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                                                  Unit 9
                                      CHEMICAL EQUILIBRIUM
 Exercise: Short Questions.
 Q1. Define chemical equilibrium with two examples.
 Ans: Definition: “The state of a reversible chemical reaction in which the rate of forward reaction
 becomes equal to the rate of backward reaction in called chemical equilibrium/Dynamic equilibrium”.
 OR The state in which both reactants and products are presents in concentration in which no further
 tendency to change with time.
 Examples
 i. Change of a liquid to gas in a closed container:
 when liquid water is placed in a closed container at constant temperature, part of liquid evaporates. As
 water begins to evaporate, at the same time, some of the vapours also begin to condense. Although in
 the beginning rate of evaporation is faster than the rate of condensation but with the passage of time
 the rate of evaporation becomes equal to the rate of condensation and thus a state of equilibrium is
 established
           𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
 Water              ⇄              Vapours
          𝑐𝑐𝑒𝑒𝑒𝑒𝑐𝑐𝑒𝑒𝑒𝑒𝑐𝑐𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
 Equilibrium state is also in the following chemical reactions
 ii. Reaction of sulphur dioxide with oxygen:2SO2 (g)+ O2 (g)⇌ 2SO3(g)
Q3. Compare the different macroscopic characteristics of forward and reverse reactions?
            Forward reaction                                Reverse reaction
Characteristics
ii. It takes place from left to right ii. It takes place from right to left.
iii. At the beginning, rate of forward reaction is iii. At the beginning, rate of reverse reaction is
very fast.                                        very slow.
Which means when we increase the Molar concentration of reacting substances the rate of reaction also
increases and vice versa.
Q6: At equilibrium a mixture of N2, H2, and NH3 gas at 500°C is determined to consist of
0.602mol/dm3 of N2, 0.420mol/dm3 of H2 and 0.113mol/dm3 of NH3. What is the equilibrium
constant for the reaction for the reaction at this temperature?
N2 (g) + 3H2 (g) ⇌ 2NH3 (g)
Given: Concentrations of Nitrogen (N2) = 0.602mol/ d𝑚𝑚3
Concentration of hydrogen (H2) = 0.420mol/ d𝑚𝑚3
Concentrations of NH3 = 0.113mol/ d𝑚𝑚3
Required:
Value of Kc =?
Solution:
Writing equilibrium constant expression for the above reaction:
           [𝑁𝑁𝑁𝑁3 ]2
    Kc=
          [𝑁𝑁2 ][𝑁𝑁2 ]3
               [0.113𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]2
=
     [0.602𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ][0.420𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]3
               [0.012769𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]2
=
    [0.602𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ][0.074088𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]3
               [0.012769𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]2                  [0.012769𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]2
=                                               −3 3 =
    [0.602𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ][0.074088𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚      ]       [0.0446𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]4
                              -3 -2
   = 0,28630[mol/dm ]
   0.28630𝑚𝑚𝑒𝑒𝑙𝑙 −2 𝑐𝑐𝑚𝑚6
Q7: State conditions necessary for the chemical equilibrium.
Ans: Conditions necessary for equilibrium:
Followings are the conditions necessary for chemical equilibrium:
i. Closed container:
Equilibrium state can only be attained in closed container. It cannot be attained in open container
because in open container the gaseous reactants and products will escape due to which there will be no
possibility of equilibrium.
ii. Constant concentrations:
When equilibrium state is attained by a reaction in a closed container then the concentrations of various
species in the reaction become constant. These concentrations are called equilibrium concentrations.
iii. Effect of catalyst on equilibrium:
A catalyst cannot change the equilibrium point, it only speedy up the rate forward and backward
reactions. Thus, it helps to attain the equilibrium in short time.
Q8. Write equilibrium constant expression for the following reactions.
a. N2 (g) + 3H2 (g) ⇌ 2NH3 (g)
Kc for this reaction is
          [𝑁𝑁𝑁𝑁3 ]2
Kc =
       [𝑁𝑁2 ][𝑁𝑁2 ]3
b. 2H2 (g) + O2 (g) ⇌ 2H2O (g)                         Kc for this reaction is:
        [𝑁𝑁2𝑂𝑂]2
Kc =
       𝑁𝑁2]2 [𝑂𝑂2 ]
         [𝑁𝑁𝑂𝑂]4 [𝑁𝑁2𝑂𝑂]6
Kc =
         [𝑁𝑁𝑁𝑁3]4 [𝑂𝑂2]5
Q9. A reaction between gaseous sulphur dioxide and oxygen gas to produce gaseous sulphur
trioxide take place at 6000C. At this temperature, the concentration of SO2 is found to
1.50mol/dm3. Using the balanced chemical equation, calculate the equilibrium constant for this
system.                                                   Ans: balanced chemical equation:
2SO2(g) + O2(g) →2SO3
Solution:
 Concentration of SO2 =1.50mol/dm3
Concentration of O2 =1.250mol/dm3
Concentration of SO3 =3.50mol/dm3
  Find Kc =?
                              [𝑆𝑆𝑂𝑂3] 2
  Formula = Kc=
                            [𝑂𝑂2 ][𝑆𝑆𝑂𝑂2 ]2
                  [3.50𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]2
  =
         [1.250𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ][1.50𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]2
        12.25[𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]2
  =
         2.8125[𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]3
  4.353[mol/dm3]-1
Explanation:
i. Change of a liquid to gas in a closed container:
Consider a closed container which is partially filled with a liquid at a given temperature. At the start, and
the vapour molecules are collected at the liquid surface. At the passage of time the collected gas
molecules over the liquid surface converts to liquid back (condensation starts). In the beginning rate of
evaporation is faster than the rate of condensation but with the passage of time the rate of evaporation
becomes equal to the rate of condensation and thus a state of dynamic equilibrium is established.
v. Graphical representation:
Thus, it is clear that rate of reaction is proportional to the concentration of the reactants.
Q3. Derive an expression for the equilibrium constant and explain its units.
Ans: equilibrium constant:
The ratio of the mathematical product of the concentration of reacting substances called equilibrium
constant”. OR
“The ratio of the product of the concentration of the products to the product of the concentration of
reactants at equilibrium is called equilibrium constant
Denotation
It is denoted by Kc where the subscript c indicates the equilibrium concentrations of various species in
term of mole/om3 or litre.
Derivation of expression for the equilibrium constant:
Consider the following general reaction
          A+B⇌ C+D
Then Kc for this reaction will be
                K𝑜𝑜       [𝐶𝐶][𝐷𝐷] K𝑜𝑜
                      =                     = Kc
                K𝑐𝑐 [𝐴𝐴][𝐵𝐵] K𝑐𝑐
                 [𝐶𝐶][𝐷𝐷]
       Kc = [𝐴𝐴][𝐵𝐵]
  For more general reaction Kc is written as:
        aA +bB ⇌ cC+dD
                           [𝐶𝐶]𝑐𝑐 [𝐷𝐷]𝑎𝑎
                Kc =
                            [𝐴𝐴]𝑎𝑎 [𝐵𝐵]𝑏𝑏
Where [A], [B],[C] and [D] represents molar concentration of reactants and products while a,b,c and d
represent the number of moles from the balanced chemical equation.
Units of equilibrium constant:
 The unit of equilibrium constant is independent of pressure, concentration and catalyst depends the
equilibrium constant expression.
Q3.b How Can you predict direction of reaction for the Kc value.
Ans: Prediction of direction of reaction:
The direction of reaction can be predicted by means of [product]/ [reactants] ratio.
      [𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑖𝑖𝑐𝑐𝑐𝑐𝑖𝑖𝑖𝑖𝑐𝑐𝑖𝑖𝑖𝑖𝑖𝑖𝑐𝑐𝑖𝑖 𝑐𝑐𝑜𝑜 𝑝𝑝𝑐𝑐𝑐𝑐𝑝𝑝𝑝𝑝𝑐𝑐𝑖𝑖𝑝𝑝]
Qc=
        [𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑖𝑖𝑐𝑐𝑐𝑐𝑖𝑖𝑖𝑖𝑐𝑐𝑖𝑖𝑖𝑖𝑖𝑖𝑐𝑐𝑖𝑖 𝑐𝑐𝑐𝑐𝑖𝑖𝑐𝑐𝑖𝑖𝑖𝑖𝑐𝑐𝑖𝑖𝑝𝑝]
By comparing the ratio with Kc we have three possibilities:
When the ratio is less than Kc:
if the ratio is less then Kc. The system is not at equilibrium and more product are required to reach the
equilibrium. Therefore, reaction will proceed in forward direction to form products.
Case-1:
It may have no units if the number of moles of reactants and products are equal in the balanced chemical
equation.
Example:
          H2(g) + I2(g) ⇌ 2HI (g)
      Kc for this reaction is:
       [𝑁𝑁𝐻𝐻]2                     [𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚3 ]2
Kc=                  =                                        = No units
      [𝑁𝑁2 ][𝐻𝐻2 ]        [𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ][𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]
Case-2:
It may have units for the reaction in which the number of moles of product is greater than the reactants in
a balanced chemical equation.
Example:
   N2O4(g) ⇌ 2NO2(g)
Kc for this reaction is:
         [𝑁𝑁𝑂𝑂2 ]2        [𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]2
 Kc=                  =
         [𝑁𝑁2 𝑂𝑂4 ]            [𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]
   = mol.dm-3
Case-3:
It may have units for the reaction in which the number of moles of product is less than the reactants in a
balanced chemical equation.
Example:
   N2(g) + 3H2(g) ⇌ 2NH3 (g)
Kc for this reaction is:
           [𝑁𝑁𝑁𝑁3 ]2                      [𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]2
 Kc=                     3 =
         [𝑁𝑁2 ][𝑁𝑁2 ]             [𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ][𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]3
     [𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]2                    1                      1
 =                         =                        = [𝑚𝑚𝑐𝑐𝑖𝑖2
     [𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]4          [𝑚𝑚𝑐𝑐𝑖𝑖 𝑝𝑝𝑚𝑚−3 ]2                𝑝𝑝𝑚𝑚−6 ]
Q4.(b). How can you predict the extent of reaction from the value of Kc value?
Ans: Extent of chemical reaction:
 The value of Kc tells us about the extent of reaction from which quantities of reactants or products can
 also be predicted.
 Consider the general reaction:
         aA +bB ⇌ cC+dD
                          [𝐶𝐶]𝑐𝑐 [𝐷𝐷]𝑎𝑎
               Kc =
                          [𝐴𝐴]𝑎𝑎 [𝐵𝐵]𝑏𝑏
The extent of reaction depends upon the magnitude of Kc , so when
i. Kc value very small:
When the concentration of [A]and [B] is large and that of [C] and [D] is small, the equilibrium, mixture
will contain large number of reactants and small number of products. It reflects that the reaction does not
proceed appreciably in the forward direction.
ii. Kc value very large:
When the concentration of [A] and [B] is small and that of [C] and [D] is large, the equilibrium, mixture
will contain large number of products and small number of reactants. It indicates that the reaction is
completed in the forward direction
iii . Kc value neither very small nor very large (moderate):
If Kc value is neither very small nor very large then neither the forward nor the reverse reaction goes to
completion. Thus, equilibrium mixture will contain appreciable amount of product and reactants.
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When the concentration of [A} and [B] is large and that of [C] and [D] is small, the equilibrium, mixture
will contain large number of reactants and small number of products. It reflects that the reaction does not
proceed appreciably in the forward direction.
ii. Kc value very large:
When the concentration of [A} and [B] is small and that of [C] and [D] is large, the equilibrium, mixture
will contain large number of products and small number of reactants. It indicates that the reaction is
completed in the forward direction
iii. Kc value neither very small nor very large (moderate):
If Kc value is neither very small nor very large then neither the forward nor the reverse reaction goes to
completion. Thus, equilibrium mixture will contain appreciable amount of product and reactants.
Example:
           A few examples of irreversible reactions are given below:
           i. Reaction of carbon with oxygen:
                   C(S)+O2 (g)      C O2 (g)
           Ii.Reaction of magnesium with HCl:
                   Mg(g) + HCl(g)        MgCl2(s)+ H2(g)
Reversible reactions:
Those chemical reactions which proceed both in forward and backward directions are called reversible
reactions”.
Characteristics of reversible reactions:
A reversible reaction has the following characteristics:
i. These reactions proceed in two directions i.e. forward and backward.
ii. These reactions never go to completion i.e. the reactants are not fully converted into products.
iii. Their reversibility can exist in closed containers only.
iv. A state of dynamic equilibrium is established in all reversible reactions.
v. These reactions are represented by double arrow sign (⇌ ).
Examples:
Following are a few examples of reversible reactions:
   i. Reaction of nitrogen with hydrogen:
                   N2 (g)+3H2(g)⇌ 2NH3(g)
At the start reaction goes in forward direction but as soon as soon amount of ammonia is formed, the
ammonia molecules dissociate to nitrogen and hydrogen. Thus, the reaction is reversed.
   Some other examples are:
   ii. Reaction of Sulphur dioxide with oxygen:
                   2SO2(g)+ O2(g)⇌ 2SO3(g)
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                                                  UNIT 10
                                      ACIDS, BASES AND SALTS
                                          SHORT QUESTIONS:
Q1. When a clear liquid is placed in a beaker. How can you identify whether it as an acid, base or
neutral?
Ans: Different methods is used to determine the whether the given solution/ liquid is acidic, basic or
neutral.
Litmus paper:
One of them is litmus paper test. In this method litmus paper is dipped in a beaker. If it turns the blue
litmus paper to red then the given liquid will be acidic. If it turns red litmus paper then the given liquid
will be basic. If the litmus paper remains unchanged it will be neutral liquid.
pH scale:
We can also use pH scale to measure the acidity or basicity of a solution. pH scale is a number from 0 to
14. From 0 to 7 are acids. From 7 to 14 are bases while if a liquid has a pH of 7. It will be neutral
Q2.justify H+ ion as a Lewis acid?
Ans: H+ ion as a Lewis acid:
According to Lewis concept a positively charged ions that can accept an electron pair can act as Lewis
acid.
As H+ is positively charged ion and it has tendency to accept lone pair of electrons so it acts as a Lewis
acid.
H+ + :NH3 ⟶ NH4+
(Acid) (Base)          (Ammonium ion)
Q3. Distinguish strong acids from weak acids? Give two example of each.
Strong acid:
Those acids which ionizes completely in aqueous solutions and give higher concentration of H+ ions are
called strong acids”. A strong acid is strong electrolyte.
HCI, H2SO4 and HNO3 are the examples of strong acids because they completely ionize in water.
HCI (aq)⟶ H+ + CI-
H2SO4 (aq)⟶ H+ + SO4-
HNO3 (aq)⟶ H+ + NO3-
Weak acids:
“Those acids which do not completely ionize in aqueous solutions and give lower concentration of H+
ions are called weak acids”. A weak acid is weak electrolyte.
Examples:
CH3COOH (aq) ⇌ H++ CH3COO-
Acetic acid (CH3COOH) is a weak acid because when it is added into water, very few molecules of
CH3COOH are dissociated. Some other weak acids are H2S, H2CO3, H2SO3 and HNO2 etc.
Q4. Compare the physical properties of acid and bases.
Ans:
                       Acids                                              Bases
   iv.     Acids solution have PH values less       iv.     Basics solution have PH values
           than 7.                                          greater than 7.
  Q5 A carbonated drink has [H+] = 3.2 × 10-3 M, classify the drink as neutral acidic or basic with
  reason.
Ans: Solution:
Data: hydrogen ion concentration = [H+] = 3.2 × 10-M
  Determination of PH
  PH = -log [H+]
  Putting the value of [H+] we get:
  PH = - log [3.2x10-3]
  =PH=-(log3.2+log10-3) ∵ log mn = logm + logn
  =PH=-[log3.2-log10-]3
  =PH= - (0.5051) –(-3) log 10 ∵ logmn = n log m
  =PH = - 0.5051 + 3log10
  =PH= -.0.5051 +3 (1) ∵ log10=1
    =PH= 0.5051 +3
  =PH= 2.49
  As the pH is less than 7 so the carbonated drink is acidic solution.
  OR
  We can also solve this problem by comparing the given [H] + of the solution with that of neutral water.
  As we know that [H]+ = [OH]- = 1×10-7
  If [H ]+ ˃ 1×10-7 the solution will be acidic.
  If [OH] - ˃ 1×10-7 the solution will be basic.
  As in the given solution [H] += 3.2×10-3 M ˃ 1×10-7 the solution is acidic.
Q6. Write the chemical name of an acid present in the following.
Ans: (a). Apple juice: Malic acid
(b).Grape: Tartaric acid
(c). Lemon juice: Citric acid
(d). Sour milk: Lactic acid.
Q7.What determine the strength of a base? Give one example of each solution of strongly acidic
and weakly acidic.
Ans: Strong bases:
“Those bases which completely dissociate in aqueous solution and give a higher concentration of OH-
are called strong bases”.
Strong bases completely ionize in water and almost no unionized molecule is left behind.
Examples:
Examples of some of the strong bases are given below:
NaOH (aq)⟶ Na+ OH-
KOH (aq) ⟶ K+ + OH-
LiOH (aq)⟶ Li + OH-
Weak bases:
“Those bases which do not dissociate completely in aqueous solution and give a lower concentration of
OH- are called weak bases”.
Examples:
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Some bases which do not completely ionize in water are given below:
NH4OH (aq)⇌ NH4+ + OH-
Ba(OH)2(aq) ⇌ Ba+2+2OH-
Mg(OH)2(aq)⇌ Mg+2+20H-
Q8. Calculate the pH and pOH of O.5 M solution of HCl.
 Calculate the PH and POH of 0.5M HCL solution.
 Given:
 Molarity of hydrochloric acid (HCI) solution= 0.05M
 Required:
 To calculate the PH and PH
 Solution:
 Chemical equation for ionization of hydrochloric acid is:
 HCI → H+ + CI-
 0.5M 0.05M
                                             5
 Hydrogen ion concentration [H+] = 0.5 =          = 5x10-1 M
                                             10
 Determination of PH
 PH = -log [H+]
 Putting the value of [H+] we get:
 PH = - log [5x10-1]
 =PH=-(log5+log10-1) ∵ log mn = logm + logn
 =PH=-log5-log10-1
 =PH= - (.06989) – (-1) log 10 ∵ logmn = n log m
 =PH = - 0.6989+1log10
 =PH= -.06989 +1 (1) ∵ log10=1
  =PH= -.06989 +1
 =PH= 0.301
 Determination of POH:
 We know that:
 PH + POH = 14
 POH = 14 - PH
 Putting the value of PH we get:
 POH = 14-0.3o1
 ⟹POH = 13.
 Result:
 PH = 0.301 and POH = 13.69
 Q9. Calculate the PH and POH of 0.005M H2SO4 solution.
 Given:
 Molarity of H2SO solution= 0.005M
 Required:
 To calculate the PH and PH
 Solution:
 Chemical equation for ionization of nitric acid is:
 H2SO → 2H+ + SO4-2
 0.005M 0.005M
                                                  5
 Hydrogen ion concentration [H+ ]:0.005 =             = 5x10-3 M
                                              1000
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Determination of PH
PH = -log [H+]
Putting the value of [H+] we get:
PH = - log [5x10-3]
=PH=-(log5+log10-3) ∵ log mn = logm + logn
=PH=-log5-log10-3
=PH= - (.06989) – (-3) log 10 ∵ logmn = n log m
=PH = - 0.6989+3og10
=PH= -.06989 +3 (1) ∵ log10=1
=PH= -.06989 +3
=PH= 2.301
Determination of POH:
We know that:
PH + POH = 14
POH = 14 - PH
Putting the value of PH we get:
POH = 14- 2.301
⟹POH = 11.699
Result:
PH = 2.301 and POH = 11.699
10. Calculate the PH of 0.2M NaOH solution?
Given:
Concentration of NaOH solution = 0.2M
Required:
POH =?
Solution:
Chemical equation for ionization of sodium hydroxide is:
NaOH → Na+ + OH-
0.2M          0.2M
                                            1
Hydroxide ion concentration [OH-]: 0.2M =        = 10-1M
                                            10
Determination of POH
POH = -log [OH-]
Putting the value of [OH-] we got:
POH = - log [2x10-1]
POH =-(log2+log10-1) ∵ log mn = logm + logn
POH =-log5-log10-3
POH = - (0.301) –(-1) log 10 ∵ logmn = n log m
POH = - 0.301+ 1 log10
POH = -.0.301+1 (1) ∵ log10=1
POH = 0.699
Determination of PH:
We know that:
PH + POH = 14
PH = 14 – POH
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                                             LONG QUESTIONS:
   Q1, (a). What is salt?
Salt:
“The substance obtained due to neutralization reaction of an acid base reaction is called salt”.
Composition of salt:
A salt consists of positive ions combined with negative ions. Positive ions come from a base while
negative ions come from an acid i.e. In NaCI Na+ is from NaOH while CI- is from HCI.
Examples of some salts:
Examples of salts are given below:
ii. Silver bromide (AgBr)
iii. Potassium sulphate (K2SO4)
iv. Ferric phosphate (FePO4).
Q1. (b). write down the different types of salts with example?
Ans: Types of salts:
There are three types of salts i.e. neutral salt, acidic salts and basic salts their detail is given below:
i. Neutral salts:
“The salts formed when hydrogen atom of an acid is completely replaced by a metal ion or group of
atoms behaving like metal ion are called normal salts”.
Neutral salts are formed when a strong acid react with strong base.
Examples:
NaOH + HCI →NaCI + H2O
Base Acid         Normal Salt
Some other examples of normal salts are given below:
i. Potassium sulphate (K2SO4)
ii. Sodium phosphate (Na3PO4)
iii. Ammonium sulphate [(NH4)2SO4]
iv. Sodium carbonate (Na2CO3)
ii. Acidic salts:
“The salts formed when hydrogen atom of a polyprotic acid is partially replaced by a metal ion or group
of atoms behaving like metal ion are called acidic salts”.
Examples:
           H2SO4 + KOH →KHSO4 + H2O
These acids can further react with bases forming neutral salt
           KHSO4 + KOH →K2SO4 + H2O
Some other examples of acidic salts are given below:
i. Ammonium bi phosphate [NH4) H2PO4]
ii. Potassium bicarbonate (KHCO3)
ii. Sodium bicarbonate [(NaHCO3] etc.
iii. Basic salts:
“The salts formed when hydroxide ions (OH-) of a base are partially neutralized by an acids are called
basic salts”.
Basic salts are formed by poly acid bases only e.g.
Examples:
Pb(OH)2, Mg(OH)2, AI(OH)3 etc.
Pb(OH)2 + HCI →Pb(OH)CI + H2O.
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   https://tehkals.com/                                                                                       17
Q2. (a). Define the auto-ionization of water. How can you find the pH of Water?
Ans: Auto-ionization of water:
The reaction in which two water molecules produce ions is called as the self-ionization or auto-
ionization of water.
Explanation:
In order to understand the concept of self-ionization or auto ionization of water, we take one molecule of
water and its dissociation at 250C, as
H2O(l) →H+(aq) + OH-(aq)
Equilibrium constant expression for this reaction is:
      �𝑁𝑁 + ] [𝑂𝑂𝑁𝑁 − ]�
Kc=       [𝑁𝑁2 𝑂𝑂]
As concentration of H2O remains constant and the above equation can be written as:
Kc [H2O] = [H+] [OH-]
Kw = [H+] [OH-]
where Kw= Kc [H2O]
Dissociation constant of water (Kw):
Kw is called dissociation constant or ionization constant of water. It is defined as “ the product of molar
concentration of H+(H3O+) and OH- ions is called dissociation constant of water”. The value of Kw of
water at 250C is 1.0x10-14 i.e.
  Kw = [H+] [OH-] =1x10-14 mol/dm3 of water at 250C
As one molecule of water produces one H+ and oneOH- ion on dissociation.
Therefore, we can say that,
  [H+] [OH-] = 1x10-14
  [H+] = [OH-]
  Or [H+] [H+] = 1x10-14
  (H+)2 = 1x10-14
  (H+)2 = 1x10-14
  Therefore
  (H+)= 1x10-7
  And [OH-] = 1.0× 10-7
  In water at 250C, (H+)= 1x10-7Mand [OH-] = 1.0× 10-7M
 Kw = [H+] [OH-]
Kw = 1x10-7M × 1x10-7M
  Kw = 1x10-14 M2
pH Definition:
  PH can also be defined as:
“The negative logarithm of molar concentration of H+ ions is called PH”.
PH = -log [H+]
PH of Water:
According to this scale, pH of water is calculated as,
pH = -log [H+]
putting values of [H+],
pH = -log[1.0× 10-7]
pH = - (-7.0) log10      log10= 1
  pH = 7.0
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   https://tehkals.com/                                                                              18
Q.2(b). Why some acids are called monoprotic, diprotic and polyprotic acids. Explain your answer
with suitable examples.
Ans: Acids can be classified in term of number of protons that can be given by per molecule of an acid
when added to water. Acids are classified as,
i.      Mono-protic acids
ii.     Poly-protic acids
Monoprotic or monobasic acids:
“Those acids which give one proton per molecule are called monoprotic acids or monobasic acids”.
Examples:
HCI, HNO3, CH3COOH, HBr, HCN etc are some of the examples of monoprotic acids because they give
only one portion per molecule.
HCI (aq)⟶ H+ + CI-
HNO3 (aq) ⟶ H+ + NO-
CH3COOH (aq)⟶ H++CH3COO-
HBr (aq)⟶ H++ Br-
HCN (aq)⟶ H+ + CN-
Polyprotic acids:
“Those acids which give more than one proton per molecule are called polyprotic acids or polybasic
acids”.
These are further divided into the following subgroups:
i. Diprotic acids:
“Those acids which give two protons per molecule are called diprotic acids or dibasic acids”.
Examples:
H2SO4H2CO3 etc. are the examples of diprotic acids because they give two protons per molecule.
H2SO4 (aq)⟶ 2H+ + SO-2
H2CO3 (aq) ⟶ 2H+ + CO3-2
ii. Triprotic acids:
“Those acids that give three protons per molecule are called Triprotic acids or tribasic acids”.
Example:
Phosphoric acid is an example of triprotic acid as clear from the following chemical equation:
H3PO4(aq)⟶3H+ + PO4-2
Q3. (a). Discuss the concept of lewis acids and bases with examples.
 Ans: The Lewis concept:
 Introduction:
G.N Lewis presented his own concept of acids and bases in 1923.
Definitions of acids and bases:
Acids:
According to this concept an acid is a species that can accept an election pair.
Base: Base is a species which can donate electron pair.
An acid is electrophile (electron loving) while a base is a nucleophile (nucleus loving).
Species which acts as Lewis acids:
Those compound in which central atom has less than eight electrons in valance shell or positively
charged ions that can accept an electron pair act as Lewis acids. e.g. BF3, AICI3,H+ etc.
Species which acts as Lewis Base:
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   https://tehkals.com/                                                                               19
Those compounds in which central atom has lone pair of electrons in valance shell or negatively charged
ions that can donate an electron pair can act as Lewis bases. act as Lewis bases e.g. NH3, H2O,CN-,CI-
Example:
Q3. (b) Give the bronsted-lowery definition of acids and bases. Write equation that explain the
definition.
 Ans: Introduction:
 Bronsted and Lowry presented a broader concept about acids and bases in 1923.
 Definitions of acids and bases:
 According to Bronsted-Lowry concept.
 Acids: Acids are defined as the substances which donate or tend to donate protons (H+ ions).
 Bases: Bases are defined as the substances which accept or tend to accept protons.
 Examples:
 i. When ammonia (NH3) is added to water, the following reaction occurs:
                          +
 HCl (aq) + NH3 (aq)⟶ N𝑯𝑯𝟒𝟒(𝒂𝒂𝒂𝒂) + Cl-(aq)
 (Acid)                (Base)
In the above example the ammonia accepts a proton from HCI, therefore, it acts as a base while HCI
donates a proton and therefore it acts as an acid.
ii.HCl + H2O ⟶ H3O+ + Cl-
Acid      Base
In this example water act as a bronsted-lowery base and HCl act as bronsted-lowery acid?
                         +
iii. H2O (l) + NH3 ⟶ N𝑯𝑯𝟒𝟒(𝒂𝒂𝒂𝒂) + OH
 In the above example the ammonia accepts a proton from H2O, therefore, it acts as a base while H2O
 donates a proton and therefore it acts as an acid.
Q4. Below are two equations showing how two alkalis react with water.
NaOH (aq) + H2O (i) ⇌ Na+ (aq) + OH-
NH3 (aq) + H2O ⇌NH4+ (aq) + OH-(aq)
A). name both alkalis.
Ans: NaOH = sodium hydroxide
NH3 = ammonia
B. which is classified as weak alkali and why?
NaOH is a strong base while NH3 is weak base.
c. What is the likely pH of each alkali?
PH of NaOH is13 while that of NH3 is 11.6
Q5. Write the balanced neutralization reaction of,
  It is also called as baking soda because it is used in baking of cakes and other confectionaries. It is also
  as antacids in medicines and in toothpastes.
  v. Copper sulphate (CuSO4.5H2O) (Blue vitriol):
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It is used as electrolyte in copper electroplating process. It is also used to kill algae in water reservoirs
and in agriculture spray.
vi. Magnesium sulphate (MgSO47H2O) (Epsom Salt):
It is used as antacid and laxative in medicines. It is also used in dye industries.
vii. Potash alum (K2SO4.AI2 (AO4)3, 24H2O):
It is used for water purification to remove suspended impurities. It is also used in textile industry and
as blood coagulant in small injuries.
viii. Potassium nitrate (KNO3):
It is used as a fertilizer and for the manufacture of flint glass.
ix. Calcium Carbonate (CaCO3):
It is used in the preparation of cement and in ceramics industry.
x. Sodium sulphate:
It is used in the manufacture of paper, detergents and glass etc.
CHEMISTRY
Class 10th (KPK)
NAME: __________________________
F.NAME: _________________________
ADDRESS: ___________________________________
__________________________________________
SCHOOL: _____________________________________
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                                                 UNIT 11
                                       ORGANIC CHHEMISTRY
                                          SHORT QUESTIONS
   Q1. Define functional group? Give example of functional groups containing oxygen.
   Ans: Functional group:
   A functional group is an atom or group of atoms attached with R that is responsible for the specific
   properties of an organic compound.
Explanation: A functional group is the active part of an organic compound. Most of the organic
compounds consist of two parts i.e.
i. The hydrocarbon part which is an alkyl group/
ii. The functional group part.
For example:
In methanol (CH3-OH),-CH3 is the alkyl group (R) while –OH is the functional group part.
Example of functional groups containing oxygen:
Examples:
The functional groups containing oxygen are given in the table below:
      Functional group Name of the classes          Examples             Name of
                                                                         compound
Q2. How can we obtain the organic compounds from natural sources?
Ans:
Q3. What are cycloalkanes?
Ans: Cyclic alkanes:
“The type of alkanes in which the carbon atoms are arranged in cyclic form are called cyclic alkanes or
cyclo-alkanes.
General formula:
Cycloalkanes have two less hydrogen atoms than in corresponding alkanes.
These have the general formula similar to alkenes i.e. CnHn2
Examples:
Q4. Write down examples of three unsaturated hydrocarbons with structural and condensed
formulae?
Ans: Unsaturated hydrocarbons:
“The hydrocarbons containing at least one carbon-carbon double or triple bond are called unsaturated
hydrocarbons”.
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   https://tehkals.com/                                                                                   26
b. Alkenes CnH2n
c. Alkynes CnH2n-2
                                            LONG QUESTONS
   Q1.List the different characteristics of organic compounds.
Ans: Characteristic properties of organic compounds:
Organic compounds have the following general properties:
i. Origin: The main source of organic compounds is plants and animals.
ii. Composition: Carbon is an essential component of all organic compounds. However, beside carbon
they also contain hydrogen as essential part. They may also contain some other elements like sulphur,
nitrogen, oxygen and halogen.
iii. Thermal instability: Many organic compounds are thermally unstable and decompose to simple
substances on heating. This property is of great commercial importance e.g. as in the cracking of
petroleum.
iv. Low melting points and boiling points: Organic compounds have generally low melting points and
boiling points due to weak intermolecular forces. They can be easily broken down and are generally
volatile in nature.
v. Bonding: Organic compounds are generally covalent in nature.
vi. Solubility: As most of the organic compounds are non-polar therefore, they are soluble in non-polar
solvent like benzene, acetone, and ether and less soluble or insoluble in polar solvents like water.
vii. Electrical conductivity: Most of the covalent compounds are non-polar therefore, poor conductors
of electricity in molten or solution form.
viii. In flammability: Most of the organic compounds are inflammable. They burn out to give carbon
dioxide, water vapours and energy.
ix. Reactivity: The reactions of organic compounds are much slower than the inorganic compounds.
x. Isomerism: The compounds having same molecular formula but different structures are called
isomers and this phenomenon is called isomerism. Isomerism is common in most organic compounds.
For example, butane has two isomers n-butane and iso-butane both have the molecular formula C4H10
but they have different structures.
b. Which of these is not an unsaturated molecule?
i. C6H6 ii. C6H6 iii. C8H18 iv. C3H6
C. Define destructive distillation of coal. Name the different types of products obtained by the
destructive distillation of coal.
 Ans: Destructive distillation:
The heating of a compound in the absence of air is called destructive distillation.
Destructive distillation of coal:
The process in which coal is heated in the absence of air is called destructive distillation of coal.
Products of destructive distillation of coal:
During this process, the coal is converted into coal gas, coke and coal tar and ammmonical liquior,
which are the sources of other organic compounds.
Q2. What is catenation?
Ans: Catenation:
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   https://tehkals.com/                                                                                   29
The self-linking ability of carbon atoms to covalently bond with other carbon atom to form straight
chain, branched chain and rings is called catenation.
b. How does catenation contribution to the diversity of organic compound?
Ans: Catenation contribute to the diversity of organic compound by allowing carbon atoms to bond
together in many possible arrangement .Due to catenation variety of molecules with different structures
including chains and rings of many shapes and sizes are formed which have different properties.
Examples:
Q4. (A). what do term saturated and unsaturated mean when applied to hydrocarbons?
Ans: Hydrocarbons:
 The compounds of carbon and hydrogen are called hydrocarbons”.
 Classification of hydrocarbons: Hydrocarbons are classified into two main groups:
       i.     Saturated hydrocarbons (alkanes)
       ii.    Unsaturated hydrocarbons (alkenes and alkynes)
  i. Saturated hydrocarbons: “The hydrocarbons which contain all carbon-carbon single bonds are
  called saturated hydrocarbons or alkanes”.
  General formula: They have the general formula of CnH2n+2, where n is the number of carbon atoms.
  Examples:
  Methane (CH4), Ethane (C2H6), propane (C3H8) and butane (C4H10) etc are the examples of saturated
  hydrocarbons.
  ii. Unsaturated hydrocarbons: “The hydrocarbons containing at least one carbon-carbon double or
  triple bond are called unsaturated hydrocarbons”. They are further classified to:
  a. Alkenes: “The hydrocarbons which contain at least one carbon-carbon double bond are called
  alkenes”.
  General formula: They have the general formula of CnH2n, where n is the number of carbon atoms.
  Example:
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   https://tehkals.com/                                                                               30
 Ethene (C2H4), propene (C3H6), Butene (C4H8) and pentene (C5H10) are the examples of alkanes.
 b. Alkyne: The hydrocarbons which contain at least on carbon─carbon triple bonds are called
 alkynes.”
 General formula: They have the general formula of CnH2n-2, where n is the number of carbon atoms.
 Examples: Ethyne (C2H2), propyne (C3H4), butyne (C4H6), and pentyne (C5H8) etc are the examples of
 alkynes.
 b. What other meanings do these term have in chemistry?
 Ans: Saturated:
 In chemistry term saturated also refer to chemical solutions. A solution which can’t dissolve more
 solute is called saturated solution.
 Unsaturated:
 A solution which can dissolve further amount of solute to form a saturated solution is called
 Unsaturated solution.
 c. Classify alkenes, alkanes, alkynes, and aromatic hydrocarbons as either saturated or
 unsaturated.
 Ans: Alkanes: The hydrocarbons which contain all carbon-carbon single bonds are called saturated
 hydrocarbons. As alkanes contain all C-C singles bonds so alkanes are saturated hydrocarbons.
 Unsaturated hydrocarbons: The hydrocarbons containing at least one carbon-carbon double or triple
 bond are called unsaturated hydrocarbons. Alkenes and alkynes contain double and triple bonds and
 aromatic hydrocarbons contain alternate single and double bonds. So, alkenes, alkynes and aromatic
 hydrocarbons are unsaturated.
         Chemical
                         Number of         Greek numerals
          formula                                              Full Name
                          C-atoms            (Prefixes)
         (CnH2n+2)
C. name the straight chain alkane with the molecular formula C8H18.
Ans: As the number of carbon atom is eight and the prefix Greek numerals oct is used for 8. So, the
name of given compound C8H18 is octane.
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 Condensed formula:
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   The condensed formula of a molecule is the formula where the groups of atoms are shown in order as
   they appear in the structural formula with no bonds or dashes.
   Example:
   Hexane has six carbon and fourteen hydrogen atoms with molecular formula of C6H14. The condensed
   formula of hexane is CH3 (CH2)4CH3.
   Dot and cross formula:
    A structural formula in which electrons are shown as dots and cross between various atoms in one
   molecule of a compound are called dot and cross formula.
   Example:
   In methane molecule, the four electrons of carbon is represented by dots (●) and cross (×) is used to
   represent the electrons of four hydrogen atoms. The molecules of methane and of propane is shown
   below.
   Q4. Explain the classification of organic compound?
Ans: Classification of Organic compound: On the basis of structure of carbon chain, organic
compound are classified into the following two groups:
i. Open chain alkanes
ii. Cyclic alkanes
i. Open chain organic compound or aliphatic organic compound: “The type of organic compound
which consist of open chain of carbon atoms are called open chain or aliphatic organic compound”.
Open chain organic compound are further classified into two types which are given below:
a. Straight chain organic compound s: “The type of open chain organic compound which contain
straight chain of carbon in their molecules are called straight chain organic compound”.
A carbon atom in straight chain organic compound is not directly bonded to more than two carbon
atoms. They are commonly named as n-alkanes (normal-alkanes).
Examples:
Some examples of straight chain alkanes are given below:
CH3-CH2-CH2-CH3 (n-butane)
CH3-CH2-CH2-CH2- CH3 (n-pentane)
CH3-CH2-CH2-CH2- CH2-CH3 (n-hexane)
b. Branched chain organic compound: “The type of open chain organic compound which do not
contain straight chain of carbon in their molecules are called straight chain organic compound”.
At least one carbon atom in branched chain alkanes is directly bonded to more than two carbon atoms.
They are commonly named as iso-alkanes.
Examples:
ii. Cyclic alkanes: “The type of organic compound in which the carbon atoms are linked together to
form a close chain structure
Are called cyclic alkanes or cyclo-alkanes”.
These have the general formula similar to alkenes i.e. CnHn2
The closed chain alkanes are further divided into two groups. These are,
i.      Homocyclic or Carbocyclic organic compound
ii.     Heterocyclic organic compound
i. Homocyclic or Carbocyclic organic compound:
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   https://tehkals.com/                                                                                 34
Organic compounds which are in close chain structure are called Homocyclic compounds. The ring is
composed of only carbon atoms.
Homocyclicorganiccompound are further divided into two types.
a. Ali-cyclic organic compound
b. Aromatic organic compound
a.Ali-cyclic organic compound:
The cyclic organic compound which is composed of only carbon atom is called Ali-cyclic organic
compound.
These organic compounds have properties similar to open chain organic compound but differ in their
structure and formulae.
Example:
The important member of this class is benzene and the derivatives of benzene i.e. Naphthalene,
Anthracene.
ii. Heterocyclic organic compound:
The organic compound which contain one or more atoms other than carbon such as sulphur, oxygen or
nitrogen in the ring are called heterocyclic organic compound.
Example:
Examples are furan, thiophene and pyridine.
General formula: The general formula of alkanes is CnH2n+2 where n is the number of carbon atoms e.g.
if n=1 then the formula will be CH4 which is called methane. Similarly, if n=2 then the formula will be
C2H6 called ethane etc.
Example:
  S. No.        Name of alkane           Chemical formula (CnH2n+2)
1 Methane CH4
2 Ethane C2H6
3 Propane C3H8
4 Butane C4H10
5 Pentane C5H12
ii.      Functional groups containing C, H and N: functional groups containing carbon hydrogen and
         nitrogen are called amines. Functional group of amines is -NH2. The general formula is R-NH2.
         For examples methyl-amine (CH3-NH2)
iii.      Functional groups containing C,H and X:
   Functional groups containing carbon hydrogen and halogens are called alkyl halides. Functional group
   of alkyl halides is halogens represented by ‘X’. The general formula is R-X. X may be F, Cl, Br, or I for
   examples methyl-chloride (CH3-Cl)
CHEMISTRY
Class 10th (KPK)
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F.NAME: _________________________
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                                                 Unit 12
                                             HYDROCARBONS
Short Questions:
Q1. How would you test that alkenes undergoes an addition reaction?
 Ans: Addition reaction of alkenes: alkenes are unsaturated hydrocarbons and can be converted to
 saturated hydrocarbon. Addition reaction is characteristics properties of alkenes. One of the addition re
 action is addition of halogens (Halogenation).
 Halogenation of alkene:
 When bromine is added to ethene in the presence of an inert solvent like carbon tetra chloride (CCl4),
 the double bond of it is converted into single bond. The red colour of bromine is disappear and
 produces 1, 2-dibromoethane.
Q2. Which one is more reactive between alkane and alkene? Explain.
  Ans: Alkenes are more reactive than the corresponding alkanes due to the following reason:
  Reason: Alkenes contain C-C double bonds and double bonds are weaker and can easily be broken.
  Therefore, alkenes are more reactive. On the other hand alkanes contain all C.C single bonds which are
  stronger and cannot be easily broken that is why alkenes are more reactive than the corresponding
  alkanes.
For example, ethane does not react with bromine solution while ethane reacts easily with bromine
solution decolourizing its red colour.
Q3. Justify alkenes and alkynes as unsaturated hydrocarbons.
Ans: Unsaturated hydrocarbons:
The hydrocarbons containing at least one carbon-carbon double or triple bond are called unsaturated
hydrocarbons. Alkenes contain C=C double and alkynes contains C≡C triple bonds. Therefore, alkenes
and alkynes are termed as unsaturated hydrocarbons.
  Q4. Why alkane is inert in nature?
  Ans: Alkanes are chemically inert in nature due to the following reason:
  Reason: Alkanes are chemically inert in nature because they contain carbon-carbon single bonds. The
  single bonds are very strong and stable and high energy is required to break them; therefore, alkanes
  are chemically inert as compared to alkynes and alkenes because they contain double or triple bonds
  which can easily be broken.
  Q5. What happened when alkyl halide is reduced?
  Ans: Reduction of alkyl halides:
  Reduction means addition of hydrogen. Alkyl halides on reaction with nascent hydrogen in the
  presence of Zinc dust and HCl form the corresponding alkane’s i.e. methyl halide will form methane,
  ethyl halide will form ethane etc.
  General reaction of alkyl halide and nascent hydrogen to form alkane is:
  R─X + 2[H] → R─H + HX
  Methyl iodide gives methane and hydrogen iodide HI.
  CH3─I+ 2[H] → CH3─H+ HI
  Similarly, ethyl bromide gives ethane and hydrogen bromide HBr.
  CH3─ CH2 ─Br + 2[H] → CH3─ CH2 ─H+ HBr
  Q6. Can you predict the products if KMnO4 solution reacts with alkene?
  Ans: Oxidation of alkanes by KMnO4 (Baeyer’s test):
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   https://tehkals.com/                                                                                    39
 Alkenes react with cold dilute potassium permanganate solution to form glycol. Glycols are the
 alcohols containing two hydroxyl groups (-OH) on two adjacent carbon atoms.
 The reaction of potassium permanganate with ethane is given below:
         3CH2=CH2+2KMnO4+4H2O → 3CH2-CH2+2MnO2+2KOH
                                          │    │
                                              OH OH
                                          (Ethylene glycol).
 Q7. Why colour of bromine water discharges on addition to ethene?
 Ans: when bromine water having red brown colour is added to ethene in the presence of inert solvent
 like carbon tetra chloride, its colour is discharged. Because during this reaction bromine water reacts
 with ethene in carbon tetra chloride to form ethylene bromide which is colourless compound.
                                        LONG QUESTIONS
Q1. (i). write down the equations for the preparation of alkanes, alkenes and alkynes.
Ans: Preparation of alkanes:
1. Hydrogenation of alkenes:
Alkanes can be prepared by the hydrogenation of alkenes. During this reaction alkenes are
hydrogenated in the presence of Nickel (Ni) catalyst at 250- 300 0C temperature.
H2C═ CH2 + H─H →H3C ─CH3
Ethene                     Ethane
2. Hydrogenation of alkynes:
Alkynes are hydrogenated to alkenes in the presence of Nickel (Ni) catalyst at 250- 300 0C temperature
in the first step and in the second step, alkene is further hydrogenated and converted into alkanes.
HC ≡ CH + H─H → H2C═ CH2
H2C═ CH2 + H─H →H3C ─CH3
3. Reduction of alkyl halides:Alkyl halides on reaction with nascent hydrogen in the presence of Zinc
dust and HCl form the corresponding alkanes i.e. methyl halide will form methane, ethyl halide will
form ethane etc.
General reaction of alkyl halide and nascent hydrogen to form alkane is:
R─X + 2[H] → R─H + HX
Methyl iodide gives methane and hydrogen iodide HI.
CH3─I+ 2[H] → CH3─H+ HI
Preparation of alkenes:
1. Dehydrogenation of alcohols:
In this reaction water molecule is removed from alcohol and double bond is formed. This reaction is
carried out in the presence of concentrated Sulphuric acid at 180oC.
Preparation of alkynes:
1. Dehalogenation of adjacent Dihalides:
The organic compound in which dihalides are attached to the adjacent carbon atoms are called vicinal
dihalides.
The reaction occurs in two steps.
Step 1.
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Step 2. In the next step another molecule of hydrogen bromide is removed and the double bond is
converted into triple bond.
Q1.(ii) Draw the molecular, dot, and cross, condensed and structural formula of each of the
following.
Propene, Butyne, Pentane, Heptyne, butane.
Q2. (i). the general formula of alkanes is CnH2n+2. Determine the general formula of cycloalkane?
Ans: Cycloalnaes: The alkanes in which carbon atoms are arranged in a ringorcyclic structure is called
cycloalkanes.The general formula of Cycloalnaes: The general formula of cycloalkanes is CnH2n.
Cycloalkanes have less than two hydrogen atom than in corresponding straight chain alkanes.
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(ii). write a balanced equation for the complete combustion of each of the following.
   a. Methane b. Ethane c. Ethyne
a.Combustion of Alkanes: Alkanes burn completely in the presence of excess of air (oxygen) to
produce a lot of heat, carbon dioxide and water. The reaction is highly exothermic and thus alkanes is
used as a fuel.
CH4 + 2O2 → CO2 + 2H2O + heat
Combustion of Alkenes: On complete combustion, alkenes produce carbon dioxide and water with
the release of high amount of energy.
C2H4 + 3O2 → 2CO2 + 2H2O + heat
C.Combustion of Alkynes:On complete combustion, alkynes produce carbon dioxide and water with
the release of high amount of energy,
2C2H2 + 5O2 → 4CO2 + 2H2O + heat.
(iii). Explain briefly that why ethyne (a) undergoes addition rection in two steps?
Answer: In addition reaction of ethyne first triple bond is converted into double bond and then double
bond is converted into single bond. That is the reason ethyne undergoes addition rection in two step.se
Example: hydrogenation of ethyne:
Step 1.HC ≡ CH + H─H → H2C═ CH2 Step 2. H2C═ CH2 + H─H →H3C ─CH3.
Q3. (i). Give an example reaction of that would yield the following products. Name the organic
reactants and products in each reaction.
Ans: a). Alkanes:
Hydrogenation of alkenes: alkanes can be prepared by the hydrogenation of alkenes. During this
reaction alkenes are hydrogenated in the presence of Nickel (Ni) catalyst at 250- 300 0C temperature.
H2C═ CH2 + H─H →H3C ─CH3
Ethene                  Ethane
b). Mon halogenated alkanes: Alkanes react with halogens in the presence of UV light and produce
alkyl halide and hydrogen halide.
H3C ─H + Cl ─ Cl → H3C─Cl + HCl
Methane              methyl chloride                                                            c). Di
halogenated alkanes:
Cl─CH2─H + Cl ─ Cl → Cl─CH2─Cl + H─Cl
methyl chloride         Dichloromethane
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2. Baeyer’s test:
This test is used for determining the presence of double bond in compounds. Alkenes is reacted with
acidified aqueous solution of potassium permanganate (KMnO4) and form ethylene glycol (1, 2-
ethanediol). During this reaction purple colour of KMnO4.
 (b). Ethene with hydrogen, name the catalyst used. Which industrial process uses a similar
reaction?
Hydrogenation of ethene:
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When hydrogen (H2) is added to ethene, the double bond is converted into single bond and produces
ethane. The reaction occur in the presence of nickel as a catalyst at 250-3000C.
ii. Combustion: Alkanes react with oxygen or air at higher temperature to form carbon dioxide, water
vapours with evolution of large amount of heat.
Examples:
        CH41g+O21g→ CO21g+H2Og+Heat
         2C2H +7021g 4CO21g+6HOg+ Heat
In limited supply of oxygen, incomplete combustion of alkanes takes place and produce carbon
monoxide, water and carbon.
3CH41g+ 4O21g→ 2CO+ 6H2O + C
iii. Meting points and boiling points: The melting points and boiling points of alkenes increases with
the increase in molecular masses because generally higher molecular masses alkenes have stronger
intermolecular forces.
iv.Combustion: on complete combustion, they produce carbon dioxide and water with the release of
high amount of energy,.
Chemical properties Alkynes:
Because of the unsaturated nature of alkynes, they easily undergoes addition reaction and in this way,
they are converted into saturated compounds.
Step 2. In the second step another halogen molecule, i.e. chlorine (Cl2) or bromine (Br2) is added to
double bond. thw double bond is converted into single bond and forms 1,1,2,2-tetra chloro ethene.
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                                              UNIT 13
                                           BIOCHEMISTRY
Q1. Differentiate between glucose and fructose?
Ans: Difference between glucose and fructose:
Glucose and fructose are hexoses having same molecular formula C6H12O6. The main difference
between these two compounds is that glucose is pentahydroxy aldehyde having aldehydic functional
group, while fructose is pentahydroxy ketone having ketonic functional group.
Vitamins which dissolve in water are called water soluble vitamins. These vitamins are B-complex and
Vitamin c. Vitamin B–complex include eight vitamins i.e. B1, B2, B3, B5, B6, Boltin, Folic acid and
B12.
Q6. Vitamin are vital for us?
Ans:Vitamins are vital for us because it plays an important role in the healthy development and growth
of our body. Our body needs a small amount of vitamins and minerals every day to remain healthy,
function properly and prevent in future from health problems. It also play important roles in bodily
function such as metabolism, immunity and digestion.
Q7.Write down the of sucrose hydrolysis
Ans: Hydrolysis of sucrose:                                                                        On
hydrolysis sucrose produce one unit of Glucose and one unit of Fructose.
C12H22O11 + H2O → C6H12O6 + C6H12O6
Sucrose + H2O → Glucose + Fructose
Q8. Draw the general formula of amino acid. Identify the different parts and functional groups
present in it.
Ans: General formula of amino acid:Amino acid is an organic compound having central carbon
atom called alpha carbon, which is attached to basic amino group (NH2), acidic carboxyl group
(COOH), a side chain alkyl group (R) and hydrogen atom.
                                           LONG QUESTIONS
Q1. What function do carbohydrates perform in living organisms?
Ans: Function of carbohydrates:1. Source of energy:
Carbohydrates are used by our body as a rich source of energy.
2. Source of food:
Carbohydrates are used as a source of food. For example, many vegetables fruits and cereals contain
carbohydrates. It is also used as a common sweetener for food.
3. Structural material:
Carbohydrates serves as a structural material in living organism. For example, plant cell wall is made
up of cellulose similarly exoskeleton of insects is made up of chitin.
4. Energy reservoir:
Glucose is stored in animal muscles and liver cell in the form of glycogen. It serves as a long term
energy reservoir in the body, converted back into glucose when needed. Plants store excess energy in
the form of starch.
5. Medicinal Uses: Constipation and diarrhea are mainly controlled bythe use of carbohydrates fibers.
Q2. (a). Distinguish between monosaccharaides, disaccharides and polysaccharides and also give
example of each one.
Ans: Carbohydrates:
Definition: “Polyhydroxy ketones or polyhydroxy aldehydes are called carbohydrates”.
Polyhdroxy means having many hydroxyl groups.
Common name: Carbohydrates are commonly known as sugars.
Composition: Carbohydrates are poly functional (alcohol+aldehydes or ketones) organic compounds
which mainly consist of carbon, hydrogen and oxygen some time along with nitrogen (chitin) or
sulphur (keratin sulphate).
Classification of carbohydrates: Carbohydrates are classified into the following three groups:
1. Monosaccharides 2. Oligosaccharides 3. Polysaccharides
Their detail is given below:
1. Monosaccharides:
Definition: “The simplest carbohydrates which cannot be further hydrolyzed into smaller units are
called monosaccharides”.
General formula: They have the general formula of Cn (H2O) n where n=3, 4, 5 and so on.
Composition: Monosaccharides are either aldoses having aldehydic functional group (for example
glucose) or ketoses having ketonic functional group for examples glucose) or ketoses having ketonic
functional group for example fructose.
These may be trioses (C3H12O6), tetroses (C4H8O4), pentoses (C5H10O5) and hexoses (C6H12O6).
General properties of monosaccharides: Some general properties of monosaccharides are:
i. Solubility: They are soluble in water.
ii. Physical state: They are crystalline solid at room temperature.
iii. Taste: They have sweet taste.
iv. Colour: They are colourless.
v. hydrolysis: They cannot be hydrolyzed further.
Examples: Some examples of carbohydrates along with their structures are given below:
In straight forms these structures can be written as following
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3. Structural material:
Carbohydrates serves as a structural material in living organism. Forexample plant cell wall is made up
of cellulose similarly exoskeleton of insects is made up of chitin.
                                                                                              4.
Energy reservoir:
Glucose is stored in animal muscles and liver cell in the form of glycogen. It serves as a long term
energy reservoir in the body, converted back into glucose when needed. Plants store excess energy in
the form of starch.
5. Medicinal Uses:
Constipation and diarrhea are mainly by the use of carbohydrates fibers.
Q3. a. How are proteins important to living organism?
Ans: Importance of proteins: Proteins are very important to living organisms because it performs
many important function in body. Some important function of protein are given below.
i. As Oxygen carrier: Haemoglobin is a protein which carries oxygen to all cells of body.
ii. Body structure: Skin, nail, hair, hoofs, horns and feather are composed of proteins.
iii. Growth: Proteins are essential for the physical and mental growth especially in children.
iv. Enzymatic action: Enzymes are proteins, which are produced by the cells of living organisms.
Enzymes catalyze the biochemical reactions going on in our bodies.
v. As Body regulators: Hormones and enzymes are the chemical regulators of the body.
vi. Body Defence System: The antibodies that help us to fight against diseases are large proteins
molecules.
3. b. Explain the nature of bonding in protein?
Ans: Nature of bonding in Protein:
In protein three types of bonding may occur.
1. Peptide Bonding 2. Hydrogen bonding 3. Disulphide bridges
1. Peptide Bonding: The bonding between carbonyl group (-C=O) and NH group are called peptide
bonding. Amino acid are linked together through peptide bonding to form proteins.
2. Hydrogen bonding:
The hydrogen bonding is formed between O of carbonyl group (-C=O) and h of –NH- group in
proteins.
3. Disulphide bridges/bonding:
The disulphide bonds occur only in sulphur containg protein. In these –SH- group are bonded to form
disulphide linkage,–H-S-S-H
Q4. a.Define the term Lipid?
Ans: Lipid:
Origin of word lipid: the word lipid is derived from Greek word lipose which means fat.
Definition: lipids are defined as biological molecules obtained from plant and animal tissue which are
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not soluble in water but are soluble in non-polar organic solvents, such as alcohol, chloroform, ether
etc.
Composition of lipids: lipids are generally composed of carbon, hydrogen and oxygen. Generally,
lipids are esters of long chain fatty acids and alcohols. These esters are made up of three fatty acids,
therefore they are called triglycerides.
Fats Oils
Those triglycerides which contain higher              Those triglycerides which contain higher
proportion of saturated fatty acidscomponents are     proportion of unsaturated fatty acids components
called fats.                                          are called oils.
They are solid or semi-solid at room temperature They are liquid at room temperature
Fats are mainly present in animals. Fats are mainly present in plants and fish.
Fats have high melting point Oils have low melting point
These molecules are translated and expressed by synthesis of specific proteins. These proteins perform
various function according to the direction which are given by the codes present in DNA.
DNA RNA
In DNA the pentose sugar is deoxyribose In RNA the pentose sugar is ribose
In DNA the nitrogen bases are adenine, cytosine,     In RNA the nitrogen bases are adenine, cytosine,
guanine and thymine                                  guanine and uracil
Q3. Write down the sources of Lipids. Sources of Lipids: Lipids obtained from animals, plants and
marine organisms. For example, salmon and whales are rich source of cod liver oil.
Animal’s sources: Milk is a rich source of animal fat from which butter, ghee, cheese etc. are
obtained.
Plants Sources: seeds of many plants such as sunflower, corn, cotton, peanut, ground nut, walnut,
coconut etc. are good source of lipids.
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 Q4. What is nucleic acid? Describe its composition, types and uses.
Ans: Nucleic acid:
Discovery: nucleic acid were first discovered in the nuclei of white blood cells in 1868 and in sperm
head in 1872 by Johannes Friedrich Miescher, a Swiss physician and biologist.
Definition: Complex organic compound consisting of nucleotides which synthesize proteins and
transmit characters from parents to offsprings are called nucleic acid.
Explanation: Nucleic acid are found in each and every living cell as well as in viruses. They are
essential part of genes.
Composition of Nucleic acid: Nucleic acid are composed of units called nucleotides. These
nucleotides are composed of three distinct parts:
i. Sugar: nucleotides consists of five carbon sugar which may be ribose (RNA) or deoxyribose (DNA).
ii. A nitrogen-containing bases of five types, uracil, cytocine, thymine, adenine and guanine.
iii. Phosphate group: Third part is phosphate group. .
Types of nucleic acid: Naturally occurring nucleic acid are of two types.
i. Deoxyribonucleic acid (DNA) ii. Ribonucleic acid (RNA)
i. Deoxyribonucleic acid (DNA):
Discovery: DNA was first discover by Watson and Crick in 1953.
Definition: DNA is a double-stranded structure composed of deoxyribose sugar, phosphate group and
four nitrogenous bases. The four nitrogenous bases present in DNA are adenine, cytocine, thymine,
and guanine.
Explanation: pentose sugar and phosphate group make the back bone of each strand. Two strand are
linked together through bases. The strands are held together by hydrogen bonds.
Function of Nucleic acid:
Function of DNA:
Genetic information: DNA has the ability to store and transmit genetic information. The genetic
information for the cell are present in DNA. The genetic information for the cell are present in the
form of special codes.
Protein synthesis: DNA also instructs how to synthesize a particular protein from a particular amino
acids. These instruction are known as genetic code.
Mutation: Mutation is a sudden chemical change in a DNA molecule that can lead to the synthesis of
protein with different amino acid sequence a DNA molecule that can lead to the synthesis of protein
with different amino acid sequence. Changes in DNA molecules. Changes in DNA molecules may be
caused by mutagens like radiations, chemical agents or viruses.
Function of RNA:
RNA is responsible for directing the synthesis of new proteins. RNA receives, reads, decodes and uses
genetic information from DNA to synthesize
Q5. Briefly describe the sources and deficiency symptoms of Vitamin A.
Ans: Vitamin A.
Vitamin A is fat soluble which is important for growth, vision and immune system.
Sources:
Vitamin A is found in milk, butter, fish oils, eggs, fresh green vegetables and fruits.
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Deficiency Symptoms:
Deficiency of Vitamin A cause night blindness, dry skin, burning and irritation of eyes.
Vitamin B Complex.
Vitamin B Complex is water soluble and consist of Vitamin B1, B2, B3, B5, B6, Boltin, Folic acid and
B12. They are important for energy production, nerves and cells,
Sources: Vitamin B is found in bread, rice, yeast, liver, milk, meat, fish, eggs, soybean oil, and fresh
green vegetable.
Deficiency Symptoms:
Deficiency of Vitamin B cause skin diseases, tongue/lips inflammation, anemia, bleeding gums and
beriberi.
Vitamin C.
Vitamin C is water soluble which is important for blood vessels, gums, healing wounds and preventing
cold.
Sources:
Vitamin C is found in citrus fruits (oranges, lemons) tomatoes, fresh green vegetables
Deficiency Symptoms:
Deficiency of Vitamin C scurvy and pain in joints.
Vitamin D.
Vitamin D is fat soluble which is important for bones and teeth.
Sources:
Vitamin D is found in milk, butter, eggs, fish oils, vegetables.
Deficiency Symptoms:
Deficiency of Vitamin D cause Rickets, osteomalacia.
Vitamin E.
Vitamin E is fat soluble which acts as an antioxidant.
Sources:
Vitamin E is found in bread, rice, eggs, liver, butter, fish, soybean oil, and fresh green vegetables.
Deficiency Symptoms:
Deficiency of Vitamin E causes anemia and sterility.
Vitamin K.
Vitamin K is fat soluble which is important for blood clotting.
Sources:
Vitamin K is found in liver, egg yolk, meat, cheese, fresh green vegetables.
Deficiency Symptoms:
Deficiency of Vitamin D cause Rickets, osteomalacia.
CHEMISTRY
Class 10th (KPK)
NAME: __________________________
F.NAME: _________________________
ADDRESS: ___________________________________
__________________________________________
SCHOOL: _____________________________________
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                                            CHAPTER 14
                                 ENVIRONMENTAL CHEMISTRY
 Q1: How ozone layer is depleted in the atmosphere?
Ans: Ozone layer depletion:
Ozone is present in the stratosphere layer helps to absorb ultraviolet ray. when ozone depleting
substance such as chlorofluorocarbons (CFCs), Hydro chlorofluorocarbons (HCFCs) and volatile
organic compound goes to stratosphere layer react with uv-ray .This reaction breakdown the depleting
substances and release the free cl atom. This free cl atom reacts with ozone gas and deplete the ozone
layer.
Q2: Differentiate between primary and secondary pollutant. Write the name and sources of
greenhouse gases to atmosphere.
Primary pollutant is an air pollutant which are   Secondary pollutant is not produced directly.
released directly into air.                       Secondary pollutant are formed from primary
                                                  pollutants.
Example:                                          Example:
The gases like SO2 released from burning
                                                  Photochemical oxidants, secondary
fossil fuel. Carbon Ammonia etc
                                                  particulate matter
The product which escape from the chimney         Acid rain, like SO2 mix with water in
of industrial unit. Exhaust of automobiles.       atmosphere and cause acid rain.
                                                  Nitrogen oxide and hydrocarbon react with
                                                  sunlight cause ozone
Q4. Sketch and briefly discuss the composition of atmosphere, by mentioning the percentage of
each component.
Ans: Sketch of atmosphere:
Nitrogen N2 78.084
Oxygen O2 20.95
Argon Ar 0.93
Neon Ne 0.0018
Helium He 0.000524
Ozone O3 0.00006
Krypton Kr 0.000114
Hydrogen H2 0.01
Q5. How the acid rain effect the building material or statues?
Ans: Acid rain: The rain having PH less than 5.6is called acid rain.
Effect of acid rain on building materials: Acid rain damages building material such as steel, paint,
plastic, cement sculptural work, material especially of marble and limestone.
CaCO3 +H2SO4 → CaSO4 +CO2 + H2O
The calcium sulphate is soluble in water and washed away with rain water.
The historical statues in Greece and Italy and The Taj Mahal in India were damaged due to acidic rain.
Q6: Increase in concentration of CO2 cause greenhouse effect, justify it.
Ans: As we know that CO2 is produced by burning fuels like oil, natural gas , diesel and organic
petroleum.CO2 is released to the atmosphere where it remain for many years. This leads to an
increasing concentration of CO2 in our atmosphere. Which in turn cause average temperature on earth
to raise. The CO2 is greenhouse gas which emission causes global warming.
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Q7: Briefly discuss how the acid rain affect the aquatic life.
Ans: Effect of Acid rain on aquatic life:
The rain which PH is less than 5.6 is called acid rain. The aquatic life like fish, plants and
microorganism are very sensitive to acidity .i.e. at PH-5, most fish eggs cannot hatch. Some adult fish
die. Ia fish or animal survive in acidic water but food they eat might not be available as food. i.e., frog
can tolerate the condition at around PH-4 but mayflies the food of frog are more sensitive and not
survive below PH-5.6.
Q8: List the sources which produce CO and CO2 to the atmosphere.
Ans: The following are sources which produced CO and CO2 to atmosphere.
CO: Natural sources: CO is one of the gaseous pollutant .its natural sources are Volcanic eruption,
Natural gas emission and forest fires.                           .
Anthropogenic sources: Most of carbon monoxide released to atmosphere by human activities.
Automobile release 75% of CO to the environment. Besides this incomplete combustion of fossil fuels,
smoking, forest fires and steel industries are the main sources of carb on monoxide.
Sources of CO2:
the main sources of carbon dioxide gas emission are our everyday activities such Cooking and baking
etc. vehicle and industrial emission, petroleum production and thermal power plant are also the source
of CO.
Q9: what are the adverse effects of global warming?
Ans: Global Warming: The gradual increase in the average temperature of the earth due to emission
of greenhouse gases is called global warming
Effect of Global warming:
Global warming is a phenomenon of climate change characterized by general increase in average
temperature. It is adversely effecting sea level, ozone layer, crop yield, precipitation (rain and snow
fall) and health.
Global warming is harming the environment in several ways include
Desertification increase in temperature around the world changes the water cycle and rainfall patterns
which causes desertification of certain areas.
Increases melting of ice and snow:
Snow and ice are melting at a faster pace due to increase in temperature.
Sea level rise:
Increase in temperature cause ocean waters to expand on other hand glaciers and ice are melting due to
rise in temperature which rise the sea level.
Strong storms and cyclone:
Rise of temperature also increase the frequency of strong cyclone and strong storms
Q10:What is the importance of ozone?
Ans: Ozone : Ozone is an allotropic form of oxygen, consisting of three chemically bonded oxygen
atoms
Importance of ozone:
The ozone acts as a protective layer in the atmosphere. it save earth and living thing from harmful UV
rays from sun.it also helps to remove the pollutant from the surface of the earth.
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                                           Long questions
Q1: Sketch and identify the different layer of atmosphere.
A: Definition:
The layer of gases surrounding the earth is called atmosphere.it extends up to about 500km from the
earth’s surface.
Layers of Atmosphere:
The atmosphere is divided into four layers based on the variation in temperature in each layer.
(i) Troposphere
(ii) Stratosphere
(iii) Mesosphere
(iv) Thermosphere / Ionosphere
Troposphere :
The first layer of the atmosphere which is closest to Earth’s surface is called troposphere .we live in
this layer.it contains approximately 75% of atmosphere mass and 99% of total mass of water vapor
aerosol.
Characteristics of Troposphere :
The few characteristics of Troposphere are given below:
a. Height:
The average height of this layer from the earth’s surface is about 11km.It height is depends upon the
latitude and season and pressure. It is lowest over the poles and highest at equator and by season it is
lower in winter and higher in summer. The pressure, moisture content, density of air also decrease with
height
b. Main components:
The major components of this layer are N2, O2, and water vapours.70 to75% of the atmosphere gases
are present in this layer. Dust, particles are also present here.
phenomenon occurring in Troposphere:
All the weather phenomenon like cloud formation, winds, rainfall, snowfall takes place in this layer.
Lapse rate:
The change in temperature of atmosphere with increase in height is called lapse rate. It may be positive
or negative
Positive lapse rate:
If the temperature increase with height it is called positive lapse rate.
Negative lapse rate:
If the temperature decrease with height it is called negative lapse rate.
The air of this layer is warmer than other layers because it is heated from the earth’s surface below.
D. Temperature:
The temperature range of this layer is from 15℃to -56. ℃. It shows negative laps rate.
ii. Stratosphere:
The layer which is above the Troposphere is called Stratosphere.
It is second layer of the atmosphere. It is warmer at top than bottom. The lower portion has nearly
constant temperature with height but upper part temperature increase with latitude.
Characteristics of Stratosphere
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a. Height:
The height of this layer is from 11km to 50 km.
b. Temperature:
The temperature of this layer is increases from -56℃ to -2℃ with height.
c. Main component:
The major component of this layer is ozone O3.
Ozonosphere:
Stratosphere contains ozone, at height of about 30 km, therefore it is called ozonosphere.
Production of ozone:
when the oxygen molecules absorbs ultraviolet radiation, it decompose to form oxygen atom. This
oxygen atom react with another oxygen and ozone is produced.
O2(g) + O●(g) → O3(g))
Phenomenon occurring in Stratosphere:
The ozone layer is present in this layer which absorbs harmful UV radiation .Thus protects the living
organism from harm full UV radiation.
The CFCs and ultra violet radiation from the sun, break down the Ozone molecule into mono atomic
oxygen and diatomic oxygen, resulting in damage to ozone layer.
O3(g)) ) →O2(g) + O●(g)
Troposphere Stratosphere
The main component of this layer is N2, O2          The main component of this layer is ozone.
andCO2and water vapor
The air density is more in this laye The air is less dense in this layer.
 Clouds formation, rain, snowfall, and winds        No clouds formation, rain, snowfall and wind
occur here.
Ans: Air pollution: According to world health organization (WHO), air pollution is defined as:
the substances released to air either by human activities or by natural activities, in sufficient
concentration to cause harmful effects to human beings, plants and other living thing is called Air
pollution.
Explanation: Air pollution change the physical, biological and chemical characteristics of airthat
cause adverse effects on human and other organism.
Air pollution in Pakistan: The big cities of Pakistan like Karachi, Lahore, Faisalabad and Islamabad
faces the air pollution due to following reasons.
1. Urbanization and population growth
2. Increase in number of vehicles used daily
3. Deforestation for making house and other uses.
4. Inefficient use of energy
5. Burning of garbage and plastic increase the smog and other air pollution due to worst air quality.
Q2.b.Enlist the main source of air pollution.
Ans: Pollutants: The substances that are responsible for causing air pollution are called air pollutant.
Sources of air pollution:
A: There are two main sources of air pollution.
i. Nature sources
ii. Anthropogenic/ man made sources
Natural sources: Natural sources of air pollutants are as follow,
A. Particulate:
The particulate pollutant is produced naturally by volcanic eruption, soil erosion by wind, dust storms ,
natural forest fires and salts spray from oceans . The contribution of the natural sources towards the
particulate emission is greater than man made sources. It has been estimated that natural sources
releases millions of tones particulate matter in to air.
B. Oxides of carbon (Coax):
CO and CO2 are the oxides of carbon.
I. Carbon monoxide;
The natural concentration of carbon monoxide in air is around 0.2 parts per million (ppm), and that
amount is not harmful to humans. Natural sources of carbon monoxide include volcanic eruption,
decomposition of organic matter, and electrical discharge during storms, seed germination and natural
gas emission etc.
ii. Carbon dioxide:
The natural sources of CO2 are animal and plants respiration, decomposition of organic matter, forest
fires and emissions from volcanic eruption.
iii. Oxides of nitrogen (NOx)
Bacteria and microorganisms are the main sources for emission of oxides of nitrogen into air. They
convert nitrates present in soil into oxides of nitrogen. Nitrogen oxides are produced naturally by
lightning, and atmospheric nitrogen gas into oxides of nitrogen and ammonia.
Iv. Oxides of Sulphur dioxides
The major oxides of Sulphur are Sulphur dioxide and Sulphur trioxide. They are produced by volcanic
eruptions, rock weathering and biological activities.
v. volatile organic compound:
Volatile organic compounds or VOCs, are chemicals that have a strong tendency to vaporize from
liquids or solids into gaseous states e.g. Methane. Wetlands are the largest natural source of methane.
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They contribute 78% of natural methane in the environment. Animals like sows, sheep and goats
during their normal digestion process produce large amount of methane.
Anthropogenic/man made sources:
The burning of different types of fuels by man that produced air pollution is called Anthropogenic
/man made sources.
I. Particulates:
Particulate matter is the sum of all solid and liquid particles suspended in air. The man-made sources
of particulate matter are combustion of fuels wood burning, construction, mining and industrial
process.                                                                                                ii.
Oxides of carbons;
a. Carbon dioxide:
Carbon dioxide is also added to the atmosphere through human activities, such as the burning of fossil
fuels and forests and the production of cement. Also thermal power generation and our everyday
activities such as cooking and baking released CO2.
Carbon monoxide:
CO is produced by incomplete combustion, i.e. when there isn't enough oxygen to make CO2,
Automobiles exhausts, smoking forest fires and steel industries are also the main sources of CO gas.
Oxides of nitrogen (NOx):
Automobiles exhausts, industrial activities, furnaces, thermal power generation. , jet airplanes are the
main sources of nitrogen oxides.
Sulphur oxides (SOx):
Burning of fossil fuels such as coal, oil and natural gas are the main source of Sulphur dioxide
emissions. Coal fired power stations, in particular, are major sources of Sulphur dioxide, other sources
of Sulphur dioxide are industrial processes and transportation.
Volatile organic compounds (VOCs):
Volatile organic compound e.g. methane is produced by burning fossil fuel.
Common sources of these organic compounds are:
Furniture polish and other wood finishing products
Solvents and thinners, e.g. nail polish remover with acetone or paint thinner.
Aerosols, such as air fresheners and other cleaners
Smoke from burning stoves or candle, as well as cigarettes.
Automobile exhaust and factories are also the main sources of VOCs.
Chlorofluorocarbons (CFCs) :
It is anthropogenic compounds that have been released into the atmosphere from various applications
such as in air-conditioning, refrigeration, blowing agents in foams, insulations and packing materials.
Q2.c. Prove that oxides of nitrogen and Sulphur cause air pollution.Oxides of Sulphur:
The major oxides of Sulphur are SO2 and SO3.
These oxides are produced by the burning of Sulphur containing fossil fuel. Sulphur dioxide is a gas. It
is invisible and has a nasty, sharp smell. It reacts easily with other substances to form harmful
compounds, such as sulfuric acid, sulphurous acid and sulphate particles. The main source of Sulphur
dioxide in the air is industrial activity e.g. the generation of electricity from coal, oil or gas that
contains sulphur. Sulphur dioxide is also present in motor vehicle emissions and air become
polluted.When sulphur dioxide combines with water and air, it forms sulfuric acid, which is the main
component of acid rain. Acid rain can:
Cause deforestation.
Effect the aquatic life and other organism life.
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More frequent and intense heat waves: Dangerously hot weather is already occurring more
frequently than it did 60 years ago heat waves to become more frequent and severe as global warming
intensifies. This increase in heat waves creates serious health risks, and can lead to heat exhaustion and
heat stroke. An increase in extreme weather events: Global warming is increasing certain types of
extreme weather events, including heat waves, coastal flooding, extreme precipitation events, and
more severe.
Topic wise:
Aging of skin: Exposure to UV rays can lead to acceleration of the aging process of our skin. We look
older than our actual age. It can also lead to photo allergy that result in outbreak of rashes in fair
skinned people.
Other effect on human: In humans, exposure to UV rays can also lead to difficulty in breathing, chest
pain, and throat irritation and can even lead to curb of lung function.
Effect on amphibians: UV rays affect other life forms too. It adversely affects the different species of
amphibians and is one of the prime reasons for the declining numbers of the amphibian species. It
affects them in every stage of their life cycle; from hampering the growth extreme cases.
and development in the larvae stage, deformities and decreases immunities in some species.
Effect on marine life: UV rays also have adverse effect on the marine ecosystem. It adversely affects
the planktons which plays a vital role in the food chain and oceanic carbon cycle. Affecting
phytoplankton will in turn affect the whole ocean ecosystem.
Effect on plants: UV rays will also affect the plants. UV radiations can alter the time of owering in
some plant species. It can also directly affect the plant growth by altering the physiological and
developmental processes of fl the plants.
Effect on material:Ozone depletion will cause many materials to degrade faster. These materials
include PVC (used in window and doorframes, pipes and gutters), nylon and polyester.
CHEMISTRY
Class 10th (KPK)
NAME: __________________________
F.NAME: _________________________
ADDRESS: ___________________________________
__________________________________________
SCHOOL: _____________________________________
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UNIT: 15
Q: Identify the factor which are responsible for dissolving polar substance in water?
A: Factors affecting solubility:
Water because of its polarity and ability to form hydrogen bonds, water makes an excellent solvent,
meaning that it can dissolve different kinds of molecules.
Explanation: Water molecules have a polar arrangement of the oxygen and hydrogen atoms—one side
(hydrogen) has a positive electrical charge and the other side (oxygen) had a negative charge. This
allows the water molecule to become attracted to many other different types of molecules. Water can
become so heavily attracted to a different molecule, like salt (𝑁𝑁𝑒𝑒𝑁𝑁𝑙𝑙) that it can disrupt the attractive
forces that hold the sodium and chloride in the salt molecule together and, thus, dissolve it.
apart. This produces a crystal lattice, which is commonly known as ice. When water is cooled down
below 4℃, its density decreases. At 0℃, the density of water becomes 0.91 g/cm. Thus, ice is lighter
than liquid water and therefore floats on the surface of water.
compounds carries a positive charge, while the chlorine part carries a negative charge. The two ions
are connected by an ionic bond. The hydrogen and oxygen in the water, on the other hand, are
connected by covalent bonds. Hydrogen and oxygen atoms from different water molecules are also
connected via hydrogen bonds. When salt is mixed with water, the water molecules orient so that the
negative charge oxygen anions faces the sodium ion, while the positive-charged hydrogen cations face
the chloride ion. Although ionic bonds are strong, the net effect of the polarity of all the water
molecules is enough to pull the sodium and chlorine atoms apart. Once the salt is pulled apart, its ions
become evenly distributed, forming a homogeneous solution.
A: Disadvantages:
1).The major disadvantage of detergent is that they are non - biodegradable. Microorganisms like
bacteria cannot decompose detergent, while these microorganisms can easily decompose soap.
2) They causes soil and water pollution
3) Excessive alkalis used in some detergent can damage the fabric.
4) Colour may run out while using cheaper variety of detergent.
5) More amount of water is required for rinsing to remove foam, otherwise it will damage the fabric
10: Identify the different toxic substances in household wastes?
A: Household waste products that contain corrosive, toxic, ignitable, or reactive ingredients are
considered to be "household hazardous waste". It is hazardous to human health and the environment.
The following are the toxic household waste substances: Aerosols, Batteries, Automotive, Chemicals,
Fertilizer, Fluorescent light bulbs, Hair color, Kerosene, Lawn and garden pesticides Nail polish,
Shampoo and Thermometers with mercury
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LONG QUESTIONS:
1. Oceans (97.5%)
2: Water is an excellent solvent. Explain how this property is beneficial for life but sometimes
harmful for us?
A: Water is the best-known solvent. It can dissolve more substances than any other solvent. Due to this
fact, it is termed as a universal solvent. This property is very beneficial for us but sometimes become
harmful for us, because mostly salts (pollutants) dissolve in water and cause water pollution. This
characteristic property of water is due to the following reasons;
i: Polarity of water molecules
ii: Hydrogen bonding in water
iii: Dielectric constant
i: Polarity of water molecules: The water molecule has polar structure. The hydrogen on one end of
the water molecule is partially positive.while the oxygen on the other end is partially negative. It is due
to the electronegativity difference between oxygen (O=3.5) and hydrogen (H=2.1) atoms. When an
ionic compound is added into water, oppositely charged ions are surrounded by water molecules.
These oppositely charged ions of ionic compounds are pulled by the water molecules and it becomes
soluble in water. For example,𝑁𝑁𝑒𝑒𝑁𝑁𝑙𝑙,𝐾𝐾𝑁𝑁𝑙𝑙, are soluble in water. When an ionic compound such as
sodium chloride (𝑁𝑁𝑒𝑒𝑁𝑁𝑙𝑙) is added to water, the sodium ion (Na+) of NaCl is attracted towards the partial
negative pole (O-δ) of water molecule while the chloride ion (Cl+) is attracted towards the partial
positive pole (H+𝛿𝛿 ) of water molecule. The ionic bond break down in NaCl, which causes the sodium
chloride to split in water and dissolves. The water molecules orient in such a way that the negative
poles are towards the positive ions. Similarly, the positive poles of water molecules orient themselves
around the negative ions. A hydration shell is formed around the ions, which prevents Na+ and Cl -
from attracting each other.
ii: Hydrogen bonding in water: Water molecule is composed of oxygen and hydrogen atoms. The
hydrogen on one end of the water molecule is partially positive .while the oxygen on the other end is
partially negative. Because of the presence of two nonbonding electrons (lone pairs of electrons) on
oxygen, water molecule forms four hydrogen bonds with other H2O molecule. These water molecules
are arranged in a tetrahedral manner. Hydrogen bonding makes water unique. Covalent compounds,
which have polar ends also form hydrogen bond with water. This behavior enables water to dissolve
some of the covalent compounds having hydroxyl group (OH˗) very easily. For example, sugar
(C12H22O11), alcohol like methyl alcohol (CH3OH) etc. will dissolve easily. iii: Dielectric constant:
Dielectric constant is based on the coulomb law in which the force of attraction between two
oppositely charged bodies “x” and “y” is directly proportional to the product of charges and inversely
proportional to the square of distance between them r2 .Mathematically it can be written as:
               (𝒙𝒙)(𝒚𝒚)                  (𝒙𝒙)(𝒚𝒚)
 𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭∞              𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭 =             Where D is the proportionality constant and is called the dielectric
                 𝑭𝑭𝒓𝒓                     𝑫𝑫𝑭𝑭𝒓𝒓
constant of water. Greater the value of D the smaller will be the force of attraction and vice versa.
Water has a high dielectric constant of 80 at 18℃. Thus, the positive and negative ions of a polar salt
dissolved in water will have less force of attraction and would remain soluble, while other liquids have
small value of dielectric constant compared to water and therefore, these are not good solvents.3a):
Classify temporary hard water and permanent hard water?
A: There are two types of hard water;
1: Temporary hard water2: Permanent hard water
1: Temporary hard water: The temporary hardness of water is due to the dissolved Calcium
bicarbonate and Magnesium bicarbonate. These salts are soluble in water and are present in the form of
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The kidneys and liver use it to help flush out these wastes from our body.
Water is used in cooking and washing.
Running water is used to generate electricity.
Water in lakes, rivers and oceans are used as a means of transportation.
Fish and other aquatic animals and many plants live in water.
Agriculture needs large amount of water, to cultivate fruits, vegetables, and other food.
It is required for irrigating crops, as seeds cannot germinate without water.
Many industries such as petroleum, fertilizers, dye and drugs industries require large quantities of
water for various processes.
b): Write the disadvantages of hard water?
A: The following are the disadvantages of hard water;
Hard water consumes large amount of soap in washing process.
Hard water is unfit to use in steam engines and boilers. When hard water is used in the boiler, calcium
and magnesium salts settles down at the bottom as hard insulating scale. As a result, more fuel is
consumed in producing steam. If these are not removed, they block the tubes, which lead to the
engines. These make a constant threat to the explosion of boiler. This deposition of scales inside the
boiler causes overheating and reduces the life of boiler.
Use of hard water for drinking purposes for a long time causes dysentery, intestinal and stomach
diseases. If magnesium sulphate is present in the hard water, it weakens the stomach function.
c): Enlist the advantages of wastewater treatment?
A: Wastewater treatment is defined as the process of removing any harmful contaminants from water
no longer needed
Advantages of wastewater treatment:
1)It prevents disease:
It removes harmful bacteria and chemicals from water that cause disease. Thus makes the water safe
for living organisms.
It provides clean water: It filtering out harmful contaminants, and leaving a water source clean and safe
for everyone. This removes the fear of droughts, water shortages.
5a): How would you evaluate the effects of water pollution?
A: Effects of water pollution:
The effects of water pollutants are not only disturbing to people but also to plants, animals, fishes and
birds. Polluted water is unsuitable for drinking, agriculture and industry.
The major effect of water pollutants are as under;
i: Infectious disease: Water pollution is the major cause of infectious diseases in human beings. These
diseases include typhoid, cholera, dysentery, amoebiasis, ascariasis and hepatitis etc.
ii: Nutrient pollution: Nutrient pollution is a form pf water pollution. In this process, large amount of
nutrients such as nitrogen, phosphorus, nitrates etc. become part of water as a runoff from agricultural
fields or weathering of rocks. The enrichment of nutrients in water bodies is called eutrophication.
These excessive amounts of nutrients create problems such as excessive growth of algae, decrease in
dissolved oxygen in water etc. This in turn, can kill fish, carbs, oysters, and other aquatic animals.
iii: Chemical contamination: Some of the major effects of chemical contamination are as under;
Pesticides affect and damage the nervous system, liver, reproductive system, endocrine glands and
DNA etc.
Oil and petrochemical can alter the ecology of aquatic habitats and the physiology of marine
organisms. In human beings, it causes gastro-intestinal irritation, liver and kidney damage and nervous
system effects.            Mercury and its compounds are used in many industries. It finds its way into
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the water bodies primarily through air pollution and industrial wastes. Mercury gets into the body
through food especially seafood. In children, it causes brain damage, learning defects and incomplete
mental development. In adults, mercury causes Parkinson’s disease, Alzheimer’s disease etc.
iv: Thermal water pollution: Hot water from industrial processes is directly allowed to become the
part of the environment. This hot water effects the aquatic life in two ways;
a. The hot water decrease the solubility of oxygen as a result aquatic organisms will die due to the
shortage of oxygen. Many aquatic animals especially young cannot survive in water above 30℃ and
will die.
Topic wise:
Composition of water: water is a compound of hydrogen and oxygen. Henry Cavendish proved that
water is a compound. Composition of water can be determined,
1) Volume             2) Mass .
Composition of water by volume: Water molecule composed of two Hydrogen atoms and one
Oxygen atom. Its molecular formula is H2O.Composition of water can be determined by using an
apparatus called "Hofmann voltammeter". When an electric current is passed through water containing
a little salt, water decomposes into hydrogen gas and oxygen gas.it shows that water composed of two
Composition of water by mass: Water is composed of one part of hydrogen to eight parts of oxygen
by mass. Joseph Proust determined the ratio of water by mass I.e. 1:8.
Explanation: One mole of H2O is made up of 2 moles of Hydrogen atoms and 1 mole of Oxygen atom
Mass of 1 mole of Hydrogen atoms= 1 g /mol
Mass of 1mole of Oxygen atoms = 16 g/mol
Mass of two moles of Hydrogen atoms = 2x 1 g/mol = 2 g/mol.
Mass of one mole of Oxygen atoms = 1 x 16 g /mol
Mass of one mole of water = 2 g/mol + 16 g/mol = 18 g/mol.
% of hydrogen = molecular mass of hydrogen / molecular mass of H2O ×100
%of hydrogen= 2/18 ×100=11.11
% of oxygen =         Atomic mass of oxygen/molecular mass of H2O ×100
% of oxygen =           16/18=88.88
Ratio           = 1 : 8
Water pollution:
Definition: water pollution occurs when undesirable foreign substance are introduced into natural
water. The substances that cause water pollution is called pollutants. Water pollution is one the main
environmental issues that we are facing, as more than 70% of the Earth’s surface is water-covered.
Sources of water pollution: The major sources of water pollution are following;
Industrial waste: waste from factories, refineries, waste treatment plants etc. that emit fluids of
varying quality directly into urban water supplies cause water pollution sources include contaminants
that enter the water supply from soils/groundwater systems and from the atmosphere via rain water.
Soils and groundwater contain the residue of human agricultural practices (fertilizers, pesticides, etc.)
and improperly disposed of industrial wastes. Atmospheric contaminants are also derived from human
practices (such as gaseous emissions from automobiles, factories and even bakeries).
Waterborne diseases: The diseases that spread due to polluted water or eating those foods that are
prepared from polluted water are called waterborne diseases. They bacterial, parasitic or viral diseases.
(1)Bacterial infections:
cholera: The symptoms are watery diarrhea and vomiting. It effects intestine and causes dehydration.
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Dysentery: It is the large intestinal disease, causing abdominal cramp, anal pain and bloody stool.
Typhoid: it is bacterial disease and person is suffer from typhoid fever.
(2) Viral infections:
Hepatitis A and E: Hepatitis-A virus infection (HAV) and Hepatitis E virus infection (HEV) are due
to contaminated water: it attack liver and cause jaundice.
Polio: Polio, also called poliomyelitis or infantile paralysis, is an infectious disease caused by the
poliovirus.The polio virus usually enters the environment in the feces of someone who is infected. In
areas with poor sanitation.
Protozoal infections: In this disease Amoeba enters through unsanitary food or contaminated water
effect the gastrointestinal illness.
Parasitic infections: Guinea worm and pinworm infection are common parasitic infection. In
contaminated water the egg of these worms are swallowed cause ulcers, fever, and vomiting.
Arsenicosis and fluorosis:
Fluorosis: It is an abnormal condition caused by excessive intake of water having fluorine.it effects
the bones and teeth.
Arsenicosis: Drinking water rich in arsenic over a long period leads to arsenic poisoning or arsenicos
is Most waters in the world have natural arsenic concentrations of less than 0.01 mg/dm3.The
symptoms of this disease are excess of saliva, vomiting, nausea and blood in urine.
CHEMISTRY
Class 10th (KPK)
NAME: __________________________
F.NAME: _________________________
ADDRESS: ___________________________________
__________________________________________
SCHOOL: _____________________________________
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                                                UNIT 16
                                               EXERCISE
                                           SHORT QUESTIONS
Q1. How could you convert NaHCO3 into Na2CO3?
Ans: Calcination: sodium bicarbonate (NaHCO3) is converted into sodium carbonate (Na2CO3) by the
process called                    calcination, in this process NaHCO3 is heated in rotatory funnel call
calcinatory, to give anhydrous sodium carbonate.
                                2NaHCO3(s) → Na2CO3(s) + CO2 (g) + H2O
Q2. Enlist the different uses of urea?
Ans: Uses of urea:
it is a white crystalline organic compound. It is important due to the following usage.
1. Fertilizer: about 86% of urea is used as solid fertilizer.
2. Resins:Urea-formaldehyde resins are used as a polywood adhesive/glues.
3. Use as explosive:Urea can be used to make urea nitrate, which is highly explosive.
4. Chemical Industry:Urea is used as a raw material for manufacture of many important chemical
compounds like plastics, resins, and various adhesives etc.
6. Flame proofing agent:Urea is used as a flame proof in agent.
7. Cosmetics:it is used as an ingredient in hair conditioners, facial cleaners and lotions.
8.Repellent to Corrosion:It is used as an alternative to rock salt in the deicing roadways and runways.
It does not promote metal corrosion to extent that salt does.
9. Cigarette:It is also used as flavor enhancing additive for cigarette.
10. Medicinal uses:urea containing creams are used as tropical dermatological products to promote
rehydration of skin.
Q3. Differentiate between minerals and ore.
Ans: Minerals: The naturally combined state of metal is called is called mineral.
Ores: An aggregate of mineral and other impurities is known as ore.
Q4. What is metallurgy? What are its types?
Ans: Metallurgy: The art and science of making of making metals and alloys from their ores with
properties suitable for practical uses is called metallurgy. OR
The science that deals with the procedures used in extracting metals from their ores, purifying, alloying
metal sand creating useful objects from metal is called metallurgy.
Types of metallurgical operations: In metallurgy the ores are mined and subjected to various
mechanical and chemical processes. There is no single method for extracting metals from their ores,
But certain basic operation are ,              required that is,
i. Concentration of ores
ii. Extraction of metal (roasting and reduction)
iii. Refiningof metal.
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removed as anode mud. The copper obtained by this method is 99.9% pure.
Reaction at anode: Cu(s) → Cu+2(aq) + 2e-
Reaction at Cathode: Cu+2(aq) + 2e- → Cu(s)
                                           LONG QUESTIONS
Q1. How could you convert the concentration ore to its oxide?
Ans: Concentration ore to its oxide: Following methods are used to convert the concentration ore to
its oxide form.
   1. Roasting: It is the process in which concentrated ore is heated alone or in the presence of some
      other materials in excess of air in a process.
      Procedure: the ore of metals such a copper and nickel when roasted in a furnace in the
      presence of air between 50000C and 70000C are converted into their oxide i.e.
      CuCO3(s) → CuO(s) + CO2 (g)
      Cu (OH) 2(s) → CuO(s) + H2O (g)
      2CuFeS2(s) → Cu2S(s) + 2FeS(s) + SO2 (g)
   2. Smelting: It is the process in which the oxide ore in the fused state is reduced with reducing
      agents such as coke to get the metal is called roasting.
      Procedure: The roasted ore is mixed with coke and sand and smelted into blast furnace.
      During smelting iron sulphide (FeS) get oxidized to iron oxide (FeO). The iron oxide then react
      with silica (SiO2) forming iron silicate (FeSiO3).
      2FeS(s) + 3O2 (g) → 2FeO(s) + 2SO2 (g)
      FeO(s) + SiO2(s) → FeSiO3
      Cuprous sulphide (Cu2S) is also oxidizes and form copper oxide (Cu2O).
      2Cu2S + 3O2 → 2Cu2O + 2SO2 (g)
      Cu2O then reacts with un-reacted FeS and form Cu2S and FeO.
      Cu2O(s) + FeS(s) → Cu2S(s) FeO(s)
      TheCu2Sand FeS is called matte and is removed through slag hole.
   3. Bessemerization:
   4. History: Bessemer process was invented by Henry Bessemer.
Bessemer converter: This process is carried out in a special kind of egg shaped or pear shaped
furnace. This furnace is called Bessemer.
Procedure: In Bessemerization matte is reacted with sand. Iron sulphide (FeS) oxidized to Iron oxide
(FeO). This iron oxide (FeO) reacts with sand (SiO2) forming (FeSiO3), slag which is float on the
surface.
2FeS + 3O2 → 2FeO + 2SO2
FeO + SiO2 → FeSiO3
Similarly Cuprous sulphide (Cu2S) is converted to cuprous oxide (Cu2O). This Cu2O reacts with Cu2S
to produce copper (Cu) in molten form and sulphur dioxide (SO2). This copper is known as blister
copper (95-97% pure).
2Cu2S + 3O2 → 2Cu2O + 2SO2
2Cu2O + Cu2S → 6Cu + SO2
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NH2. . Urea is one of the most important nitrogenous fertilizers. Its chemical formula is NH2CONH2.
The formula shows that urea is composed of nitrogen, hydrogen. Carbon and oxygen.
Raw material used in the manufacture of urea material used in the manufacture of urea are
following.
    •   Ammonia (NH3)
    •   Carbon dioxide (CO2)
Ammonia (NH3): ammonia is prepared by Haber process. In this process, nitrogen and hydrogen react
when they are passed over iron catalyst at 4500C and 200 atmosphere pressure. It produces ammonia.
N2 (g) + 3H2 (g) → 2NH3 (g)Carbon dioxide: it can be prepared from natural gas (CH4).
CH4 (g) + H2O (g) → CO2 (g) + 4H2 (g)
Q3.b. What basic reactions would you use to support the manufacture of urea?
Ans: Basic reaction:
Manufacture of urea involves the following steps.
1. Reaction of Ammonia and carbon dioxide: ammonia and carbon dioxide are heated at 17o –
2000C and 100-200 atmospheric pressure to form ammonium carbamate.
2NH3 (g) + CO2 (g) → NH2COONH4
2.Urea formation: when ammonium carbamate is heated it decomposes and produce water and urea.
NH2COONH4 → NH2CONH + H2O
3. Evaporation and granulation of liquid: in this step, liquid urea is concentrated in vacuum
evaporators. It is sprayed from top of tower under pressure and hot current of air in introduced from
the bottom in opposite direction. It evaporates water from urea. It is rapidly cooled and sent to the
granules. This urea is stored to be marketed.
Q3.c. What is the advantage of recycling of untreated compound in manufacture of urea?
Ans: Advantages of Recycling: manufacture urea contain untreated ammonia and carbon dioxide and
ammonium carbamate. Ammonium carbamate is removed by reduction the pressure. When heating
ammonia and carbon dioxide is separated. The advantage of this process is that ammonia and carbon
dioxide can be recycle back to the process. Which increases urea yield.
Q3.d. Sketch the flow sheet diagram of urea manufacture process.
Composition of petroleum: petroleum is the mixture of various hydro carbons. It includes petroleum
gas such as methane, ethane, propane and butane, naphtha petrol which range from C4 – C1. Kerosene
oil ranges from C12 – C16, diesels oils from C14 – C25, lubricating oil C20 – C70 and residue which is
above C70.
Q4.b. What are the two theories about origin of petroleum?
Ans: There are two theories about origin of petroleum.
Inorganic origin: This theory was put forward by Russian chemist Mendeleev. He proposed that
steam reacted with metallic carbides at high temperature and pressure under the surface of the earth
produced petroleum. This theory did not gain popularity. According to this theory petroleum is
originated from non-living source.
Organic origin: according to this theory, the remain of plants and animals were buried under the soil
millions of year ago. These dead organic matter were converted into petroleum by the action of
bacteria under the influence of temperature and pressure of the earth. These deposits trapped between
the layers of nonporous rocks. The oil and gas thus formed could not come out and collected
underground. This theory gain popularity.
Q.4.c. write a detail note on fractional distillation of petroleum.
Ans: Fractional distillation: The type of distillation in which different fractions of petroleum are
separated according to difference in their boiling point is called Fractional distillation.
Basic principle of Fractional distillation: The fractions are separated according to difference in their
boiling point. The substances having less boiling points boils out first leaving behind others.
Fractional Column: Fractional distillation is carried out in fractionating column. It has different
compartment to collect different fractions.
Fractional distillation of petroleum: First of all crude oil is heated up to 4000C under high pressure
in a furnace. The heated mixture then passed through the fractional distillation column. The
fractionating g column is divided into different compartment. Each compartment has definite range of
temperature as hot vapours move up, they condense according to their boiling point into various
fractions. Compound with high boiling point will condense first near bottom, while those compounds
having low boiling points move to the top of column. Thus crude oil is separated into different
fractions.
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2. Hydraulic washing:
Basis: This method is based on difference in densities of the ore and gangue.
Procedure: In this process, the ore particles are poured over a hydraulic classifier which is vibrating
inclined able with grooves and a jet of water is allowed to flow over it. The denser are settled in the
grooves while the lighter gangue particles are washed away.
Forth floatation: A separating method of the mineral particles of ore from the gangue that depends on
the wetting of the minerals pieces.
Function of forth floatation: It is a process for selectively separating hydrophobic materials from
hydrophilic. In this process the ore is wetted with water and pine oil. And the water insoluble
impurities float on the surface and thus separated from the ore.
Steps:
1. Floatation tank: concentration of copper ore is carried out in floatation tank.
2. Crushing and grinding: In this step the ore is reduced to fine powder through crushing and
grinding.
 3. Froath formation: the powdered ore is suspended in water, soap or pine oil is added and a blast of
air is bubbled through the suspension to produce froth.
4. Particle of ore: the particle are wetted by oil and float at the top of the mixture in container from
which it is collected.
5. Concentrated Ore: the froth is washed with water and then filtered to obtain concentrated ore.
While undissolved particles settle down at the bottom.
ii. Extraction of metal (roasting and reduction)
See L.Q 1 (part b and c).
iii. Refining of metal.
See L.Q 1 part d.
Q5.b. Explain the process of smelting and Bessemerization with reference to copper extraction.
Ans: Smelting: It is the process in which the oxide ore in the fused state is reduced with reducing
agents such as coke to get the metal is called roasting.
Procedure: The roasted ore of copper is mixed with coke and sand and smelted into blast furnace.
During smelting iron sulphide (FeS) get oxidized to iron oxide (FeO). The iron oxide then react with
silica (SiO2) forming iron silicate (FeSiO3).
2FeS(s) + 3O2(g) → 2FeO(s) + 2SO2(g)
FeO(s) + SiO2(s) → FeSiO3
Cuprous sulphide (Cu2S) is also oxidizes and form copper oxide (Cu2O).
2Cu2S + 3O2 → 2Cu2O + 2SO2(g)
Cu2O then reacts with un-reacted FeS and form Cu2S and FeO.
Cu2O(s) + FeS(s) → Cu2S(s) FeO(s)
TheCu2Sand FeS is called matte and is removed through slag hole.
Bessemerization:
History: Bessemer process was invented by Henry Bessemer.
Bessemer converter: This process is carried out in a special kind of egg shaped or pear shaped
furnace.
This furnace is called Bessemer.
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Procedure: In Bessemerization matte is reacted with sand. Iron sulphide (FeS) oxidized to Iron oxide
(FeO). This iron oxide (FeO) reacts with sand (SiO2) forming (FeSiO3), slag which is float on the
surface.
2FeS + 3O2 → 2FeO + 2SO2
FeO + SiO2 → FeSiO3
Similarly Cuprous sulphide (Cu2S) is converted to cuprous oxide (Cu2O). this Cu2O reacts with Cu2S
to produce copper (Cu) in molten form and sulphur dioxide (SO2). This copper is known as blister
copper (95-97% pure).
2Cu2S + 3O2 → 2Cu2O + 2SO2
2Cu2O + Cu2S → 6Cu + SO2
Q.5.c. Enlist the different uses of urea.
Ans: Uses of urea:
it is a white crystalline organic compound. It is important due to the following usage.
1. Fertilizer: about 86% of urea is used as solid fertilizer.
2. Resins: Urea-formaldehyde resins are used as a poly-wood adhesive/glues.
3. Use as explosive: Urea can be used to make urea nitrate, which is highly explosive.
4. Chemical Industry: Urea is used as a raw material for manufacture of many important chemical
compounds like plastics, resins, and various adhesives etc.
5. Flame proofing agent: Urea is used as a flame proof in agent.
6. Cosmetics: it is used as an ingredient in hair conditioners, facial cleaners and lotions.
7. Repellent to Corrosion: It is used as an alternative to rock salt in the deicing roadways and
runways. It does not promote metal corrosion to extent that salt does.
8. Cigarette: It is also used as flavor enhancing additive for cigarette.
9. Medicinal uses: urea containing creams are used as tropical dermatological products to promote
rehydration of skin.
Artificially: in case of artificial mining, mines are bored. If the natural gas is present with the
petroleum its pressure forces the petroleum to come out. If there is no natural gas then the air pressure
is applied to force the oil from the well.