Odpowiedzi

2010-04-15T20:00:53+02:00
A) Al= (27/132)*100%= 20.5%
Cl3= (35*3)/132*100%=79.5%
b) Na= 23/119*100%=19%
S=32/119*100%=27%
O4=64/119*100%=54%
c) Pb=207/333*100%=62%
N2= 28/333*100%=8%
O6=96/333*100%=30%
2010-04-15T20:01:37+02:00
A)AlCl3
m AlCl3 = 27u + 3*35u = 27u + 105u = 132
m Al = 27u
132u - 100%
27u - x
x = 27u * 100%/ 132
x = 20,45%

mCl3 = 105u
132u - 100%
105u - x
x = 105*100%/132u
x = 79,54%

b)NaSO4
mNaSO4 = 23u + 32u + 4*16u = 119u
mNa = 23u
119u - 100%
23u - x
x = 23u*100%/119u
x = 19,32%

mSO4 = 96u
119u - 100%
96u - x
x = 96u*100%/119u
x = 80,67%

c)Pb(NO3)2
mPb(NO3)2 = 207u + (14u + 3*16u)*2 = 207u + 124u = 331
mPb = 207u
331u - 100%
207u - x
x = 207u*100%/331u
x = 62,53%

m(No3)2 = 124u
331u - 100%
124u - x
x = 124u*100%/331u
x = 38,47%
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