Odpowiedzi

2010-04-15T15:59:51+02:00
Kwas węglowy: H₂CO₃
m = 2*1u + 12u + 3*16u = 14u + 48u = 62u
m(wodoru) = 2u
2u/62u * 100% = 3%
kwas fosforowy V: H₃PO₄
m=3u + 4*16u + 31u = 98u
m(wodoru) = 3u
3u/98u * 200% = 3%
2010-04-15T16:01:06+02:00
H2CO3

MmolH2CO3=62g/mol
MmolH=1g/mol 1g*2=2g

100%-62g
x% -- 2g

x=3,22% - wodoru

H3PO4

MmolH3PO4=98g/mol
MmolH=1g/mol 1g*3=3g

100%- 98g
x% -- 3g

x=3,06%- wodoru
2010-04-15T16:09:52+02:00
Masa h2co3=2*1+16*3+12=62g

62g-100%
2g-x%
x=2g*100%/62g=3.2%

masa H3PO4=3*1+31+16*4=98

98g-100%
3g-x&

x=300/98g=3.06%