1.

Maximum percentage of available Cl_(2) as per formula CaOCl_(2). H_(2)O is x xx 7. Where the value of 'x' ?

Answer»


Solution :`underset( 145 G m)(CaOCl_(2).H_(2)O) RARR underset( 71 g m ) ( Cl_(2)) , ` Valu of x `xx 7 = 49.``:. x = ( 49)/( 7) = 7`


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