The oxidation states of transition metal atoms in K₂Cr₂O₇, KMnO₄ and K₂FeO₄, respectively, are x,y and z. The sum of x,y and z is [numerical value].
The oxidation states of transition metal atoms in K₂Cr₂O₇, KMnO₄ and K₂FeO₄, respectively, are x,y and z. The sum of x,y and z is [numerical value].
4 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
A
Answered by
5 months ago
K? Cr? O? (x=+6), KMnO? (y=+7), K? FeO? (z=+6). x+y+z=19.
Similar Questions for you
K2Cr2O7 + H2O2 + H2SO4->
Potassium permanganate in alkaline medium oxidise lodide to lodate.
Compound A is
KMnO4 decomposes upon heating at 513 K and forms K2MnO4 and MnO2.
2KMnO4
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else.
On Shiksha, get access to
66K
Colleges
|
1.2K
Exams
|
6.9L
Reviews
|
1.8M
Answers
Learn more about...

Chemistry Ncert Solutions Class 12th 2023
View Exam DetailsMost viewed information
SummaryDidn't find the answer you were looking for?
Search from Shiksha's 1 lakh+ Topics
or
Ask Current Students, Alumni & our Experts
Have a question related to your career & education?
or
See what others like you are asking & answering



