Odpowiedzi

2010-01-15T18:02:00+01:00
K₂O
39*2+16=94
100%-94
x=16
1600/94≈17% a K 83%
Al₂O₃
27*2+16*3=54+48=102
4800/102≈47% O 53%Al
H₂O
1*2+16=18
200/18≈11% H 89% O
H²SO₄
2+32+16*4=98
200/98≈2% H
3200/98≈32,5% S 65,5% O
H₃PO₄
3+16*4+31=98
300/98≈3% H
3100/98≈31,5% P 65,5% O
H₂CO³
2+12+16*3=62
200/62≈3% H
1200/62≈19% C 78% O
2010-01-15T18:03:34+01:00
MK₂O=94u %K= 7400%/94=78,7% %O=100-78,7=21,3%

mAl₂O₃=102u %Al=5400%/102=ok.52,9% %O=100-52,9=47,1%

mH₂O=18u %H=200%/18=11,1% %O=100-11,1=88,9%

mH²SO₄=98u %H=200%/98=ok.2% %S=3200%/98=32,7% %O=100-34,7=65,3%

mH₃PO₄=98u %H=300%/98=3% %P=3100/98=31,7% %O=100-34,7=65,3%

mH₂CO³=62u %H=200/62=3,2% %C=1200/62=19,4% %O=100-22,6=77,4%

2010-01-15T18:06:45+01:00
K₂O
K-83
O-17

Al2O3
Al-53%
O-47%

H2O
H-14%
O-86%

H2SO4
H-2
S-33%
O-65%

H3PO4
H-3%
P-32%
O-65%


H2CO3
H-3,25%
C-19,35%
O-77,4%