An organic compound undergoes first decompoistion. The time taken for its decompoistion to `1//8` and `1//10` of its initial concentration are `t_(1//8)` and `t_(1//10)`, respectively. What is the value of `([t_(1//8)])/([t_(1//10)]) xx 10`? `(log_(10)2 = 0.3)`

An organic compound undergoes first order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are t_(1/8) and t_(1//10) respectively. What is the value of ([t_(1//8)])/([t_(1//10)])xx10 ? (log_(10)2=0.3)

An organic compound undergoes first-order decomposition. The time taken for decomposition to 1/8 and 1/10 of its initial concentration are t_(1//8) and t_(1//10) respectively. What is the value of ([t_(1//8)])/([t_(1//10)]) xx 10 ? (take log_10 2 = 0.3 )

The decomposition of a compound follows the rate law of a first order reaction. For the initial concentration of the compound to drop to (1)/(8) and (1)/(10) th of this value , times required are t_(1//8)th and t_(1//10) respectively. Find the value of (t_(1//8)/(t_(1//10))xx10). ["given":log"10^(2)=0.3]

For a first order reaction, t_(7//8) = n xx t_(1//2) . The value of n will be .......... .