In amides, the carbonyl carbon is bonded to a nitrogen. The nitrogen in an amide can be bonded either to hydrogens, to carbons, or to both. ... Another way of thinking of an ester is that it is a carbonyl bonded to an alcohol. Thioesters are similar to esters, except a sulfur is in place of the oxygen.
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i think thats right
Answer: Li is the reducing agentg and O is the oxidizing agent.
Explanation:
1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.
2) The given reaction is:
4Li(s) + O₂ (g) → 2 Li₂O(s)
3) Determine the oxidation states of each atom:
Li(s): oxidation state = 0 (since it is alone)
O₂ (g): oxidation state = 0 (since it is alone)
Li in Li₂O (s) +1
O in Li₂O -2
That because 2× (+1) - 2 = 0.
4) Determine the changes:
Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.
O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.
Answer:Rubidium and Strontium
Explanation:
