Answer:
450. g of 0.173 % KCN solution contains 779 mg of KCN.
Explanation:
Mass of the solution = m
Mass of the KCN in solution = 779 mg
Mass by mass percentage of KCN solution = 0.173%



1 mg = 0.001 g
m = 450,289 mg × 0.001 g = 450.289 mg ≈ 450. g
450. g of 0.173 % KCN solution contains 779 mg of KCN.
Answer:
chemical
Explanation:
because heat is being taken to the egg
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.

= 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)
= 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)
= 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.
Answer:
43.05 moles of Al needed to react with 28.7 moles of FeO.
Explanation:
Given data:
Moles of FeO = 28.7 mol
Moles of Al needed to react with FeO = ?
Solution:
Chemical equation:
2Al + 3FeO → 3Fe + Al₂O₃
Now we will compare the moles of Al with FeO.
FeO : Al
2 : 3
28.7 : 3/2×28.7 = 43.05 mol
Thus 43.05 moles of Al needed to react with 28.7 moles of FeO.