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Compiled​ ​by​ ​Aaron​ ​Hui
...

1
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What​ ​is​ ​the​ ​pressure​ ​of​ ​the​ ​gas​ ​(in​ ​mmHg)​ ​if​ ​the​ ​volume​ ​is​ ​reduced​ ​at​ ​constant
temperature​ ​to​ ​154​ ​mL?

2
...
20​ ​L​ ​at​ ​125℃
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54​ ​L​ ​if​ ​the​ ​pressure​ ​remains​ ​constant
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​ ​An​ ​aerosol​ ​spray​ ​deodorant​ ​can​ ​with​ ​a​ ​volume​ ​of​ ​350
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2​ ​g​ ​of​ ​propane
gas​ ​(C​3​H​8​)​ ​as​ ​propellant
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​ ​A​ ​gas​ ​initially​ ​at​ ​4
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2​ ​atm,​ ​and​ ​66℃​ ​undergoes​ ​a​ ​change​ ​so​ ​that​ ​its​ ​final​ ​volume
and​ ​temperature​ ​are​ ​1
...
​ ​ ​What​ ​is​ ​the​ ​final​ ​pressure?​ ​Assume​ ​number​ ​of
moles​ ​remains​ ​constant
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​ ​Argon​ ​is​ ​an​ ​inert​ ​gas​ ​used​ ​in​ ​lightbulbs​ ​to​ ​retard​ ​the​ ​vaporization​ ​of​ ​the​ ​filament
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20​ ​atm​ ​and​ ​18℃​ ​is​ ​heated​ ​to​ ​85℃​ ​at​ ​constant
volume
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​ ​What​ ​is​ ​the​ ​density​ ​(in​ ​g/L)​ ​of​ ​uranium​ ​hexafluoride​ ​(UF​6​)​ ​at​ ​779​ ​mmHg​ ​and​ ​62℃?

7
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10​ ​L​ ​vessel​ ​contains​ ​4
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00​ ​atm​ ​and​ ​27
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​ ​ ​What​ ​is​ ​the​ ​molar
mass​ ​of​ ​the​ ​gas?

8
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​ ​ ​She​ ​measures​ ​the​ ​volume​ ​as​ ​1
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54​ ​g​ ​at​ ​STP
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​ ​Calculate​ ​the​ ​density​ ​of​ ​CO​2​​ ​at​ ​STP
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​ ​A​ ​gaseous​ ​compound​ ​is​ ​78
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86%​ ​H
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3​ ​mL​ ​of​ ​the​ ​gas
exerted​ ​a​ ​pressure​ ​of​ ​1
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​ ​ ​If​ ​the​ ​mass​ ​of​ ​the​ ​gas​ ​was​ ​0
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​ ​What​ ​is​ ​the​ ​volume​ ​of​ ​CO​2​​ ​produced​ ​at​ ​37
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00​ ​atm​ ​when​ ​5
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​ ​If​ ​34
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​ ​mmHg,​ ​how​ ​many​ ​grams​ ​of​ ​NaN​3
were​ ​used​ ​in​ ​the​ ​reaction:
2N aN 3 (s) → 2N a (s) + 3N 2 (g)

13
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3​ ​g​ ​of​ ​NaN​3​​ ​are​ ​used​ ​in​ ​the
reaction:
2N aN 3 (s) → 2N a (s) + 3N 2 (g)

14
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9​ ​L​ ​of​ ​butane
(C​4​H​10​)​ ​at​ ​STP
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​ ​Calculate​ ​the​ ​liters​ ​of​ ​CO​2​​ ​produced​ ​during​ ​the​ ​complete​ ​combustion​ ​of​ ​27
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2C 8 H 18 (g) + 25O2 (g) → 16CO2 (g) + 18H 2 O (g)

16
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14​ ​L​ ​sample​ ​of​ ​HCl​ ​gas​ ​at​ ​2
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​ ​Calculate​ ​the​ ​molarity​ ​of​ ​the​ ​acid​ ​solution
assuming​ ​no​ ​change​ ​in​ ​volume
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​ ​Carbon​ ​dioxide​ ​inside​ ​a​ ​submarine​ ​cabin​ ​has​ ​a​ ​volume​ ​of​ ​2
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9​ ​x​ ​10​-3​​ ​atm​ ​at​ ​312​ ​K
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​ ​Eventually​ ​the​ ​pressure​ ​fell​ ​to​ ​1
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​ ​How​ ​many​ ​grams​ ​of​ ​lithium
carbonate​ ​are​ ​formed​ ​by​ ​this​ ​process?
2LiOH (aq) + C O2 (g) → Li2 CO3 (aq) + H 2 O (l)

18
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0​ ​L​ ​of​ ​C​4​H​10​​ ​and​ ​25
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25​ ​atm?
2C 4 H 10 (g) + 13O2 (g) → 8CO2 (g) + 10H 2 O (g)

19
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5​ ​g​ ​of​ ​C​4​H​10​​ ​and​ ​14
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​ ​36​ ​mL​ ​of​ ​gas​ ​is​ ​collected​ ​over​ ​water​​ ​at​ ​25℃​ ​and​ ​an​ ​atmospheric​ ​pressure​ ​of​ ​757
mmHg
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​ ​A​ ​sample​ ​of​ ​natural​ ​gas​ ​contains​ ​8
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421​ ​moles​ ​of​ ​C​2​H​6​​ ​and​ ​0
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​ ​ ​If​ ​the​ ​total​ ​pressure​ ​of​ ​the​ ​gases​ ​is​ ​1
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​ ​Dry​ ​air,​ ​which​ ​is​ ​made​ ​up​ ​of​ ​21
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0%​ ​nitrogen,​ ​and​ ​less​ ​than​ ​0
...
​ ​mmHg
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​ ​Hydrogen​ ​gas​ ​is​ ​generated​ ​when​ ​Ca​ ​reacts​ ​with​ ​water
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​ ​ ​The​ ​volume​ ​of​ ​gas​ ​collected
was​ ​641​ ​mL
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82​ ​mmHg​ ​at​ ​30℃
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​ ​What​ ​is​ ​the​ ​pressure​ ​of​ ​the​ ​gas​ ​in​ ​the​ ​drawing​ ​if​ ​T=22℃​ ​and​ ​P​atm​​ ​=​ ​743
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​ ​If​ ​CO​ ​gas​ ​effuses​ ​at​ ​a​ ​rate​ ​that​ ​is​ ​1
...
​ ​Which​ ​of​ ​the​ ​following​ ​can​ ​form​ ​Hydrogen​ ​bonds​ ​among​ ​themselves?
H 2 S , C 6 H 6 , CH 3 OH , CH 3 OCH 3

27
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​ ​What​ ​types​ ​of​ ​intermolecular​ ​forces​ ​exist​ ​between​ ​each​ ​of​ ​the​ ​following​ ​species
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H Br , CH 4 , P H 3 ,​ ​ N H 3 ,​ ​ LiCl , Al3+ + H 2 O

29
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​ ​ ​Why?

30
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​ ​Which​ ​substance​ ​will​ ​have​ ​a​ ​larger​ ​boiling​ ​point:
● F​2​​ ​or​ ​HCl,​ ​Xe​ ​or​ ​Ar,​ ​NaCl​ ​or​ ​H​2​O

32
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33
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5​ ​mL​ ​of​ ​0
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0​ ​mL​ ​of​ ​0
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34
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154​ ​nm​ ​are​ ​diffracted​ ​from​ ​a​ ​crystal​ ​at​ ​an​ ​angle​ ​of​ ​14
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Assuming​ ​that​ ​n=1,​ ​what​ ​is​ ​the​ ​distance​ ​(in​ ​pm)​ ​between​ ​tlayers​ ​in​ ​the​ ​crystal?
● 1000​ ​pm​ ​=​ ​1​ ​nm,​ ​ 2 d sinθ = nλ

35
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​ ​Arrange​ ​the​ ​following​ ​in​ ​order​ ​of​ ​increasing​ ​melting​ ​point
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​ ​A​ ​sample​ ​of​ ​0
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6​ ​g​ ​of​ ​water
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​ ​What​ ​is​ ​the​ ​molality​​ ​of​ ​a​ ​solution​ ​containing​ ​7
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​ ​What​ ​is​ ​the​ ​molarity​ ​and​ ​molality​ ​of​ ​a​ ​solution​ ​made​ ​by​ ​dissolving​ ​2
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0​ ​mL​ ​(assume​ ​density​ ​of​ ​solution​ ​is​ ​1
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​ ​What​ ​is​ ​the​ ​molarity​ ​of​ ​a​ ​2
...
976
g/mL?

41
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86​ ​M​ ​ethanol​ ​(C​2​H​5​OH)​ ​solution​ ​whose​ ​density​ ​is​ ​0
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​ ​Calculate​ ​the​ ​molarity​ ​and​ ​molality​ ​of​ ​a​ ​35
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45​ ​g/mL
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​ ​Calculate​ ​the​ ​molarity​ ​and​ ​molality​ ​of​ ​a​ ​44
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19​ ​g/mL
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​ ​Calculate​ ​the​ ​ppm​ ​of​ ​a​ ​solution​ ​that​ ​is​ ​5
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​ ​Calculate​ ​the​ ​%​ ​weight​ ​of​ ​a​ ​solution​ ​that​ ​is​ ​32
...


46
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22​ ​atm
(Henry’s​ ​law​ ​constant​ ​for​ ​oxygen​ ​is​ ​1
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47
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4​ ​g​ ​of​ ​urea​ ​(molar
mass=60
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​ ​What​ ​is​ ​the​ ​change​ ​in​ ​vapor​ ​pressure?
The​ ​vapor​ ​pressure​ ​of​ ​water​ ​at​ ​35ºC​ ​is​ ​42
...


48
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01​ ​g​ ​and​ ​the​ ​K​f
for​ ​water​ ​equals​ ​1
...


49
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85​ ​g​ ​sample​ ​of​ ​a​ ​compound​ ​with​ ​an​ ​empirical​ ​formula​ ​C​5​H​4​​ ​is​ ​dissolved​ ​in​ ​301​ ​g
of​ ​benzene​ ​and​ ​depresses​ ​the​ ​freezing​ ​point​ ​by​ ​1
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​ ​What​ ​is​ ​the​ ​molar​ ​mass​ ​and
molecular​ ​formula​ ​of​ ​this​ ​compound?​ ​The​ ​Kf​ ​of​ ​benzene​ ​is​ ​5
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50
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85​ ​g​ ​of​ ​an​ ​organic​ ​compound​ ​in​ ​100
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16ºC
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50ºC
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12​ ​ºC/m
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​ ​Which​ ​solution​ ​would​ ​have​ ​the​ ​lowest​ ​freezing​ ​point?
0
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20 m HC 2 H 3 O2 ,​ ​ 0
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20 m N a3 P O4

ANSWER​ ​KEY
1
...
​ ​

V1
T1

=

P 1V 1
V2

=

(726 mmHg)(946 mL)
154 mL

= 4460 mmHg = 4
...
15)K = 398 K
3
...
54 L
T2

=

3
...
54 L)
T 2 = 192 K
T 2 = (192 − 273
...
0℃
L
1000 mL

= 0
...
2 g C 3 H 8 *

mol C 3 H 8
44
...
​ ​ V = 350
...
0726 mol

T = (20
...
15)K = 293 K
P =

4
...
0726 mol)(0
...
350 L

=

P 2V 2
T2

T 1 = (66 + 273
...
15)K = 315 K
(1
...
0 L)
339 K

P 2 (1
...
6 atm
5
...
15)K = 291 K
T 2 = (85 + 273
...
20 atm
291 K

P

= 3582K
P 2 = 1
...
0 atm

6
...
15)K = 335 K
P = 779 mmHg *

1 atm
760 mmHg

= 1
...
00​ ​mol
nRT
P

V =

=

L*atm )(335 K)
(1
...
08206 mol
*K
1
...
0
352
...
8 L

D=

= 13
...
0 + 273
...
15 K

(1
...
10 L)
L*atm )(300
...
08206 mol
*K

molar mass =

4
...
0853 mol

mol gas
22
...
​ ​ 1
...
​ ​ molar mass =
n=

= 26
...
0853 mol

= 54
...
0513 mol

1
...
0513 mol

= 30
...
​ ​ 44
...
414 L

10
...
14 g B *

= 1
...
81 g B

21
...
228 mol B

mol H
1
...
69 mol H

B 7
...
69 = B H 3
7
...
228

P V = nRT
n=

PV
RT

=

T = (27 + 273
...
15 K

(1
...
0743 L)
L*atm )(300
...
08206 mol
*K

molar mass =

0
...
003379 mol

27
...
83
= 2d
g
=​ ​ (BH 3 )2 =​ ​ B 2 H 6

= 0
...
6 g/mol

11
...
60 g C 6 H 12 O6 *

mol C 6 H 12 O6
180
...
186 mol CO2

T = (37
...
15)K = 310
...
15 K)
(0
...
08206 mol
*K
1
...
73 L CO2

12
...
00 atm)(34
...
15)K
(0
...
21 mol N 2 *

2 mol N aN 3
3 mol N 2

*

= 1
...
02 g N aN 3
mol N aN 3

= 52
...
​ ​ 52
...
3 g N aN 3 *

mol N aN 3
65
...
​ ​ 14
...
9 L O2

15
...
2 L C 8 H 18 *

16 L CO2
2 L C 8 H 18

= 218 L CO2

16
...
414 L N 2
mol N 2

= 27
...
15)K = 301
...
61 atm)(2
...
15 K)
(0
...
226 mol HCl
0
...
226 mol HCl

= 0
...
​ ​ V 1 P 1 T (CO2 ) ⇒ mol (CO2 ) ⇒ mol (Li2 CO3 ) ⇒ g Li2 CO3
n=

PV
RT

=

−3

(7
...
8 mol CO2 *

−3

5

atm − 1
...
4*10 L)
L*atm )(312 K)
(0
...
89 g Li2 CO3
mol Li2 CO3

= 62
...
6 * 103 g Li2 CO3

18
...
0 L C 4 H 10 *

13 L O2
2 L C 4 H 10

= 32
...
5 L but only have 25 L
25
...
0 L C 4 H 10 *

8 L CO2
13 L O2

8 L CO2
2 L C 4 H 10

25
...
4 L CO2

= 20
...
4 L CO2

O2 is the limiting reagent
19
...
5 g C 4 H 10 *
14
...
12 g C 4 H

*

mol O2
32
...
172 mol CO2

= 0
...
172 mol CO2 *

22
...
9 L CO2

20
...
76)mmHg
P O2 = 733 mmHg *

1 atm
760 mmHg

P O2 = 0
...
15)K = 298 K
nO2 =

PV
RT

=

(0
...
036 L)
L*atm )(298 K)
(0
...
42 * 10−3 mol O2 *

32
...
42 * 10−3 mol O2
= 0
...
​ ​ P C 3 H 8 = X C 3 H 8 P T
X C 3H 8 =

0
...
24+0
...
116) mol

= 0
...
0132)(1
...
0181 atm
22
...
210(790
...
780(790
...
009(790
...
​ ​Part​ ​1:
C a + 2H 2 O → C a(OH)2 + H 2
P H 2 = P T − P H 2 O = (988 − 31
...
26 atm

T = (30 + 273
...
26 atm)(0
...
08206 mol
*K

0
...
016 g H 2
mol H 2

Part​ ​2:
0
...
0325 mol

= 0
...
08 g Ca
mol Ca

= 1
...
​ ​ ​ P gas = P atm − P H 2 O − P height dif f erence
P h = 154 mmH 2 O *

1 mmHg
13
...
3 mmHg

P gas = (743
...
5 − 11
...
9 mmHg
25
...
48 rX
rCO
rX

√

= (1
...
48)2 =

MX 2
)
M CO

MX
28
...
4 g/mol
26
...

C H 3 OCH 3 no,​ ​H​ ​is​ ​not​ ​attached​ ​to​ ​FON
...
​ ​ C H 3 OCH 3 ​ ​yes,​ ​water​ ​can​ ​bind​ ​to​ ​lone​ ​pairs​ ​on​ ​O
C H 4 no,​ ​there​ ​is​ ​no​ ​FON
F − ​ ​yes,​ ​water​ ​can​ ​bind​ ​to​ ​F​ ​(hydrogen​ ​bond​ ​and​ ​ion-dipole)
H COOH ​ ​yes,​ ​water​ ​can​ ​bind​ ​to​ ​lone​ ​pairs​ ​on​ ​O
N a+ ​ ​no,​ ​there​ ​is​ ​no​ ​FON
28
...


○

Al3+ + H 2 O
■ Dispersion,​ ​Ion-Dipole​ ​force

29
...

30
...

Higher​ ​molar​ ​mass​ ​=​ ​stronger​ ​IMF​ ​=​ ​lower​ ​VP
...
​ ​F​2​​ ​(38
...
46​ ​g/mol)
○ F​ ​has​ ​dispersion,​ ​HCl​ ​has​ ​dipole​ ​dipole​ ​AND​ ​dispersion
■ HCl​ ​is​ ​stronger​ ​IMF​ ​so​ ​higher​ ​BP
● Xe​ ​or​ ​Ar
○ Both​ ​dispersion,​​ ​but​ ​Xe​ ​has​ ​higher​ ​MM​ ​so​ ​it​ ​is​ ​stronger​ ​IMF​ ​with
higher​ ​BP
● NaCl​ ​or​ ​H​2​O
○ NaCl​ ​has​ ​ion-ion
○ Water​ ​has​ ​H-bond,​ ​d-d,​ ​disp
■ NaCl​ ​is​ ​a​ ​solid​ ​(with​ ​a​ ​higher​ ​melting​ ​point)​ ​and​ ​ion-ion​ ​is
significantly​ ​stronger​ ​than​ ​h-bonds
32
...
​ ​ (N H 4 )2 SO4 + K OH → 2N H 4 OH + K 2 SO4 ​ ​(no​ ​need​ ​to​ ​write​ ​aqueous)
○

2N H 4 + + S O4 2− + 2K + + 2OH − → 2N H 4 OH + 2K + + S O4 2−

N H 4 + + OH − → N H 4 OH

■
●

(N H)4 SO4 : (87
...
40
K OH : (225
...
20
mmol

N H 4+

I nitial

(35 * 2) = 70
...
5

= 35 mmol

= 45 mmol

+ OH −

→ N H 4 OH
0

35

45

− 45
...


0

45

35

45


...
14M

=
...
5

=
...
5 mL + 225
...
5 mL

34
...
154 nm * 1 nm )
2sin(14
...
​ ​SO​2​​ ​-​ ​Molecular​ ​solid,​ ​dipole​ ​dipole,​ ​dispersion
KI​ ​-​ ​Ionic​ ​solids,​ ​ion​ ​ion
Zn​ ​-​ ​Metallic​ ​solid,​ ​metallic​ ​bonding
NaBr​ ​-​ ​Ionic​ ​solid,​ ​ion​ ​ion
CH​4​ -​​ ​Molecular​ ​solid,​ ​dipole​ ​dipole,​ ​dispersion
36
...
​ ​ (0
...
892
100% = 1
...
6 g) soln *

38
...
78 g urea *

mol urea
60
...
130 mol urea

0
...
203 kg H 2 O

39
...
638 m urea

mol H 2 SO4
L soln

2
...
0241 mol H 2 SO4
0
...
08 g H 2 SO4

mol H 2 SO4
kg H 2 O

= 0
...
481 M H 2 SO4

50
...
00 g soln
mL soln

= 50
...
0 g − 2
...
64 g H 2 O
0
...
04764 kg H 2 O

40
...
506 m H 2 SO4

2
...
976 g soln
mL soln

⇒

mol CH 3 OH
L soln

Assume​ ​ 2
...
73 mol CH 3 OH *

32
...
5 g CH 3 OH

1000 g H 2 O + 87
...
5 g soln
1087
...
73 mol CH 3 OH
1
...
976 g soln

*

L
1000 mL

= 1
...
45 M CH 3 OH
0
...
​ ​ 5
...
86​ ​mol​ ​ethanol
1
...
86 mol ethanol *

*

0
...
07 g ethanol
mol ethanol

= 927 g soln

= 270
...
g ethanol = 657 g H 2 O = 0
...
86 mol ethanol
0
...
​ ​

35
...
92 m ethanol

3
...
4 g H 3 P O4
100 g soln

*

mol H 3 P O4
97
...
45 g soln
mL soln

*

1000 mL
L

= 12
...
4 g H 3 P O4 → 100 g soln
35
...
99 g H 3 P O4

= 0
...
4 g = 64
...
0646 kg H 2 O
0
...
0646 kg H 2 O

= 5
...
​ ​ 44
...
19 g soln
mL soln

44
...
44 g N aCl

*

(molarity) ⇒
2
...
7 M N aCl

Assume​ ​ 44
...
6 g N aCl *

mol N aCl
58
...
763 mol N aCl

100 g − 44
...
4 g H 2 O = 0
...
763 mol N aCl
0
...
​ ​

5
...
23 g N O2
106 g soln

*

104
104

g solute
106 g soln

= 5
...
​ ​ 32
...
7 m N aCl

10−4
10−4

3
...
21 * 10−5 %

46
...
3 * 103 Lmol
)(0
...
9 * 10−4 M
2
...
00 g soln
mL soln

⇒

mol O2
kg H 2 O

Assume​ ​ 2
...
0 L soln
32
...
9 * 10−4 mol O2 *
1
...
00928 g O2

1
...
00928 g ≈ 1000 g H 2 O
2
...
0 kg H 2 O

= 2
...
​ ​ P 1 = X 1 P 0 1
82
...
06 g urea

212 mL H 2 O *
X1 =

= 1
...
00 g H 2 O
mL H 2 O

11
...
8+1
...
02 g H 2 O

= 11
...
896

P 1 = (0
...
18 mmHg) = 37
...
896)(42
...
39 mmHg
or

●

ΔP = P 0 1 − P 1
ΔP = 42
...
8 mmHg
ΔP = 4
...
​ ​

478 g antif reeze *

mol
62
...
202 kg H 2 O

= 2
...
86℃
)(2
...
48℃
m

0℃ − 4
...
48 (freezing​ ​point​ ​of​ ​soln)
49
...
05℃ = ( 5
...
205 m =

0
...
301 kg benzene *
7
...
0617 mol

n=

0
...
0617 mol (C 5 H 4 )n

= 127 g/mol

127 g/mol
64
...
​ ​ ΔT f = K f m
5
...
16℃ = ( 5
...
066 m

0
...
85 g
0
...
066 mol org
...
2 * 102 g/mol

51
...
20 m KN O3
● i​ ​=​ ​2
●

= 0
...


2 * 0
...
40 m

0
...
20 m < x < 0
...
20 m

0
...
20 m

0
...
80 m

● THIS​ ​HAS​ ​THE​ ​LOWEST​ ​FP

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