Answer:
D. C14 appears at C‑6 of fructose 6‑phosphate
Explanation:
Conversion of glucose B-phosphate via 6-phospho-
gluconate to pentose phosphate and CO2 followed by conversion
of the pentose phosphate to fructose 6-phosphate and glyceralde-hyde phosphate followed by conversion of the fructose 6-phos-
phate to glucose g-phosphate .
The answer is: <span>The principal idea here is how r they obtained:
for example: sodium u put equal molar concentrations of sodium (Na) and sodium ion (Na+) together in a beaker , then dip in this solution a platinium wire (zero potential) which is connected to a normal hydrogen electrode (electrode with zero potential) then u see the reading of the whole circuit
if it is negtive, this means negative potential which means that the reducing property predominates where Na(reducing agent) is oxidized and electrons r accumulated on the platinum which gives it negative charge
This means that (Na) is a reducing agent, its strength depends on the value of the potential obtained, and here the table can help you
If u want to know if it's strong red. agent, look for it in the table, see if it has higher reduction potentail (or lower oxidation potential, same idea) than most other substances then it is reducing agent
and vice versa
So if we look at ur examples, u will find that MnO4- is the very strong oxidizing agent (has highest oxidation potential) (lowest reduction potential)
H+ and H2 are both with zero potential, no redox properties
And lastly Na and Na+:
This u can know from ur knowledge in chemistry, that sodium is very rarely found in elemental form and always in the form of ion so u can deduce that Na is the very strong reducing agent
or u can see the value of its standard oxi or red potetial and deduce which is the predominating form of them.
I hope this helps</span>
Answer:
E. None of the above statements are true.
Explanation:
