The reaction between mercury (Hg) and sulfur (S) to form HgS is:
Hg + S ------------- HgS
Therefore: 1 mole of Hg reacts with 1 mole of S to form 1 mole of HgS
The given mass of Hg = 246 g
Atomic mass of Hg = 200.59 g/mol
# moles of Hg = 246 g/ 200.59 gmol-1 = 1.226 moles
Based on the reaction stoichiometry,
# moles of S that would react = 1.226 moles
Atomic mass of S = 32 g/mol
Therefore, mass of S = 1.226 moles*32 g/mole = 39.23 g
39.2 g of sulfur would be needed to react completely with 246 g of Hg to produce HgS
Answer:
Add G and H to determine the atomic mass for helium
Explanation:
It's simple, just follow my steps.
1º - in 1 L we have

of

2º - let's find the number of moles.



3º - The concentration will be

But we have this reaction

This concentration will be the concentration of

![K_{sp}=\frac{[Ba^{2+}][CO_3^{2-}]}{[BaCO_3]}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cfrac%7B%5BBa%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D%7D%7B%5BBaCO_3%5D%7D)
considering
![[BaCO_3]=1~mol/L](https://tex.z-dn.net/?f=%5BBaCO_3%5D%3D1~mol%2FL)
![K_{sp}=[Ba^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BBa%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
and
![[Ba^{2+}]=[CO_3^{2-}]=5.07\times10^{-5}~mol/L](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D%3D%5BCO_3%5E%7B2-%7D%5D%3D5.07%5Ctimes10%5E%7B-5%7D~mol%2FL)
We can replace it


Therefore the

is: