определить массовою долю кислорода в оксиде углерода
Mr(CO2)=12+16*2=44
w(O2)=32/44=0.72%
w(C)=12/44=0.27%
0.72+0.27=0.99%
C O2
12:32
3:8
W(C)=(100%:11)x8=72,73%
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Mr(CO2)=12+16*2=44
w(O2)=32/44=0.72%
w(C)=12/44=0.27%
0.72+0.27=0.99%
C O2
12:32
3:8
W(C)=(100%:11)x8=72,73%