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givi [52]
3 years ago
11

The half-life is the amount of time it takes for one-half of an isotope sample to decay into a different element. The half-life

of 238U is 4.5 billion years. Other radioactive isotopes decay in a much shorter time. By comparison, the half-life of 14C (carbon 14) is approximately 5,700 years. Radiometric dating has calculated the age of the earth at approximately 4.6 billion years. Using a circle represent the amount of an isotope before decay has begun
Chemistry
1 answer:
katovenus [111]3 years ago
5 0

Answer:

idk but this is what i know only

Explanation:

he half-life is the amount of time it takes for one-half of an isotope sample to decay into a different element. The half-life of 238U is 4.5 billion years. Other radioactive isotopes decay in a much shorter time. By comparison, the half-life of 14C (carbon 14) is approximately 5,700 years. Radiometric dating has calculated the age of the earth at 215 approximately 4.6 billion years. 1. Which circle in Figure 13-2 represents the amount of the original isotope before decay began?A 2. Assume that Figure 13-2 represents the half-life of 238U. Color in the area of circles B, C, and D that represents the amount of 238U remaining in the rock layer as each half-life passes. 3. If Figure 13-2 represented the half-life of 238U, which circle that you colored would represent a rock layer with the greatest concentration of lead? _A__ 4. Which circle that you colored shows the amount of the original isotope remaining after two half-life periods have expired? 5. IT Figure 13-2 represented the half-life of 14C, how many years would have passea to reach letter D? years 70 um ens ghe ting Fie ni boe nibs D rlt to got no oA teB Lgoc velFIGURE 13-2. Half-Life of a Radioactive Isotope C teob

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Answer:

0.800 mol

Explanation:

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Step 1. <em>Gather all the information</em> in one place.

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n/mol:  4.00    4.00

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

Step 2. Identify the <em>limiting reactant </em>

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<em>From C₃H₈:</em>

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

Step 3. Calculate the <em>moles of C₃H₈ consumed</em>.

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