In the following reaction (unbalanced), equivalent weight of `As_(2)S_(3)` is related to molecular weight `M` by
`As_(2)S_(3)+H+NO_(3)^(-) rarr NO+H_(2)O+AsO_(4)^(3-)+SO_(4)^(2-)`

In the following reaction (unbalanced) equivalent wt. of As_2S_3 is related to molecular wt. M By As_2S_3+H^(+)NO_3^(-)rarrNO+H_2O+AsO_4^(3-)+SO_4^(2-)

Equivalent weight of H_(3) PO_(2) is ( M rarr molecular weight)

Equivalent weight of H_(3) PO_(2) is ( M rarr molecular weight)

The equivalent weights of oxidising and reducing agents can be calculated by the number of electrons gained or lost. The equivalent weight of an oxidising agent is the number of parts by weight of the substance which gains one electron. Thus, it is equal to the molecular weight of the substance divided by the number of electrons gained in the balanced chemical equation. Similarly, equivalent weight of a reducing agent is equal to the molecular weight divided by the number of electrons lost as represented in the balanced chemical equation The equivalent weght of As_(2),S_3 in the following reaction As_(2)S_(3) + H^(+) + NO_(3)^(-) rarr NO + H_(2)O + AsO_(4)^(3-) + SO_(4)^(2-) is related to its molecular weight as

Balance the equation : As_(2)S_(3) + HNO_(3) rarr H_(2)AsO_(3) + H_(2)SO_(4) + NO

In the given reaction determine the equivalent weight of of As_(2)S_(3) : [Assume that M.W of As_(2)S_(3)=M ] As_(2)S_(3)+7CIO_(3)^(-)7OH^(-)to2AsO_(4)^(3-)+7CIO^(-)+3SO_(4)^(2-)+6H_(2)O

Suppose molecular weight of AS_(2)S_(3) is "M" and the equivalent weight of AS_(2)S_(3) in the reaction As_(2)S_(3)+H^(+)+NO_(3)^(-)rarr NO+H_(2)O+ASO_(4)^(-3)+SO_(4)^(-2) is (M)/((14times p)) .Then "p"=

The equivalent weight of hypo in the reaction [M is molecular weight] 2Na_(2)S_(2)O_(3)+I_(2)rarr2NaI+Na_(2)S_(4)O_(6) is

The equivalent weight of Hypo in the reaction [M = molecular weight] 2Na_(2)S_(2)O_(3)+I_(2)rarr2NaI+Na_(2)S_(4)O_(6) is