Molarity = moles of solute/volume of solution in liters.
The solute here is NaCl, of which we have 46.5 g. To calculate the molarity of an NaCl solution, we need to know the number of moles of NaCl. To convert from grams to moles, we divide the mass by the molar mass of NaCl. The molar mass of NaCl is the sum of the atomic masses of Na and Cl: 23 amu + 35 amu = 58 amu. For our purposes, we can regard amu as equivalent to grams/mole.
(46.5 g)/(58 g/mol) = 0.8017 moles NaCl.
Now that we know both the number of moles of our NaCl solute and the volume of the solution, we can calculate the molarity:
(0.8017 moles NaCl)/(2.2 L) = 0.364 M.
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Explanation:
A matchstick has a lot of chemical energy stored in it. When the match is struck, it burns and the chemical energy in it produces heat energy and light energy.
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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.
