The answer is A because it gets caught in clouds then is let down in rain
The scientific method is similar to superstition, guessing, intuition, or hope because it is a form of finding the answer. It is different because it is a more effective and efficient way of finding an answer to a question.
The electron bit would be 2 and you could colour in 4 boxes:) I had the same question as this and I got it correct
Answer:
a)
fraction of carbon−14 in a piece of charcoal remains after 14.0 years.
b)
fraction of carbon−14 in a piece of charcoal remains after ![1.900\times 10^4 years](https://tex.z-dn.net/?f=1.900%5Ctimes%2010%5E4%20years)
c)
fraction of carbon−14 in a piece of charcoal remains after
.
Explanation:
The fraction of a radioactive isotope remaining at time t is given by:
![[A]=\frac{(\frac{1}{2})^t}{t_{\frac{1}{2}}}](https://tex.z-dn.net/?f=%5BA%5D%3D%5Cfrac%7B%28%5Cfrac%7B1%7D%7B2%7D%29%5Et%7D%7Bt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D)
Taking log both sides:
![\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}]](https://tex.z-dn.net/?f=%5Clog%20%5BA%5D%3Dt%5Clog%5B%5Cfrac%7B1%7D%7B2%7D%5D-%5Clog%20%5Bt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%5D)
[A] = fraction at given time t
= half life of the carbon−14 =5,730 years
a)When , t = 14 years
![\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}]](https://tex.z-dn.net/?f=%5Clog%20%5BA%5D%3Dt%5Clog%5B%5Cfrac%7B1%7D%7B2%7D%5D-%5Clog%20%5Bt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%5D)
![\log [A]= 14 years\times (-3010)-\log [5,730 years]](https://tex.z-dn.net/?f=%5Clog%20%5BA%5D%3D%2014%20years%5Ctimes%20%28-3010%29-%5Clog%20%5B5%2C730%20years%5D)
![[A]=1.065\times 10^{-8}](https://tex.z-dn.net/?f=%5BA%5D%3D1.065%5Ctimes%2010%5E%7B-8%7D%20)
b)When , t = ![1.900\times 10^4 years](https://tex.z-dn.net/?f=1.900%5Ctimes%2010%5E4%20years)
![\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}]](https://tex.z-dn.net/?f=%5Clog%20%5BA%5D%3Dt%5Clog%5B%5Cfrac%7B1%7D%7B2%7D%5D-%5Clog%20%5Bt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%5D)
![\log [A]= 1.900\times 10^4 years\times (-3010)-\log [5,730 years]](https://tex.z-dn.net/?f=%5Clog%20%5BA%5D%3D%201.900%5Ctimes%2010%5E4%20years%5Ctimes%20%28-3010%29-%5Clog%20%5B5%2C730%20years%5D)
![[A]=0.000\times 10^{-3} [/tex](https://tex.z-dn.net/?f=%5BA%5D%3D0.000%5Ctimes%2010%5E%7B-3%7D%20%5B%2Ftex%20)
c)When , t = ![1.0000\times 10^5 years](https://tex.z-dn.net/?f=1.0000%5Ctimes%2010%5E5%20years)
![\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}]](https://tex.z-dn.net/?f=%5Clog%20%5BA%5D%3Dt%5Clog%5B%5Cfrac%7B1%7D%7B2%7D%5D-%5Clog%20%5Bt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%5D)
![\log [A]= 1.0000\times 10^5 years\times (-3010)-\log [5,730 years]](https://tex.z-dn.net/?f=%5Clog%20%5BA%5D%3D%201.0000%5Ctimes%2010%5E5%20years%5Ctimes%20%28-3010%29-%5Clog%20%5B5%2C730%20years%5D)
![[A]=0.0000\times 10^{-4}](https://tex.z-dn.net/?f=%5BA%5D%3D0.0000%5Ctimes%2010%5E%7B-4%7D%20)