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3 years ago

The half-life of I-37 is 8.07 days. If 25 grams are left after 40.35 days, how many grams were in the original sample?

Chemistry

1 answer:

andreev551 [17]3 years ago

3 0

Answer:

800 g

Explanation:

We can express the decay of I-37 using the formula:

Final Mass = Initial Mass * $0.5^{\frac{Time}{Half-Life} }$

We input the data:

25 g = Initial Mass * $0.5^{(\frac{40.35}{8.07})}$

And solve for Initial Mass:

25 g = Initial Mass * $0.5^{5}$

25 g = Initial Mass * 0.03125

Initial Mass = 800 g

Meaning that out 800 grams of I-37, only 25 will remain after 40.35 days.

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A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

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Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

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Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

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Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

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$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

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