Zad w załączniku
1. tlenek żelaza (II) - FeO
m Feo = 56u + 16 u = 72u
% O = 16u:72u * 100% = 22 %
m Fe = 100% - 22% = 78%
22:78 = 11:39
siarczan (IV) potasu - K2SO3 = 39u*2 + 32u + 16u*3 = 168u
% K2 = 78u:168u * 100% = 46%
% SO3 = 100% - 46% = 54%
46:54 = 23:27
2. 500g * 10% = 480g * x
x = (500g * 10%) : 480g = 10.4 %
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Zad w załączniku
1. tlenek żelaza (II) - FeO
m Feo = 56u + 16 u = 72u
% O = 16u:72u * 100% = 22 %
m Fe = 100% - 22% = 78%
22:78 = 11:39
siarczan (IV) potasu - K2SO3 = 39u*2 + 32u + 16u*3 = 168u
% K2 = 78u:168u * 100% = 46%
% SO3 = 100% - 46% = 54%
46:54 = 23:27
2. 500g * 10% = 480g * x
x = (500g * 10%) : 480g = 10.4 %