During an electrolysis of conc. `H_(2)SO_(4)`, peroxydisulphuric acid `(H_(2)S_(2)O_(8))` and `O_(2)` form in an equimolar amount. The moles of `H_(2)` that will be formed simultaneously will be

During an electrolysis of conc H_(2)SO_(4) , perdisulphuric acid (H_(2)S_(2)_(8)) and O_(2) are formed in equimolar amount. The moles of H_(2) that will be formed simultaneously will be a. Thrice that of O_(2)" "b. Twice that of O_(2) c. Equal to that of O_(2)." "d. Half of that of O_(2)

During an electrolysis of conc H_(2)SO_(4) , perdisulphuric acid (H_(2)S_(2)_(8)) and O_(2) are formed in equimolar amount. The moles of H_(2) that will be formed simultaneously will be a. Thrice that of O_(2)" "b. Twice that of O_(2) c. Equal to that of O_(2)." "d. Half of that of O_(2)

During the electrolysis of conc H_(2)SO_(4) , it was found that H_(2)S_(2)O_(8) and O_(2) liberated in a molar ratio of 3:1 . How many moles of H_(2) were found of moles of H_(2)S_(2)O_(8) ? ( Express your answer as :3xx mol e s of H_(2), integer answer is between 0 and 5 0

During the electrolysis of conc H_(2)SO_(4) , it was found that H_(2)S_(2)O_(8) and O_(2) liberated in a molar ratio of 3:1 . How many moles of H_(2) were found of moles of H_(2)S_(2)O_(8) ? ( Express your answer as :3xx mol e s of H_(2), integer answer is between 0 and 5 0

H_(2)S burns in O_(2) to form

H_(2)S burns in O_(2) to form

During electrolysis of H_(2)O , the molar ratio of H_(2) and O_(2) formed is