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          A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

  Chemistry 7






Previous page


1
H
2
He
3
Li
4
Be
5
B
6
C
7
N
8
O
9
F
10
Ne
11
Na
12
Mg
13
Al
14
Si
15
P
16
S
17
Cl
18
Ar
19
K
20
Ca
21
Sc
22
Ti
23
V
24
Cr
25
Mn
26
Fe
27
Co
28
Ni
29
Cu
30
Zn
31
Ga
32
Ge
33
As
34
Se
35
Br
36
Kr
37
Rb
38
Sr
39
Y
40
Zr
41
Nb
42
Mo
43
Tc
44
Ru
45
Rh
46
Pd
47
Ag
48
Cd
49
In
50
Sn
51
Sb
52
Te
53
I
54
Xe
55
Cs
56
Ba
57-
71
72
Hf
73
Ta
74
W
75
Rn
76
Os
77
Ir
78
Pl
79
Au
80
Hg
81
Tl
82
Pb
83
Bi
84
Po
85
At
86
Rn
87
Fr
88
Ra
89-
103
104
Rf
105
Db
106
Sg
Bh
108
108
Hs
109
Mt
110
Ds
111
Rg
112
Cn
113
Uut
114
Fl
115
Uup
116
Lv
117
Uus
118
Uuo

Oxygen



Nitrogen


Let us talk for the main part, about oxygen and also a bit about nitrogen. Both have certain specific features, which can answer our question: How can a petrol driven car, e.g., the Mercedes S-Class, have a consumption of 8,6 liters per 100 kms and an exhaust emission of 200g/cm≥ of CO2? After all, the 200 grams mean 20 kilograms per 100 kms, the 8,6 liters of premium petrol however, at the most, 6,8 kgs. Where then, is the huge additional mass of 13,2 kgs coming from?

Let's assume that the fuel is made up of uniform C8H16-molecules, this would be expressed by the following formula:

C8H16 + 12 O2 = 8 CO2 + 8 H2O

The oxygen atoms however, can only be had together with air and, only as exactly 23,16 percent of the mass. That means that they are bringing in an additional 76,84 percent of mass, most of it with 75,53 percent nitrogen.

Rounded off atomic masses
O16 u*
C12 u*
H1 u*
* The 'u' is explained here.

C8H16+12 O2=8 CO2 + 8 H2O
112 u+384 u=352 u + 144 u

So, between the fuel and the pure oxygen, there is a mass ratio of 112 to 384. The 384 however, make up a portion of 23,16 percent. Thus, using simple cross-multiplication, this portion of 384 can be divided by 23,16 and then, multiplied by 100, the result is 1659. If you put this result in relation to the 112, mathematically speaking, it will produce the:

Stoichiometrical mixing ratio
14,8 : 1 (Mass proportions)

However, during the combustion process, the carbon atoms have joined forces with the heavy oxygen taken from the air. All you need to do, is to divide the 352 (8 CO2) by the 112 (C8H16) and you'll arrive at roughly the same ratio as the 20 kgs of CO2 divided by the 6,8 kgs of fuel, which is approx. 3 to 1.

What do we learn from this? The combustion engine must, by no means, carry everything around that it needs for the combustion process. In terms of figures, it takes the much larger part of the oxygen from the air. This explains once again, the enormous amount of basic energy found in a liter of fuel and just how difficult it will be, to find a substitute for this form of energy.

How much larger the tanks would have to be if one had to carry along oxygen tanks, is made perfectly clear if we look at the field of space travel. When, in this case, the re-usable shuttle is spoken of, only the point of the rocket is meant. The gigantic rest is newly built for each start, then filled with fuel and oxygen, once in flight, the sections are then ejected, stage by stage. 10/13

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Translator: Don Leslie - Email: lesdon@t-online.de

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