90ml of pure dry O2 is subjected to silent electric discharge. If only 10%of it is converted to O3 volume of mixture of gases ( O2 and O3) after the reaction will be........ and after passing through turpentine oil will be ........ .

Question

Vikas TU · Answer

3O2 ----> 2O3 Now for 90 ml of O2 volume of Ozone => 90*2/3 => 60 ml. Now only 10% is gets converted hence 10 % of 60 => 6 ml. Hence, AFter tge reaction O3 has 6 ml remaining and O2 has90 – 6 = 84 ml.

anjali gupta · Answer

302 ----203given vol.of o2 which is converting into o3 is 10 %Now,we should convert 90 ml in 10 % =90×10/100 =9ml.After reaction (90-9)=81 ml 2/3×9 = 6 ml of o3vol.of the mix .of gases (o2&o3) =81 + 6 = 87 mlturpentine oil vol.will be =87 ml

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90ml of pure dry O2 is subjected to silent electric discharge. If only 10%of it is converted to O3 volume of mixture of gases ( O2 and O3) after the reaction will be........ and after passing through turpentine oil will be ........ .

90ml of pure dry O2 is subjected to silent electric discharge. If only 10%of it is converted to O3 volume of mixture of gases ( O2 and O3) after the reaction will be........ and after passing through turpentine oil will be ........ .

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90ml of pure dry O2 is subjected to silent electric discharge. If only 10%of it is converted to O3 volume of mixture of gases ( O2 and O3) after the reaction will be........ and after passing through turpentine oil will be ........ .

90ml of pure dry O2 is subjected to silent electric discharge. If only 10%of it is converted to O3 volume of mixture of gases ( O2 and O3) after the reaction will be........ and after passing through turpentine oil will be ........ .

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