Sikander Aqeel

ATMOSPHERE OF CARBON ON EARTH

May 20th 2017, 11:44 am
Posted by aqeelsika
483 Views

ENERGY LAW OF SIKANDER AQEEL = E = m2c 

 

Ketene 

        Carbon = 12.0107-C  

        Carbon = 12.0107 

        Carbon = 6 proton 

        Carbon = 6 neutron 

 

        a2 + b2 = c2  

        a2 (neutron)     +     b2 (proton) = c2  

        a2 (6)           +     b2 (6)   

 

        a2 (36)          + b2 (36) = c2

 

        a2 + b2 = 72 = c2 

        a2 + b2 = 72 / 12.0107-C = 5.99

        a2 + b2 = 5.99 

        Carbon proton = 6 - 5.99 + H0.01

 

 

Carbon energy atmosphere = 42.5% * H0.01 of fusion = 0.425 hour 

 

                       Because 

                       Hour = 0.425 * 17 proton of Cl = 7.225-Nitrogen   

                            = 7.225 – 6 proton of C = 1.225-Hydrogen 

 

Nitrogen = 7 proton   

Hydrogen = 1.25 proton 

 

    Carbon = 7.225 * 1.225 = 8.850625

           = 8.850625 + 33.649375 (CO + H4.5 = 33.625) = 42.5

           = 42.5

 

Chemical = 42.5 – 16 (O1) A Ketene atmosphere of earth = 26.5

         = 26.5 – 2.5 (H2) = 24  

         = 24 = C2 = Carbon   

 

 

Chemical formula from 0.425 hours = CH2CO

                                  = CH2CO, Ketene 

 

ENERGY IN ACT BY C1     

Light speed meter seconds = 300000000 / 72 (m2) = 4166666.66

                          = 4166666.66 / 24 hour = 173611.11

                          = 173611.11 / 60 minute = 2893.51

                          = 2893.51 / 68.08(CO)2 + C = 68) = 42.5

                          = 42.5 = CH2CO = E = m2c,         

 

 

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