pH of the solution for 6 .14 × 10 − 3 M HI has to be calculated and the solution is acidic, basic or neutral also has to be identified. Concept introduction: The concentration of H 3 O + is calculating by using following formula, K w = K a × K b K w = [H 3 O + ][OH - ] = 1 .0 × 10 − 14 The Relationships Among pH, pOH, and pKw pH = - log (H 3 O + ) pOH = - log (OH − ) pK w = pH + pOH = 14 .00 The pH is a unit of measure which describe the degree of alkanity or acidity of a solution. pH = − log [ H + ] A solution may be acidic and basic depends on relative strength of anions as an acid or base. Anions and cations can be acidic, basic and netural. Acidic have pH < 7 , basic have pH > 7 and netural have pH=7 .
pH of the solution for 6 .14 × 10 − 3 M HI has to be calculated and the solution is acidic, basic or neutral also has to be identified. Concept introduction: The concentration of H 3 O + is calculating by using following formula, K w = K a × K b K w = [H 3 O + ][OH - ] = 1 .0 × 10 − 14 The Relationships Among pH, pOH, and pKw pH = - log (H 3 O + ) pOH = - log (OH − ) pK w = pH + pOH = 14 .00 The pH is a unit of measure which describe the degree of alkanity or acidity of a solution. pH = − log [ H + ] A solution may be acidic and basic depends on relative strength of anions as an acid or base. Anions and cations can be acidic, basic and netural. Acidic have pH < 7 , basic have pH > 7 and netural have pH=7 .
pH of the solution for 6.14×10−3 M HI has to be calculated and the solution is acidic, basic or neutral also has to be identified.
Concept introduction:
The concentration of H3O+ is calculating by using following formula,
Kw = Ka × Kb
Kw = [H3O+][OH-] = 1.0×10−14
The Relationships Among pH, pOH, and pKw
pH = -log(H3O+)pOH = -log(OH−)
pKw = pH + pOH = 14.00
The pH is a unit of measure which describe the degree of alkanity or acidity of a solution.
pH = −log[H+]
A solution may be acidic and basic depends on relative strength of anions as an acid or base. Anions and cations can be acidic, basic and netural. Acidic have pH < 7, basic have pH > 7 and netural have pH=7.
(b)
Interpretation Introduction
Interpretation:
pOH of the solution for 2.55MBa(OH)2 has to be calculated and the solution is acidic, basic or neutral also has to be identified.
Concept introduction:
The concentration of H3O+ is calculating by using following formula,
Kw = Ka × Kb
Kw = [H3O+][OH-] = 1.0×10−14
The Relationships Among pH, pOH, and pKw
pH = -log(H3O+)pOH = -log(OH−)
pKw = pH + pOH = 14.00
The pH is a unit of measure which describe the degree of alkanity or acidity of a solution.
pH = −log[H+]
A solution may be acidic and basic depends on relative strength of anions as an acid or base. Anions and cations can be acidic, basic and netural. Acidic have pH < 7, basic have pH > 7 and netural have pH=7.
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
Part IV. C6H5 CH2CH2OH is an aromatic compound which was subjected to Electron Ionization - mass
spectrometry (El-MS) analysis. Prominent m/2 values: m/2 = 104 and m/2 = 9) was obtained.
Draw the structures of these fragments.
Chapter 18 Solutions
Chemistry: The Molecular Nature of Matter and Change - Standalone book
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