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{:(,"Column-I",,"Column-II",),((A),"Law ...

`{:(,"Column-I",,"Column-II",),((A),"Law of conservation of mass",(p),CH_(4) "has carbon and hydrogen in" 3:1 "mass raio",),((B),"Law of multiple proportion",(q),10mL N_(2) "combines with" 30mL "of" H_(2) "to form" 20mL "of" NH_(3),),((C),"Law of definite proportion",(r),"S and" O_(2) "combine to form" SO_(2) "and" SO_(3),),((D),"Law of reciprocal proportion",(s),"In" H_(2)S "and" SO_(2) "mass ratio of H and O w.r.t. sulphure is" 1:16 "hence in" H_(2)O "mass ratio of H and O is" 1:8.,),((E),"Gay Lussac's Law",(t),4.2 g MgCO_(3) "gives" 2.0 g "residue on heating",):}`

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Nitrogen and oxygen combine to form N_(2)O , NO and NO_(2) . This is in accordance with law of reciprocal proportions.

The compounds H_2O and D_2O follows Law of multiple proportions.

Knowledge Check

  • If 30mL of H_(2) and 20mL of O_(2) , reacts to form H_(2)O , what is left at the end of the reaction?

    A
    10mL of `H_(2)`
    B
    5mL of `H_(2)`
    C
    10mL of `O_(2)`
    D
    5mL of `O_(2)`

  • If 30mL of H_(2) and 20mL of O_(2) react to form form water, what is left at the end of the reaction?

    A
    10mL of `H_(2)`
    B
    5 mL of `H_(2)`
    C
    10mL of `O_(2)`
    D
    5 mL of `O_(2)`

  • In the reaction, 2H_(2)S+SO_(2) to 3S+2H_(2)O, H_(2)S is

    A
    Reducing agent
    B
    oxidizing agent
    C
    precipitating agent
    D
    an acid

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    Laws Of Chemical Combination|Law Of Conservation Of Mass|Law Of Definite Proportion|Law Of Multiple Proportion|OMR

    {:(,"Column I",,"Column II",),((A),"In 1 M aqueous NaoH, weight % of NaOH is " (d_("solution") = 2 gm//ml),(P),20,),((B),"Molarity of '22.4 V' " H_(2)O_(2) " solution" ,(Q),10,),((C ),"Molality of 20 ppm aqueous NaOH solution",(R ),3,),((D),"Mass % of " SO_(3) "in 102.25 % oleum sample",(S),2,),(,,(T),5 xx 10^(-4),):}

    Match the term given in Column I withthe equation given in Column II {:("Column I " , "Column II") , ("(A) Enthalpy of formation", (p) CuSO_(4) (s) + 5 H_(2) O (l) to CuSO_(4) 5H_(2)O), ("(B) Enthalpy of combustion" , (q) CuSO_(4) (s) + nH_(2) O (l) to CuSO_(4) (aq)),("(C) Enthalpy of solution " , (r) C(s) + O_(2) (g) to CO_(2) (g)) , ("(D) Enthalpy of hydration" , (s) CH_(4) + 2 O_(2) to CO_(2) + 2 H_(2)O):}

    10 mL of H_(2) combine with 5 mL of O_(2) to form H_(2)O when 200 mL of H_(2) at S.T.P. is passed through heated CuO, latter loses 0.144 g of its weight. Does the above data correspond to the law of constant composition ?

    1 mol of SO_(2) and 1 mol of H_(2)S react completely to form H_(2)O and S as follows: SO_(2)+2H_(2)S to 2H_(2)O+3S (At. mass S = 32, O = 16) The mass of S obtained is:

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