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4 years ago

If it take 3.4 years for a radioactive isotope to decay to 1/16 of its initial value, what is the value of k for the isotope?

1 answer:

4 0

In this question, k stands for the decay constant. The half-life and decay constant of a radioactive isotope are related by the following equation:

Half Life = ln(2) / Decay Constant
ln(2) is the natural log of 2.

We are provided that in 3.4 years the isotope decays to 1/16 of its initial value.
$\frac{1}{16}= \frac{1}{ 2^{4} }$
This means, in 3.4 years 4 half lifes are passed.

So, 4 Half lifes = 3.4 years
Half Life = 3.4/4 = 0.85 years.

Now we can find k by plugging in values in above equation:

$k= \frac{ln(2)}{0.85}= 0.815$

So the value of k for the isotope is 0.815 per year.

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4 years ago

What volume did a helium-filled balloon have at 18.1 °C and 2.61 atm if its new volume was 59.9 mL at 1.92 atm and 12.5°C?

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a. V=44.9mL

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For a gas undergoing all of these changes, it will be important to combine Boyle's Law and Charles' Law to form the following equation (if it isn't already known):

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

Where the Temperatures must be measured in Kelvin.

Recall that to convert Celsius to Kelvin, one must add 273 or use the equation $T_C+273=T_K$ .

Thus, $T_1=(18.1+273)[K]=291.1[K]$ and $T_2=(12.5+273)[K]=285.5[K]$

To solve for the requested quantity, note that all of the other units match between beginning and end, so we substitute and solve:

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

$\dfrac{(2.61[atm])V_1}{(291.1[K])}=\dfrac{(1.92[atm])(59.9[mL])}{(285.5[K])}$

$\dfrac{(2.61[atm] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----})\bold{V_1}}{291.1[K] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}*\dfrac{291.1[K] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}{2.61[atm]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}=\dfrac{(1.92[atm]\!\!\!\!\!\!\!\!\!\!\!{--})(59.9[mL])}{285.5[K\!\!\!\!\!{-}]}*\dfrac{291.1[K\!\!\!\!\!{-}]}{2.61[atm]\!\!\!\!\!\!\!\!\!\!\!{--}}$

$V_1=44.928677576[mL]$

Accounting for significant digits, $V_1=44.9[mL]$

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