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34kurt
1 year ago
7

Carbon from the remains of an extinct Australian marsupial, called Diprotodon, has a specific activity of 0.61 pCi/g. Modern car

bon has a specific activity of 6.89 pCi/g. How long ago did the Diprotodon apparently become extinct?
Chemistry
1 answer:
Alika [10]1 year ago
5 0

The Diprotodon apparently become extinct <u>20,000 years</u> ago.

<h3>Briefly explained</h3>

This question is in reference to the use of carbon 14 dating. We know the specific activity of a Australian marsupial. Based on the carbon 14 present. It's .61 PICO curie program. Modern carbon has a specific activity of 6.89 PICO curie program. So all we need to do then is use the first order integrated rate law to solve for the time that has elapsed with Both of these activities as part of the equation.

But before we can do that we need the decay constant for carbon 14, which we can get by dividing the half life Of carbon 14 into natural log of two. And we get 1.21 times 10 to give 14, 1 over years. So using the first order integrated rate law, the natural log of the activity at time, T divided by the activity Um at time zero represented by present day Will be equal to negative K. multiplied by T.

Little Algebraic rearrangement gives us an age of 20,000 years.

Learn more about decay

brainly.com/question/1898040

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Answer:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

\Delta G=-RTln(Ksp)\\\\\Delta G=\Delta H-T\Delta S

Thus, by combining them, we obtain:

-RTln(Ksp)=\Delta H-T\Delta S\\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{T\Delta S}{RT} \\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{\Delta S}{R}

Which is related to the general line equation:

y=mx+b

Whereas:

y=ln(Ksp)\\\\m=-\frac{\Delta H}{R} \\\\x=\frac{1}{T} \\\\b=\frac{\Delta S}{R}

It means that we answer to the blanks as follows:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Regards!

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