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abruzzese [7]
3 years ago
12

The half-life of potassium-40 is 1.3 billion years. If a rock that formed 2.6 billion years ago originally contained 5 grams of

potassium-40, about ____ grams of potassium-40 would be left in that rock today
Chemistry
1 answer:
pochemuha3 years ago
8 0

Answer: 1.25 grams of Potassium-40

Explanation: When the amount of a compound is consumed by half, the time passed is called "the half life". This term is particularly important in nuclear chemistry. Potassium-40 is an isotope of potassium element and it is radioactive. As a result, the consumption time of this radioactive isotope is valuable information since radioactive isotopes are detrimental to health and nature.

For example, in this question, within 1.3 billion years, half amount of the potassium-40 disappears. 2.6 billion years ago, there were 5 grams of potassium-40 and when 1.3 billion years pass, half of the 5 grams of potassium-40 will disappear and there will will be 2.5 grams of potassium-40 left (5 grams of potassium-5 grams of potassium*(1/2)[half amount]=2.5 grams of potassium-40 left). As a result, 2.5 grams of potassium-40 is the amount that is present 1.3 billion years ago (2.6 billion years-1.3 billion years [half-life]=1.3 billion years). There will be 1.3 billion years left until today, so we still have to calculate the amount disappeared in 1.3 billion years. Thus, half of the 2.5 grams must disappear in 1.3 billion years left (2.5 grams - 2.5 grams*(1/2)[half amount]=1.25 grams). Finally, we have 1.25 grams of potassium-40 left since 2.6 billion years ago.

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Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:

d=\frac{m}{V}

Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:

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So that we are now able to calculate the density in g/mL first:

d=\frac{87.7g}{27.8mL}=3.15g/mL

Now, we proceed to the conversion to lb/in³ by using the following setup:

d=3.15\frac{g}{mL}*\frac{1lb}{453.6g}*\frac{1in^3}{16.3871mL}\\\\d=4.24x10^{-4}\frac{lb}{in^3}

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