The order of the intermolecular forces and increasing boiling, melting points and enthalpy of vaporization are should be determined. Concept Introduction: Intermolecular forces: The forces involving within the molecule is called intermolecular forces thus, Ion-dipole forces: The attraction forces between anion charge end with opposite partial charge is called Ion-dipole forces. Dipole- Dipole forces: The attraction forces between ends of opposite partially charge in polar molecules is called Dipole- Dipole forces. Hydrogen bond: The weak bond between highly electronegative atom with partially positively charged Hydrogen is called Hydrogen bonding. London Dispersion Forces: The London dispersion force is the weaker force within the non polar molecules and which can be used to explain the state of the non polar molecules. The intermolecular forces are higher means the boiling point, melting points and enthalpy of vaporization of the compound is all so higher.
The order of the intermolecular forces and increasing boiling, melting points and enthalpy of vaporization are should be determined. Concept Introduction: Intermolecular forces: The forces involving within the molecule is called intermolecular forces thus, Ion-dipole forces: The attraction forces between anion charge end with opposite partial charge is called Ion-dipole forces. Dipole- Dipole forces: The attraction forces between ends of opposite partially charge in polar molecules is called Dipole- Dipole forces. Hydrogen bond: The weak bond between highly electronegative atom with partially positively charged Hydrogen is called Hydrogen bonding. London Dispersion Forces: The London dispersion force is the weaker force within the non polar molecules and which can be used to explain the state of the non polar molecules. The intermolecular forces are higher means the boiling point, melting points and enthalpy of vaporization of the compound is all so higher.
Solution Summary: The author explains the order of intermolecular forces and increasing boiling, melting points and enthalpy of vaporization of given compounds.
LTS
Solid:
AT=Te-Ti
Trial 1
Trial 2
Trial 3
Average
ΔΗ
Mass water, g
24.096
23.976
23.975
Moles of solid, mol
0.01763
001767
0101781
Temp. change, °C
2.9°C
11700
2.0°C
Heat of reaction, J
-292.37J -170.473
-193.26J
AH, kJ/mole
16.58K 9.647 kJ 10.85 kr
16.58K59.64701
KJ
mol
12.35k
Minimum AS,
J/mol K
41.582
mol-k
Remember: q = mCsAT (m = mass of water, Cs=4.184J/g°C) & qsin =-qrxn &
Show your calculations for:
AH in J and then in kJ/mole for Trial 1:
qa (24.0969)(4.1845/g) (-2.9°C)=-292.37J
qsin =
qrxn =
292.35 292.37J
AH in J = 292.375 0.2923kJ
0.01763m01
=1.65×107
AH in kJ/mol =
=
16.58K
0.01763mol
mol
qrx
Minimum AS in J/mol K (Hint: use the average initial temperature of the three trials, con
Kelvin.)
AS=AHIT
(1.65×10(9.64×103) + (1.0
Jimai
For the compound: C8H17NO2
Use the following information to come up with a plausible structure:
8
This compound has "carboxylic acid amide" and ether functional groups.
The peaks at 1.2ppm are two signals that are overlapping one another.
One of the two signals is a doublet that represents 6 hydrogens; the
other signal is a quartet that represents 3 hydrogens.
Vnk the elements or compounds in the table below in decreasing order of their boiling points. That is, choose 1 next to the substance with the highest bolling
point, choose 2 next to the substance with the next highest boiling point, and so on.
substance
C
D
chemical symbol,
chemical formula
or Lewis structure.
CH,-N-CH,
CH,
H
H 10: H
C-C-H
H H H
Cale
H 10:
H-C-C-N-CH,
Bri
CH,
boiling point
(C)
Сен
(C) B
(Choose
Chapter 9 Solutions
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