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Answer:
.0556 L
Explanation:
First, convert the 1.35 M to 1.35 mol/L in order for the units to correctly cancel out.
Then, multiply (0.0725 moles Na2CO3/1) times (L/ 1.35 mol).
Finally, the answer will be .0556 L.
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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.
It is A lubricant that reduces frictions creation