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.
Answer:
Molality = 7.5 mol/kg
Explanation:
Given data:
Mass of NH₄Cl = 6.30 g
Mass of water = 15.7 g (15.7/1000 =0.016 kg)
Molality = ?
Solution:
Formula of molality:
Molality = Moles of solute / mass of solvent in gram
Now we will first calculate the number of moles of solute( NH₄Cl )
Number of moles = mass/ molar mass
Molar mass of NH₄Cl = 53.491 g/mol
Number of moles = 6.30 g/ 53.491 g/mol
Number of moles = 0.12 mol
Now we will calculate the molality.
Molality = Moles of solute / mass of solvent in gram
Molality = 0.12 mol / 0.016 kg
Molality = 7.5 m
or (m=mol/kg)
Molality = 7.5 mol/kg
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20 g O2 x 1 mol O2/32 g O = 0.625 mol O2
