If a I-order reaction is completed to the extent of `60%` and `20%` in time intevals, `t_(1)` and `t_(2)`, what is the ratio, `t_(1) : t_(2)`?

Two first order reaction have half-life in the ratio 3:2 . Calculate the ratio of time intervals t_(1):t_(2) . The time t_(1) and t_(2) are the time period for 25% and 75% completion for the first and second reaction respectively:

Two first order reactions have half-lives in the ratio 8:1, calculate the ratio of time intervals t_(1):t_(2) . The time t_(1) and t_(2) are the time period foe ((1)/(4))^(th) and ((3)/(4))^(th) completion respectively:

Two I-order reaction have half-lives in the ratio 3:2 .Calculate the ratio of time intervals t_(1),t_(2) . Where t_(1) is the time period for 25% completion of the first reaction and t_(2) , for 75% completion of the second reaction.

Two 1^(st) order reactions have half - lives in the ration 3:2 . Then the ratio of time intervals t_(1):t_(2) , will be gt Where t_(1) is the time period for 25% completion completion of the first reaction and t_(2) is time required for 75% completion of the second reaction . [log 2=0.3 , log 3=0.477]

A first order reaction is completed by 75% in time t, completed by 99.9% in time t'. Ratio (t')/(t) will be equal to

20% first order reaction is completed in "50" minutes.Time required for the completion of 60% of the reaction is

20% first order reaction is completed in 50 minute. Time required for the completion of 60% of the reaction is

Find the ratio of t_(75%) and t_(50%) of first order reaction?

Show that t_(99%) = 2 xx t_(90%)

Two first order freaction have half-lives in the ratio 8:1 Calculate the ratio of time intervals t_(1):t_(2) and t_(1) and t_(2) are the time period for ((1)/(4))^(th) and ((3)/(4))^(th) completion.