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cricket20 [7]
3 years ago
11

americium-241 has a half-life of 430 years. how much of a 10.0g sample of americium-241 remains after 1720 years?

Chemistry
1 answer:
Arlecino [84]3 years ago
6 0

Answer:

(a) What are the half-lives of these two isotopes? (b) Which one decays at a faster rate? (c) How much of a 1.00-mg sample of each isotope remains after 3 half- ...

Explanation:

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In a particular mass of kau(cn)2, there are 6.66 × 1020 atoms of gold. What is the total number of atoms in this sample?
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Total number of atoms in the sample is the sum of number of atoms of all the elements present in the sample.

Number of gold, Au atoms in KAu(CN)_2 = 6.66\times 10^{20} atoms     (given)

From the formula of compound that is KAu(CN)_2 it is clear that the number of potassium and gold are same whereas those of carbon and nitrogen are 2 times of them.

So, the number of atoms of each element is:

Number of potassium, K atoms in KAu(CN)_2 = 6.66\times 10^{20} atoms    

Number of carbon, C atoms in KAu(CN)_2 = 2\times6.66\times 10^{20} atoms = 13.32\times 10^{20}

Number of nitrogen, N atoms in KAu(CN)_2 = 2\times6.66\times 10^{20} atoms = 13.32\times 10^{20}

Total number of atoms in KAu(CN)_2 = Number of gold, Au atoms+Number of potassium, K atoms +Number of carbon, C atoms + Number of nitrogen, N atoms

Total number of atoms in KAu(CN)_2 = 6.66\times 10^{20}+6.66\times 10^{20}+13.32\times 10^{20}+13.32\times 10^{20}

Total number of atoms in KAu(CN)_2 = 39.96\times 10^{20} atoms

Hence, the total number of atoms in KAu(CN)_2 is 3.996\times 10^{21} atoms.

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