Answer:
Explanation:
The formula of the reaction:
KClO₂ → KCl + O₂
To assign oxidation numbers, we have to obey some rules:
- Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
- The charge on simple ions signifies their oxidation number.
- The algebraic sum of all the oxidation number of all atoms in a neutral compound is zero. For radicals with charges, their oxidation number is the charge.
The oxidation number of K in KClO₂:
K + (-1) + 2(-2) = 0
K-5 = 0
K = +5
The oxidation number of K in KCl:
K + (-1) = 0
K = +1
The oxidation number Cl in KClO₂ is -1
For Cl in KCl, the oxidation number is -1
For O in KClO₂, the oxidation number is (2 x -2) = -4
For O in O₂, the oxidation number is 0
K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.
O moves from an oxidation state of -4 to 0. This is a loss of electrons and O has undergone oxidation. We say O is oxidized.
I would say b and d but I wouldn’t really know :/
Answer:
All three are present
Explanation:
Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: 
.
- Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
- Secondly, addition of liquid ammonia would form a precipitate with silver:
![AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq)](https://tex.z-dn.net/?f=AgCl%20%28s%29%20%2B%202%20NH_3%20%28aq%29%20%2B%20H_2O%20%28l%29%5Crightarrow%20%5BAg%28NH_3%29_2%5DOH%20%28s%29%20%2B%20HCl%20%28aq%29)
; Silver hydroxide at higher temperatures decomposes into black silver oxide: ![2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g)](https://tex.z-dn.net/?f=2%20%5BAg%28NH_3%29_2%5DOH%20%28s%29%5Crightarrow%20Ag_2O%20%28s%29%20%2B%20H_2O%20%28l%29%20%2B%204%20NH_3%20%28g%29)
. - Thirdly, we also know we have
in the mixture, since addition of potassium chromate produces a yellow precipitate: 
. The latter precipitate is yellow.
