Question

Situation:

Answer 1

$\\ \rm\Rrightarrow N=N_oe^{-kt}$

• We need half-life
• put N=N_o/2

$\\ \rm\Rrightarrow \dfrac{N_o}{2}=N_oe^{-kt}$

• Cancel N_o

$\\ \rm\Rrightarrow \dfrac{1}{2}=e^{-kt}$

$\\ \rm\Rrightarrow 2^{-1}=e^{-kt}$

• Cancel negative

$\\ \rm\Rrightarrow 2=e^{kt}$

• Apply natural log

$\\ \rm\Rrightarrow ln2=lne^{kt}$

$\\ \rm\Rrightarrow ktlne=ln2$

$\\ \rm\Rrightarrow kt=ln2$

$\\ \rm\Rrightarrow kt=0.693$

$\\ \rm\Rrightarrow T=\dfrac{0.693}{k}$

• Where T denotes half-life

Put k from the question

$\\ \rm\Rrightarrow T=\dfrac{0.693}{0.125}$

$\\ \rm\Rrightarrow T=8(0.693)$

$\\ \rm\Rrightarrow T=5.54days$

Note:-

• This equation is popularly known as Arrhenius equation
• The decay constant for first order reactions and half-life don't depend upon the initial concentration so 11g we haven't used

Answer 2

Answer:

5.5 days (nearest tenth)

Step-by-step explanation:

Given formula:

$\sf N=N_0e^{-kt}$

• $\sf N_0$ = initial mass (at time t=0)
• N = mass (at time t)
• k = a positive constant
• t = time (in days)

Given values:

• $\sf N_0$ = 11 g
• k = 0.125

Half-life:  The time required for a quantity to reduce to half of its initial value.

To find the time it takes (in days) for the substance to reduce to half of its initial value, substitute the given values into the formula and set N to half of the initial mass, then solve for t:

$\begin{aligned}\sf N & = \sf N_0e^{-kt}\\\\\implies \sf \dfrac{11}{2} & = \sf 11e^{-0.125t}\\\\\sf \dfrac{1}{2} & = \sf e^{-0.125t}\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf \ln e^{-0.125t\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t \ln e\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t(1)\\\sf t & = \sf \dfrac{\ln \left(\dfrac{1}{2}\right)}{-0.125}\\\\\sf t & = \sf 5.545177444...\\\\\sf t & = \sf 5.5\:\:days\:\:(nearest\:tenth)\end{aligned}$

Therefore, the substance's half-life is 5.5 days (nearest tenth).

Learn more about solving exponential equations here:

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