Mam prośbe trzy zadania w załączniku wydaje mi sie ze nie beda dla was trudne.
1. kwas oktanowy
C7H15COOH
H H H H H HH
| | | | | | |
H-C-C-C-C-C-C-C-C=O
| | | | | | | |
HH H HH H H OH
2. kwas nonanowy
H H H HH HH H
H-C-C-C-C-C-C-C-C-C=O
| | | | | | | | |
H H H H H HHH OH
3.C2H5COOH
c=12u czyli 3*12*100%/3*12+6*1+2*16=3600/74=48,65%
o=16u czyli 2*16*100%/3*12+2*16+6*1=3200/74=43,24%
48,65%+43,24%=91,89%
czyli
h=100%-91,89%=8,11%
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1. kwas oktanowy
C7H15COOH
H H H H H HH
| | | | | | |
H-C-C-C-C-C-C-C-C=O
| | | | | | | |
HH H HH H H OH
2. kwas nonanowy
H H H HH HH H
| | | | | | | |
H-C-C-C-C-C-C-C-C-C=O
| | | | | | | | |
H H H H H HHH OH
3.C2H5COOH
c=12u czyli 3*12*100%/3*12+6*1+2*16=3600/74=48,65%
o=16u czyli 2*16*100%/3*12+2*16+6*1=3200/74=43,24%
48,65%+43,24%=91,89%
czyli
h=100%-91,89%=8,11%