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125 2.303 (3.465 $465\times 10^{1}$ min ) J08 1205 0 $\left(log\left(1$ $-2log^{2}=2\times 0.3010=0.6an0$ $2.303$ $\times 0.6020=4$ min. $\left(3.46\bar{5\times 10-1} $ min-) (b) Now time required to reach $10%$ of initial concentration of the reaction: $α=1006$ $\left(a-x\right)=10%$ $2.303$ Therefore $t10=\dfrac {2.303} {k}log\dfrac {a} {a-x}=$ $-$ $-$ $\left(3.465\times 10-1$ min ') $log\dfrac {100} {10}$ $\left(3.465\times 10-1$ min-1). $-\times 1=6.65$ min. $2.303$ $99.9%$ T : For first order reaction. t%3D $2.303$ $log _{\dfrac {a} {\left(a2}}$
10th-13th grade
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Qanda teacher - Raghav sir
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