Answer:
-) Oxidizing agent: 
-) Reducing agent: 
Explanation:
The first step is separate the reaction into the <u>semireactions</u>:
A.
B.
If we want to balance in <u>basic medium </u>we have to follow the rules:
1. We adjust the oxygen with 
2. We adjust the H with 
3. We adjust the charge with 
Lets balance the first semireaction A. :
Now, lets balance semireaction B:
Finally, we have to add the two semireactions:
_________________________________________
_________________________________________
Cancel out the species on both sides:
Simplifying the equation :
The is <u>oxidized</u> therefefore is the <u>reducing agent</u>. The is<u> reduced</u> therefore is the <u>oxidizing agent</u>.
I believe the closest possible answer to this question are:Gas A effuses faster than gas B.The molar mass is directly proportional to the rate of effusion.Thank you for your question. Please don't hesitate to ask in Brainly your queries
Al, since it is the only metal in that list
Answer: (a) The solubility of CuCl in pure water is 
.
(b) The solubility of CuCl in 0.1 M NaCl is 
.
Explanation:
(a) Chemical equation for the given reaction in pure water is as follows.
Initial: 0 0
Change: +x +x
Equilibm: x x
And, equilibrium expression is as follows.
![K_{sp} = [Cu^{+}][Cl^{-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCu%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D)
x =
Hence, the solubility of CuCl in pure water is .
(b) When NaCl is 0.1 M,
,
, 
Net equation: 
= 0.1044
So for,
Initial: 0.1 0
Change: -x +x
Equilibm: 0.1 - x x
Now, the equilibrium expression is as follows.
K' = 
0.1044 = 
x =
Therefore, the solubility of CuCl in 0.1 M NaCl is .
It makes it thinner to calcite the rhythm
