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GenaCL600 [577]
3 years ago
9

A lab technician had a sample of radioactive Americium. He knew it was one of the isotopes listed in the table below but did not

know which one. The sample originally contained 0.008 g of the isotope, but 120 minutes later it contained 0.002 g.
a. How many half-lives have passed?
b. What is the half-life of the sample?
c. Which isotope did he have?

Chemistry
2 answers:
Zarrin [17]3 years ago
6 0
2 half lives. .008>.004>.002

the half life is 1 hour

he had Am-237
ladessa [460]3 years ago
3 0

Answer: a. 2

b. 1 hour

c. Am-237

Explanation:

Half life is the time taken by a reactant to reduce to its original concentration. It is designated by the symbol t_\frac{1}{2}.

Formula used :

a=\frac{a_o}{2^n}

where,

a = amount of reactant left after n-half lives = 0.002

a_o = Initial amount of the reactant = 0.008 g

n = number of half lives  = ?

Putting values in above equation, we get:

0.002=\frac{0.008}{2^n}

2^n=4\\\2^n=2^2

n=2

Thus two half lives have passed.

2. If two half lives have passed in 120 minutes

one half life will be passed in =\frac{120}{2}\times 1=60 minutes or 1 hour

3. As half life is characteristic of a particular isotope.

Given : the half life of isotope Am-237 is 1 hour or 60 minutes. thus the isotope was Am-237.

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Answer:

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From the given information:

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However, Since the amount of sulfur trioxide gas to be 1.6 mol.

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For O₂; we have 2.6 - x

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Recall that; the equilibrium constant for the reaction 2SO_2 + O_2 \to 2SO_3 = 0.8325;

If we want to find:

SO_2 + \dfrac{1}{2}O_2 \to SO_3

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K_c = (0.8325)^{1/2}

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Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.

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