Question #4bc86

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Question #4bc86

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Answer 1 · 2017-05-28T14:24:04

$C_{f} = 90.5 %$ $C_f=90.5%$

Explanation:

Given 1st order reaction => ?% of original sample remains after 1 hour?

Use the 1st order decay equation $C_{f} = C_{o} e^{- k t}$ $C_f = C_oe^(-kt)$
$C_{f}$ $C_f$ = Final Concentration = ?
$C_{o}$ $C_o$ = Initial Concentration = 100%
$t = t i m e$ $t=time$ = 1 hour
$k = \frac{0.693}{t_{\frac{1}{2}}}$ $k=0.693/t_(1/2)$ = $(0.693)(1/(10min))((60min)/(hr)) = 4.158hr^(-1)$
$C_f$ = $(100%)e^-[(4.158hr^-1)(1.00hr)]$ = $1.56%$ Remaining

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