The net equations are obtained from the double displacement of the cations and anions, then balance.
NH3(aq) + HC2H3O2 (aq) = NH4+(aq) + C2H3O2-(aq<span>)
</span><span>H+(aq) + C2H3O2-(aq) + NH3(aq) -> NH4+(aq) + C2H3O2-(aq)</span><span>
</span><span>2NaOH(aq) + H2SO4 (aq) = Na2SO4 (s)+ 2H2O (aq)
</span>H2S (aq) + Ba(OH)2 (aq) = BaS (s)+ 2H2O (aq)
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.
A solution that has the capacity to hold additional solute at a certain temperature is called Under-saturated.
Answer:
1218.585
Explanation:
Looking at the subscripts we know there are 2 atoms of Fe, 3 atoms of C, and 6 of O.
Take the molar mass of each atom (from the periodic table) and multiply by the # of atoms
Fe: 55.845×2= 111.69
C: 12.011×3= 36.033
O:15.999×6=95.994
Add the values together: 243.717 g/mol
That is 1 mole of the molecule. Multiply by 5 for the final answer.
243.717×5=1218.585
