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Romashka-Z-Leto [24]
3 years ago
13

The activity of a certain isotope dropped from 3200 ci to 800 ci in 24.0 years. what is the half-life of this isotope (in years)

? show your work.
Chemistry
1 answer:
Alexxx [7]3 years ago
3 0
Ln(800/3200) = - kt
t = 24 years.
ln(0.25) = -k*24
(- 1.3863) = -k*24
1.3863  / 24 = k
0.05776 = k

ln(0.5) = -k*t
-0.6931 = - 0.05776 t
12 = t

I don't know if you can just look at the question and know the answer. If 24 years is a quarter life then is it obvious that the 1/2 life is 12 years? It might be, but the method I've used works for sure. 


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What mass of carbon dioxide gas occupies a volume of 81.3L at 204 kPa and a temperature of 95.0°C?
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Answer:

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Explanation:

Given parameters:

Volume of gas  = 81.3L

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temperature of gas = 95°C

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Mathematically, it is expressed as;

                    PV = nRT

the unknown here is n which is the number of moles;

P is the pressure, V is the volume, R is the gas constant and T is the temperature.

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Convert temperature to Kelvin;   95 + 273  = 368K

       2 x 81.3 = n x 0.082 x 368

             n = \frac{2 x 81.3}{0.082 x 368}   = 5.38moles

Since the unknown is mass;

 Mass  = number of moles x molar mass

      Molar mass of carbon dioxide  = 12 + 2(16)  = 44g/mol

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