A. Combustion since it can’t be nuclear….?
Chemistry. More specifically, analytical chemistry.
My guess is b for the question
Answer:
Zero-Nine
Explanation:
this is becasue these numbers are rather small and if you plug these numbers into an equation you will most likely get 0.
<span>Average oxidation state = VO1.19
Oxygen is-2. Then 1.19 (-2) = -2.38
Average oxidation state of V is +2.38
Consider 100 formula units of VO1.19
There would be 119 Oxide ions = Each oxide is -2. Total charge = -2(119) = -238
The total charge of all the vanadium ions would be +238.
Let x = number of of V+2
Then 100 – x = number of V+3
X(+2) + 100-x(+3) = +238
2x + 300 – 3x = 238
-x = 238-300 = -62
x = 62
Thus 62/100 are V+2
62/100 * 100 = 62%
</span>62 % is the percentage of the vanadium atoms are in the lower oxidation state. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.
