A compound C is produced on an industrial scale bt oxidation of 2-ethylanthraquinol by air. Compound C decolourises an acidic solutio of `KMnO_(4)` with the evolution of `O_(2)` Compound C produces a brown precipitate when it reacts with `MnSO_(4)` in alkaline solution. In idustrial preparation of compound C,2-ethyl anthraquinone is also produced which can be converted back 2-tehyl anthraquinol byA. Addintion of strong acidB. Addition of strong baseC. Catalytic oxidationD. Catalytic reduction suninh `H_(2)//Pd`

A compound C is produced on an industrial scale bt oxidation of 2-ethy

1.	A compound C is produced on an industrial scale bt oxidation of 2-ethylanthraquinol by air. Compound C decolourises an acidic solutio of `KMnO_(4)` with the evolution of `O_(2)` Compound C produces a brown precipitate when it reacts with `MnSO_(4)` in alkaline solution. In idustrial preparation of compound C,2-ethyl anthraquinone is also produced which can be converted back 2-tehyl anthraquinol byA. Addintion of strong acidB. Addition of strong baseC. Catalytic oxidationD. Catalytic reduction suninh `H_(2)//Pd`
Answer» Correct Answer - D

Discussion

No Comment Found

Related InterviewSolutions

Strength of the sample of `H_(2)O_(2)` is generally expressed in terms of volume strength . It means the volume of oxygen liberated at NTP by heating one volume of `H_(2)O_(2)` . The concentration of `H_(2)O_(2)` in a solution can also expressed as percentage of `H_(2)O_(2)` in solution . Normality of this solution can be calculated if the equivalent mass of `H_(2) O_(2)` is known . 25 mL of `H_(2)O_(2)` solution was added to the excess of acidified KI solution . The iodine so liberated required 40 mL of `0.1 N Na_(2)S_(2)O_(3)` solution . what is normality of `H_(2)O_(2)` solution ?A. `0.08`B. `0.02`C. `0.16`D. `0.20`
Strength of the sample of `H_(2)O_(2)` is generally expressed in terms of volume strength . It means the volume of oxygen liberated at NTP by heating one volume of `H_(2)O_(2)` . The concentration of `H_(2)O_(2)` in a solution can also expressed as percentage of `H_(2)O_(2)` in solution . Normality of this solution can be calculated if the equivalent mass of `H_(2) O_(2)` is known . Percentage strength of the above `H_(2)O_(2)` solution isA. `0.126`B. `0.272`C. `0.544`D. `0.136`
Strength of the sample of `H_(2)O_(2)` is generally expressed in terms of volume strength . It means the volume of oxygen liberated at NTP by heating one volume of `H_(2)O_(2)` . The concentration of `H_(2)O_(2)` in a solution can also expressed as percentage of `H_(2)O_(2)` in solution . Normality of this solution can be calculated if the equivalent mass of `H_(2) O_(2)` is known . Volume strength of above `H_(2)O_(2)` solution is ?A. `0.448`B. `0.632`C. `0.896`D. `0.556`
What is the molarity of `H_2O_2` of the `11.2V` (volume strength)?
What is the molarity of `H_(2)O_(2)` of the 11.2 V (volume strength) ?
The volume of perhydrol which on decomposition gives 1.5 Iit of `O_(2)` gas at STP isA. 25mlB. 15mlC. 10mlD. 0ml
Perhydrol isA. `10%(w//v)H_(2)O_(2)`B. `30%(w//v)` of `H_(2)O_(2)`C. `3%(w//v)H_(2)O_(2)`D. `100%(w//v)H_(2)O_(2)`
The volume of perhydrol which on decompostion gives 1L of `O_(@)` at STP isA. 10mLB. 1mLC. 100mLD. 10L
The lead of the lead paintings becomes black due to formation of……… `H_(2)O_(2)` converts the black colour to white lead is…..A. `H_(2)S,PbSO_(4)`B. `PbS,PbSO_(4)`C. `PbSO_(4),PbS`D. `PbS,H_(2)S`
Molarity and volume strength of perhydrol respectivelyA. 100M and 8.9 VB. 8.9 M and 100VC. 8.9M and 30VD. 30M and 8.9 V

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment