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.
Cu ions plus EDTA2- ->cu(EDTA)2- plus 2H-
number of moles of CU ions used which is equal to molarity multiplied by volume in litres
that is 50xo.o2 divided by 1000 that is 0.001moles
Since ratio is 1:1 the moles of EDTA is also0.001moles
volume of EDTA is 0.001 divided by 0.0510m which is 0.0196 litres in ml is 0.0196x1000 which is 19.61ml
Answer:- Molarity of the acid solution is 0.045M.
Solution:- The balanced equation for the reaction of given acid and base is:
From the balanced equation, they react in 1:1 mol ratio. So, we could easily solve the problem using the equation:
where, 
is the molarity of acid, 
is the molarity of base, 
is the volume of acid and 
is the volume of base.
Let's plug in the given values in the equation:
on rearranging the above equation:
= 0.045M
So, the molarity of the acid solution is 0.045M.
Its condensation
the vapour has a lot of kinetic energy but if it cools down it loses that energy and condenses into a liquid
hope that helps
Answer:
4.5moles
Explanation:
First, let us balance the equation given from the question. This is illustrated below:
KClO3 —> KCl + O2
There are 2 atoms of O on the right side and 3 atoms on the left. It can be balance by putting 2 in front of KClO3 and 3 in of O2 as shown below
2KClO3 —> KCl + 3O2
Now, we have 2 atoms each of K and Cl on the left side and 1atom each of K and Cl on the right. It can be balance by putting 2 in front of KCl as shown below:
2KClO3 —> 2KCl + 3O2
Now the equation is balanced.
From the balanced equation,
2 moles of KClO3 produced 3 moles of O2.
Therefore, 3 moles of KClO3 will produce = (3 x 3) /2 = 4.5moles of O2.
Therefore 3 moles of KClO3 will produce 4.5 moles of O2
