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mezya [45]
2 years ago
9

An artifact contains one-fourth as much carbon-14 as the atmosphere. how old is the artifact

Chemistry
2 answers:
Nikolay [14]2 years ago
4 0
<h3><u>Answer;</u></h3>

= 11,460 years

<h3><u>Explanation;</u></h3>
  • <em><u>The half life of Carbon-14 is 5,730 years . Half life is the time taken by a radioactive material to decay by half of its original mass.  Therefore, it  would take a time of 5730 years for a sample of 100 g of carbon-14 to decay to 50 grams</u></em>

<em>The initial amount of carbon-14 in this case was 1 whole; thus; </em>

<em>1 → 1/2 →1/4</em>

<em>To contain 1/4 of the value, 2 half-lives have passed. </em>

<em>But, 1 half life = 5,730 years</em>

<em>Therefore; The artifact is is therefore: 2 x 5,730 </em>

<em>          = 11,460 years </em>

Blizzard [7]2 years ago
3 0
The half life of Carbon-14 is 5,730 years
To contain 1/4 of the value, 2 half-lives must have passed.
The artifact is: 2 x 5,730
= 11,460 years older than the atmosphere.
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Use <em>Raoult’s Law</em> to calculate the vapour pressure:  

<em>p</em>₁ = χ₁<em>p</em>₁°  

where  

χ₁ = the mole fraction of the solvent  

<em>p</em>₁ and <em>p</em>₁° are the vapour pressures of the solution and of the pure solvent  

The formula for vapour pressure lowering Δ<em>p</em> is  

Δ<em>p</em> = <em>p</em>₁° - <em>p</em>₁  

Δ<em>p</em> = <em>p</em>₁° - χ₁<em>p</em>₁° = p₁°(1 – χ₁) = χ₂<em>p</em>₁°  

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<em>Step 1</em>. Calculate the <em>mole fraction of glucose </em>

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Δ<em>p</em> = χ₂<em>p</em>₁° = 0.018 62 × 23.8 torr = 0.4430 torr  

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