<span>w(C)=81,8% )</span>w(H=18,2% Dh=22 найти Сx Hy
<span>Mr( Сx Hy )=2*22 =44</span>
<span>n(C)=W(C)*Mr( Сx Hy )</span>
<span> Ar( C )*100% </span>
<span>n(C)=81,8*44</span>
<span> 12*100%</span>
n(C)=3
n(H)=w(H)*Mr( Сx Hy )
Ar( H )*100%
n(H)= 8
Ответ:С2Р8-пропан
CH3CHO + Ag2O = CH3COOH + 2Ag
n(Ag) = 21.6/M(Ag)
m(CH3CHO) = n(Ag)*M(CH3CHO)/2
%(CH3CHO) = m(CH3CHO)/0.5*100
1)Ca(OH)2
2)CaSO3
3)<span>Cao+2HCl=</span>CaCl2+H2O
4)H2SO4
5)<span>SO3+2KOH= </span>K2SO3+H2O
6)BaSO3+H2O
Mr(C2H6O2)=2*12+6*1+2*16=62
W(C)=2*12/62=0.387=38.7%