When the equation `Sn^(2+)` (aq) `+IO_(3)^(-)(aq)+H^(+)(aq)rarrSn^(4+)(aq)+I_(2)(aq)+H_(2)O(l)` is balanced, what is the `Sn^(2+)(aq)//IO_(3)^(-)(aq)` mole ratio?

2Hg^(2)(aq)+Sn^(2)+(aq)toHg_(2)^(2+)(aq)+Sn^(4+)(aq)

__ Sn^(2+) (aq)+__ MnO_(4)^(-) (aq)+__ H^(+) (aq) rarr __ Sn^(4+) (aq) +__ MnO^(2+) (aq) __ H_(2)O(l) When this equation for the reaction of Sn^(2+) (aq) and MnO_(4)^(-) (aq) is balanced correctly, what is the ratio, Sn^(2+)//MnO_(4)^(-) ?

Complete the chemical equations : MnO_4^(-) (aq) + H_2O_2(aq) + H^(+)(aq) to

For the balanced equation : 8H^(+)(aq)+5Fe^(2+)(aq)+MnO_(4)^(-)(aq)rarr5Fe^(3+)(aq)+Mn^(2+)(aq)+4H_(2)O(l) Which statement is correct ?

Complete the chemical equations : MnO_4^(-) (aq) + S_2O_3^(2-)(aq) + H_2O (l) to

The reaction H_(2)O_(2)(aq)+3I^(-)(aq)+2H^(+)(aq)rarrI_(3)^(-)(aq)+2H_(2)O(l) has a rate law : rate = k[H_(2)O_(2)][I^(-)] . What is the order of the reaction with respect to H^(+) , and what is the overall order of the reaction?

Iodate ions (IO_(3)^(-)) can be reduced to iodine by iodide ions. The half equations which represent the redox reaction are: IO_(3)^(-)(aq.)+6H^(+)(aq.)+5e^(-) to (1)/(2)I_(2)(s)+3H_(2)O(l) . . . (i) I^(-)(aq.) to (1)/(2)I_(2)(s)+e^(-) . . . (ii) How many moles of iodine are produced for every mole of iodate ions consumed in the reaction?