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 
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 = 
![\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 = 
![\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)
In the systemic arteries, the partial pressure of carbon dioxide is 40 mm Hg. Partial pressure of a gas is the contribution of one gas to the total pressure exerted by all gases. Partial pressure of carbon dioxide in arterial blood is the portion of total blood gas pressure that is exerted by carbon dioxide. It decreases during heavy exercise, during rapid breathing, or in association with severe diarrhea, uncontrolled diabetes or the diseases of the kidney. It increases with chest injuries and respiratory disorders. In the systemic arteries, the partial pressure of oxygen is 100 mm Hg.
The element is TELLURIUM. Tellurium has atomic number 52 with the electronic configuration of [Kr] 4d10 5s2 5p4. The element belongs to group 16 elements on the periodic table. It exhibits various oxidation states: +2, +4, +6, and -2. +4 is its most common oxidation state.
Answer:
boiling point and melting point.