Oblicz masy czasteczkowe nastepujacych zwiazkow: H2O ; Fe(OH)2 ; H2SO4 ; Al2(CO3)3

h2o-18u ;fe(oh)2- 90u; H2SO4 -98u;Al2(CO3)3 -234u

mH2O=2*1u+16u=18u mFe(OH)2=55,8u+(16u+1u)*2=89,8u mH2SO4=(2*1u)+(32,1+16u*4)=98,1u mAl2(CO3)3=(27u*2)+(12+16*3)*3=54u+36u+144u=234u Powinno być wszytko dobrze :) Pozdrawiam!

H2O = m(na dole)H2O = 2 m(na dole)H + 1 × m(na dole)O = 2 × 1 u + 1 × 16 u = 2 u + 16 u = 18 u Fe(OH)2 = m(na dole)FE(OH)2 = 1 × m(na dole)Fe + 2 × m(na dole)O + 2 × m(na dole)H = 1 × 56 u + 2 × 16 u + 2 × 1 u = 56 u +32 u + 2 u = 90 u H2SO4 = m(na dole)H2SO4 = 2 × m(na dole)H + 1 × m(na dole)S + 4 × m(na dole)O = 2 × 1 u + 1 × 32 u + 4 × 16 u = 2 u + 32 u + 64 u = 98 u Al2(CO3)3 = m(na dole)Al2(CO3)3 = 2 × m(na dole)Al + 3 × m(na dole)C + 9 × m(na dole)O = 2 × 27 u + 3 × 12 u + 9 × 16 u = 54 u + 36 u + 144 u = 234 u