Odpowiedzi

2010-04-06T14:56:15+02:00
M ca(oh)2 =74g/mol

40=x %
74=100
x=54% ca

74=100%
2-x
x=2,7% wodor

100-2,7-54=43,3% tlen
2010-04-06T15:02:46+02:00
Ca(OH)₂

mCa(OH)₂ = mCa + 2mO + 2mH
mCa(OH)₂ = 40u + 2 * 16u + 2 * 1u
mCa(OH)₂ = 40u + 32u + 2u
mCa(OH)₂ = 74u

%Ca
74u ----------- 100%
40u ----------- x%


40 * 100%/74 = 54 ²/₃₇% ≈ 54%

%O
74u ----------- 100%
32u ----------- x%

32 * 100%/74 = 43 ⁹/₃₇% ≈ 43%

%H
74u ----------- 100%
2u ----------- x%

2 * 100%/74 = 2 ²⁶/₃₇% ≈ 3%