oblicz masy cząsteczek podanych zwiasków chemicznych.
mCuO= mH2SO4=
m Fe2O3= mCa(OH)2=
m Na2C3= mFe2(SO3)3=
CuO
masa : 64+16=80u
Fe2O3
masa; 56+56+(16u x 3)= 160
Na2CO3
masa: 23+23+12+(16u x 3)=106u
H2SO4
masa: 1+1+32+(16u x 4)= 98u
Ca(OH)2
masa : 40+16+16+1+1=74u
Fe2(SO3)3
56+56+(32u x 3) + (16u x 9) = 352u
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CuO
masa : 64+16=80u
Fe2O3
masa; 56+56+(16u x 3)= 160
Na2CO3
masa: 23+23+12+(16u x 3)=106u
H2SO4
masa: 1+1+32+(16u x 4)= 98u
Ca(OH)2
masa : 40+16+16+1+1=74u
Fe2(SO3)3
56+56+(32u x 3) + (16u x 9) = 352u