Odpowiedź :
a) mBi2O3=2*209u+3*16u=466 u
418u---x%
466u---100%
x=89,7%Bi
b) mMn2O7=2*55u+7*16u=222 u
110u---x%
222u---100%
x=49,55%Mn
100%-49,55%=50,45%O
c) mSbCl5=122u+5*35,5u=299,5 u
122u---x%
299,5u---100%
x=40,73%Sb
100%-40,73%=59,27%Cl
a) mBi2O3=2*209u+3*16u=466 u
418u---x%
466u---100%
x=89,7%Bi
b) mMn2O7=2*55u+7*16u=222 u
110u---x%
222u---100%
x=49,55%Mn
100%-49,55%=50,45%O
c) mSbCl5=122u+5*35,5u=299,5 u
122u---x%
299,5u---100%
x=40,73%Sb
100%-40,73%=59,27%Cl