Chemistry-
General
Easy

Question

A 0.200 M KOH solutions is electrolyzed for 1.5 h using a current of 8.00 A How many moles of O2 were produced at the anode?

  1. 0.48    
  2. 0.224    
  3. 0.112    
  4. 2.24 cross times 10 to the power of negative 2 end exponent    

The correct answer is: 0.112

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Conductors allow the passage of electric current through them Metallic and electrolytic are the two types of conductors Current carriers in metallic and electrolytic conductors are free electrons and free ions respectively Specific conductance of conductivity of the electrolyte solution is given by the following relation :
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logical and subscript m end subscript equals K x fraction numerator 1000 over denominator M end fraction logical and subscript e end subscript equals K x fraction numerator 1000 over denominator blank N end fraction where, M and N are the molarity and normality of the solution respectively. Molar conductance of storage electrolyte depends on concentration : logical and subscript m end subscript equals logical and subscript m end subscript superscript 0 end superscript minus b square root of c where, logical and subscript m end subscript superscript 0 end superscript equals molar conductance at infinite dilution
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The degrees of dissociation of weak electrolytes are calculated as: alpha equals fraction numerator logical and subscript m end subscript over denominator logical and subscript m end subscript superscript 0 end superscript end fraction equals fraction numerator logical and subscript e end subscript over denominator logical and subscript e end subscript superscript 0 end superscript end fraction
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