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%m1=0,038%
%m2=0,205%
%m3=99,757%
A1=17
A2=18
A3=16
Wzór: m.at.= %m1*A1+%m2*A2+%m3*A3
100%
Obliczenia:
m.at.O=0,038%*17+0,205%*18+99.757%*16=1600.448=16.004u
100% 100
Odpowiedź:Masa atomowa tlenu wynosi 16.004u.
%m 1 = 99,757 %
%m2 = 0,038 %
% m 3 = 0,205 %
A1 = 16
A2= 17
A3 = 18
Podstawiamy do wzoru :
m at = 16 * 99,757 + 17 * 0,038 + 18 * 0,205 / 100% = 1596,112 + 0,646 + 3,69 / 100 = 16,00448
ODP. Masaatomowa to 16,00448