Given :
0.00072 M solution of 
at 
.
To Find :
The concentration of 
and pOH .
Solution :
1 mole of gives 2 moles of ions .
So , 0.00072 M mole of gives :
![[OH^-]=2 \times 0.00072\ M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D2%20%5Ctimes%200.00072%5C%20M)
![[OH^-]=0.00144\ M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.00144%5C%20M)
![[OH^-]=1.44\times 10^{-3}\ M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.44%5Ctimes%2010%5E%7B-3%7D%5C%20M)
Now , pOH is given by :
![pOH=-log[OH^-]\\\\pOH=-log[1.44\times 10^{-3}]\\\\pOH=2.84](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D%5C%5C%5C%5CpOH%3D-log%5B1.44%5Ctimes%2010%5E%7B-3%7D%5D%5C%5C%5C%5CpOH%3D2.84)
Hence , this is the required solution .
Answer:
Explanation:
NaCl = Na⁺ + Cl⁻
6120 ions of NaCl will contain 3060 ions of Na⁺ and 3060 ions of Cl⁻ , forming 3060 molecules of NaCl .
6.02 x 10²³ molecules of NaCl = 1 mole
3060 molecules of NaCl = 3060 x 10⁻²³ / 6.02 moles
= 508.30 x 10⁻²³ moles
= 5.08 x 10⁻²¹ moles
Answer : 5.08E-21 moles .
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