(Electron or photon-Atomic-Theory)
Cerium = 163
Cerium Boride
It is an chemical bonding through the math, where donor and acceptor system contain on mathematics,
CeB6 cerium boride
CERIUM
A
One lines value = 0.008 photon or electron
Metals Ratio = 50
Cerium Boride = CeB6
Cerium Boride = [Ce-163 + B6- 66 = 206]
Protons = [Ce-58 + B6-30 = 88]
Cerium Boride = 206 * 0.0085 = 1.751
= 1.75 + 0.01 = 1.76
= 1.76 * 50 ratio = 88
= 88 protons of Cerium Boride = CeB6
About one line = (0.008)3
Proton donor or acceptor
Because line = 0.0085 / 0.008 = 1.0625 density (Cm3) per square gram
Because line = 0.0085 / 0.008 = 1.0625
= 1.0625 - 0.3585 = 0.704
= 0.704 / 0.008 = 88
= 88, proton of (CeB6)
Cerium Boride = CeB6 by 17.92 % reaction (0.3585 * 50 ratio = 17.92 % reaction)
Cerium Boride = 206 * 0.048 = 9.888
= 9.888 * 50 ratio = 494.4
= 494.4 / 206 = 2.4 = H1
= Cerium Boride = CeB6
Because = 2.4 * 2.5 (H2) = 6 = B6
About one line = 0.008
Proton donor or acceptor
Because line = 0.048 / 0.008 = 6 donor or acceptor = CeB6
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CeBr3 cerium(III) bromide
CERIUM
A
One lines value = 0.008 photon or electron
Metals Ratio = 50
Cerium (III) Bromide = CeBr3
Cerium (III) Bromide = [Ce-163 + Br3- 240 = 403]
Protons = [Ce-58 + Br3- 105 = 163]
Cerium (III) Bromide = 403 * 0.0080 = 3.224
= 3.22 + 0.04 = 3.26
= 3.26 * 50 ratio = 163
= 163 protons of Cerium (III) Bromide = CeBr3
About one line = (0.008)3
Proton donor or acceptor
Because line = 0.0080 / 0.008 = 1 density (Cm3) per square gram
Because line = 0.0080 / 0.008 = 1
= 1 + 0.304 = 1.304
= 1.304 / 0.008 = 163
= 163 proton of (CeBr3)
Cerium (III) Bromide = CeBr3 (by 15.2 % reaction (0.304 * 50 ratio = 15.2 % reaction)
Cerium (III) Bromide = 403 * 0.024 = 9.672
= 9.672 * 50 ratio = 483.6
= 483.6 / 403 = 1.2
= Cerium (III) Bromide = CeBr3
Because = 1.2 * 2.5 (H2) = 3 = O3
About one line = 0.008
Proton donor or acceptor
Because line = 0.024 / 0.008 = 3 donor or acceptor = CeBr3
WATER
A
One lines value = 0.008 photon or electron
Metals Ratio = 50
Water = H2O
Water = [H2-2.5 + O- 16 = 18.5]
Protons = [H2-2 + O- 8 = 10]
Water = 18.5 * 0.0108 = 0.1998
= 0.19 - 0.01 = 0.2
= 0.2 * 50 ratio = 10
= 10 protons of Water = H2O
About one line = 0.008
Proton donor or acceptor
Because line = 0.0108 / 0.008 = 1.35 density (Cm3) per square cm
Density of warm water at 0-C = 0.998
Density of water = 1.35
Density of water = 1.35 / 0.998 = 1.35
= 1.35
Water = 18.5 * 0.016 = 0.296
= 0.296 * 50 ratio = 14.8
= 14.8 / 18.5 = 0.8 = H2
= Water = H2O
Because = 0.8 * 2.5 (H2) = 2 = H2
About one line = 0.008
Proton donor or acceptor
Because line = 0.016 / 0.008 = 2 donor or acceptor = H2O
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