Answer : The amount of carbon-14 remaining will be, 2.5 grams.
Explanation :
Half-life = 5730 years
First we have to calculate the rate constant, we use the formula :
![k=\frac{0.693}{t_{1/2}}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B0.693%7D%7Bt_%7B1%2F2%7D%7D)
![k=\frac{0.693}{5730\text{ years}}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B0.693%7D%7B5730%5Ctext%7B%20years%7D%7D)
![k=1.21\times 10^{-4}\text{ years}^{-1}](https://tex.z-dn.net/?f=k%3D1.21%5Ctimes%2010%5E%7B-4%7D%5Ctext%7B%20years%7D%5E%7B-1%7D)
Now we have to calculate the time passed.
Expression for rate law for first order kinetics is given by:
![t=\frac{2.303}{k}\log\frac{a}{a-x}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B2.303%7D%7Bk%7D%5Clog%5Cfrac%7Ba%7D%7Ba-x%7D)
where,
k = rate constant = ![1.21\times 10^{-4}\text{ years}^{-1}](https://tex.z-dn.net/?f=1.21%5Ctimes%2010%5E%7B-4%7D%5Ctext%7B%20years%7D%5E%7B-1%7D)
t = time passed by the sample = 11460 years
a = initial amount of the reactant = 10 g
a - x = amount left after decay process = ?
Now put all the given values in above equation, we get
![11460=\frac{2.303}{1.21\times 10^{-4}}\log\frac{10}{a-x}](https://tex.z-dn.net/?f=11460%3D%5Cfrac%7B2.303%7D%7B1.21%5Ctimes%2010%5E%7B-4%7D%7D%5Clog%5Cfrac%7B10%7D%7Ba-x%7D)
![a-x=2.5g](https://tex.z-dn.net/?f=a-x%3D2.5g)
Therefore, the amount of carbon-14 remaining will be, 2.5 grams.