The “Chemistry in Focus” segment The Beetle That Shoots Straight discusses the bombardier beetle and the chemical reaction of the decomposition of hydrogen peroxide. m:math> H 2 O 2 ( a q ) → H 2 O ( l ) + O 2 ( g ) e balanced equation given in the segment is m:math> 2 H 2 O 2 ( a q ) → 2 H 2 O ( l ) + O 2 ( g ) y can’t we balance the equation in the following way? m:math> H 2 O 2 ( a q ) → H 2 ( g ) + O 2 ( g ) e molecular-level pictures like those in Section 6.3 to support your answer.
The “Chemistry in Focus” segment The Beetle That Shoots Straight discusses the bombardier beetle and the chemical reaction of the decomposition of hydrogen peroxide. m:math> H 2 O 2 ( a q ) → H 2 O ( l ) + O 2 ( g ) e balanced equation given in the segment is m:math> 2 H 2 O 2 ( a q ) → 2 H 2 O ( l ) + O 2 ( g ) y can’t we balance the equation in the following way? m:math> H 2 O 2 ( a q ) → H 2 ( g ) + O 2 ( g ) e molecular-level pictures like those in Section 6.3 to support your answer.
Solution Summary: The author explains the reason why the given balanced equation of decomposition of hydrogen peroxide cannot be balanced in a differently mentioned way.
The “Chemistry in Focus” segment The Beetle That Shoots Straight discusses the bombardier beetle and the chemical reaction of the decomposition of hydrogen peroxide.
m:math>
H
2
O
2
(
a
q
)
→
H
2
O
(
l
)
+
O
2
(
g
)
e balanced equation given in the segment is
m:math>
2
H
2
O
2
(
a
q
)
→
2
H
2
O
(
l
)
+
O
2
(
g
)
y can’t we balance the equation in the following way?
m:math>
H
2
O
2
(
a
q
)
→
H
2
(
g
)
+
O
2
(
g
)
e molecular-level pictures like those in Section 6.3 to support your answer.
Definition Definition Transformation of a chemical species into another chemical species. A chemical reaction consists of breaking existing bonds and forming new ones by changing the position of electrons. These reactions are best explained using a chemical equation.
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
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell