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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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A sample of water is heated from 60.0 °C to 75.0°C by the addition of 140 j of
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<h3>Furter explanation</h3>

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Answer:

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Explanation:

                      Amount of energy required by known amount of a substance to raise its temperature by one degree is called specific heat capacity.

The equation used for this problem is as follow,

                                                 Q  =  m Cp ΔT   ----- (1)

Where;

           Q  =  Heat  =  640 J

           m  =  mass  =  125 g

           Cp  =  Specific Heat Capacity  =  <u>??</u>

           ΔT  =  Change in Temperature  =  43.6 °C  -  22 °C  =  21.6 °C

Solving eq. 1 for Cp,

                                Cp  =  Q / m ΔT

Putting values,

                                Cp  =  640 J / (125 g × 21.6 °C)

                                Cp  =  0.237 J.g⁻¹.°C⁻¹

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