Answer:
The age of the sample is 4224 years.
Explanation:
Let the age of the sample be t years old.
Initial mass percentage of carbon-14 in an artifact = 100%
Initial mass of carbon-14 in an artifact = ![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
Final mass percentage of carbon-14 in an artifact t years = 60%
Final mass of carbon-14 in an artifact = ![[A]=0.06[A_o]](https://tex.z-dn.net/?f=%5BA%5D%3D0.06%5BA_o%5D)
Half life of the carbon-14 = 

![[A]=[A_o]\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-kt%7D)
![[A]=[A_o]\times e^{-\frac{0.693}{t_{1/2}}\times t}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-%5Cfrac%7B0.693%7D%7Bt_%7B1%2F2%7D%7D%5Ctimes%20t%7D)
![0.60[A_o]=[A_o]\times e^{-\frac{0.693}{5730 year}\times t}](https://tex.z-dn.net/?f=0.60%5BA_o%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-%5Cfrac%7B0.693%7D%7B5730%20year%7D%5Ctimes%20t%7D)
Solving for t:
t = 4223.71 years ≈ 4224 years
The age of the sample is 4224 years.
C5H10O5 would weight a total of 2.312 kilograms.
I think it’d be C. I’m not 100% if it is tho
The rate constant of a reaction : 8.3 x 10⁻⁴
<h3>Further explanation</h3>
Given
rate = 1 x 10⁻² (mol/L)/s, [A] is 2 M, [B] is 3 M, m = 2, and n = 1
Required
the rate constant
Solution
For aA + bB ⇒ C + D
Reaction rate can be formulated:
![\large{\boxed{\boxed{\bold{r~=~k.[A]^a[B]^b}}}](https://tex.z-dn.net/?f=%5Clarge%7B%5Cboxed%7B%5Cboxed%7B%5Cbold%7Br~%3D~k.%5BA%5D%5Ea%5BB%5D%5Eb%7D%7D%7D)
the rate constant : k =
![\tt k=\dfrac{rate}{[A]^m[B]^n}\\\\k=\dfrac{1.10^{-2}}{2^2\times 3^1}\\\\k=8.3\times 10^{-4}](https://tex.z-dn.net/?f=%5Ctt%20k%3D%5Cdfrac%7Brate%7D%7B%5BA%5D%5Em%5BB%5D%5En%7D%5C%5C%5C%5Ck%3D%5Cdfrac%7B1.10%5E%7B-2%7D%7D%7B2%5E2%5Ctimes%203%5E1%7D%5C%5C%5C%5Ck%3D8.3%5Ctimes%2010%5E%7B-4%7D)