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Sulfuric acid is used with KMnO4 during oxidation that is why we write 'acidified' KMnO4Hold on, the oxidizing agent should be the KMnO4 and not sulfuric acid
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Sulfuric acid is used with KMnO4 during oxidation that is why we write 'acidified' KMnO4Hold on, the oxidizing agent should be the KMnO4 and not sulfuric acid
No it is DDo you mean that the answer was B and not D?
Sulfuric acid is used with KMnO4 during oxidation that is why we write 'acidified' KMnO4
Idk why I am confusing this all but can u plz write an equation to show the behaviour of sulfuric acid as an acid in the reaction with ethanol?
It is C
H2O is also formed in this reaction right?It is C
YeahH2O is also formed in this reaction right?
That's pretty obvious...they said that X is the only CHLORINE CONTAINING COMPOUND, didn't say that it is the only productBut, why didn't they mention it? :/
B??Apart from general guessing, can this question be solved arithmetically?
Yes
Look 1 mol is produced by cracking the ethene, methane n propene 0.5 mol of ethene already given in the question so we need to find the ratio of other products mol so like 0.5+0.25+0.25= 1 mole to convert it into whole number we just multiply with 4 so by adding them we will get 8 carbon altogetherApart from general guessing, can this question be solved arithmetically?
Thanx bro!Look 1 mol is produced by cracking the ethene, methane n propene 0.5 mol of ethene already given in the question so we need to find the ratio of other products mol so like 0.5+0.25+0.25= 1 mole to convert it into whole number we just multiply with 4 so by adding them we will get 8 carbon altogether
thankyou so much ))View attachment 63764
Ester breakage of the compound will help you identify it.
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