<h2>
→
</h2>
Explanation:
Ethanol can be oxidized to ethanal or acetaldehyde which is further oxidized to acid that is acetic acid.
→ [oxidation by loss of hydrogen]
-
An oxidizing agent potassium dichromate(VI) solution is used to remove the hydrogen from the ethanol.
- An oxidizing agent used along with dilute sulphuric acid for acidification.
Acetaldehyde can also be reduced back to ethanol again by adding hydrogen to it by using a reducing agent that is sodium tetrahydro borate, NaBH4.
- The oxidation of aldehydes to carboxylic acids can be done by the two-step process.
- In the first step, one molecule of water is added in the presence of a catalyst that is acidic.
- There is a generation of a hydrate. (geminal 1,1-diol).
→ 
[reduction by the gain of electrons]
Here, the oxidizing agent used is
in the presence of acetone.
DR is the correct answer
hope this helps
Answer is: <span>the molarity of this glucose solution is 0.278 M.
m</span>(C₆H₁₂O₆<span>) = 5.10 g.
n</span>(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆<span>) .
</span>n(C₆H₁₂O₆) = 5.10 g ÷ 180.156 g/mol.
n(C₆H₁₂O₆<span>) = 0.028 mol.
</span>V(solution) = 100.5 mL ÷ 1000 mL/L.
V(solution) = 0.1005 L.
c(C₆H₁₂O₆) = n(C₆H₁₂O₆) ÷ V(solution).
c(C₆H₁₂O₆) = 0.028 mol ÷ 0.1005 L.
c(C₆H₁₂O₆<span>) = 0.278 mol/L.</span>
Answer:
b the delocalized
Explanation:
it could be the right answer
Answer : The rate of consumption of oxygen = 0.245 mol/s
Solution : Given,
Rate at which Hydrogen burns = 0.49 mol/s
The Reaction is,
In this reaction, 2 moles of hydrogen react with the 1 mole of oxygen.
The rate at which oxygen burns is equal to the half of rate at which hydrogen burns.
Rate at which Oxygen burns = 
× 0.49 mol/s
= 0.245 mol/s
