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2010-02-02T18:22:27+01:00
A)
AlCl3 = 27u + 3 *35u = 27u + 105u = 132u

%Al = 27/132*100%=20,45%
%Cl = 105/132*100%=79,55%
b)
Na2SO4=23u*2 + 32u+4*16u = 46u + 32u + 64u = 142u
%Na= 46u/142u*100%=32,39%
%S=32u/142u*100%=22,54%
%O=64u/142u*100%=45,07

c) Pb(NO3)2=207u+2*14u+6*16u=207u+28u+96u=331u
%Pb= 207u/331u*100%=62,54%
%N=28u/331u*100%=8,46%
%O=96u/331u*100%=29,00%
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