It's highly reactive and contains only one valence electron
Answer:
8. 2.75·10^-4 s^-1
9. No, too much of the carbon-14 would have decayed for radiation to be detected.
Explanation:
8. The half-life of 42 minutes is 2520 seconds, so you have ...
1/2 = e^(-λt) = e^(-(2520 s)λ)
ln(1/2) = -(2520 s)λ
-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1
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9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...
6.5·10^7/5.73·10^3 ≈ 11344
half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.
Answer:
The second trumpeter will be playing at frequency = 515 Hz
Explanation: Given that the note sounds lower and they can hear 20 beats in 4.0 s.
Beat frequency = 20/4 = 5 Hz
Beat frequency = F2 - F1
5 = 520 - F1
F1 = 520 - 5
F1 = 515 Hz
Since the note sound lower, the second trumpeter will be playing at 515 Hz frequency
Your answer is c. homologous structures
