Answer:
![[Cl^-]=232.3\frac{mgCl^-}{L}](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D232.3%5Cfrac%7BmgCl%5E-%7D%7BL%7D)
Explanation:
Hello,
In this case, we can represent the chemical reaction as:
![Cl^-(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NO_3^-(aq)](https://tex.z-dn.net/?f=Cl%5E-%28aq%29%2BAgNO_3%28aq%29%5Crightarrow%20AgCl%28s%29%2BNO_3%5E-%28aq%29)
In such a way, since the mass of the obtained silver chloride is 93.9 mg, we can compute the chloride ions in the ground water by using the following stoichiometric procedure whereas the molar mass of chloride ions and silver chloride are 35.45 g/mol and 143.32 g/mol respectively:
![m_{Cl^-}=93.3mgAgCl*\frac{1mmolAgCl}{143.32mgAgCl}*\frac{1mmolCl^-}{1mmolAgCl} *\frac{35.45mgCl^-}{1mmolCl^-} =23.23mgCl^-](https://tex.z-dn.net/?f=m_%7BCl%5E-%7D%3D93.3mgAgCl%2A%5Cfrac%7B1mmolAgCl%7D%7B143.32mgAgCl%7D%2A%5Cfrac%7B1mmolCl%5E-%7D%7B1mmolAgCl%7D%20%2A%5Cfrac%7B35.45mgCl%5E-%7D%7B1mmolCl%5E-%7D%20%3D23.23mgCl%5E-)
Finally, for the given volume of water in liters (0.100L), we compute the required concentration:
![[Cl^-]=\frac{23.2mgCl^-}{0.100L}\\](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D%5Cfrac%7B23.2mgCl%5E-%7D%7B0.100L%7D%5C%5C)
![[Cl^-]=232.3\frac{mgCl^-}{L}](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D232.3%5Cfrac%7BmgCl%5E-%7D%7BL%7D)
Best regards.
So, you need to have same ammount of atoms on the left and on the right side of the equation. You need to count the ammount of attoms of every substance on the left, and make sure that on the right side the ammount is same. For example in the 1st one it’s 6Sn+2P4=2Sn3P4, so that you have 6atoms of Sn on the left and 6 atoms of Sn on the right, same with the P
As the moon like orbits earth,it changes phases in an orderly way.