CO2 + Ba(OH)2 = BaCO3 + H2O
n(BaCO3)=m/M=19,7/197=0,1моль
n(BaCO3)=n(CO2)=0,1моль
V(CO2)=0,1*22,4=2,24л
4Al+3О2=2Al2O3
<span>Масса Al2O3 равна: 74/(4*27)*(2*(2*27+3*16))=140 г</span>
1.) 3
2) 3
3) 3
4) 4
5) 2
6) 1
7) 1
8) 4
9) 1
10) 4
Ответ во вложении..............................
Ω(C)=83,34%
ω(H)=100-83,34=16,66%
Dвозд=2,483
CxHy
Dвозд = Mr/29
Mr=29*D
Mr=29 * 2,483 = 72
<span>x = 72*83,34/12*100 = 5
5*12 = 60
72-60=12
y=12
</span>C5H12