Exercises-WS22-23-A. Casini
Tutorial I
e
138
a)
-
=
protons
138-56
1s0Ba:
-> 82
neutrans
+
16 neutrons,
an-15=16_,
b)
atomic
82
electrons & 56 protans
56
neutrons,
=
15&15p
amu
weight: 0,9223.27.97693 amu
+
0,0468.
28,97649
97377
+0,0309. 29,
=)
8. 3.203
28,0884
amu
+0.7205
0.9998. 1+0.000142
it's the
value
9
9 neutrons
protons, 102,
13p; son;
530;
-
-
->
FiT; I;
Th
ton;
op; non;
sse
54ei
so ei
same
given
204.4
=
+000000.3=
0.5069.79+ 0.4931.81
-,
amu
mass
in the
=
amu
1,000083
79.9862
as
the
periodic table
amo
am u
Tutorial z
a) emits
b)absabs
energy
energy
Tutorial 3
+
2
+
au
4
G
02
02
2
-
(2 0
2H2
->
+
302
+
2
->
+6HC
2 A
CH1206
+
2CuO
->
+
CH4
4 Fe
Hz
2 HcO->2NaOH
2 Na
FezOs
2A213
3Hz
->
602
+
+
6202
->
6H20
+
C3H8
502
+
4549
n 2 10.3md
m
nc
=
->
5 mal
=
+4Hz0
3CO2
nsmal n=mal
=
=
mn
=
M(0)
m=
-
02
2H
n
m
=
m
>2
+
1509
=
m
ngM(O2
=niMCHic
H20
1500g
=
n
46,875md
n
=
n(Hz)
=
m n
=
mol
=75
-
c
Grams
Mass
0.561502.
=
H 0.306
=
of
0
gH20.
mass
=
=0.255
=0.1539c
motoma
=0.0343gH
(sample)
-
(mass
20.153 0.8343)
+
-
0.06890
=
0.0043
7.9;0
=
Exercise 5
[PtC]2 +
m
n
2NHy
->
[PELNAS), Cr]
15.5gm 1,45g
n
=
0,05ma
=
n
=
->
0,09
=
m
limiticg
reactant:
0.0912
0.045 0.5
=
+2C-
2n(Nis)
=
n
=
-
M
12,1545g
=
-
n
>
2
=
n
0.0128
=
0.0340md
-
n 0.0043
=
H
=
i
=
ma
1
=
O
=>
be
mal
+
=
H 0.0043
=3.0;
-
can
mass (H))
(C)
0.0340
0.0128
C:
Minis
1358gH20
reactant
3
Exercise
grams
limiting
M
=
C3H80
formed
Tutorial 4
Me
E
m
has
->
be it
number
·
3b)
3a3
P:
152522p4
0:
the
impossible
1525p63573pe
T
T
3-NV
2
2
-
IV
1V
+
1
--
1
s
·
-
T
l
-
TV
444
TVTrII
Ir
s
·
l
energy,
greatest
electrons
of parallel
n
n
lowest
has the
Tutorial
st
removed
are
Na":
Felt:
152s32p6
p5: 1s'252p6353p6
1s"25'2p63513p6453d"-Zei
is orbital because
metal
a transition
0.5
+
3HC
->
mol
n=3ne
m=
0,500
·25am
Arlg)
BEs
+0,347
+
5, 87 mal Bi.
9,8089 Bi.
G
Inar Bi
209,
lams
6,022. 1843
1
atams
mor Bi
=
ogBi
Bi
=
5,87 mal
3,53.18"
0.70rg
=
1,
0,70n=1,
(Ne) 35"3p64s/
Bi.
+
Rg=Ml
=0,00s4
0 855
AUz
3H2O
9
=
[Ne) 3923p6
it is
Alst: 1s2s22p6
m=
+->
the
-Is'25"2p63573p63d6
RCOHSs
Arcg)
from
Bi
atoms
Bi
202
con
m
=
4;M
Dr Rz
=
m=
=0,8559
Sarethe
same
=
ne
0,347 g
ecoiserved
Tutorial 6
diffe
a) isotopes
b) ir
1
numbers
in
of
c)
hieran acupit
+
shell--2s&ept
-
I
15
!
-
-
core
electrons
average
the
c
->
b)
valece
4
->
é
I
-
20
therefore
=>
-
2
[r
0
-
c
->
node, whereas
the
is orbital
has
0
-
formal changes of
therefore
expectany
we
-==8] [.=-ä
10 Al
e
=
1
difference
electronegativity
-
↳ places
-
0
-
oxygen,
on
negativechange
el
18
0,
1
=
conic
conic
no
nodes
1
magnitude.
the
smallest
negative formal change
0:)
=
Si
1
one
-
left
rules outthe
reside
to
on
C
electronegative
the
most
the
Natur
ineine
elementin the
Fe
=
18
-Öl
j
Li
covalent
in
-Li
-
-
-
ionic
conic
values
-
-
H
covalent
ionic bonds:are
non metal
metal
->
by
formed
the
transgel
atoms
↳ have
one
dominantone!
is the
Lewis structure
middle
the
, not sureig
correct
->
greate
F
produces
generally
Is e
than thatof the
is
e
⑧
than cors->
the
es
nucleus
e
configuration
abital
have sé
both
-
100
+
electronegative
more
valence
5
I
3) the
dominantLewis structure
wis
c)
energy
the
from
the
exactsame, since
Is
the
has
Is orbital
2 valence e
->
orbital)
Is
than
in
Cs
ofthe
orbital
distance
their electronic
so
-
is lower
orbital
the
is
1
3
a)
the
nucleus
neutrons in
is
between
of
one
or
two
on
more
more
electrons
atoms
between
atoms are bound
covalentbond:the
shared electrons
->
values
differentelectronegativity
electronegativity
non metal
values
be close
need to
nonmetal
+
by
Fe
al
=
Tutorial
7
AH-TAS
Ap=
-176.165)
=
91,
=-
+
298284, S
13kJ
spontaneous
question
is
at
250
-
need
we
so
atroom Temp
spontaneous
is
the reaction
if
asking
know
to
reaction:
BH-ASj T
=
Atty
(-
=
=BHg
2500
+
27.34)
(21,90
=
=
+
2.58,64)
callkmol
21.12
= 5,476 kcallmol
reaction
is
>C
·
Exercise
2
calculate
EnAHjproduct
AH:
-
Emstigreactant
=AH=(-132.51) +(-205.00)
=>
BH
calculate
·
5:
=>
=
=A5
=>
=
Dp=
188.72
+
*m
-
Asgreactant
146.40)
-
As
0
=,
by
=
=
-
241,82
188,
28.05.183-298.nos
so
-
is
reaction
at room
->
285,53
=)
=
8-70, 0
=
=
+
At
T=
DH
D5
=
=
44,
oh
nflmol
118,88/kmal
(151.08)
ZNV
AI-T.A5
=54=- 4,134
=
19=O
equilibrium,
is at
process
AHO-TAS
At
AS:
(13.40
=
=>
H2Olg
lf
28.05
EnA5gproduct
15
(-365.56)
-
-
this
when
direction
other
4
HOces
the reaction
for
All
for formation of
Brelg) and non-sportal
spontaneous
Caris) and
for the
Exercise
133.2)
-
-30.12
x
-
of the
11.77 Ucal/mol
=
A5g
either
Gibbs energy for
calculate
AG
D9
spontaneous
temperature
will dissolve
NHyNO3(s)
water at room
temperature
in
T=
44,01183
118,
s
=
378,45K
=
97,452
Tutorial
S
16e
~>
er
-
1
mit
e
--
IV
or
3
1S
Oz
es
---------.
.....
Na -> 14e
molecular orbital
The
eegy
Oxygen's paramagnetism
electrons in the
b) the
bond
roo
first
owing
ist
-
E
ofNe
order
excited state
the
in
Ei
impaired
2
two
e
unpaired
abirat.
a
state
bond order
bond ander.
reduction
the
to
ground
the
in
ithas
in
er
Nein
3;
is
of
2.
excited state
the
first
has
a
N-Nbord.
weaken
ein
1
IV..
26
---
.....
AO
Analyze:we
of
N
will
828......
TV
25
1..------
-
Esen
AuofO
determine
electrons
comsine
populate
we
bond to
exist,
the
icon
londing
abital and
bond
thus, the
third
order
is
Because
the
bond order
is
and
Het. Formation
of
use
like thatfor He
on
Hez.
is stable.
you
energy-level diagram
the
solve:the
and
in
predictwether
Heation
the
is
of
of
He
is
molecule
electrons in the
are
in
bond
If the
shown. This ion has
antifonding
the
ion
to
it
The
valence electrons
MOs.
diagram
MO
an
ion
number
mustdetermine the
available
the
giveand
to
He
for the
oder
bond
the
bond order,
plan:To determine the
He
predicts
paramagnetic
a)
1
15
is
2S
N:
3
O2
diagram for the
explained
by presence of
antibonding ITzpy i2px) molecular
re
-
s
is
the
oder
is
orbital,
there
and how
orbitals
is
and the
greater than
electrons. Two
three
stable.
am
0,
we
peaced
a
expect
in the
orbital.
(2-1) I
=
greate
than
Heatin
0,
we
gas
the
Heation
predictthe
phase has been
the
separated
rlative to
be stable
to
demonstrated
in
laboratory
experiments.
a
Tutorial
elements
pean:
En: relement
↓
to
compare the relative
and
plan:
solve:
An
common
we
Ga:
As:
2:
semiconductor
n-type
must
look
more
same
->
we
-
that
means
if
need
of
an
element
of
and
have
must
determine the number
doped
As is
si,
into
it
would
table
ig:
gap"on
band
difference
case
compounds
atoms
dopant
periodic
difference for
En
and
se
Gapgapins
and
table,
where
higher up inand periodic orbitals
antizonding
banding
the
enhanced
elements
separation between the
bond
with
element
increases, which beads
from group
valence
of
yield
valence
more
an
than the
associated
host
with
material.
si,Ga,
Silicon-host material
As and c.
the elements with
in
this
more
than silicon
are the ones that will
produce
an
semiconductor
n-type
FALSE
a) Ag
b)En
imic solids
covalent-network solids
3)
2
(H)
(1)
+ X
+X
+352)
x
+
2(1)
+
x
2Ct)
+
3(t)
2+
11-2)
2x
2x
2x
so
(23
+4
+372)
2x
=
=
=
=
+374=0
+1(2)=0
0
=
0
0
0
=
0
=
0
=
x
=
=
x
X
=
=
=
3:
6:
3:
·
x
=Z
=
=
x
=
3:
@3t
x
=
1:
A
via
pSt
4
t
5
Mnst
Felt
Mg?" -hydrid-generalM
='*Es: "Es
=,
are
become electrical conductors
can
=
0
an
increase in the
electronegativity
case
valence
FALSE
X
to
gap
T RVE
->
a
periodic table:
in the
belows
is
(fexample
As
to
>bandgap gas
than si
valence
gewer
5t-Bor
leads
polarity
as sie
value
orbital overlap
between
gap
4Ensband
the
the
in
located
the horizontal
each
in
elements
the
electronegativity
in
band
both
si
si:
valence than
number
to
gap
periodic table
at the
less valence than
right of
Sione element to the
Inseband
Analyze:
positions of
the elements
4-GaNbandgap >Gap
than
is
zn
Ba
and horizontal positions
vertical
on
left of go;
the
higher
positioned
is
gap depends
periodic table
look at the
solve:
band
of
size
Anale:
are
langer speitting
much more
likely
to
chemical substitution
dissolve
in
water
type
material
upon
doping
Tutorial
10
FR
A
T
-
In
X
260
pp
360
BP: 360K
FP:
·
·
260 K
gas
solid
:liquid
the triple
point
is
approximately
at
185K
0.45 atm
i) Consider 1
-
H=
Cu=409;
nicul=
in)
kg of
Al-4utoCh:
960g
= 0,63; ni)=
=
Arich
35.6 mal
solid
single phase
Al-4Wt Cur
composition: alloy composition
in) at 250: two solid phases:
at
·
5502:
Cessentially) pure
+
compound
->
->
-
increase
Freezing=melting point
increase: the resistance
->
53%wtdu
increase
increase
->
composition
decrease
->
->
Al
Cullz
slowey
increase
->
increase
it
of
flows
a
liquid flow
is
called
visiosity. The
greate
a
liquid's viscosity, the
more
*
A
230
⑭
-A
we
130(
*
Pb-YowtYosn,
liquid+Pb-rick
↳liquid
solid
at
to
temp:
230°C: liquid
at const.point
the phases
approx.
pb-rich solid).
at
183°:
to
the
liquid
at
+
phases
changes
field
2(Liquid
+
4b-ich solid
ii)
at
approx
150°,
1302:
&
initial
changes
phace
to
is
changes
constpants auch solid
->
1) Pb-5w+Yosn;
3000
Sn-rich solie
Pb-rich solid
final phases: abrich solchsolid
·
solid (nzwosn
250c: ph-rich
(34wt%oSm>
and Liquid
1602: ob-rich
and
solid
diquid (57utySn)
solid Gowtososn
150C abrick
phase compositions
maintained by
changes
As the
change
in the
with
temperature,
proportions of
the overall
the
places
composition
is
Snswt%osn
Tutorial e
Given:
Vn
>n==
1
net
VeiRz
=
ideal
gas
Pc
=
V2=?
Find:
=
1,92;
ne
0.0210 mal
=
mol
0.1130
1.92.0913Md
=
2.33 L
Pil: Pil,
law:
=
166, 67
~immig
Vn
=
c.001;
PV
=
the amount
P2
Pitz=5600j00atm
=
mal
0,920
=
L
n
=
760.0
=102.6
=
mmty;
4= (MxIntor) Prot
·
T
=
MRetrosoon
->
=
dene
·ana
men
Verlinder
T
=
=
8.10-
de?
=nie
internal
ideal
gas:
Sro-L
300K
-Oseckcamal"
MNe
=
=nie.Mue
energy:
=
Ot e
or=Mupa.1.4Pa.mol
300K
8.314
=
3.21.10-5 mal
6.4818Yg
Eit
=
1.5NwKBT
P.
AT
20c
=
·But
->
PNs
PNcINN
=
nacRTIV
=
not
change,
the same
293V
Dr
=
Ra
0,432
1.5KBT
RTIV
=
=
RT/v]
=
1,54Ne'R T
=
0133
=
cPNs=(RNeIntt)Pot
DN2=4NPotIRtt
=
PotInot
=
1.5
RT/NA
18
J
=0.055 el
an
same
remains
40, +4Nc
=5.65.
as
the
bal
Pot: RotR.T/V
Enin
behave
is
volume does
430000 Pa
Ptot
We
PiVte,
the
0.135 bar
=
Assuming
substance
at m
re_242=K Peret
=
=
of
=
19.4.3/100
=
3.4 bar
=
3.418 Pa
Tutorial 12
if
a)
compare experiments 122,
we
how SB]
when SB]
affects
doubled,
is
experiments
in
the
rate,
b)
rate
Using the
=
[B]
Because
(4.0
notpart
is
constant,
a)
part
10
-
CNOz
(g)
>
2NOCy1
02
+
(g)
=>
Es +k.t
rate
k.
=
[NOz]
[N0,33600
=>
we
law, iti s
solve:To
can
plotbe SNOC and MINO
graph
en [NOz] and
M/[NO2]
time.
against
againsttime, we
other
one on the
will
will be
linean, indicating whether
firstprepare following from
the
.
.
table
reaction
the
From
the
slope
the
data
obeys
of
this
a
faction
the
is
given:
rate
second-order
istor
and
straight-line graph,
a
know
we
(wo
mi
from
+
=
[A]
while
Holding [B] constantis seconddoubling
ander in Ab
(thatis, the
reaction
affectsr ate.
to
[A]
part b),
is
if there
=
40x40-3ms
have
we
atl eastsome
is
with
A.
react
presentto
(10.54m -151)(36001]
(NO2].
and
(NO] 0.1[NO2] =0.1
=
N100056M)
(1/0.050M)
-
=
0.54M
under
very rapidly;
-
15
-
1
conditions,
these
(0.056M)
3.0.102
5.0
=
the
reaction
is
0.0056M
=
=
90%
min.
complete
in
only
5.0 min.
k[NO]
can
ms-1
0.543
determine that:x
could also
=
obtain Ea
from
disappearance of
for the
the
slope
data.
given
est:conventtemperature
solve:a)
inverse
and:take the
->
Ea
=-=
scope
(510pe)(R)
=
-
=
-
of
a
graph of
enk
versus
1/5.
Once we know
Ea, we
K.
a to
from
each
of
NOC.
temperature, 15 and
the
natural
log of
each rate
= onas?"ons=-n.9 no"
Slope:
(
-
1.9.104k) (8.314
mak)()
=
in
E E(
=
rate
b) To determine the
Ea
160h]/mol
(Ea/RT)
k A.e
->
entb
of
2.5210-55-1
and
mCe5c.eris):
40 5g
2.52
-
use
k
=
(sigm)(an ior) ()--
e
=
-
I 462.9
1
318
4.15.10
-
2
=
-
= ( (4.15.10
=
-
4 (2.52.10 55 1)
-
-
with
equation (A)
temperatures from the
given
can
we
the
and
r a te constants
=
=
=>
a
(n)
constant, by, atT 430.0K,
such as:
Rz
)
=
and one
=
data,
-
1.0.10
=
-
25
-
1
38
graph of
i räugt
line.
can
calculate
constant, ent.
=1.6.104] /mol 160h] /mal
↳
d
=
1.0.18-5MIs
rate
the
irrelevantt o
whattime
at
the
at4 30.Ok with
=
n
=
plan:We
ent
(thatis,
order
law:Rate
we
in B
=
0056m
+kt
=
Nos decompose
Blan.We
experimentsshows
orden
5.1.18 4M
know
wantt o
t
therefore
zero
=
B) In this care,
3
of
s am e
mate remains the
Because the
k==a
have
we
constant
and the
rate
l
Swa]ssoo"(Norlo
Ab
proportional
is
experiment 1,
from
M-15-1) (0.050M)2
rate
of the
[A]
show how
data
these
so
is
[A] [B] =k . [A]2
k.
=
x
=
to
B.
effecton
has no
i
of
Thus, this pair
doubled.
is
the
rate law with
respect
of the
rate law
raction rate the
rate
resultindicates thatthe
from
law
[A]
Rate k
held constanta nd (B)
is
oder
deduce the
data
the
law and
that( A)
see
we
to
held
is
is:Rate
the
rate
c) Using
[B]
3,
fourfold.This
law
rate
Hence, the
us
concentration
the
1 and
rate
the
increases
allowing
This
gives
the rate constant
us
this table.
Tutorial 13
a)
c)
SO27"
b)
k=
[03]
r
kc
[NOC]"
(00] [cer]
=
e
=
partial
aleeWeare
given balanced etvationandquilibrium
a
kp
p(NHz)
=
p(Nz).
kp kc.
=
p(Hz)
0.1662
pressures
and
are
2.79.18-5
=
=
2.46.7.383
(RT)*
3
-
ke
AG/RI
=
-
=e
(- 33,300]/mol)/8.314]1molk. 298k
13.4
=e
7.105
=
[H+] [C10]
=
pH
=
-
0.040M
=
log[H
Ht
0.20M
change
-
Ka
CHCN]
0
+xM
xM
xM
[H+) [CN]
x
2
=4.9.16
=
0.20
-
-
=1x
-
18
X
9.9.10 6M
=
1.40
=
+CN-
xM
+
xM
-
=
(0.040]
0
(0.20 x)M
equilibrium
Rog
-
=
->
HCA
initial
]
+
[H 1]
=
-
pH
=
-
log[H+]
5.00
=
are an