Saturday 22 June 2013

Numerical Based On Square ( Mathematics )

Q. Find the perimeter of a square with the following area:
  1. 81 m
  2. 36 cm
  3. 49 mm
  4. 16 dm
  5. µm
  6. 9 km
  7. 25 inches
  8. 64 nm
Sol.
1. Area of the Square = 81 m         
    Also, Area of the Square = ( side )2
      ∴   ( side ) = 81 m
               side   =   81 m
                        =  9 m
    ∴   Perimeter of the Square = 4 x ( side ) = 4 x 9 m = 36 m


2. Area of the Square = 36 cm         
    Also, Area of the Square = ( side )2
      ∴   ( side ) = 36 cm
               side   =   36 cm
                        =  cm
    ∴   Perimeter of the Square = 4 x (side) = 4 x 6 cm = 24 cm


3. Area of the Square = 49 mm
        
    Also, Area of the Square = ( side )2
      ∴   ( side ) = 49 mm
               side   =   49 mm
                        =  7 mm
    ∴   Perimeter of the Square = 4 x side = 4 x 7 mm = 28 mm


4. Area of the Square = 16 dm
         
    Also, Area of the Square = ( side )2
      ∴   ( side ) = 16 dm
               side   =  16 dm
                        = 4 dm
    ∴   Perimeter of the Square = 4 x side = 4 x 4 dm = 16 dm


5. Area of the Square = µm
         
    Also, Area of the Square = ( side )2
      ∴   ( side ) = 4 dm
               side   =   4 dm
                        = 2 dm
    ∴   Perimeter of the Square = 4 x side = 4 x 2 dm =8 dm


6. Area of the Square = 9 km
       
    Also, Area of the Square = ( side )2
      ∴   ( side ) = 9 km
               side   =  √ 9 km
                        =  3 km
    ∴   Perimeter of the Square = 4 x side = 4 x 3 km = 12 km


7. Area of the Square = 25 inches
         
    Also, Area of the Square = ( side )2
      ∴   ( side ) = 25 inches
               side   =   25 inches
                        =  5 inches
    ∴   Perimeter of the Square = 4 x side = 4 x 5 inches = 20 inches


8. Area of the Square = 64 nm
         
    Also, Area of the Square = ( side )2
      ∴   ( side ) = 64 nm
               side   =  √ 64 nm
                        =  8 nm
    ∴   Perimeter of the Square = 4 x side = 4 x 8 nm = 32 nm


Q. Find the area of a square with the following perimeter:
  1. 4 m
  2. 8 cm
  3. 12 mm
  4. 16 dm
  5. 36 µm
  6. 40 km
  7. 44 inches
  8. 48 nm
Sol. 
1. Perimeter of the Square = 4 m 
    Also, Perimeter of the Square = 4 x ( side ) 
      ∴  4 x ( side ) = 4 m
                   side  =  ( 4/4 ) m
                            =  1 m
    ∴    Area of the Square = ( side )= ( 1 x 1 ) m = 1 m

2. Perimeter of the Square = 8 cm 
    Also, Perimeter of the Square = 4 x ( side ) 
      ∴  4 x ( side ) = 8 cm
                   side  =  ( 8/4 ) cm
                            =  2 cm
    ∴    Area of the Square = ( side )= ( 2 x 2 ) cm = 4 cm

3. Perimeter of the Square = 12 mm
    Also, Perimeter of the Square = 4 x ( side ) 
      ∴  4 x ( side ) = 12 mm
                   side  =  ( 12/4 ) mm
                            =  3 mm
    ∴    Area of the Square = ( side )= ( 3 x 3 ) mm = 9 mm

4. Perimeter of the Square = 16 dm
    Also, Perimeter of the Square = 4 x ( side ) 
      ∴  4 x ( side ) = 16 dm
                   side  =  ( 16/4 dm
                            =  4 dm
    ∴    Area of the Square = ( side )= ( 4 x 4 ) dm = 16 dm

5. Perimeter of the Square = 36 µm
    Also, Perimeter of the Square = 4 x ( side ) 
      ∴  4 x ( side ) = 36 µm
                   side  =  ( 36/4 ) µm
                            =  9 µm
    ∴    Area of the Square = ( side )= ( 9 x 9 ) µm = 81 µm

Monday 10 June 2013

REDOX REACTIONS

REDOX REACTIONS
Reactions which involve change in oxidation number of the interacting species.

OXIDATION
Addition of oxygen/electronegative element to a substance or removal of hydrogen /electropositive element from a substance.
                                                                       OR
Loss of electron(s) by any species
                                                                       OR
An increase in the oxidation no. of the element in the given substance.


OXIDISING AGENT
Acceptor of electron(s)
                                                                       OR
A reagent which can increase the oxidation number of an element in a given substance are called as oxidants.

REDUCTION
Removal of oxygen/electronegative element from a substance or addition of hydrogen/electropositive element to a substance.
                                                                       OR      
Gain of electron(s) by any species
                                                                       OR
An decrease in the oxidation no. of the element in the given substance.

REDUCING AGENT
Donor of electron(s)
                                                                       OR
Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that electron pair in a covalent bond belongs entirely to more electronegative elements.

8 Rules for calculation of the oxidation number

1.  In elements, in the free or the uncombined state, each atom bears an oxidation number of zero
  • Ex:.  O2  ,  CL2  O3  ,  P4  ,  S8  , Na , Mg , Al
2.  For ions composed of only one atom, the oxidation number is equal to the charge on the ion
  • Ex:.  oxidation number of Na+ is +1
  • Ex:.  oxidation number of Mg2+  is +2
3 All alkaline earth metals have an oxidation number of +2
4 Aluminium is regarded to have an oxidation number of +3 in all of its compounds.
5.  The oxidation number of oxygen in most of the compounds is -2. 
     In peroxides ( e.g., H2O2, Na2O) each oxygen atom is assigned an oxidation number of -1
     In super-oxides ( e.g., KO2, RbO2) each oxygen atom is assigned an oxidation number of - (1/2)
     When oxygen atom is bonded to fluorine ( e.g., OF2O2F) the oxygen is assigned to have a oxidation number of +2 and +1 respectively.
6.  The oxidation number of hydrogen is +1, except when it is bonded to metals in binary compounds. For example, in LiH, NaH, CaH2 its oxydation number is -1
7.  In all its compounds, fluorine has an oxidation number of -1
     Other halogens ( Cl, Br and I) also have an oxidation number of -1 as halide ions in their compounds
8.  The algebraic sum of the oxidation number of all the atoms in a compound must be zero.
     In poly-atomic ion, the algebraic sum of all the oxidation numbers of atoms of the ion must equal the charge on the ion.
     Thus, the sum of oxidation number of three oxygen atoms and one carbon atom in the carbonate ion, ( CO)2-  must equal -2.

Saturday 1 June 2013

CH 1 Reproduction In Organisms

1. Life Span
The period from birth to the natural death of an organism is called its life span.
Examples : 1. Butter Fly ( 1-2 weeks )  2. Tortoise ( 100-150 years )

2. Clone
The individual which are morphologically and genetically similar to each other and are exact copies of their parents are called clones. 
Examples : 1. Amoeba ( Produced by binary fission ), 2. Yeast ( Produced by budding )

3. Reproduction
The biological process in which in which an organism gives rise to young ones ( offspring ) similar to itself, is called reproduction.
→ it enables the continuity of the species, generation after generation.
→ factors affecting reproduction
     1. Organism Habitat
     2. Internal Physiology