Review Session-2

Review Session-2 - REVIEW
SESSION‐
2
 GAS
LAWS


Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: REVIEW
SESSION‐
2
 GAS
LAWS
 INTERMOLECULAR
FORCES
 10.23)
If
the
atmospheric
pressure
is
0.985
atm,
what
is
the
pressure
of
 
the
enclosed
gas
in
each
of
the
three
cases
depicted
in
the
drawing?
 10.23)
If
the
atmospheric
pressure
is
0.985
atm,
what
is
the
pressure
of
 
the
enclosed
gas
in
each
of
the
three
cases
depicted
in
the
drawing?
 P
gas
=
Patm
+
Ph
 10.23)
If
the
atmospheric
pressure
is
0.985
atm,
what
is
the
pressure
of
 
the
enclosed
gas
in
each
of
the
three
cases
depicted
in
the
drawing?
 P
gas
=
Patm
+
Ph
 P
gas
=
0.985
–(52cm/76cm)
 












=
0.985‐
0.684

 








=
0.31
atm
 10.23)
If
the
atmospheric
pressure
is
0.985
atm,
what
is
the
pressure
of
 
the
enclosed
gas
in
each
of
the
three
cases
depicted
in
the
drawing?
 P
gas
=
Patm
+
Ph
 P
gas
=
0.985
–(52cm/76cm)
 












=
0.985‐
0.684

 








=
0.31
atm
 10.23)
If
the
atmospheric
pressure
is
0.985
atm,
what
is
the
pressure
of
 
the
enclosed
gas
in
each
of
the
three
cases
depicted
in
the
drawing?
 P
gas
=
Patm
+
Ph
 P
gas
=
0.985
–(52cm/76cm)
 P
 
=
0.985
+
(67mm/760mm)
 gas 












=
0.985‐
0.684

 








=
0.31
atm
 












=
0.985
+
0.088

 








=
1.073
atm
 10.23)
If
the
atmospheric
pressure
is
0.985
atm,
what
is
the
pressure
of
 
the
enclosed
gas
in
each
of
the
three
cases
depicted
in
the
drawing?
 Pgas
=(10.3cm/76cm)
 










=0.136
atm
 P
gas
=
Patm
+
Ph
 P
gas
=
0.985
–(52cm/76cm)
 P
 
=
0.985
+
(67mm/760mm)
 gas 












=
0.985‐
0.684

 








=
0.31
atm
 












=
0.985
+
0.088

 








=
1.073
atm
 10.31)
suppose
you
are
given
two
1‐L
flasks
and
told
that
one
contains
 a
gas
of
MW
30,
the
other
a
gas
of
MW
60,
both
at

same
Temperature.

 The
Pressure
in
flask
A
is
X
atm
and
mass
of
gas
is
1.2g.
The
Pressure
in

 flask
B
is
0.5
atm
and
mass
of
the
gas
is
1.2g.
Which
flask
contains
the
 gas
of
MW
30
and
which
contains
the
gas
of
MW
60?

 10.31)
suppose
you
are
given
two
1‐L
flasks
and
told
that
one
contains
 a
gas
of
MW
30,
the
other
a
gas
of
MW
60,
both
at

same
Temperature.

 The
Pressure
in
flask
A
is
X
atm
and
mass
of
gas
is
1.2g.
The
Pressure
in

 flask
B
is
0.5
atm
and
mass
of
the
gas
is
1.2g.
Which
flask
contains
the
 gas
of
MW
30
and
which
contains
the
gas
of
MW
60?

 PV
=
nRT
 P/n
=
RT/V
 P/n
=
Constant
 P
α
n
 P
α
Mass/MWt
 P
α
1/MWt
 
Lower
the
molecular
weight,
higher
will
be
the
P
(under
the
above
menZoned
condiZons)
 10.31)
suppose
you
are
given
two
1‐L
flasks
and
told
that
one
contains
 a
gas
of
MW
30,
the
other
a
gas
of
MW
60,
both
at

same
Temperature.

 The
Pressure
in
flask
A
is
X
atm
and
mass
of
gas
is
1.2g.
The
Pressure
in

 flask
B
is
0.5
atm
and
mass
of
the
gas
is
1.2g.
Which
flask
contains
the
 gas
of
MW
30
and
which
contains
the
gas
of
MW
60?

 PV
=
nRT
 P/n
=
RT/V
 P/n
=
Constant























You
can
get
this
answer
without
doing
this
calculaZon,
 
 
 
 
 







Think
about
the
number
of
gas
molecules
in
each
container..
 P
α
n
 P
α
Mass/MWt
 P
α
1/MWt
 
Lower
the
molecular
weight,
higher
will
be
the
P
(under
the
above
menZoned
condiZons)
 Calculate
the
raZo
of
rates
of
effusion
of
Helium
and
SO2
 Calculate
the
raZo
of
rates
of
effusion
of
Helium
and
SO2
 r
He
/
r
SO2
=
(M
SO2/
M
He)
½

 r
He
/
r
SO2
=
(64/4)
½

 r
He
/
r
SO2
=
4
 r
He
=
4
r
SO2

 Samples
of
HCl
and
(C2H5)NH2
were
placed
into
opposite
ends
of
a
 100
cm
glass
tube.
At
what
distance
from
the
HCl
end
of
the
tube
will
 the
two
gases
meet
and
react?
 HCl
 (C2H5)NH2
 Samples
of
HCl
and
(C2H5)NH2
were
placed
into
opposite
ends
of
a
 100
cm
glass
tube.
At
what
distance
from
the
HCl
end
of
the
tube
will
 the
two
gases
meet
and
react?
 (C2H5)NH2
 HCl
 Rate
α
1/
√MWt

 r
HCl
/
r
Amine
=
(M
amine/
M
HCl)
½

 r
HCl
/
r
Amine
=
(46/36.5)
½

 r
HCl
/
r
Amine
=
1.122
 r
HCl
=
1.122
X
r
Amine
 Samples
of
HCl
and
(C2H5)NH2
were
placed
into
opposite
ends
of
a
 100
cm
glass
tube.
At
what
distance
from
the
HCl
end
of
the
tube
will
 the
two
gases
meet
and
react?
 (C2H5)NH2
 HCl
 Rate
α
1/
√MWt

 r
HCl
/
r
Amine
=
(M
amine/
M
HCl)
½

 From
here
it
is
simple
math!
 r
HCl
/
r
Amine
=
(46/36.5)
½

 r
HCl
/
r
Amine
=
1.122
 r
HCl
=
1.122
X
r
Amine
 100
cm
distance
should
be
divided
into
 1.122
:
1
raZo
 r
HCl
=
(1.122/2.122)
X
100
=
52.9
cm
 r
amine
=
(1/
2.122)
X
100
=
47.1
cm
 Samples
of
HCl
and
(C2H5)NH2
were
placed
into
opposite
ends
of
a
 100
cm
glass
tube.
At
what
distance
from
the
HCl
end
of
the
tube
will
 the
two
gases
meet
and
react?
 100
cm
 HCl
 52.9
cm
 47.1
cm
 (C2H5)NH2
 Rate
α
1/
√MWt

 r
HCl
/
r
Amine
=
(M
amine/
M
HCl)
½

 From
here
it
is
simple
math!
 r
HCl
/
r
Amine
=
(46/36.5)
½

 r
HCl
/
r
Amine
=
1.122
 r
HCl
=
1.122
X
r
Amine
 100
cm
distance
should
be
divided
into
 1.122
:
1
raZo
 r
HCl
=
(1.122/2.122)
X
100
=
52.9
cm
 r
amine
=
(1/
2.122)
X
100
=
47.1
cm
 10.43)
Oxygen
consumpZon
for
male
cockroaches:
In
1
hr,
the
roach
 running
at
a
0.08
kmph
consumed
0.8mL
of
O2
at
1
atm,
24oC
per
 gram
of
insect
weight.

(a)
how
many
moles
of
O2
consumed
by
5.2g
 cockroach
moving
at
this
speed
in
1hr?
(b)
If
it
placed
in
a
1qt
jar,
 assuming
the
same
level
of
acZvity
as
in
research,
will
the
cockroach
 consume
more
than
20%
of
available
O2
in
48hr
period?
(air
is
21
 mole
percent
O2)
 10.43)
Oxygen
consumpZon
for
male
cockroaches:
In
1
hr,
the
roach
 running
at
a
0.08
kmph
consumed
0.8mL
of
O2
at
1
atm,
24oC
per
 gram
of
insect
weight.

(a)
how
many
moles
of
O2
consumed
by
5.2g
 cockroach
moving
at
this
speed
in
1hr?
(b)
If
it
placed
in
a1qt
jar,
 assuming
the
same
level
of
acZvity
as
in
research,
will
the
cockroach
 consume
more
than
20%
of
available
O2
in
48hr
period?
(air
is
21
 mole
percent
O2)
 (a) 
0.8ml
of
oxygen
in
1
hr
per
1gram
weight

 For
the
same
Zme
period,

cockroach
weighs
5.2g,

 the
O2
consumpZon
is

5.2X0.8
=

4.16mL
 How
many
moles
of
O2?
 PV=nRT
 
n=
PV/RT
=
(1atmX
0.00416L)/(0.0821L‐atm/mol.K)(297K)
 n=
1.71
X
10‐4Moles
 10.43)
Oxygen
consumpZon
for
male
cockroaches:
In
1
hr,
the
roach
 running
at
a
0.08
kmph
consumed
0.8mL
of
O2
at
1
atm,
24oC
per
 gram
of
insect
weight.

(a)
how
many
moles
of
O2
consumed
by
5.2g
 cockroach
moving
at
this
speed
in
1hr?
(b)
If
it
placed
in
a
1qt
jar,
 assuming
the
same
level
of
acZvity
as
in
research,
will
the
cockroach
 consume
more
than
20%
of
available
O2
in
48hr
period?
(air
is
21
 mole
percent
O2)
 (b)
In
1
hr,
cockroach
needs
1.71X10‐4
moles
of
O2
 In
48
hrs,
it
needs
48
X
1.71X10‐4
moles
=
8.2X10‐3
moles
of
O2
 Now
calculate
the
number
of
moles
in
jar:

 1
qt
=0.946L
of
air
(air
is
21
mole
percent
O2)
 Means,
available
oxygen
=
(0.946)
(0.21)
=
0.199
L
of
Oxygen
 Number
of
moles
of
Oxygen:
 PV=nRT
 n=
(1atmX
0.199L)/(0.0821L‐atm/mol.K)(297K)=
8.16X10‐3
moles
of
O2
 Cockroach
consumes
all
the
Oxygen
within
48
hrs
 10.56)
Both
Charles
and
Guy‐Lussac
were
avid
balloonists.
In
his
 original
flight
in
1783,
Charles
used
a
balloon
that
contained
31,150
L
 of
H2.
He
generated
H2
using
the
reacZon
between
Fe
and
HCl
 Fe(s)
+
2HCl
(aq)

FeCl2(aq)
+H2
(g).
How
many
kg’s
of
iron
were
 needed
to
produce
this
volume
of
H2,
if
the
temperature
was
22oC
 10.56)
Both
Charles
and
Guy‐Lussac
were
avid
balloonists.
In
his
 original
flight
in
1783,
Charles
used
a
balloon
that
contained
31,150
L
 of
H2.
He
generated
H2
using
the
reacZon
between
Fe
and
HCl
 Fe(s)
+
2HCl
(aq)

FeCl2(aq)
+H2
(g).
How
many
kg’s
of
iron
were
 needed
to
produce
this
volume
of
H2,
if
the
temperature
was
22oC
 Number
of
moles
of
Fe
reacted
=
Number
of
moles
of
H2
produced
 n
=
PV/RT
=
(1X
31150)/(295X
0.0821)
=
1.287X103
moles
of
H2
 1.287
X103
mol
Fe
X
55.845g/mol
=71872g
=
71.87
kg
of
Fe
 10.80)
Enriched
Uranium
can
be
produced
by
gaseous
diffusion
of
 UF6.Suppose
a
process
were
developed
to
allow
diffusions
of
gaseous
 Uranium
atoms,
U(g).
Calculate
the
raZo
of
diffusion
for
235U
and
238U
 and
compare
it
to
the
raZo
for
UF6
given
in
the
essay.
 10.80)
Enriched
Uranium
can
be
produced
by
gaseous
diffusion
of
 UF6.Suppose
a
process
were
developed
to
allow
diffusions
of
gaseous
 Uranium
atoms,
U(g).
Calculate
the
raZo
of
diffusion
for
235U
and
238U
 and
compare
it
to
the
raZo
for
UF6
given
in
the
essay.
 Isotopic
mass
of
235U:
235.04
amu
 Isotopic
mass
of
238U:
238.08
amu
 r
(235U)/
r
(238U)
=
(238.08/235.04)
½
 r
(235U)
=
1.0064
r
(238U)

 10.80)
Enriched
Uranium
can
be
produced
by
gaseous
diffusion
of
 UF6.Suppose
a
process
were
developed
to
allow
diffusions
of
gaseous
 Uranium
atoms,
U(g).
Calculate
the
raZo
of
diffusion
for
235U
and
238U
 and
compare
it
to
the
raZo
for
UF6
given
in
the
essay.
 Isotopic
mass
of
235U:
235.04
amu
 Mwt
of

235UF6:
349.04
amu
 Isotopic
mass
of
238U:
238.08
amu
 MWt
of
238UF6:
352.08
amu
 r
(235U)/
r
(238U)
=
(238.08/235.04)
½
 r
(235UF6)/
r
(238UF6)
=


 































(352.08/349.04)
½
 r
(235U)
=
1.0064
r
(238U)

 r
(235UF6)
=
1.0043
r
(238UF6)

 10.105)
You
have
a
sample
of
gas
at
‐33o
C.

You
wish
to
increase
the
 rms
speed
by
10.0%.

To
what
temperature
should
the
gas
be
heated?
 10.105)
You
have
a
sample
of
gas
at
‐33o
C.

You
wish
to
increase
the
 rms
speed
by
10.0%.

To
what
temperature
should
the
gas
be
heated?
 If
IniZal
velocity
is
100
the
final
velocity
should
be
110

 
 
 
 
(10%
Increase
in
velocity)
 V1
=
100
 V2
=110
 T1
=
‐33oC
=
240.15
K
 T2
=?
 10.105)
You
have
a
sample
of
gas
at
‐33o
C.

You
wish
to
increase
the
 rms
speed
by
10.0%.

To
what
temperature
should
the
gas
be
heated?
 If
IniZal
velocity
is
100
the
final
velocity
should
be
110%

 
 
 
 
(10%
Increase
in
velocity)
 u1
=
100
 u2
=110
 T1
=
‐33oC
=
240.15
K
 T2
=?
 u1/u2
=
(T1/T2)
½
 100/110
=
(240.15/T2)
½
 T2
=
240.15
(1.1)2

=
290.58
K
=
17.4oC
 10.84)
The
planet
Jupiter
has
a
surface
temperature
of
140K
and
a
 mass
318
Zmes
that
of
Earth.
Planet
Mercury
has
a
surface
 temperature
about
700K
and
a
mass
0.05
Zmes
that
of
earth.
On
 which
planet
the
atmosphere
more
likely
obey
Ideal
Gas
law?
Explain
 10.84)
The
planet
Jupiter
has
a
surface
temperature
of
140K
abd
a
 mass
318
Zmes
that
of
Earth.
Planet
Mercury
has
a
surface
 temperature
about
700K
and
a
mass
0.05
Zmes
that
of
earth.
On
 which
planet
the
atmosphere
more
likely
obey
Ideal
Gas
law?
Explain
 ‐ 
Ideal
gas
behavior
is
most
like
to
occur
at
high
T
and
Low
P
 ‐ 
High
T

KE
in
high

intermolecular
interacZons
have
low
influence!
 ‐ 
Low
mass
of
planet

low
gravitaZonal
force

atmospheric
pressure
on
Mercury
will
be
 


lower!
 10.88)
Calculate
the
pressure
that
CCl4
will
exert
at
40oC
if
1.00
mol
 occupies
28
L,
assuming
that
(a)
CCl4
obeys
the
ideal‐gas
equaZon
(b)
 Van
der
waals
equaZon
(c)
which
would
you
expect
to
deviate
more
 from
ideal
behavior
under
these
condiZons
Cl2
or
CCl4?
Why?
 a=
20.4
L2‐atm.mol2,

b=
0.1383
L/mol

 10.88)
Calculate
the
pressure
that
CCl4
will
exert
at
40oC
if
1.00
mol
 occupies
28
L,
assuming
that
(a)
CCl4
obeys
the
ideal‐gas
equaZon
(b)
 Van
der
waals
equaZon
(c)
which
would
you
expect
to
deviate
more
 from
ideal
behavior
under
these
condiZons
Cl2
or
CCl4?
Why?
 a=
20.4
L2‐atm.mol2,

b=
0.1383
L/mol

 P=
nRT/V
 P
=
(1
mol)
(0.0821
L‐atm/mol.K)
(313K)/
28L
=
0.917
atm
 10.88)
Calculate
the
pressure
that
CCl4
will
exert
at
40oC
if
1.00
mol
 occupies
28
L,
assuming
that
(a)
CCl4
obeys
the
ideal‐gas
equaZon
(b)
 Van
der
waals
equaZon
(c)
which
would
you
expect
to
deviate
more
 from
ideal
behavior
under
these
condiZons
Cl2
or
CCl4?Why?
 a=
20.4
L2‐atm.mol2,

b=
0.1383
L/mol

 P=
nRT/V
 P
=
(1
mol)
(0.0821
L‐atm/mol.K)
(313K)/
28L
=
0.917
atm
 P=
[nRT/V‐nb]
–
[an2/V2]
 P
=
[(1
X
0.0821
X
313)/
(28‐
(1X
0.1383))]
–
[
20.4
(1)2/
(28)2]=
0.896
atm
 Note: van der Waals result indicates that the real Pressure will be less than the ideal pressure . That is, the intermolecular forces reduce the effective number of particles and the real Pressure. This is reasonable for 1 mole of a gas at relatively Low T and P 10.88)
Calculate
the
pressure
that
CCl4
will
exert
at
40oC
if
1.00
mol
 occupies
28
L,
assuming
that
(a)
CCl4
obeys
the
ideal‐gas
equaZon
(b)
 Van
der
waals
equaZon
(c)
which
would
you
expect
to
deviate
more
 from
ideal
behavior
under
these
condiZons
Cl2
or
CCl4?
Why?
 a=
20.4
L2‐atm.mol2,

b=
0.1383
L/mol

 P=
nRT/V
 P
=
(1
mol)
(0.0821
L‐atm/mol.K)
(313K)/
28L
=
0.917
atm
 P=
[nRT/V‐nb]
–
[an2/V2]
 P
=
[(1
X
0.0821
X
313)/
(28‐
(1X
0.1383))]
–
[
20.4
(1)2/
(28)2]=
0.896
atm
 Note: van der Waals result indicates that the real Pressure will be less than the ideal pressure . That is, the intermolecular forces reduce the effective number of particles and the real Pressure. This is reasonable for 1 mole of a gas at relatively Low T and P a,
b
values
of
Cl2
are
lower
than
the
a,
b
values
of
CCl4.
Means,
CCl4

 experiences
stronger
intermolecular
forces.
(larger
volume
of
CCl4).

 CCl4
deviate
more
from
ideal
behavior
than
Cl2
under
these
condiZons.

 11.14)
Based
on
your
understanding
on
intermolecular
forces,
would
 you
say
that
maser
is
fundamentally
asracted
or
repulsed
by
other
 maser?
 IMF’s
are
based
on
forces
of
asracZon
and
repulsion.
While
 repulsions
occur,
the
net
forces
are
asracZve
because
the
asracZve
 forces
lower
the
overall
energy
of
the
system
(gas,
liq
or
solid)
and
 maser
tends
to
exist
in
lowest
possible
energy
state
 11.16)
What
type
of
IMF’s
accounts
for
the
following
differences
in
 each
case.
 (a)  CH3OH
boils
at
65oC
and
CH3SH
boils
at
6oC
 (b)  
Xe
is
liquid
at
1
atm,
120K,
where
as
Ar
is
a
gas
 (c)  
Kr,
atomic
wt
84
boils
at
121K,
where
as
Cl2,
MWt:
71
boils
at
 238K
 (d)  Acetone
boils
at
56oC
and
2‐methyl
propane
boils
at
‐12oC
 O C H3C CH3 CH CH3 Acetone H3C CH3 2-Methylpropane 11.16)
What
type
of
IMF’s
accounts
for
the
following
differences
in
 each
case.
 (a)  CH3OH
boils
at
65oC
and
CH3SH
boils
at
6oC
:
Hydrogen
Bonding
 (b)  
Xe
is
liquid
at
1
atm,
120K,
where
as
Ar
is
a
gas
 (c)  
Kr,
atomic
wt
84
boils
at
121K,
where
as
Cl2,
MWt:
71
boils
at
 238K
 (d)  Acetone
boils
at
56oC
and
2‐methyl
propane
boils
at
‐12oC
 O C H3C CH3 CH CH3 Acetone H3C CH3 2-Methylpropane 11.16)
What
type
of
IMF’s
accounts
for
the
following
differences
in
 each
case.
 (a)  CH3OH
boils
at
65oC
and
CH3SH
boils
at
6oC
:
Hydrogen
Bonding
 (b)  
Xe
is
liquid
at
1
atm,
120K,
where
as
Ar
is
a
gas:
London
 Dispersion
forces‐
Heavier
the
gas,
stronger
the
interacZon
 (c)  
Kr,
atomic
wt
84
boils
at
121K,
where
as
Cl2,
MWt:
71
boils
at
 238K
 (d)  Acetone
boils
at
56oC
and
2‐methyl
propane
boils
at
‐12oC
 O C H3C CH3 CH CH3 Acetone H3C CH3 2-Methylpropane 11.16)
What
type
of
IMF’s
accounts
for
the
following
differences
in
 each
case.
 (a)  CH3OH
boils
at
65oC
and
CH3SH
boils
at
6oC
:
Hydrogen
Bonding
 (b)  
Xe
is
liquid
at
1
atm,
120K,
where
as
Ar
is
a
gas:
London
 Dispersion
forces‐
Heavier
the
gas,
stronger
the
interacZon
 (c)  
Kr,
atomic
wt
84
boils
at
121K,
where
as
Cl2,
MWt:
71
boils
at
 238K:
London
Dispersion
forces:
Diatomic,
larger,
more
polarizable

 (d)  Acetone
boils
at
56oC
and
2‐methyl
propane
boils
at
‐12oC
 O C H3C CH3 CH CH3 Acetone H3C CH3 2-Methylpropane 11.16)
What
type
of
IMF’s
accounts
for
the
following
differences
in
 each
case.
 (a)  CH3OH
boils
at
65oC
and
CH3SH
boils
at
6oC
:
Hydrogen
Bonding
 (b)  
Xe
is
liquid
at
1
atm,
120K,
where
as
Ar
is
a
gas:
London
 Dispersion
forces‐
Heavier
the
gas,
stronger
the
interacZon
 (c)  
Kr,
atomic
wt
84
boils
at
121K,
where
as
Cl2,
MWt:
71
boils
at
 238K:
London
Dispersion
forces:
Diatomic,
larger,
more
polarizable

 (d)  Acetone
boils
at
56oC
and
2‐methyl
propane
boils
at
‐12oC
 O C H3C CH3 CH CH3 Acetone H3C CH3 2-Methylpropane Similar
London
Dispersion
forces‐
but
Acetone
experience
 dipole‐dipole
interacZons
too!
 11.18)
True
or
False:
 a)  The
more
polarizable
the
molecules,
the
stronger
the
dispersion
 forces
between
them
 b)  The
boiling
point
of
noble
gases
decreases
as
you
go
down
the
 column
in
the
periodic
table
 c)  In
general
smaller
the
molecule,
the
stronger
the
dispersion
forces
 d)  All
other
factors
being
the
same,

dispersion
forces
between
 molecules
increase
with
the
number
of
electrons
in
the
molecule
 11.18)
True
or
False:
 a)  The
more
polarizable
the
molecules,
the
stronger
the
dispersion
 forces
between
them:
TRUE
 b)  The
boiling
point
of
noble
gases
decreases
as
you
go
down
the
 column
in
the
periodic
table
 c)  In
general
smaller
the
molecule,
the
stronger
the
dispersion
forces
 d)  All
other
factors
being
the
same,

dispersion
forces
between
 molecules
increase
with
the
number
of
electrons
in
the
molecule
 11.18)
True
or
False:
 a)  The
more
polarizable
the
molecules,
the
stronger
the
dispersion
 forces
between
them:
TRUE
 b)  The
boiling
point
of
noble
gases
decreases
as
you
go
down
the
 column
in
the
periodic
table:
FALSE
 c)  In
general
smaller
the
molecule,
the
stronger
the
dispersion
force
 a)  All
other
factors
being
the
same,

dispersion
forces
between
 molecules
increase
with
the
number
of
electrons
in
the
molecule
 11.18)
True
or
False:
 a)  The
more
polarizable
the
molecules,
the
stronger
the
dispersion
 forces
between
them:
TRUE
 b)  The
boiling
point
of
noble
gases
decreases
as
you
go
down
the
 column
in
the
periodic
table:
FALSE
 c)  In
general
smaller
the
molecule,
the
stronger
the
dispersion
force
 






FALSE
 a)  All
other
factors
being
the
same,

dispersion
forces
between
 molecules
increase
with
the
number
of
electrons
in
the
molecule
 11.18)
True
or
False:
 a)  The
more
polarizable
the
molecules,
the
stronger
the
dispersion
 forces
between
them:
TRUE
 b)  The
boiling
point
of
noble
gases
decreases
as
you
go
down
the
 column
in
the
periodic
table:
FALSE
 c)  In
general
smaller
the
molecule,
the
stronger
the
dispersion
force
 






FALSE
 a)  All
other
factors
being
the
same,

dispersion
forces
between
 molecules
increase
with
the
number
of
electrons
in
the
molecule
 






TRUE
 11.20)
Which
member
of
the
following
pairs
has
the
strongest
 intermolecular
dispersion
forces?
 (a)  Br2
or
O2
 (b)  
CH3CH2CH2CH2SH

or

CH3CH2CH2CH2CH2SH
 (c)  CH3CH2CH2Cl


or
(CH3)2CHCl
 11.20)
Which
member
of
the
following
pairs
has
the
strongest
 intermolecular
dispersion
forces?
 (a)  Br2
or
O2

 (b)  
CH3CH2CH2CH2SH

or

CH3CH2CH2CH2CH2SH
 (c)  CH3CH2CH2Cl


or
(CH3)2CHCl
 11.20)
Which
member
of
the
following
pairs
has
the
strongest
 intermolecular
dispersion
forces?
 (a)  Br2
or
O2

 (b)  
CH3CH2CH2CH2SH

or

CH3CH2CH2CH2CH2SH
 (c)  CH3CH2CH2Cl


or
(CH3)2CHCl
 11.20)
Which
member
of
the
following
pairs
has
the
strongest
 intermolecular
dispersion
forces?
 (a)  Br2
or
O2

 (b)  
CH3CH2CH2CH2SH

or

CH3CH2CH2CH2CH2SH
 (c)  CH3CH2CH2Cl


or
(CH3)2CHCl
 ...
View Full Document

Ask a homework question - tutors are online