During complete oxidation of the fatty acid CH3(CH2)14COOH, molecules of acetyl-CoA are produced, and the fatty acid goes through the β-oxidation cycle times

During complete oxidation of the fatty acid CH3(CH2)14COOH, eight molecules of acetyl-CoA are produced, and the fatty acid goes through the β-oxidation cycle seven times.

Explanation:

In the β-oxidation of fatty acids, an acetyl-CoA molecule is removed from the fatty acid chain after every β-oxidation cycle that the fatty acid undergo, leaving behind a fatty acyl-CoA molecule shortened by two cabon atoms..

The removal of the acetyl-CoA molecule starts from the carboxyl end and shortens the fatty acid molecule by two carbon units. Successive β-oxidation cycles results in the complete oxidaton of the fatty acid molecle to acetyl -CoA molecules.

The compound CH3(CH2)14COOH, is a 16-carbon saturated fatty acid molecule known as palmitic acid.. It undergoes seven passes through the β-oxidation cycle to yield eight molecules of acetyl-CoA with each cycle yielding an acetyl-CoA molecule and a fatty acyl-CoA shortened by two carbon atoms. Finally, the seventh step yields two acetyl-CoA molecules.

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-

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