`Cl_(2)+NaOHoverset(cold // dil) to A+NaCl+H_(2)O` ` ; ` ` Cl_(2)+NaOHoverset(hot//conc) to B+NaCl+H_(2)O`. Sum of oxidation numbers of chlorine in A and B is

Oxidation number of O in H_(2)O_(2) will be

Consider the following reaction: 6NaOH+3Cl _2 →5NaCl+A+3H_ 2O . What is the oxidation number of chlorine in A?

NH_(4)Cl+NaNO_(2) overset(Delta)toN_(2)uarr+NaCl+H_(2)O

NH_(4)Cl+NaNO_(2) overset(Delta)toN_(2)uarr+NaCl+H_(2)O

When Cl_(2) gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from

Cl_(2) overset(NaOH)(rarr)NaCl+NaClO_(3)+H_(2)O The equivalent mass of Cl_(2) in the above reaction is

Cl_(2)+NaoH to NaCl+NaOCl

Cl_(2)+NaoH to NaCl+NaOCl

Consider the following series of reactions : Cl_(2)+2NaOH to NaCl+NaClO+H_(2)O 3NaClO to 2NaCl+NaClO_(3) 4NaClO_(3) to 3NaClO_(4)+NaCl How many moles of NaCl will be formed by using 1 mole Cl_(2) and other reagents in excess ?

In [Cr(O_(2))(NH_(3))_(4)H_(2)O]Cl_(2) oxidation number of Cr is +3 then oxygen will be the form: