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Find the two-thirds life (t(2//3)) of a ...

Find the two-thirds life `(t_(2//3))` of a first order reaction in which `k=5.48xx10^(-1) "sec"^(-1) ` ( log 3 = 0.4771 , log 2 = 0.3010)

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The correct Answer is:
2.0005 s
`(t_(2//3))=(2.303)/(k)log_(10).(a)/(a-(2)/(3)a)`
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