ATMOSPHERE OF LITHIUM ON EARTH

ENERGY LAW OF SIKANDER AQEEL = E = m2c 

Lithium ortho-molybdate,  

        Lithium = Li-7  

        Lithium = 7 

        Lithium = 3 proton 

        Lithium = 4 neutron 

        a2 + b2 = c2  

        a2 (neutron)     +     b2 (proton) = c2  

        a2 (4)           +     b2 (3)   

        a2 (16)          + b2 (9) = c2

        a2 + b2 = 25 = c2 

        a2 + b2 = 25 / 7-Li = 3.57

        a2 + b2 = 3.57 = 3 Li proton + H0.57

Lithium energy atmosphere = 174 % * H0.57 of fusion = 99.18 hour 

                       Because 

                       Hour = 99.18 / 6 proton of C = 16.53-Chlorine

                            = 16.53 – 3 proton of Lithium = 13.53-Silicon

Chlorine = 17 proton 

Silicon = 14 proton

  Lithium = 16.53 * 13.53 = 223.6509

          = 223.6509 - 49.6509% (O3 + H = 49.25) = 174

          = 174

 Chemical = 174 - 96 (Mo) molybdate atmosphere of earth = 78

          = 78 – 64 (O4) = 14

          = 14 = Li2 = Lithium 

Chemical formula from 99.18 hour = Li2MoO4  

                                 = Li2MoO4, lithium ortho-molybdate 

ENERGY IN ACT BY C1     

Light speed meter seconds = 300000000 / 25 (m2) = 12000000

                          = 12000000 / 24 hour = 500000

                          = 500000 / 60 minute = 8333.33

                          = 8333.33 / 47.89% (O3 = 48) = 174

                          = 174 = Li2MoO4 = E = m2c,