a giant molecule (polymer) having the same empirical formula. (e.g. Eghylene glycol)
Condensation of two different monomers to give a new monomer called copolymer which undergoes polymerization (e.g. Dacron)
2- Chemical equilibrium
Ionic equilibrium
It is a state of equilibrium in the reversible reaction at which the rate of forward reaction equals the rate of backward reaction.
It is the equilibrium arising between molecules of a weak electrolyte and the ions resulting from it .
37- Compare between each of the following:
1. Irreversible reactions and reversible.
2. Alkaline hydrolysis of esters and acidic hydrolysis of esters
1. Complete (irreversible) reactions
Incomplete (reversible) reactions
1.The reactions which proceed in one direction (forward)
1. The reactions which proceed in both directions; forwaed and backward.
2. One of the products escapes from the system as evolving of a gas or forming a precipitate.
2. Both the reactants and products are always found in the reaction medium
Example:
NaCl + AgNO3 NaNO3 + AgCl
Mg + 2HCl MgCl2 + H2
Example :
CH3COOH + C2H5OH
CH3COOC2H5 + H2O
2NO2 N2O4 , DH = -ve
Alkaline hydrolysis of esters
Acidic hydrolysis esters
1. Hydrolysis of esterin in presence of aqueos alkali e.g. (aqueous NaOH)
1. Hydrolysis of ester in presence of dilute mineral acid e.g. (H2So4)
2.Gives an alcohol and the salt of the acid.
2. Gives an alcohol and organic acid.
3.Example:
CH3COOC2H5 + NaOH
CH3COONa+C2H5OH
3. Example:
CH3COOC2H5 + H2O
C2H5OH + CH3COOH
38- The following reaction has tow equilibrium constant value at two different temperatures:
H2(g) + I2(g) 2HI (g)
Kc at 850 °C = 67 and at 448 °C = 50
Is this reaction exothermic or endothermic? Why?
The reaction is endothermic
As the Value of Kc incrases, concentration of products, concentration of products increases as the temperature increases, therefore the reaction is endothermic.
39- How each of the following changes affects the concentration of hydrogen in the following equilibrium system:
H2(g) + CO2(g) H2O(g) + CO(g) ; H = + 41.1 kj
a. Addition of more CO2 gas,
b. Addition of more water vapor.
c. Addition of a catalyst.
d. Increase the temperature.
e. Decrease the vessel volume.
a) Decreases the concentration of hydrogen.
b) Increases the concentration of hydrogen.
c)Does not affect the concentration of hydrogen.
d) Decreases the concentration of hydrogen.
e) Dose not affect the concentration of hydrogen.
FeCl3 + 3NH4SCN Fe(SCN)3 + 3NH4CI
1- The red colour of the solution increase indicating to the formation of more iron (III) thiocyanate.
2- The intensity of the red colour decreases because the equilibrium shifts towards backward reaction to the formation of iron (III) chloride.
40- In the balanced reaction :
CH3COOH + C2H5OH CH3COOC2H5 + H2O
What happens to the equilibrium of this reaction in the following cases ?
- Adding excess of water .
- Adding drops of cone sulphuric acid .
- The equilibrium shifts in the opposite direction . (backward)
- The equilibrium shift in the forward direction .
41- Consider the reaction at equilibrium :
CH3COOH + H2O CH3COO- + H3O+
Explain the effect of each of the following changes on the concentration of \
acetate ion :
1) Adding drops of hydrochloric acid.
2) Adding drops of sodium Hydoxide solution .
1) Adding drops of HCl leading to increase the concentration of Hydronium ion in the Solution as a result of combination between the Hydrogen ion of the acid and water molecules in the solution, according to Le Chatelier’s principle the reaction shifts in the direction which decreases the concentration of Hydronium ion then the concentration of acetate ion decrease.
2) Adding drops of Sodium Hydroxide neutralizes acetic acid, the concentration of the acid decrease, according to Le Chatelier’s principle the reaction shifts in the direction which decreases the concentration of Hydronium ion then the concentration of acetate ion decrease .
42- Arrange the following in an ascending order as reducing agents :
1) Zn2+ / Zn [ -0.762 volt ]
2) Mg/Mg2+ [ 2.375 volt ]
3) 2Cl- / Cl2 [ -1.36 volt ]
4) K+ / K [ -2.924 volt ]
5) Pt2+ / Pt [ 1.2 volt ]
Cl < Pt < Zn < Mg < K
43- Arrange the following alcohols in ascending order according to their boiling points, give the scientific reason. (Ethylene Glycol – Sorbitol – Ethanol – Glycerol )
Ethanol < ethylene glycol < glycerol < sorbitol Due to increase of the number of polar hydroxyl groups which have the ability to form hydrogen bonds between the molecules of alcohols causing increase of its boiling point.
44- Arrange the following compounds according to the pH value of their aqueous solutions :
NaCl – CH3COONa – NH4Cl
NH4Cl < NaCl < CH3COONa
45- Arrange the following compounds in ascending order with respect to increase in their acidity :
Arrangement of the following compounds in ascending order with respect to increase in their acidity :
46- Rearrange correctly what are inside the boxes in the following diagram :
& Events of the scientisis &
1)Berzelius:
a) Classify the compounds into two types :
1- Organic compounds which are extracted from animal or plant origin.
2- Inorganic compounds which are originated from mineral sources in the earth
b) He thought that organic compounds are formed only under the influence of vital force, which is found in living cells of the body and he was the first scientist who classified the elements into metals and non-metals and non-matals.
2)Wohler :
Wohler performed an experiment which was considered to be the beginning of the end of Berzelius theory.
Wohler was able to prepare Urea, which is organic compound, by heating an aqueous solution of two inorganic compounds (Ammonium chloride and silver cyanate solution).
NH4Cl + AgCNO AgCl + NH4CNO
3)Baeyer “Baeyer’s reaction” :
He discovered reaction of alkene with potassium in an alkaline medium. This reaction is used to direct the presence of double bond.
4)Markownikoff :
He discovered a rube which controls the addition of HX or HOSO3H to unsymmetrical alkenes as propene. Where H is added to carbon rich with hydrogen and X is added to carbon poor with hydrogen.
5)Kekule:
He suggested a new idea about the bonding between carbon in the benzene molecule.
He sad carbon atoms from a ring, in which singel and double bonds are exchanged, so that, all bonds between carbon atoms similar in length,froming a hexagonal regular shaoe in which carbon are there at every angle
The ring in the figure indicates that 6 electrons are delocalized at certain carbon atoms.
6)Fridelcraft’s:
He introduced an alky1 group to a benzene ring in the presence of anhydrous AlCl3
As a catalyst.
7)Gaylussac’slaw:
The volume of the reactant gases and produced are in a fixed ration.
H) Avogadro’s number:
Equal volumes of gases contain the same number of molecules under the same temperature and pressure
8) Lechchatelier:
He formulated a principle named after him. This principle that
“When a system at equilibrtum is subjected to any change such as (Conc., temp. and pressure), the equilibrturm will shift in the direction wich will oppose this change.
9)guldbergandWaage:
He discovered the law of mass action which, the relation between the rate of chcmical reaction and the concentration of the reactants low states that.
At constant temperature, the rate of chemical reaction is dirctty propational to the product of multiplication of the reactant concentrated each is raiscd to the power of number of molecules or ions in the balanced chemical equations.
10)Ostwald:
He found the relation between the degree of ionization (a) and concentration (c)
(mole/liter) Ka
Ka =
11)Galvani:
He was the first scientist who changed the energy of the thermal energy into electrical energy.
12)Farady:
He discovered the relation between the amount of electricity is passed in the solution and the amount of matter which is released at the electrodes.
Question ( no. 4 ) Problems:
1- Calculate the volume of 0.1 of CO2 gas at S.T.P.
Volume of CO2 gas at S.T.P = 0.1 × 22.4 = 2.24 liter
2- Calculate the density of oxygen (O2) at (STP) . [O=16]
Molecular mass = 2 × 16 = 32 grams
The molecular mass = The density × volume of one mole
32 = The density × 22.4
The density = 32 ÷ 22.4 = 1.43 grams / liter
3- Calculate the volume of chlorine gas evolved at (S.T.P) during the electrolysis of sodium chloride (NaCl) solution by passing an electric current, its intensity is 10 amperes for 20 minutes. ( Cl = 35.45 )
The equivalent mass of chlorine = = 35.45 gm
The mass of chlorine = = 4.4 gm
The number of chlorine moles = = 0.062 moles
The volume of chlorine at S.T.P = 0.062 × 22.4 = 1.3888 liters
4- Potassium super oxide KO2 is a compound used to purify atmospheric air from carbon dioxide in a closed atmosphere as following equation .
4KO2 + 2CO2 2K2CO3 + 3O2
If 14.2 gm of KO2 is used, calculate the volume of oxygen formed in liters
(K=39 , O=16)
4KO2 + 2CO2 2K2CO3 + 3O2
4( 39 + 2 × 16 ) 3 × 22.4 liters
284 grams 67.2 liters
14.2 grams (X) liters
Volume of Oxygen = = 3.36 liters
5- Calculate the mass of calcium oxide produced from the thermal decomposition of 50 gram calcium carbonate, then calculate the volume of carbon dioxide gas evolve. ( Ca = 40 , O = 16 , C = 12 )
CaCO3 CaO + CO2
40 + 12 + 3 × 16 40 + 16
100 grams 56 grams
50 grams (X) grams
(X) Mass of calcium oxide = = 28 grams
CaCO3 CaO + CO2
100 grams 22.4 liter
50 grams (X) liter
Volume of CO2 = = 11.2 liter
6- Calculate the number of ions produce by dissolving 7.1 grams of sodium sulphate (Na2SO4) in water.
Na2SO4 + H2O(I) 2Na+(aq) + SO42-(aq)
Molecular mass of sodium sulphate = (23 × 2) + (32 × 1) + (16 × 4) = 142 grams
Number of moles of sodium sulphate = = 0.05 mole
Each mole of sodium sulphate gives 3 moles of ions.
Total number of moles of the produced ions = 3 × 0.05 = 0.15 mole
Total number of the produced ions = 0.15 × 6.02 × 1023 = 0.9 × 1023 ions
7- A two gram of impure sodium chloride was dissolved in water, Excess of silver nitrate was added to precipitate 4.628 grams of silver chloride. Calculate the percentage of chlorine in the sample. Given that :
(0 = 16 , N = 14 , Na = 23 , Cl = 35.5 , Ag = 108 )
NaCl + AgNO3 AgCl + NaNO3
Mole of AgCl = (1× 108) + ( 1×35.5) = 143.5 gram
143.5 gram AgCl 35.5 gram chlorine
4.628 gram AgCl (X) gram chlorine
Therefore mass of chlorine in silver chloride = mass of chlorine in sodium chloride
= ـ = 1.145 = = 57.245 %
8- 25 ml of 0.5 molar hydrochloric acid was neutralized with 20 ml of calcium hydroxide solution. Calculate the concentration of calcium hydroxide (mole/liter)
2HCl+ Ca(OH)2 CaCl2 + H2O
= ® = Concentration of calcium hydroxide (M2) = = 0.3125 moles / liter
9- 10 ml of 0.1 molar sulphuric acid added to 0.2 grams of impure sample of calcium carbonate until complete reaction takes place. Calculate the percent of calcium carbonate in The equation for the reaction is :
CaCO3 + H2SO4 CaSO4 + CO2 + H2O
The number of moles of H2SO4 = = 0.001 mole
The balanced equation shows that :
The number of moles of CaCO3 = the number of moles of H2SO4
The number of moles of CaCO3 reacting with H2SO4 = 0.001 moles
One moles of CaCO3 = 40 + 12 + ( 3 × 16 ) = 100 gm.
Mass of one CaCO3 in the sample = 100 × 0.001 = 0.1 gm .
% of CaCO3 in the impure sample = = 50 %
10- 0.2 gm of a mixture of solid substance containing sodium hydroxide and sodium chloride was titrated wih 0.1 molar of hydrochloric. The complete reaction takes place by the consumption of 10 ml of the acid. Calaulate the percentage of sodium hydroxide in the mixture. (Na = 23. O = 16. H = 1 )
Number of moles of HCI = = 0.001 mole
Number of moles of NaOH = 0.001 mole
The mass of one mole of NaOH = 23 + 16 + 1 = 40 gm
The mass of NaOH in the mixture = 40 x 0.001 = 0.04 gm
The percentage of NaOH the mixture = = 20%
11- Four grams of impure Sodium chloride was dissolved in water and an excess of silver nitrate solution was added to precipitate 9.256 gram of silver choride. Calculate the percentage of sodium chloride in the sample.
(Na = 23,Cl = 35.5,Ag = 108)
NaCl + AgNo3 NaNo3 + AgCl
The molecular mass in grams of AgCl = 108 + 35.5 = 143.5 gm
The molecular mass in grams of NaCl = 23 + 35.5 = 58.5 gm .
Each 58.5gm. of NaCl 143.5gm.ofAgCl
(X) gm. of NaCl 9.256gm.ofAgCl
The mass of NaCl in the sample = = 3.7733gm.
% NaCl = x 100 = 94.3325%
12- A sample of 1.47grams of hydrated calcium chloride salt ( CaCl2. ×H2O). was heated several times till a constant mass of 1.11 grams., Find out the number of water molecules of crystallization in the molecule of hydrated calcium chloride.
(H = 1.O = 16.Ca = 40.Cl = 35.5)
The mass of water of crystallization in the sample = 1.47 – 1.11 = 0.36 gram.
Each 1.11gram of anhydrous (CaCl2) bind with 0.36 gram water of crystallization
The molecular mass of (CaCl2) = (35.5 x2) + 40 = 111gram
1.11 gram CaCl2 bind with 0.36 gram water of crystallization
111 gram CaCl2 bind with (X) gram water of crystallization
X (mass of water of crystallization) = ( 111 x 0.36)÷ 1.11 = 36 gram
The molecular mass of (H2O)= 16 + (2×1) = 18 gram
The number of molecules of water of crystallization in the molecule
of hydrate CaCl2 = 36 ÷ 18 = 2 molecules
13- Calculate the equilibrium constant ( Kp ) of the reaction :
N2(g) + 3H2(g) 2NH3(g) DH = -92 KJ
The pressures of gases are 2.3 atmosphere for N2 , 7.1 atmosphere for H2
and 0.6 atmosphere for NH3. Comment on the value of (Kp).
KP = = = 4.4 × 10-4
The small value of Kp indicating the small amount of resulting ammonia .
14- Calculate the equilibrium constant for the reaction : 12 + H2 2HI
Provided that concentrations of iodine , hydrogen and hydrogen iodide at equilibrium are 0.1105.0.1105 and 0.7815 mole / liter respectively .
Kc = = = 50.018
15- If the equilibrium constant for the following reaction is 15.75 Cl2 + PCl3 PCl5
And the concentrations of chlorine and phosphorus trichloride were 0.3 and 0.84 mole / liter respectively. Calculate the concentration of phosphorous pentachloride.
Equilibrium constant Kc = ® 15.75 = ® 15.75 =
The concentration of PCl5 = 15.75 × 0.252 = 3.969 mole / liter
16- Calculate the degree of dissociation in 0.1 molar hydrocyanic acid (HCN) solution at 25°C. Providing that the equilibrium constant of the acid Ka = 7.20 × 10-10
HCN + H2O H3O+ + CN-
Applying Ostwald law : Ka = a2 × C
7.2 × 10-10 = a2 × 0.1
a2 = = 72 × 10-10
a = = 8.5 × 10-5
The degree of dissociation of hydrogen ion = 8.5 × 10-5
17- Calculate the concentration of hydronium ions (H3O+) in a 0.2 molar solution of acetic acid at 25°C given that equilibrium constant of the acid is (1.8 × 10-5)
[H3O+] =
[H3O+] = = 1.897 × 10-3 molar
18- Calcium fluoride has Ksp of 3.9 × 10-11 . What is the fluoride ion concentration at equilibrium ?
CaF2 (s) Ca-2 (aq) + 2F-(aq)
(X) (X) (2x)
Ksp = [Ca-2] × [2f]2
3.9 × 10-11 = (X) × (2x)2
3.9 × 10-11 = 4 x3
x3 = C x = 2.1 × 10-4 mole / liter
19- Calculate the solubility product. (Ksp) of calcium phosphate Ca3(PO)4 which is sparingly soluble in water , given that the concentration of calcium ions is
( 1× 10-4 ) mole / litre.
Ca3(PO4)2 3Ca2+ + 2PO43-
Ksp = [Ca2+]3 [PO43-]2
= [1 × 10-8]3 [0.5 × 10-3]2 = 2.5 × 10-31
20- Two metals A and B their standard oxidation potentials are – 0.3 and 0.7 volts respectively, each of them is divalent :
1- Using a diagram how can you represent the cell which can be formed from these two metals?
2- Calculate the electromotive force of this cell.
3- Does this cell produce an electric current ? Why ?
1- B B+2 A+2 A
2- e.m.f = standard oxidation potential of anode – standard oxidation potential of cathode e.m.f = 0.7 – ( -0.3 ) = 1 volt
3- An electric current is produced because the e.m.f is positive value and the reaction inside the cell is spontaneous .
21- What is the number of faradays required to precipitate gram/atom of copper according to the following reaction :
Cu2- + 2e- Cu (at cathode)
2 Faradays
22- Calculate the electric current intensity required for passing 3.7 Faraday through an electrolyte for 40 minutes.
The quantity of electricity= current strength × time
3.7 × 96500 = I × 40 × 60
I = (3.7 × 96500) + (40× 60) = 148.7 ampere
23- What is the quantity of electricity (coulomb) necessary to separate 5.6 grams of iron from solution of Iron (III) chloride ? Where the cathode reaction is :
Fe3- + 3e- Fe0 (Fe = 55.86)
Equivalent mass of iron = 18.62 grams
Quantity of electricity (coulomb) =
= 29022.5 coulomb
24- Aluminum metal is produced from electrolysis of molten aluminum oxide.
Calculate the number of aluminum moles produced when an electric current of intensity 9.6 amperes is passed for 5 minutes . the cathode reaction equation is : Al3 + 3e- Al0 (Al = 27)
Quantity of electricity = electric current intensity intensity × time
= 9.65 × 5 × 60 = 2895 coulombs
Equivalent mass of Al = atomic mass / valency = = 27 / 3 = 9 gram
96500 coulombs 9 grams Al
285 coulombs (X) grams AI
X = = 0.27 gram
27 grams Al 1 mole Al
0.27 gram Al (X) mole Al
(X) = 0.01 mole Al
Question ( no. 5 )
Various questions
1- By using balanced symbolic chemical equations, how can you obtain :
Phenol from benzene .
Ethylene glycol from ethanol .
1- Obtaining phenol from benzene :
2- Ethylene glycol from ethanol
2- Show by balanced chemical equations : (Write the conditions of reaction ):
1- Conversion of carboxylic group of an organic compound into hydroxyl group
2- Conversion of an aromatic acid into an acid amide.
3- Show by balanced chemical equations, how to obtain :
A- Methane gas from ethanoic acid.
B- Picric acid from chlorobenzene .
4- Show by balanced chemical equations, how can you obtain each of the following:
1- aAcetone from 2- Bromopropane .
2- Ethyl benzoate ester from toluene.
5- Show by balanced chemical equations what happens in each of the following cases:
1- Addition of iron (Ill) chloride solution to ammonium thiocyanate solution.
2- Addition of yeast (zymase enzyme) to glucose solution.
3- Dry distillation of sodium benzoate.
4- The reaction of ethanol with concentrated sulphuric acid at 180°C.
6- Show by balanced chemical equations, how can you obtain each of the following :
1- Benzene sulphonic acid from sodium benzoate.
2- Ethyl chloride from acetic acid.
7- By using the balanced symbolic chemical equations, how can you obtain:
1- Nitrobenzene from sodium benzoate.
2- Dihydric alcohol from monohydric alcohol.
8- Illustrate by the balanced chemical equations how you can obtain the following:
1- Cycloalkane from normal alkane.
2- Pricric acid from chloro benzene.
2- propanol
1- propanol
Picric acid
2- methyl -1- propanol
2- methyl -2- propanol
Catecol
Choose from the above table, the compound (or compounds) that considered:
1- A Phenols .
2- A Secondary alcohol .
3- Alcohol which gives aldehyde on oxidation.
4- Alcohol which gives ketone on oxidation.
5- Product of nitration of phenol.
6- Disubstituted benzene.
7- Tetra substituted benzene .
1) Catecol. 2) 2-propanol.
3) 1-propanol. 4) 2-propanol.
5) Picric acid 6) Catecol.
7) Picric acid.
10-
Acetic acid
Formic acid
Oxalic acid
Ethyl formate
Methyl acetate
Ethyl acetate
From the above table find :
1- Two isomers.
2- Two compound that give acetamide on ammonolysis.
3- Compounds that give effervescence with NaHCO3 (one enough).
4- Compound that is named according to IUPAC ethyl methanoate.
1) Ethyl formate-methyl acetate (Two isomers) .
2) Methyl acetate – ethyl acetate (Two compounds from acetamide)
3) Acetic acid- formic acid – oxalic give effervescence with sodium bicarbonate (One acid is enough)
4) Ethyl formate.
11-
Methyl acetate
Sodium acetate
Ethanoic acid
Methyl formate
Potassium acetate
Ethyl formate
From the previous table illustrate the following :
1) The esters .
2) The salts of the carboxylic acids.
3) The compounds which take a name according to IUPAC system .
4) The isomeric compounds.
12-
Formic acid
Ethanoic acid
Oxalic acid
Salicylic acid
Butyric acid
Benzoic acid
From the previous table mention the following :
1) Aromatic monocarboxylic acid .
2) Dicarboxylic acid.
3) An acid in which the number of carboxylic groups equals the number of carbon atoms .
4) An acid contains two functional groups.
1) Benzoic acid. 2) Oxalic acid. 3) Oxalic acid. 4) Salicylic acid.
13- Give reason for each of the following :
Salt bridge must be used in Galvanic cells.
Cone. Sulphuric acid is added in ester formation and nitration.
The catalyst does not affect the position of equilibrium in the reversible reactions.
density of oxygen gas is more than of nitrogen. [N = 14 , O = 16 ]
1) To neutralize of excess ions in both half cells. Thus, it prevents the formation of electric potential between two solutions.
2) To absorb the water produced and prevent the reversible reaction.
3) Catalyst decreases the activation energy required to accelerate reversible and irreversible reaction at the same time.
4) Because molecular mass of O2 which is (32) is more than of N2 which is (28).
14- Give reasons for the following :
- Using a mixture of fluoride salts of aluminum, Sodium and calcium instead of cryolite containing a little amount of fluorspar in the extraction of aluminum from bauxite.
- Because this mixture gives with bauxite a melt which is characterized by a low melthing point and a lower density compared to the melt obtained from Cryolite.
15- Mention the structural formula of :
1- Naphthaline.
2- Diphyenyl
Do you consider these two compounds isomers? Why ?
1) Structural formula for naphthaline is
2) Structural formula for diphyenyl is
They are not isomers due to different molecular formula for each of them, where (C10H8) naphthaline – C12H10 diphenyl )
16- The molecular mass of a hydrocarbon is 58 grams, its mole contains 48 grams carbon ( H= 1 , C = 12)
1- Write the molecular formula for this compound.
2- This compound has two isomers (isomerism), write the structural formula for each isomer.
Mass of hydrogen = 58 – 48 = 10 grams
Number of carbon atoms = 48 ÷ 12 = 4 atoms
Number of hydrogen atoms = 10 atoms
The molecular formula of the hydrocarbon C4H10
The structural formulas for the two isomers are :
17- Read the following statement. Then answer the questions that follow it :
When glycerol reacts with nitric acid in the presence of concentrated sulphuric acid substance (x) is produced. Substance (x) is used to widen arteries in the treatment of heart crisis .
1- What is the name of the substance (x) ?
2- Mention another use for substance (x) .
3- Write the balanced chemical equation which describes the reaction of nitric acid with benzene in the presence of hot concentrated sulphuric acid (50°C).
1) Tri-nitroglycerine.
2) It is used in the prearation of explosive substance.
18- A and B are two organic compounds having the molecular formula [C2H6O]
1- Write the structural formula for each compound.
2- If compound (A) responds to oxidation reactions and compound (B) does not. How can you convert compound (A)) to a compound that has the function group of compound (B) ?
3- How can you obtain hydrogen from one of the two compounds?
19- When calcium reacts with carbon compound (A) is formed. On dropping water on compound (A) compound (B) is formed. On adding water to compound (B) in presence of catalyst and heating a liquid (C) is formed. On passing compound (B) through a red hot nickel tube a vapour of compound (D) is formed.
From the previous information :
1- Write the balanced chemical equations that illustrate the following :
a) How to obtain dichloroethane from compound (B) ?
b) The effect of concentrated sulphuric acid on compound (D).
2- Mention one use for compound (C).
2- Preparation of ethanoic acid (acetic)
20- An experiment was carried out on two organic compound (A) and (B). It was found that :
- Compound (A) reacts with sodium metal and not with sodium hydroxide.
- Compound (A) reacts with both sodium carbonate and sodium hydroxide.
1- To which group of compounds do (A) and (B) belong? Give an example for each group.
2- Write a symbolic equation for the reaction between (A) and (B) with reference to the conditions of the reaction.
Compound (A) is ethanol (C2H5OH)
Compound (B) is acetic acid (CH3COOH)
The reaction between (A) and (B) :
21- Explain the role of each of the following :
1- Analytical chemistry in agriculture.
2- Catalysts in industry.
1- Analytical chemistry in agriculture :
It is possible to know the constituents of a soil and rocks to decide whether the soil is suitable or not for cultivation.
2- Catalysts in industry :
It is possible to accelerate most of the slow reactions by using catalysts which increase the reaction rate without raising the temperature. This tends to save energy. Catalysts are used in more than 90% of the industrial processes such as foods, petrochemicals and fertilizer industries.
22- What is the role of each of the following :
1) Potassium hydroxide in the mercury cell.
2) Soda lime in preparation of methane .
3) Potassium permanganate in Baeyer’s reaction .
4) Concentrated sulphuric acid in the reaction of ester formation.
1- It is used as an electrolyte .
2- Soda lime is a mixture of sodium hydroxide and calcium oxide.
Sodium hydroxide reacts with sodium acetate forming methane and sodium carbonate calcium oxide helps in :
- Reducing the melting point of the mixture.
- Absorbing water vapour.
3- It acts as an oxidizing agent .
4- It absorbs the produced water and prevents the reversible reaction .
23- You are provided with three glasses containing : Ethanol – phenol – acetic acid. Explain how you can identify each of them practically.
Add acidified potassium dichromate dichromate to the same amount of the three solutions , heat the mixture in a water bath for 10 minutes. If the colour changes from orange to green then the solution is ethanol.
- Add iron III chloride to the three solutions. If violet colour appears then the solution is phenol.
- Acidic test: add sodium carbonate or bicarbonate, if effervescence takes place and carbon dioxide gas is evolved which turbids lime water then the solution is acetic acid.
24-How can you differentiate practically between ?
Dilute acetic acid and pure acetic acid.
Ethyl alcohol and phenol.
Methane and ethyne.
litmus solution and phenolphthalein solution.
1) By passing an electric current in each of them.
Electric conductivity of dilute acid is more than pure acid, In case of dilute acetic acid the lamp gives illumination – in case of pure acetic acid the lamp does not illuminate .
- BY adding FeCl3 solution
- If violet colour is formed, it is phenol .
- If no violet colour is formed , it is ethanol.
3- Experiment
Methane gas
Acetylene gas
By adding bromine dissolved in (CCl4) to each of them
No effect
The red colour of bromine disappears
4- By adding acid solution to both solutions,
– If the colour turns colourless, the solution is phenolphthalein
– If the colour turns red, the solution is litmus .
25- Show by practical experiment how you can detect the presence of carbon of carbon and hydrogen elements in an organic substance.
Put a small amount of an organic substance mixed with copper oxide in a glass tube and heat the test tube strongly then pass the resulting gases over anhydrous white copper sulphate , then through lime water.
Observation :
1- The white colour of anhydrous copper sulphate turns blue.
2- Lime water turns turbid.
Conclusion : The organic compound contains carbon and hydrogen elements.
The equations:
C + 2 CuO 2 Cu + CO2
2H + CuO Cu + H2O
26- How can you differentiate practically between the following :
1- Methane and ethylene.
2- Ethanol and 2- methyl -2- butanol.
1- Experiment
Methane gas
Ethene (ethylene) gas
By adding bromine dissolved in (CCl4) to each of them
No effect
The red colour of bromine disappears
2- Experiment
Ethanol
2- Methyl -2- butanol
By adding KMnO4 solution and conc. H2SO4 to each of them
The violet colour of potassium permanganate disappears
No effect
27- How can differentiate between :
( Illustrate your answer by observation and balanced chemical equations)
1- Ethanoic acid and carbolic acid (By acidity test ).
2- Ethylene gas and methane gas (By using potassium permanganate solution)
3-
1- Experiment
Ethanoic acid
Carbolic acid
By adding Na2CO3 solution to each of them
Effervescence takes places and CO2 evolves which turbids lime water
No effect
CH3COOH + NaHCO3 CH3COONa + Ca2 + H2O
C6H3OH + NaHCO3 no effect
2- Experiment
Methane gas
Ethene (ethylene) gas
b) By adding KMnO4 in an alkaline medium to each of them
No effect
The violet colour of KMnO4disappears
C2H4 + H2O + [O] C2H4 (OH) ethylene glycol
28- How can you differentiate practically between each of the following:
1-Litmus solution and phenolphthalein solution.
2- Ethanoic acid and carbolic acid.
1- Indicator
Colour in acidic medium
Colour in basic medium
Phenolphthalein
Colourless
Red
Litmus
Red
Blue
2- Experiment
Ethanoic acid
Carbolic acid
By adding Na2Co3 solution to each of them
Effervescence takes places and CO2evolves which turbids lime water
No effect
29- Draw the apparatus used for the preparation of ethyne in the laboratory then answer the following:
1- Show by practical experiment gow you can detect ethyne gas.
2- Write the balanced chemical equation which describes the reaction of ethyne with hydrogen bromide.
30- The apparatus used for preparation of ethane (ethylene) gas in laboratory:
1- Ethyne gas + bromine dissolved in CCl4 the red colour of bromine disappears.
2- The reaction of ethyne with hydrogen bromide.
31- Draw a labeled diagram for sector in dry cell.
32- Write the balanced chemical equation for preparation of methane in laboratory then draw the apparatus used in the preparation.
? The apparatus used :
Equation of reaction : CH3COONa + NaOH CH4 + Na2CO3
33- The degree of purity of metals which are prepared in industry is usually lower than the required degree of purity which is required for definite purposes.
Explain by drawing how to obtain copper of 99.95 % Purity degree.
The steps : Dip the pure copper and the impure copper in copper sulphate solution.
connect the pure copper to the negative pole of the battery.
( The pure copper acts as the cathode ) Connect the impure copper to the positive pole of the battery.
( The impure copper acts as the solution ) pass the electric current in the solution.
Observation :
Copper sulphate solution is ionized as follows :
CuSO4 Cu2+ + SO42-
At the anode : Cu Cu2- + 2e- (Oxidation)
At the cathode : Cu2+ + 2e- Cu (Reduction)
Therefore, copper will dissolve from the impure copper (anode) as copper ions (Cu2+), whereas copper ions in the solution will deposit (precipitate) as pure copper (Cu) at the pure copper (cathode).
In this process 99.95 % pure copper can be obtained.
34- If you have a copper spoon what are the different steps you must follow it to plate it by a layer of silver. Write down the equations (reactions) which occur at each of the cathode and anode.
The electroplanting of copper spoon by a layer of silver .
1) Clean the surface of the spoon completely.
2) Dip the spoon into electrolyte solution that contains silver ions.
3) Connect the spoon with the (-ve) electrode of the battery (cathode) and a rod of silver metal with the (+ve) electrode (anode).
4) Switch on the circuit.
When the electric current passes through the circuit :
- Oxidation process takes place at anode.
Ag Ag+ + e-
- Reduction process takes place at cathode.
Ag+ + e- Ag
So, silver atoms precipitate on the surface of the spoon.
35- Compare between each of the following :
1- Acidity of alcohol and acidity of phenol.
2- Dry cell and lead acid battery.
1- Acidity of alcohol
Acidity of phenol
1- Alcohols are less acidic than phenols because the alkyl group decreases the length of bond between oxygen and hydrogen atom in (O-H) group and this makes the separation of H+ ions is more difficult.
1- Phenols are more acidic than alcohols because the aryl group increases the length of bond between oxygen and hydrogen atom in (O-H) group and this makes the separation of H+ ions is easier.
2- Alcohols react with active metals such as sodium but do not react with NaOH.
2- Alcohols react with active metals such as sodium and react with NaOH.
P.O.C
Dry cell
Lead acid battery
1- Anode
Zinc electrode
Lead electrode
2- Cathode
Graphite electrode
Lead tetra oxide electrode
3- Electolite
Ammonium chloride
Dilute sulphuric acid
4- E.M.F
1.5 volt
2.0535 volt
5- Reactions
Zn0 + 2MnO2 + 2NH4
Zn2+ + 2Mn (OH) + 2NH3
At anode (Oxidation) :
Pb + SO42- PbSO4 + 2e-
At cathode (Reduction) :
PbO2 +4H+ + SO4 + 2e-
PbSO4 + 2H2O
Total reaction :
Pb + PbO2 + 4H+ + 2SO4
2PbSO4 + 2H2O
36- Compare between each of the following :
1- Addition polymerization and condensation polymerization .
2- Chemical equilibrium and ionic equilibrium.
1- Addition polymerization
Condensation polymerization
Combination of a large number of unsaturated simple molecules (monomers) to give a giant molecule (polymer) having the same empirical formula. (e.g. Eghylene glycol)
Condensation of two different monomers to give a new monomer called copolymer which undergoes polymerization (e.g. Dacron)
2- Chemical equilibrium
Ionic equilibrium
It is a state of equilibrium in the reversible reaction at which the rate of forward reaction equals the rate of backward reaction.
It is the equilibrium arising between molecules of a weak electrolyte and the ions resulting from it .
37- Compare between each of the following:
1. Irreversible reactions and reversible.
2. Alkaline hydrolysis of esters and acidic hydrolysis of esters
1. Complete (irreversible) reactions
Incomplete (reversible) reactions
1.The reactions which proceed in one direction (forward)
1. The reactions which proceed in both directions; forwaed and backward.
2. One of the products escapes from the system as evolving of a gas or forming a precipitate.
2. Both the reactants and products are always found in the reaction medium
Example:
NaCl + AgNO3 NaNO3 + AgCl
Mg + 2HCl MgCl2 + H2
Example :
CH3COOH + C2H5OH
CH3COOC2H5 + H2O
2NO2 N2O4 , DH = -ve
Alkaline hydrolysis of esters
Acidic hydrolysis esters
1. Hydrolysis of esterin in presence of aqueos alkali e.g. (aqueous NaOH)
1. Hydrolysis of ester in presence of dilute mineral acid e.g. (H2So4)
2.Gives an alcohol and the salt of the acid.
2. Gives an alcohol and organic acid.
3.Example:
CH3COOC2H5 + NaOH
CH3COONa+C2H5OH
3. Example:
CH3COOC2H5 + H2O
C2H5OH + CH3COOH
38- The following reaction has tow equilibrium constant value at two different temperatures:
H2(g) + I2(g) 2HI (g)
Kc at 850 °C = 67 and at 448 °C = 50
Is this reaction exothermic or endothermic? Why?
The reaction is endothermic
As the Value of Kc incrases, concentration of products, concentration of products increases as the temperature increases, therefore the reaction is endothermic.
39- How each of the following changes affects the concentration of hydrogen in the following equilibrium system:
H2(g) + CO2(g) H2O(g) + CO(g) ; H = + 41.1 kj
a. Addition of more CO2 gas,
b. Addition of more water vapor.
c. Addition of a catalyst.
d. Increase the temperature.
e. Decrease the vessel volume.
a) Decreases the concentration of hydrogen.
b) Increases the concentration of hydrogen.
c)Does not affect the concentration of hydrogen.
d) Decreases the concentration of hydrogen.
e) Dose not affect the concentration of hydrogen.
FeCl3 + 3NH4SCN Fe(SCN)3 + 3NH4CI
1- The red colour of the solution increase indicating to the formation of more iron (III) thiocyanate.
2- The intensity of the red colour decreases because the equilibrium shifts towards backward reaction to the formation of iron (III) chloride.
40- In the balanced reaction :
CH3COOH + C2H5OH CH3COOC2H5 + H2O
What happens to the equilibrium of this reaction in the following cases ?
- Adding excess of water .
- Adding drops of cone sulphuric acid .
- The equilibrium shifts in the opposite direction . (backward)
- The equilibrium shift in the forward direction .
41- Consider the reaction at equilibrium :
CH3COOH + H2O CH3COO- + H3O+
Explain the effect of each of the following changes on the concentration of \
acetate ion :
1) Adding drops of hydrochloric acid.
2) Adding drops of sodium Hydoxide solution .
1) Adding drops of HCl leading to increase the concentration of Hydronium ion in the Solution as a result of combination between the Hydrogen ion of the acid and water molecules in the solution, according to Le Chatelier’s principle the reaction shifts in the direction which decreases the concentration of Hydronium ion then the concentration of acetate ion decrease.
2) Adding drops of Sodium Hydroxide neutralizes acetic acid, the concentration of the acid decrease, according to Le Chatelier’s principle the reaction shifts in the direction which decreases the concentration of Hydronium ion then the concentration of acetate ion decrease .
42- Arrange the following in an ascending order as reducing agents :
1) Zn2+ / Zn [ -0.762 volt ]
2) Mg/Mg2+ [ 2.375 volt ]
3) 2Cl- / Cl2 [ -1.36 volt ]
4) K+ / K [ -2.924 volt ]
5) Pt2+ / Pt [ 1.2 volt ]
Cl < Pt < Zn < Mg < K
43- Arrange the following alcohols in ascending order according to their boiling points, give the scientific reason. (Ethylene Glycol – Sorbitol – Ethanol – Glycerol )
Ethanol < ethylene glycol < glycerol < sorbitol Due to increase of the number of polar hydroxyl groups which have the ability to form hydrogen bonds between the molecules of alcohols causing increase of its boiling point.
44- Arrange the following compounds according to the pH value of their aqueous solutions :
NaCl – CH3COONa – NH4Cl
NH4Cl < NaCl < CH3COONa
45- Arrange the following compounds in ascending order with respect to increase in their acidity :
Arrangement of the following compounds in ascending order with respect to increase in their acidity :
46- Rearrange correctly what are inside the boxes in the following diagram :
& Events of the scientisis &
1)Berzelius:
a) Classify the compounds into two types :
1- Organic compounds which are extracted from animal or plant origin.
2- Inorganic compounds which are originated from mineral sources in the earth
b) He thought that organic compounds are formed only under the influence of vital force, which is found in living cells of the body and he was the first scientist who classified the elements into metals and non-metals and non-matals.
2)Wohler :
Wohler performed an experiment which was considered to be the beginning of the end of Berzelius theory.
Wohler was able to prepare Urea, which is organic compound, by heating an aqueous solution of two inorganic compounds (Ammonium chloride and silver cyanate solution).
NH4Cl + AgCNO AgCl + NH4CNO
3)Baeyer “Baeyer’s reaction” :
He discovered reaction of alkene with potassium in an alkaline medium. This reaction is used to direct the presence of double bond.
4)Markownikoff :
He discovered a rube which controls the addition of HX or HOSO3H to unsymmetrical alkenes as propene. Where H is added to carbon rich with hydrogen and X is added to carbon poor with hydrogen.
5)Kekule:
He suggested a new idea about the bonding between carbon in the benzene molecule.
He sad carbon atoms from a ring, in which singel and double bonds are exchanged, so that, all bonds between carbon atoms similar in length,froming a hexagonal regular shaoe in which carbon are there at every angle
The ring in the figure indicates that 6 electrons are delocalized at certain carbon atoms.
6)Fridelcraft’s:
He introduced an alky1 group to a benzene ring in the presence of anhydrous AlCl3
As a catalyst.
7)Gaylussac’slaw:
The volume of the reactant gases and produced are in a fixed ration.
H) Avogadro’s number:
Equal volumes of gases contain the same number of molecules under the same temperature and pressure
8) Lechchatelier:
He formulated a principle named after him. This principle that
“When a system at equilibrtum is subjected to any change such as (Conc., temp. and pressure), the equilibrturm will shift in the direction wich will oppose this change.
9)guldbergandWaage:
He discovered the law of mass action which, the relation between the rate of chcmical reaction and the concentration of the reactants low states that.
At constant temperature, the rate of chemical reaction is dirctty propational to the product of multiplication of the reactant concentrated each is raiscd to the power of number of molecules or ions in the balanced chemical equations.
10)Ostwald:
He found the relation between the degree of ionization (a) and concentration (c)
(mole/liter) Ka
Ka =
11)Galvani:
He was the first scientist who changed the energy of the thermal energy into electrical energy.
12)Farady:
He discovered the relation between the amount of electricity is passed in the solution and the amount of matter which is released at the electrodes.