Масса(М)(Са(NO3)2)=40+(14+16×3)×2=164грам\моль
W=n×Ar÷Mr
W(Ca)=(40÷164)×100%=24%
W(N)=(14÷164)×100%=8,5%
W(O)=(3×16÷164)×100%=29%
M(воды)=1400г
m(р-ра)=120+1400=1520г
w(в-ва)%=m(в-ва)/m(р-ра)×100% ->
120/1520=7.9% или 0.079м.д.
ОТВЕТ: 7.9%
С-<span>1s </span>2<span>2s </span>2<span>2p </span><span>2
Si - </span><span>1s </span>2<span>2s </span>2<span>2p </span>6<span>3s </span>23p<span>2
Li - </span><span>1s </span>2<span>2s </span><span>1
O - </span><span>1s </span>2<span>2s </span>2<span>2p </span><span>4</span>