(a) Interpretation: The reason for no effect of the presence of sodium carbonate in sodium hydroxide solution on the experimental results should be determined. Concept introduction: The substance that releases OH − ions when dissolved in water is known as base. The bases can be classified into two types on the basis of the strength. The two types are as follows: 1. Strong Bases 2. Weak Bases The bases that dissociate completely into the respective ions when dissolved in the solution is known as strong bases while the bases that dissociate partially into the respective ions when dissolved in the solution is known as liquid bases.
(a) Interpretation: The reason for no effect of the presence of sodium carbonate in sodium hydroxide solution on the experimental results should be determined. Concept introduction: The substance that releases OH − ions when dissolved in water is known as base. The bases can be classified into two types on the basis of the strength. The two types are as follows: 1. Strong Bases 2. Weak Bases The bases that dissociate completely into the respective ions when dissolved in the solution is known as strong bases while the bases that dissociate partially into the respective ions when dissolved in the solution is known as liquid bases.
Solution Summary: The author explains that the reason for no effect of the presence of sodium carbonate in sodium hydroxide solution should be determined.
Definition Definition Method of quantitative analysis for determining the concentration of an acid or base by exactly neutralizing it with a standard solution of base or acid having known concentration.
Chapter 5, Problem 73IAE
Interpretation Introduction
(a)
Interpretation:
The reason for no effect of the presence of sodium carbonate in sodium hydroxide solution on the experimental results should be determined.
Concept introduction:
The substance that releases OH− ions when dissolved in water is known as base. The bases can be classified into two types on the basis of the strength. The two types are as follows:
1. Strong Bases
2. Weak Bases
The bases that dissociate completely into the respective ions when dissolved in the solution is known as strong bases while the bases that dissociate partially into the respective ions when dissolved in the solution is known as liquid bases.
Interpretation Introduction
(b)
Interpretation:
The reason for effect of the presence of sodium carbonate more than 1% to 2% on the experimental results should be determined.
Concept introduction:
The substance that releases OH− ions when dissolved in water is known as base. The bases can be classified into two types on the basis of the strength. The two types are as follows:
1. Strong Bases
2. Weak Bases
The bases that dissociate completely into the respective ions when dissolved in the solution is known as strong bases while the bases that dissociate partially into the respective ions when dissolved in the solution is known as liquid bases.
3. Consider the compounds below and determine if they are aromatic, antiaromatic, or
non-aromatic. In case of aromatic or anti-aromatic, please indicate number of I
electrons in the respective systems. (Hint: 1. Not all lone pair electrons were explicitly
drawn and you should be able to tell that the bonding electrons and lone pair electrons
should reside in which hybridized atomic orbital 2. You should consider ring strain-
flexibility and steric repulsion that facilitates adoption of aromaticity or avoidance of anti-
aromaticity)
H H
N
N:
NH2
N
Aromaticity
(Circle)
Aromatic Aromatic Aromatic Aromatic Aromatic
Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic
nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic
aromatic TT
electrons
Me
H
Me
Aromaticity
(Circle)
Aromatic Aromatic Aromatic
Aromatic Aromatic
Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic
nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic
aromatic πT
electrons
H
HH…
A chemistry graduate student is studying the rate of this reaction:
2 HI (g) →H2(g) +12(g)
She fills a reaction vessel with HI and measures its concentration as the reaction proceeds:
time
(minutes)
[IH]
0
0.800M
1.0
0.301 M
2.0
0.185 M
3.0
0.134M
4.0
0.105 M
Use this data to answer the following questions.
Write the rate law for this reaction.
rate
= 0
Calculate the value of the rate constant k.
k =
Round your answer to 2 significant digits. Also be
sure your answer has the correct unit symbol.
Chapter 5 Solutions
General Chemistry: Principles And Modern Applications Plus Mastering Chemistry With Pearson Etext -- Access Card Package (11th Edition)