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horrorfan [7]
3 years ago
10

Americium-241 is used in smoke detectors. It has a first-order rate constant for radioactive decay of k = 1.6x10^−3yr−1. By cont

rast, iodine-125, which is used to test for thyroid functioning, has a rate constant for radioactive decay of k = 0.011 day−1.
(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 three half-lives?
(d) How much of a 1.00-mg sample of each isotope remains after 4 days?
Chemistry
1 answer:
Vika [28.1K]3 years ago
4 0

Answer:

A. 433 years and 63 days

B. Iodine

C. 0.125mg

D. 1.00mg

Explanation:

A. Americium

The formula of a radioactive decay constant half life is t = 0.693/k

Where k is the decay constant.

For americium, k = 0.0016

t = 0.693/0.0016 = 433.125 apprx 433 years

For iodine, k = 0.011

Half life = 0.693/0.011 = 63 days.

B. Iodine decays at a faster rate.

C. After three half lives

For both, first half life yields a mass of 0.5mg, next yields 0.25mg, next yield 0.125mg

D. 1.00mg still remains.

Half life is high for both so the decay after one day is insignificant.

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5 0
2 years ago
What is the total volume of gaseous products formed when 116 liters of butane (C4H10) react completely according to the followin
Contact [7]

<u>Answer:</u> The total volume of the gaseous products is 1044.29 L

<u>Explanation:</u>

We are given:

Volume of butane = 116 L

At STP:

22.4 L of volume is occupied by 1 mole of a gas

So, 116 L of volume will be occupied by = \frac{1}{22.4}\times 116=5.18mol of butane

The chemical equation for the combustion of butane follows:

2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

  • <u>For carbon dioxide:</u>

By Stoichiometry of the reaction:

2 moles of butane produces 8 moles of carbon dioxide

So, 5.18 moles of butane will produce = \frac{8}{2}\times 5.18=20.72mol of carbon dioxide

Volume of carbon dioxide at STP = (20.72 × 22.4) = 464.13 L

  • <u>For water vapor:</u>

By Stoichiometry of the reaction:

2 moles of butane produces 10 moles of water vapor

So, 5.18 moles of butane will produce = \frac{10}{2}\times 5.18=25.9mol of water vapor

Volume of water vapor at STP = (25.9 × 22.4) = 580.16 L

Total volume of the gaseous products = [464.13 + 580.16] = 1044.29 L

Hence, the total volume of the gaseous products is 1044.29 L

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(HELP) how can you make the metal chair less cold when you sit on it?
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Answer:

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3 years ago
Assuming 80 mg of vitamin c present 29.57 ml if oj how many milliliters of 0
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