Molarity = Moles/Liter
Use the molecular atomic mass of NaCl to convert from grams to moles.
Molecular mass of NaCl is the sum of its atomic masses. Look at the periodic table to find these. Na is 23 g/mol and Cl is 35.5 g/mol ,
so NaCl = 23 + 35.5 = 58.5 g/mol
multiply to cancel out grams
76 g NaCl * (1mol / 58.5 g NaCl) = 1.3 mol NaCl
over 1 Liter is just 1.3 M NaCl
Hope this helps!
We have been given the condition that carbon makes up 35%
of the mass of the substance and the rest is made up of oxygen. With this, it
can be concluded that 65% of the substance is made up of oxygen. If we let x be
the mass of oxygen in the substance, the operation that would best represent
the scenario is,
<span> x = (0.65)(5.5 g)</span>
<span> <em> </em><span><em>x =
3.575 g</em></span></span>
Density/Earth’s gravitational pull.
The given question is incomplete. The complete question is as follows.
Sodium sulfate is slowly added to a solution containing 0.0500 M 
and 0.0390 M 
. What will be the concentration of 
(aq) when 
begins to precipitate? What percentage of the can be separated from the Ag(aq) by selective precipitation?
Explanation:
The given reaction is as follows.
![[Ag^{+}]](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D)
= 0.0390 M
When 
precipitates then expression for 
will be as follows.
![K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5BSO%5E%7B2-%7D_%7B4%7D%5D)
![1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]](https://tex.z-dn.net/?f=1.20%20%5Ctimes%2010%5E%7B-5%7D%20%3D%20%280.0390%29%5E%7B2%7D%20%5Ctimes%20%5BSO%5E%7B2-%7D_%7B4%7D%5D)
![[SO^{2-}_{4}]](https://tex.z-dn.net/?f=%5BSO%5E%7B2-%7D_%7B4%7D%5D)
= 0.00788 M
Now, equation for dissociation of calcium sulfate is as follows.
![K_{sp} = [Ca^{2+}][SO^{2-}_{4}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BSO%5E%7B2-%7D_%7B4%7D%5D)
![4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788](https://tex.z-dn.net/?f=4.93%20%5Ctimes%2010%5E%7B-5%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%20%5Ctimes%200.00788)
![[Ca^{2+}]](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D)
= 0.00625 M
Now, we will calculate the percentage of remaining in the solution as follows.
= 12.5%
And, the percentage of that can be separated is as follows.
100 - 12.5
= 87.5%
Thus, we can conclude that 87.5% will be the concentration of when begins to precipitate.
