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USPshnik [31]
3 years ago
8

Why isn't carbon-14 dating accurate for estimating the age of materials older than 50,000 years?

Chemistry
1 answer:
soldi70 [24.7K]3 years ago
5 0
Carbon-14 is radioactive isotope of carbon.
Carbon is essential element of living cells. While the living cells are alive, the carbon contained in them are in equilibrium with the carbon in atmosphere. But, once the cell dies, the carbon-14 isotope undergoes radioactive decay. By measuring the carbon-14 in atmosphere to the carbon-14 in dead organism, we can calculate the time (or years) that organism have died.

However, carbon-14 dating technique is not accurate for estimating the age of materials older than 50,000 years old (above 40,000 years). This is because, 99% of carbon is carbon-12, 1% is carbon-13 and trace remaining is the carbon-14. This means, carbon-14 is found in very trace amount, in fact 1 part per trillion of carbon atoms present is carbon-14. The half of life of carbon-14 is 5,730 years. For dating the organism, we use the concept of half lives of the carbon-14 isotope in the dead organisms and calculate how many half life old the sample is. But as the years increases, the number of carbon-14 isotope becomes too low to detect and make accurate calculation.
This means, at some point the organism can simply run out of carbon-14.

Hence carbon-14 dating is not accurate for estimating age of materials older than 50,000 years old.

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Heat is transferred by conduction through a wall of thickness 0.87 m. What is the rate of heat transfer if the walls thermal con
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The given data is as follows.

         Thickness (dx) = 0.87 m,       thermal conductivity (k) = 13 W/m-K

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        T_{2} = 93^{o}C

According to Fourier's law,

                    Q = -kA \frac{dT}{dx}

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3 years ago
Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppo
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Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

CH_4+2O_2\rightarrow CO_2+2H_2O

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

left\ over=0g

Regards.

7 0
3 years ago
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