The Haber process is the principal method for fixing nitrogen °Converting N 2 to nitrogen compounds). N 2 ( g ) +3H 2 ( g ) → 2 NH 3 ( g ) Assume that the reactant gases are completely converted to NH 2 (g) and that the gases behave ideally. a. What volume of NH 2 (g)can be produced from 152 L N 2 (g)and 313 L of H 2 (g) if the gases are measured at 315°C and 5.25 atm? b. What volume of NH 2 (g)measured at 25 °C and 727 mmHg, can be produced from 152 L N 2 (g) 313 L H 2 (g), measured at 315 °C and 5.25 atm?
The Haber process is the principal method for fixing nitrogen °Converting N 2 to nitrogen compounds). N 2 ( g ) +3H 2 ( g ) → 2 NH 3 ( g ) Assume that the reactant gases are completely converted to NH 2 (g) and that the gases behave ideally. a. What volume of NH 2 (g)can be produced from 152 L N 2 (g)and 313 L of H 2 (g) if the gases are measured at 315°C and 5.25 atm? b. What volume of NH 2 (g)measured at 25 °C and 727 mmHg, can be produced from 152 L N 2 (g) 313 L H 2 (g), measured at 315 °C and 5.25 atm?
The Haber process is the principal method for fixing nitrogen °Converting N2 to nitrogen compounds).
N
2
(
g
)
+3H
2
(
g
)
→
2 NH
3
(
g
)
Assume that the reactant gases are completely converted to NH2(g) and that the gases behave ideally. a. What volume of NH2(g)can be produced from 152 L N2(g)and 313 L of H2(g) if the gases are measured at 315°C and 5.25 atm? b. What volume of NH2(g)measured at 25 °C and 727 mmHg, can be produced from 152 L N2(g) 313 L H2(g), measured at 315 °C and 5.25 atm?
Relative Intensity
Part VI. consider the multi-step reaction below for compounds
A, B, and C.
These compounds were subjected to mass spectrometric analysis and
the following spectra for A, B, and C was obtained.
Draw the structure of B and C and match all three compounds
to the correct spectra.
Relative Intensity
Relative Intensity
100
HS-NJ-0547
80
60
31
20
S1
84
M+
absent
10
30
40
50
60
70
80
90
100
100-
MS2016-05353CM
80-
60
40
20
135 137
S2
164 166
0-m
25
50
75
100
125
150
m/z
60
100
MS-NJ-09-43
40
20
20
80
45
S3
25
50
75
100
125
150
175
m/z
Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following:
(a) match structures of isomers given their mass spectra below (spectra A and spectra B)
(b) Draw the fragments given the following prominent peaks from
each spectrum:
Spectra A m/2 =43 and 1/2-57
spectra B m/2 = 43
(c) why is 1/2=57 peak in spectrum A more intense compared
to the same peak in spectrum B.
Relative abundance
Relative abundance
100
A
50
29
29
0
10
-0
-0
100
B
50
720
30
41
43
57
71
4-0
40
50
60 70
m/z
43
57
8-0
m/z = 86
M
90 100
71
m/z = 86
M
-O
0
10 20 30
40 50
60
70
80
-88
m/z
90
100
Chapter 6 Solutions
General Chemistry: Principles and Modern Applications (11th Edition)
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