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Anit [1.1K]
3 years ago
12

A 3.5 gram sample of a radioactive element was formed in a 1960 explosion of an atomic bomb at Johnson Island in the Pacific tes

t site. The half-life of the radioactive element is 28 years. How much of the original sample will remain in the year 2030? Show your work.
Chemistry
1 answer:
Annette [7]3 years ago
5 0
The answer is 0.62g.
Solution:
From year 1960 to year 2030, it has been 
     2030-1960 = 70 years

The half-life of the radioactive element is 28 years, then the sample will go through
     70 years * (1 half-life/28 years) = 2.5 half-lives

Starting with a 3.5 gram sample, we will have
     3.5*(1/2) after one half-life passes
     3.5*(1/2) * (1/2) = 3.5*(1/4) after two half-lives pass
     3.5*(1/4) * (1/2) = 3.5*(1/8) after three half-lives pass and so on

Therefore, we can write the remaining amount of the sample after the number n of half-lives have passed as
     mass of sample = initial mass of sample/2^n

The mass of the remaining sample for n = 2.5half-lives can be now calculated as
     mass of sample = 3.5 grams / 2^2.5 = 0.62 g
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When 10.0 grams of sulfur reacts with fluorine gas at a pressure of 2.69 atmosphere in a 5.00 L container at 0.00 degrees Celsiu
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Answer:

74.1%

Explanation:

Based on the reaction:

S₈ + 16F₂ → 8SF₄

<em>1 mole of sulfur reacts with 16 moles of F₂ to produce 8 moles of SF₄</em>

<em />

To solve this question we must find the moles of each reactant in order to find the moles of SF₄. Thus, we can find the theoretical mass produced. Percent yield is:

Percent yield = Actual yield (25.0g) / Theoretical yield * 100

<em>Moles S₈: 256.52g/mol</em>

10.0g * (1mol / 256.52g) = 0.0390 moles

<em>Moles F₂:</em>

<em>PV = nRT</em>

PV/RT = n

<em>Where P is pressure in atm, V is volume in liters, R is gas constant and T is absolute temperature (0°C = 273.15K)</em>

2.69atm*5.00L / 0.082atmL/molK*273.15K = n

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For a complete reaction of 0.600 moles F₂ are required:

0.600mol F₂ * (1mol S₈ / 8 mol F₂) = 0.075 moles S₈

As there are just 0.0390 moles, S₈ is limiting reactant.

The theoretical moles and mass of SF₄ -Molar mass: 108.07g/mol- is:

0.0390 moles S₈ * (8mol SF₄ / 1mol S₈) = 0.312 moles SF₄ * (108.07g) =

33.7g

Percent yield = 25.0g / 33.7g * 100

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