CHM 111/112 LAB MANUAL >C<
11th Edition
ISBN: 9781337310956
Author: SLOWINSKI
Publisher: CENGAGE LEARNING (CUSTOM)
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Question
Chapter 17, Problem 1ASA
Interpretation Introduction
Interpretation:
The properties of a substance that definitely establish that it is a molecular material are to be stated.
Concept introduction:
Substances are classified on the basis of the kinds of bonding present in it. There are three types of substances.
- Ionic substances
- Molecular substances
- Metallic substances
Molecular substances have covalent bonds between the atoms, ionic substances have electrostatic forces between the ions, and metallic substances have metallic bonds.
Expert Solution & Answer
Answer to Problem 1ASA
Molecular substances are formed by the sharing of the electrons. They are volatile in nature, and do not conduct electricity either in solid or in the molten state.
Explanation of Solution
The properties of a substance that is molecular material are listed below.
- Molecular substances are formed by the sharing of the electron (covalent bond) between the atoms.
- Molecular substances are liquid at room temperature.
- Molecular substances are volatile in nature.
- Molecular substances do not conduct electricity either in solid or in the molten state.
- Molecular substances are soluble in a few organic solvents.
- Some molecular substances are polar in nature and they all are soluble in water.
Conclusion
Molecular are liquid at room temperature, volatile in nature, and do not conduct electricity either in solid or in the molten state. A few molecular substances are polar in nature.
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Calculating standard reaction free energy from standard reduction...
Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG° for the following redox reaction.
Be sure your answer has the correct number of significant digits.
3Cu+ (aq) + Cro²¯ (aq) +4H₂O (1) → 3Cu²+ (aq) +Cr(OH)3 (s)+5OH˜¯ (aq)
0
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☐ x10
00.
18
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Calculating the pH of a weak base titrated with a strong acid
An analytical chemist is titrating 241.7 mL of a 0.4900M solution of methylamine (CH3NH2) with a 0.7800M solution of HNO3. The pK of methylamine is
3.36. Calculate the pH of the base solution after the chemist has added 17.7 mL of the HNO3 solution to it.
Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO3 solution added.
Round your answer to 2 decimal places.
pH
=
☑
?
18
Ar
Chapter 17 Solutions
CHM 111/112 LAB MANUAL >C<
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