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BartSMP [9]
3 years ago
9

The technetium-99 isotope has a half-life of 6.0 hours. If 100.0 mg were injected into a

Chemistry
1 answer:
BartSMP [9]3 years ago
8 0

Answer:

12.5mg of technetium-99 remain

Explanation:

When a radioactive isotope decays in the time, its concentration follows the equation:

ln[A] = -kt + ln[A]₀

<em>Where [A] could be taken as the mass of radioactive isotope after time t (18h), </em>

<em>k is rate constant = ln 2 / Half-Life = ln 2 / 6hours = 0.1155hours⁻¹ </em>

<em>[A]₀ is initial amount of the isotope = 100.0mg</em>

<em />

Replacing:

ln[A] = -0.1155hours⁻¹ *18hours+ ln[100.0mg]

ln[A] = 2.5257

[A] =

<h3>12.5mg of technetium-99 remain</h3>

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The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.500 atm? (∆Hvap = 28.5
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