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JulsSmile [24]
3 years ago
12

Thorium-234 has a half life of about 25 days. What percent of the thorium will remain after 125 days

Chemistry
1 answer:
solong [7]3 years ago
4 0

Answer:

3.125% will remain after 125 days

Explanation:

Given data:

Half life of Th-234 = 25 days

Percent of thorium remain after 125 days = ?

Solution:

Number of half lives = T elapsed / half life

Number of half lives = 125 days / 25 days

Number of half lives = 5

At time zero =100%

At 1st half life = 100%/2 = 50%

At second half life = 50%/2 = 25%

At third half life = 25%/ 2 = 12.5%

At 4th half life = 12.5% /2 = 6.25%

At 5th half life = 6.25% /2 = 3.125%

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Read 2 more answers
Calculate ∆G ◦ r for the decomposition of mercury(II) oxide 2 HgO(s) → 2 Hg(ℓ) + O2(g) ∆H◦ f −90.83 − − (kJ · mol−1 ) ∆S ◦ m 70.
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Answer:

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Explanation:

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ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

ΔHr: - 2ΔHf HgO(s) = <u><em>181,66 kJ/mol</em></u>

<u><em /></u>

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

ΔSr= 216,8 J/Kmol = <em><u>0,216 kJ/Kmol</u></em>

Thus, ΔGr at 298K is:

ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ <em>4. +117,1 kJ/mol</em>

<em></em>

I hope it helps!

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