Question 1 ( point) A 64g sample of Germanium-66 is left undisturbed for 12.5 hours. At the end of that period, only 2.0g remain. How many half-lives transpired during this time period? half-lives Answer = Blank 1:​

Answers

Answer 1

Answer:

2.5 half-lives

Step-by-step explanation:

Find [tex]t_{h}[/tex] in [tex]N(t)=N_{0}e^{-kt}[/tex] given [tex]N(t) = 2, N_{0} =64,t=\frac{25}{2}[/tex]

[tex]N(t)[/tex] is the amount after the time [tex]t[/tex], [tex]N_0[/tex] `is the initial amount, [tex]t_{h}[/tex] is the half-life

We know that after half-life there will be twice less the initial quantity:

[tex]N(t_{h})=\frac{N_{0} }{2}=N_{0}e^{kt_{h} }[/tex].

Simplifying gives [tex]\frac{1}{2} =e^{-kt_h} or k = -\frac{In(\frac{1}{2}) }{t_{h} } .[/tex]

Plugging this into the initial equation, we obtain that [tex]N(t)=N_{0}e^\frac{In(\frac{1}{2}) }{t_{h} }[/tex]or [tex]N(t)= N_{0} (\frac{1}{2})^\frac{t}{t_h}.[/tex]

Finally, just plug in the given values and find the unknown one.

From [tex]2=64(\frac{1}{2})^\frac{\frac{25}{2} }{t_h}[/tex], we have that [tex]t_h[/tex] = [tex]\frac{25 In (2)}{2 In (32)\\}[/tex]

So you should get: [tex]t_{h} =\frac{25 In (2)}{2 In (32)\\}=2.5.[/tex]

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Answer 2

The number of half-lives that transpired during this time period is equal to 2.5 hours.

Given the following data:

Time = 12.5 hours.

Initial mass = 64 grams.

Remaining mass = 2.0 grams.

How to determine the number of half-lives.

Mathematically, the amount of a radiocative element that is remaining after time (t-hours) is given by this formula:

[tex]ln(\frac{A}{A_0} )=ln(\frac{1}{2} )\frac{t}{H}[/tex]

Where:

A is the amount remaining.Ao is the initial amount.H is the half-life.t is the time.

Substituting the given parameters into the formula, we have;

ln(2/64) = ln(1/2) × 12.5/H

-3.466 = -0.693 × 12.5/H

H = 12.5 × -0.693/-3.466

H = 12.5 × 0.1999

H = 2.5 hours.

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Given

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Step-by-step explanation:

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The speed of current is [tex]\rm 2 \:km/h[/tex].

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Calculations:

Given:

[tex]\begin{aligned} \rm Distance\: traveled\: on\: motorboat &= 25\:km\\\rm Speed\:of\:motorboat\:in\:still\:water &= 12\:km/h \end[/tex]  

 

Let the speed of the current be [tex]\rm x[/tex]

 

The speed of the boat traveling against the current is [tex]{(12 -\rm x) km/h}[/tex]

 

Time spent on motorboat is 10 hours less than time spent on raft.

Therefore:

[tex]\rm \dfrac{25km}{x\:km/h} - \dfrac{25km}{(12 - x)km/h} = 10[/tex]


The equation obtained will be:

[tex]\dfrac{25}{\rm x} - \dfrac{25}{(12-\rm x)} - = 10[/tex]

 

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==========================================================

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We could stop here if you wanted.

----------------------

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-----------------------

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