Chemistry-
General
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Question

Inversion of a sugar follows first order rate equation which can be followed by noting the change in rotation of the plane of polarisation of light in a polarimeter. If 1:", li and '0 are the fotations at t = 00, t = t and t = 0, then first order reaction can be written as:

  1. k equals fraction numerator 1 over denominator t end fraction l o g subscript e end subscript invisible function application fraction numerator r subscript t end subscript minus r subscript infinity end subscript over denominator r subscript 0 end subscript minus r subscript infinity end subscript end fraction    
  2. k equals fraction numerator 1 over denominator t end fraction l o g subscript e end subscript invisible function application fraction numerator r subscript 0 end subscript minus r subscript infinity end subscript over denominator r subscript t end subscript minus r subscript 0 end subscript end fraction    
  3. k equals fraction numerator 1 over denominator t end fraction l o g subscript e end subscript invisible function application fraction numerator r subscript infinity end subscript minus r subscript 0 end subscript over denominator r subscript infinity end subscript minus r subscript t end subscript end fraction    
  4. k equals fraction numerator 1 over denominator t end fraction l o g subscript e end subscript invisible function application fraction numerator r subscript infinity end subscript minus r subscript t end subscript over denominator r subscript infinity end subscript minus r subscript 0 end subscript end fraction    

The correct answer is: k equals fraction numerator 1 over denominator t end fraction l o g subscript e end subscript invisible function application fraction numerator r subscript 0 end subscript minus r subscript infinity end subscript over denominator r subscript t end subscript minus r subscript 0 end subscript end fraction

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A substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as: A stretchy sum from k subscript 2 end subscript superscript 2 end superscript C to k subscript 1 end subscript B of   k subscript 1 end subscript equals 1.26 cross times 10 to the power of negative 4 end exponent s to the power of negative 1 end exponent text  and  end text k subscript 2 end subscript equals 3.8 cross times 10 to the power of negative 5 end exponent s to the power of negative 1 end exponentThe percentage distributions of Band Care:

A substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as: A stretchy sum from k subscript 2 end subscript superscript 2 end superscript C to k subscript 1 end subscript B of   k subscript 1 end subscript equals 1.26 cross times 10 to the power of negative 4 end exponent s to the power of negative 1 end exponent text  and  end text k subscript 2 end subscript equals 3.8 cross times 10 to the power of negative 5 end exponent s to the power of negative 1 end exponentThe percentage distributions of Band Care:

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Two reactions A rightwards arrow Products and B rightwards arrow Products, have rate constants k,4 and kB at temperature T and activation energies EA and EB respectively. If kA > kB and EA < EB and assuming that A for both the reactions is same, then:

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<img src="https://mycourses.turito.com/tokenpluginfile.php/c161933dbfaab094c54655ab71e9b8f0/1/question/questiontext/643590/1/1165511/Picture1.png" alt="" width="244" height="166"

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<img src="https://mycourses.turito.com/tokenpluginfile.php/c161933dbfaab094c54655ab71e9b8f0/1/question/questiontext/643590/1/1165511/Picture1.png" alt="" width="244" height="166"

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