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the type of isomer in which two different srructural form which are readily interconvertible and at dynamic equilibrium is known as tautomerism and the different forms are called tautomers.

 

hre the equilibrium between keto and enol ( alkene + alcohol )

KETO                        ENOL

NH2 - C - NH2  =  NH - - C - NH2

         //                          /

        O                         OH

        

 

               

the term active mass is defined as the molar concentration oer unit volume (number of moles per dm³ of a subatance

 

consider the example

 

in a reaction  vessel of one cubic metre volume, hydrogen gas reacts with gaseous iodine to form hydrogen iodide gas at 300K . at equilibrium 2.5 moles of  hydrogen 6.0 moles of iodine, 25 moles of HI were found to be present.

 

   the equation is                                    H_²        +       I₂     =      2HI

no of moles  at equilibrium                        2.5             6              25

 

since volume is 1 m³  , active mass           2.5/1          6/1             25/1 

MULLIKEN'S ELECTRONEGEVATIVITY SCALE CAN BE USED TO FIND IP (IONISATION POTENTIAL  OF A COMPOUND

 

ACCORDING TO MULLEKEN ELECTRONEGETIVITY(EN) CAN BE DEFINED ES THE AVERAGE OF IONISATION POTENTIAL  ( IE) AND ELECTRONIC AFFINITY (EA) OF AN ATOM

 

EN  = ( IP + EA )/ 56

 

IN THE ABOVE EQUATION IP AND EA ARE IN eV

 

EN = (IP+ EA )/2*62.5

 

IN THE ABOVE EQUATION IP AND EA ARE IN Kcal

 

 H = IP - EA

 

WHY DO YOU WANT THE IP 0F THE GIVEN COMPOUND

1]  A straight line through A(-2, -3) cut the lines x + 3y = 9

     and x + y + 1 = 0 at B and C respectively. Find the equation

     of the line if AB x AC = 20

 

2]  In the angle ABC, the equations of AB and AC are 2x + 3y = 29

     and x + 2y = 16. If the median through A cuts BC at (5, 6) find the

    equation of the side BC.  

ans is 1000% (a),because n-butane has two isomer , one anti and two gauche isomer . The two gauche isomer are mirror images of each other which are enantiomres .thusn-butane has a pair of enantiomer ,i would have explainede in a better manner but unable to attach the files , hope your query is solved

 

the compound contains,  1% KI therefore 2g of compound will contain 0.02g of KI

now molecular weight of KI = 39 + 127= 166

 

The % weight of I in KI  = (127/ 166 ) * 100 = x %

 

THERE FORE  weight of I in 0.02g of KI  = ( x/ 100) * 0.02

 

on calculating we get, weight of I = 0.0153g in 0.02 0f KI

 

NOW 1 MOLE OF I = 127g = 6.023 x 10^{23   ATOMS OR IONS}

 

therefore  0.153g of  I WILL CONTAIN = (6.023/127) * 10²³ 0.0153 IONS

 

THEREFOE WE GET  NO OF IONS = 7.2 X 10¹⁹

the balanced chemical reaction is

 

2C8H18  +15 O2= 16CO2 +18H2 O

 

molecular weight of C8H18 = 114G AND molecular weight of O2 =32 g

 

 therefore  228g of octane consume 480g of oxygen for complete combustion

 

therefore 1 g of octane will consume 480/228 g 0f oxygen

now the dendity of octane is 0.8 g/ ml

 

therefore  1 ml of octane wieghs 0.8g , so 1425ml (1.425 lts) will weigh =0.8 x 1425 g =1140

 

since 1g of octane will consume 480/228g

therefore 1140g of octane will consume  = 1140x480/228 g of oxygen

 

                                                           =2400

sine one mole of oxygen = 32g

 

therefore 2400g = 75 moles of oxygen

 

this the no of moles consumed