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Jak coś to ta dwójka jest na dole, tylko nie wiem jak ją napisać ;P
mCa(OH)2 = 40u + (16u*2) + (1u*2) = 40u + 32u + 2u = 74u
WAPŃ
74u --- 100%
40u --- x
74x = 4000/:74
x = 54,05%
TLEN
74u --- 100%
32u --- x
74x = 3200/:74
x = 43,24%
WODÓR
100% - (54,05% + 43,24%) = 100% - 97,29% = 2,71%
WAPŃ - 54,05%
TLEN - 43,24%
WODÓR - 2,71%