   # Your Turn Half-Life Radon-222 decays via alpha emission to Po-218 with a half-life of 3.82 days. If a house initially contains 512 mg of radon 222, and no new radon enters the house, how much will be left in 19.1 days? How many alpha emissions would have occurred within the house? ### Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692

#### Solutions

Chapter
Section ### Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692
Chapter 8, Problem 8.3YT
Textbook Problem
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## Your TurnHalf-LifeRadon-222 decays via alpha emission to Po-218 with a half-life of 3.82 days. If a house initially contains 512 mg of radon 222, and no new radon enters the house, how much will be left in 19.1 days? How many alpha emissions would have occurred within the house?

Interpretation Introduction

Interpretation:

The amount of Rn222 which is left after 19.1 days, and the number of alpha particles which are emitted within the house during the given period are to be calculated.

Concept Introduction:

Half-life time of a radioactive substance is the time in which the parent nuclei is reduced to half of its initial amount.

Alpha emission occurs when an unstable radioactive nucleus emits a particle containing 2 protons and 2 neutrons.

One mole contains 6.022×1023 atoms. It can be expressed as:

1 mol=6.022×1023 atom

The conversion factor is (6.022×10231mol).

One mole of any substance has a mass equal to its molar mass.

One milligram is equal to 103 g. The conversion factor for grams from milligrams is (10-3 g1 mg).

### Explanation of Solution

Half-life time 3.82 days is the time in which Radon-222 is reduced to half of its initial amount. Therefore, the amount that remains after 3.82 days is 256 mg. It can be expressed as:

512 mg2=256 mg

Now, the available amount is 256 mg which, after 3.82 days, is reduced to half. It can be expressed as:

256 mg2=128 mg

Now, the available amount is 128 mg which, after 3.82 days, is reduced to half. It can be expressed as:

128 mg2=64 mg

Now, the available amount is 64 mg which, after 3.82 days, is reduced to half. It can be expressed as:

64 mg2=32 mg

Now, the available amount is 32 mg which, after 3.82 days, is reduced to half. It can be expressed as:

32 mg2=16 mg

Hence, 16.0 mg of the given sample is left after 19.1 days.

Hence, the total amount of the sample of Rn222 that has decayed is calculated as:

512 mg16

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