The answer would be KMnO4, please let me know if you would like me to explain how i got this
Answer:
Kb = 

Explanation:
For a weak organic base, the formula to find
is given by:

where c is the concentration of base.
Here c= 

Substituting the above values in the formula,we get:

Hence:
Kb = 

In carbohydrates the C:H:O is 1:2:1
Answer:
Compound that has the smallest ions with the greatest charge.
Explanation:
The patient will have less hemoglobin in the blood
The patient will have a reduced oxygen supply to the cells