VCE Chemistry Units 1 and 2 Transition Booklet...VCE Chemistry Units 1 and 2 Transition Booklet...
23
St Leonard's College VCE Chemistry Units 1 and 2 Transition Booklet Name:___________________________
Transcript of VCE Chemistry Units 1 and 2 Transition Booklet...VCE Chemistry Units 1 and 2 Transition Booklet...
St Leonard's College
VCE Chemistry Units 1 and 2
Transition Booklet
Name:___________________________
3
1.Pe
riod
ic ta
ble o
f the
elem
ents
1 H
1.0
hydr
ogen
2 He
4.0
heliu
m
3 Li
6.9
lithi
um
4 Be
9.0
bery
llium
5 B
10.8
bo
ron
6 C
12.0
ca
rbon
7 N
14.0
ni
troge
n
8 O
16.0
oxygen
9 F 19.0
fluorine
10
Ne
20.2
ne
on
11
Na
23.0
so
dium
12
Mg
24.3
m
agne
sium
13
Al
27.0
al
umin
ium
14
Si
28.1
si
licon
15
P 31.0
ph
osph
orus
16
S 32.1
su
lfur
17
Cl
35.5
ch
lorin
e
18
Ar
39.9
ar
gon
19
K
39.1
po
tass
ium
20
Ca
40.1
ca
lciu
m
21
Sc
45.0
sc
andi
um
22
Ti
47.9
tit
aniu
m
23
V
50.9
va
nadi
um
24
Cr
52.0
ch
rom
ium
25
Mn
54.9
m
anga
nese
26
Fe
55.8
iro
n
27
Co
58.9
co
balt
28
Ni
58.7
ni
ckel
29
Cu
63.5
co
pper
30
Zn
65.4
zi
nc
31
Ga
69.7
ga
llium
32
Ge
72.6
ge
rman
ium
33
As
74.9
ar
seni
c
34
Se
79.0
se
leni
um
35
Br
79.9
br
omin
e
36
Kr
83.8
kr
ypto
n
37
Rb
85.5
ru
bidi
um
38
Sr
87.6
st
ront
ium
39
Y
88.9
yt
trium
40
Zr
91.2
zi
rcon
ium
41
Nb
92.9
ni
obiu
m
42
Mo
96.0
m
olyb
denu
m
43
Tc
(98)
te
chne
tium
44
Ru
101.
1 ru
then
ium
45
Rh
102.
9 rh
odiu
m
46
Pd
106.
4 pa
lladi
um
47
Ag
107.
9 si
lver
48
Cd
112.
4 ca
dmiu
m
49
In
114.
8 in
dium
50
Sn
118.
7 tin
51
Sb
121.
8 an
timon
y
52
Te
127.
6 te
lluriu
m
53
I 12
6.9
iodi
ne
54
Xe
131.
3 xenon
55
Cs
132.
9 ca
esiu
m
56
Ba
137.
3 ba
rium
57–7
1 la
ntha
noid
s
72
Hf
178.
5 ha
fniu
m
73
Ta
180.
9 ta
ntal
um
74
W
183.
8 tu
ngst
en
75
Re
186.
2 rh
eniu
m
76
Os
190.
2 os
miu
m
77
Ir
192.
2 iri
dium
78
Pt
195.
1 pl
atin
um
79
Au
197.
0 go
ld
80
Hg
200.
6 m
ercu
ry
81
Tl
204.
4 th
alliu
m
82
Pb
207.
2 le
ad
83
Bi
209.
0 bi
smut
h
84
Po
(210
) po
loni
um
85
At
(210
) as
tatin
e
86
Rn
(222
) ra
don
87
Fr
(223
) fr
anci
um
88
Ra
(226
) ra
dium
89–1
03
actin
oids
104
Rf
(261
) ru
ther
ford
ium
105
Db
(262
) du
bniu
m
106
Sg
(266
) se
abor
gium
107
Bh
(264
) bo
hriu
m
108
Hs
(267
) ha
ssiu
m
109
Mt
(268
) m
eitn
eriu
m
110
Ds
(271
) da
rmst
adtiu
m
111
Rg
(272
) ro
entg
eniu
m
112
Cn
(285
) co
pern
iciu
m
113
Nh
(280
) ni
honi
um
114
Fl
(289
) flerovium
115
Mc
(289
) m
osco
vium
116
Lv
(292
) liv
erm
oriu
m
117
Ts
(294
) te
nnes
sine
118
Og
(294
) og
anes
son
57
La
138.
9 la
ntha
num
58
Ce
140.
1 ce
rium
59
Pr
140.
9 pr
aseo
dym
ium
60
Nd
144.
2 ne
odym
ium
61
Pm
(145
) pr
omet
hium
62
Sm
150.
4 sa
mar
ium
63
Eu
152.
0 eu
ropi
um
64
Gd
157.
3 ga
dolin
ium
65
Tb
158.
9 te
rbiu
m
66
Dy
162.
5 dy
spro
sium
67
Ho
164.
9 ho
lmiu
m
68
Er
167.
3 er
bium
69
Tm
16
8.9
thul
ium
70
Yb
173.
1 yt
terb
ium
71
Lu
175.
0 lu
tetiu
m
89
Ac
(227
) ac
tiniu
m
90
Th
232.
0 th
oriu
m
91
Pa
231.
0 pr
otac
tiniu
m
92
U
238.
0 ur
aniu
m
93
Np
(237
) ne
ptun
ium
94
Pu
(244
) pl
uton
ium
95
Am
(2
43)
amer
iciu
m
96
Cm
(2
47)
curiu
m
97
Bk
(247
) be
rkel
ium
98
Cf
(251
) ca
lifor
nium
99
Es
(252
) ei
nste
iniu
m
100
Fm
(257
) fe
rmiu
m
101
Md
(258
) m
ende
levi
um
102
No
(259
) no
beliu
m
103
Lr
(262
) la
wre
nciu
m
The
valu
e in
bra
cket
s ind
icat
es th
e m
ass n
umbe
r of t
he lo
nges
t-liv
ed is
otop
e.
79
Au
197.
0 go
ld
atom
ic n
umbe
r
rela
tive
atom
ic m
ass
sym
bol o
f ele
men
t
nam
e of
ele
men
t
ELECTRONIC CONFIGURATIONS
THEORYOld • electrons existed in definite energy levels or shells
• the levels were concentric rings• the further the energy level is from the nucleus, the higher its energy• each level held a maximum number of electrons• when a level was full up you moved to fill the next level
New Instead of circulating in orbits around the nucleus, electrons were in orbitals.
ORBITAL ”A region in space where one is likely to find an electroneach orbital can hold up to two electrons, with opposite spins”.
• 3-dimensional statistical maps showing the likeliest places to find electrons• come in different shapes and sizes• hold a maximum of two electrons each (as long as they have opposite spins)
ENERGYLEVELS In the newer theory, main energy levels are split into sub-levels. Each level has
orbitals and the electrons fill the orbitals. The first four main levels (shells) are ...
Main shell Sub-shells Orbitals Electrons
n = 1 1 1s 2 = 2
n = 2 2 2s 22p 6 = 8
n = 3 3 3s 23p 63d 10 = 18
n = 4 4 4s 24p 64d 104f 14 = 32
Electronic Structure 1 F321
© KNOCKHARDY PUBLISHING 2008
S ORBITALSpherical
ONE in each main shell
P ORBITAL‘Dumb-bell’ shaped
THREE in each main shell(except the first)
Electronic Structure
RULES FOR FILLING ENERGY LEVELS
Aufbau Principle ”Electrons enter the lowest energy orbital available.”
Energy levels are not entered until those below them are filled.
Pauli’s Exclusion ”No two electrons can have the same four quantum numbers.” or Principle
Orbitals can hold a max. of 2 electrons provided they have opposite spin.
Hund’s Rule Orbitals of the same energy remain singly occupied before pairing up. This is due to the repulsion between electron pairs.
FILLING ORDER • Orbitals are filled in order of increasing energy
• Orbitals are not filled in numerical order ...1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, etc.
The 4s is filled before the 3d because it is lower in energy
• The effect can be explained by assuming the ...principal energy levels get closer together as they get furtherfrom the nucleus.
• As a result, the highest energy orbitals in one principal level may beabove the lowest in the next level
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d ...
1s 2s 3s 4s 5s 6s 7s
FILLING ORDER
2p 3p 4p 5p 6p
3d 4d 5d 6d
4f 5f
Electronic Structure
INCREASING ENERGY
The diagram helps explain why the 4s orbitals are filled before the 3d orbitals
4f 4d
4 4p
4s
3d
3 3p
4f 4d
4 4p 3d 4s
3 3p 3s 2p
3s 2 2s
2p 2s
NOT TO SCALE
1 1s
A
1 1s
B
There is plenty of evidence to explain the filling order. This will be dealt with in sections on Ionisation Energies and Periodicity.
EVIDENCE
Energy levels Periodically there was a large drop in the energy to remove electrons caused by the electrons being further from the nucleus.
Sub-levels The energy required to remove electrons was sometimes less than expected due to shielding from filled sub-levels.
2
Electronic Structure
Electronic configurations of the first 36 elements. 1s
H He Li Be B C N O F Ne Na Mg
2s
2p
3s
- -
3p
- - - - - - - - - - - - - - - - - - - - - - - - - - - 1s1
Al Si P S Cl Ar 4s K Ca 3d Sc Ti V Cr Mn Fe Co Ni Cu Zn 4p Ga Ge As Se Br Kr
The filling proceeds according to the rules ... but watch out for chromium and copper.
Cr
Cu
Electronic Structure
(g)
IONISATION ENERGY
• A measure of the energy required to remove electrons from an atom.• Value depends on the distance of the electron from the nucleus and the
effective nuclear charge (not the nuclear charge) of the atom.• There are as many ionisation energies as there are electrons in the atom.
Nuclear Charge (NC) The actual charge (relative) due to the protons in the nucleus
Effective nuclear Charge (ENC) • The effectiveness of nuclear charge after passing through filled shells
• A simple way to compare effective nuclear charges is to knock offa + for every electron in a filled inner level.
Species Protons Electron config. NC ENC
H 1 1 1+ 1+ He 2 2 2+ 2+ Li 3 2,1 3+ 1+ Be 4 2,2 4+ 2+ B 5 2,3 5+ 3+ Ne 10 2,8 10+ 8+ Na 11 2,8,1 11+ 1+ K 19 2,8,8,1 19+ 1+
1st I.E. The energy required to remove one mole of electrons (to infinity) from one mole of isolated, gaseous atoms to form one mole of gaseous positive ions.
e.g. Na(g) ——> Na+ + e¯ Mg(g) ——> Mg+ (g) + e¯
Its value gives an idea of how strongly the nucleus pulls on the electron being removed. The stronger the pull, the more energy needed to pull out the electron.
First Ionisation Energies / kJ mol-1
1 H 1310 15 P 1060 29 Cu 745 43 Tc 699 2 He 2370 16 S 1000 30 Zn 908 44 Ru 724 3 Li 519 17 Cl 1260 31 Ga 577 45 Rh 745 4 Be 900 18 Ar 1520 32 Ge 762 46 Pd 803 5 B 799 19 K 418 33 As 966 47 Ag 732 6 C 1090 20 Ca 590 34 Se 941 48 Cd 866 7 N 1400 21 Sc 632 35 Br 1140 49 In 556 8 O 1310 22 Ti 661 36 Kr 1350 50 Sn 707 9 F 1680 23 V 648 37 Rb 402 51 Sb 833
10 Ne 2080 24 Cr 653 38 Sr 548 52 Te 870 11 Na 494 25 Mn 716 39 Y 636 53 I 1010 12 Mg 736 26 Fe 762 40 Zr 669 54 Xe 1170 13 Al 577 27 Co 757 41 Nb 653 55 Cs 376 14 Si 786 28 Ni 736 42 Mo 694 56 Ba 502
Q.1 • Plot a graph of 1st I.E. v. Atomic No. for the first 56 elements.• Plot graphs of 1st I.E. v. Atomic No. for the elements H to Na and for Ne to K
Electronic Structure
Interpretation of Ionisation Energy graphs
ACROSS PERIODS
2400
2200
2000
1800
1600
1400
1200
1000
800
600
400 0 2 4 6 8 10 12 14 16 18
ATOMIC NUMBER
He > H • nuclear charge is greater - one extra proton• extra electron has gone into the same energy level• increased attraction makes the electron harder to remove.
Li < He • increased nuclear charge, but...• outer electron is held less strongly• it is shielded by full inner levels and is further away - easier to remove
Be > Li • increased nuclear chargeMg > Na • electrons in the same energy level
B < Be • despite the increased nuclear charge, the outer electron is held less stronglyAl < Mg • it is now shielded by the 2s energy sub-level and is also further away
(LED TO EVIDENCE FOR SUB LEVELS)
O < N • despite the increased nuclear charge the electron is easier to removeS < P • in N the three electrons in the 2p level are in separate orbitals whereas
• in O two of the four electrons are in the same orbital• repulsion between paired electrons = less energy needed to remove of one
Values in Period 3 are always smaller than the equivalent Period 2 value - the electron removed is further from the nucleus and has more shielding
Na < Li • despite the increased nuclear charge the electron is easier to remove• increased shielding and greater distance from the nucleus• outer electron in Na is held less strongly and easier to remove
PERIOD 2 PERIOD 3
Ist I
ON
ISAT
ION
EN
ERG
Y /
kJ m
ol -1
Electronic Structure
Be
Mg
Ca Li
Na Sr Ba
K Rb
Cs
(g)
DOWN GROUPS 1000
Ist IONISATION ENERGY / kJ mol-1
800
600
400
200 INCREASING ATOMIC NUMBER
GROUP I Value decreases down the Group
• despite the increased nuclear charge the outer s electron is easier to remove• this is due to increased shielding and greater distance from the nucleus• outer electron is held less strongly and easier to remove
GROUP II Similar trend to Group I
• Group II values are greater than their Group I neighbours• increased nuclear charge = stronger pull on electron• more energy required to remove an s electron
SUCCESSIVE IONISATION ENERGIES
2nd I.E. The energy required to remove one mole of electrons (to infinity) from one mole of gaseous unipositive ions to form one mole of gaseous dipositive ions.
e.g. Mg+
Al+———> Mg2+
———> Al2+(g) + e¯
+ e¯(g) (g)
Successive I.E. values for calcium / kJ mol-
Trends • Successive ionisation energies arealways greater than the previous one- the electron is being pulled away from
a more positive species
• Large increases occur when there isa change of shell- this can be used to predict the group
of an unknown element
Q.2 • Plot a graph of log10 I.E. of calcium v. no. of electron removed.
1 590 10 20385 2 1145 11 57048 3 4912 12 63333 4 6474 13 70052 5 8145 14 78792 6 10496 15 86367 7 12320 16 94000 8 14207 17 104900 9 18192 18 111600
Worksheet 1.1 Using nuclide symbol notation
© Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2008.
T
INTRODUCTION One way to represent a particular atom is with a nuclide symbol:
XAZ
where A = mass number, Z = atomic number and X = element symbol Often this notation is used to represent only a nucleus. For the questions below, assume that the atoms represented also contain electrons.
No. Question Answer
1 98252Cf is an isotope used to treat cervical cancer. How many of each of the following does one atom of 98
252Cf contain? a protons b neutrons
2 One isotope of nickel used in the detection of explosives has a mass number of 63. Write a nuclide symbol to represent this isotope.
3 How many protons, neutrons and electrons does each of the following ions contain? a 94
241Pu+ b 92
235U2+
4 How many electrons does the particle represented by the symbol 1531P3– have?
5 What is the nuclide symbol for carbon-12?
6 The most abundant isotope of antimony has a mass number of 121. If an atom of this isotope losttwo electrons, what would itsnuclide symbol be?
Worksheet 1.1 Using nuclide symbol notation
© Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2008.
No. Question Answer
7 How are the two isotopes of gallium, 31
69Ga and 3171Ga , similar
in terms of: a atomic structure? b chemical properties?
8 Briefly explain why it is easier to separate 38
88Sr and 3988 Y than 38
88Sr and 38
86Sr .
9 The term isoelectronic refers to species with the same number of electrons. Give the nuclide symbol for a positively charged ion that is isoelectronic with an atom of neon.
10 Give the formulas of two non-metal ions that are isoelectronic with an atom of argon.
Worksheet 1.2 Electron configurations
Copyright © Pearson Australia 2009 (a division of Pearson Australia Group Pty Ltd) ISBN 9780733993817
INTRODUCTION Electrons behave like waves, thus mathematical equations can be derived to describe electrons in terms of probability functions. The wave-mechanical view of the atom can be summarized as follows: • A shell is an energy level which electrons may occupy in an atom. It is labelled with a number
(n) where n = 1, 2, 3 etc. n is the principal quantum number.• A subshell is a sub-level of energy within an electron shell.• An orbital is a region of space around a nucleus in which the electron will probably be found.A maximum of two electrons may occupy each orbital (Pauli exclusion principle).Examples of electron configuration in terms of subshells are:
magnesium (12): 1s22s22p63s2 oxygen (8): 1s22s22p4
No. Question Answer
1 State the difference between a shell and an orbital.
2 State the difference between a subshell and an orbital.
3 Sodium is in the s-block of the periodic table. Explain how this relates to its electron configuration.
4 State the maximum number of electrons that could occupy the 3d subshell.
5 State Hund’s rule.
Worksheet 1.2 Electron configurations
Copyright © Pearson Australia 2009 (a division of Pearson Australia Group Pty Ltd) ISBN 9780733993817
No. Question Answer
6 Compare the relative energies of s, p, d and f orbitals in a single energy level.
7 Write the electron configurations for these elements. The atomic number is given in brackets. a Sulfur (16) b Nitrogen (7) c Argon (18) d Scandium (21)
8 Write the electron configurations for: a cobalt (27) b germanium (32) c rubidium (37) d tin (50)
9 Write the electron configurations for the following: a chloride ion (Cl–) b strontium ion (Sr2+) c silver ion (Ag+) d selenium ion (Se2–)
10 a Write the electron configuration for the nitride ion, N3–.
b Write the electron configuration for the sodium ion, Na+.
c Determine the order of increasing atomic radius for the following species and explain your answer: Na+, N3–, Ne
WKS001x019 © 2000 Cavalcade Publishing (http://www.cavalcadepublishing.com) All Rights Reserved
Balancing Chemical Equations
Balance the equations below:
1) ____ N2 + ____ H2 ____ NH3
2) ____ KClO3 ____ KCl + ____ O2
3) ____ NaCl + ____ F2 ____ NaF + ____ Cl2
4) ____ H2 + ____ O2 ____ H2O
5) ____ Pb(OH)2 + ____ HCl ____ H2O + ____ PbCl2
6) ____ AlBr3 + ____ K2SO4 ____ KBr + ____ Al2(SO4)3
7) ____ CH4 + ____ O2 ____ CO2 + ____ H2O
8) ____ C3H8 + ____ O2 ____ CO2 + ____ H2O
9) ____ C8H18 + ____ O2 ____ CO2 + ____ H2O
10) ____ FeCl3 + ____ NaOH ____ Fe(OH)3 + ____NaCl
11) ____ P + ____O2 ____P2O5
12) ____ Na + ____ H2O ____ NaOH + ____H2
13) ____ Ag2O ____ Ag + ____O2
14) ____ S8 + ____O2 ____ SO3
15) ____ CO2 + ____ H2O ____ C6H12O6 + ____O2
16) ____ K + ____ MgBr ____ KBr + ____ Mg
17) ____ HCl + ____ CaCO3 ____ CaCl2 + ____H2O + ____ CO2
18) ____ HNO3 + ____ NaHCO3 ____ NaNO3 + ____ H2O + ____ CO2
19) ____ H2O + ____ O2 ____ H2O2
20) ____ NaBr + ____ CaF2 ____ NaF + ____ CaBr2
21) ____ H2SO4 + ____ NaNO2 ____ HNO2 + ____ Na2SO4
Experiment – The Journey of Copper Aim x To observe some of the changes that indicate a chemical reaction has taken place. x To write balanced chemical equations based on observations x To demonstrate atoms are conserved in a chemical reaction, although there is a change in the
way they are combined with one another.
Materials x 0.5-1.0g copper foil (~0.7 g) x 10 mL 6M nitric acid x 20 mL 6M sodium hydroxide x 20 mL 3M sulfuric acid x 2-3g granular zinc (usually 1-2 pieces)x stirring rod x watch glass x 250 mL beaker x 1 × 50 mL measuring cylinder
x filter funnel x filter paper x conical flask x hot plate x bench mat x electronic balance x drying oven
Procedure 1. The series of reactions that you will carry out is summarised in the flowchart on page 3.2. As you carry out each step, write a description of the appearance of each copper-containing
substance in the appropriate box.3. For each step, make a detailed list of the changes you observed that indicate a chemical reaction has
taken place.
Part A: copper o copper(II) nitrate 1 Label with your group name an empty 250 mL beaker. Weigh the beaker and record the mass of
the beaker in the results box at the bottom of page 4. 2 To the beaker, add approximately 0.7g copper foil and reweigh.
Record the mass of copper foil used in the results box on page 4. 3 In the fume cupboard, using a dispenser bottle, add 10 mL 6M nitric acid to the copper metal in the
beaker. 4 Cover the beaker with a watch glass and allow it to stand until there is no copper metal left. 5 Feel the underside of the beaker and note any change in temperature. 6 Record all your observations on your flowchart (page 3).
Part B: copper(II) nitrate o copper(II) hydroxide 7 In the fume cupboard, to the solution from Part A, slowly add 20 mL 6M sodium hydroxide solution
while stirring. 8 Record all your observations on your flowchart (page 3).
Part C: copper(II) hydroxide o copper(II) oxide 9 Add 50 mL distilled water to the beaker and contents from Part B. 10 Using a hot plate to carefully heat the beaker and contents. STIR CONSTANTLY WHILE HEATING.
Continue heating the solution until it just begins to boil and you observe no further changes. 11 Cool the mixture on a bench mat and allow any precipitate to settle at the bottom of the beaker. 12 Carefully pour off the liquid into another beaker, try not to lose any precipitate. Dispose of the
liquid into the waste container provided, but keep the solid precipitate. 13 Add 60-100 mL water to the precipitate. Stir thoroughly and allow the solid to settle again. 14 Repeat steps 12 and 13 with 100mL washes. 15 Fill your beaker with ~200mL of water and allow it to stand until the next class. 16 Record all your observations on your flowchart.
Part D: copper(II) oxide o copper(II) sulfate 17 Do not disturb the contents of the beaker. Carefully pour off the liquid into another beaker, try not
to lose any precipitate. Dispose of the liquid into the waste container provided, but keep the solid precipitate.
18 Take your beaker to the fume cupboard. To the solid precipitate, add 20 mL of 3M sulfuric acid and stir thoroughly.
19 Record all your observations on your flowchart.
Part E: copper(II) sulfate o copper 20 In the fume cupboard, add zinc granules (2-4g) to the solution from Part D. 21 Swirl the beaker to mix the contents and then place a watch glass over it. Feel the outside of the
beaker. 22 Allow the beaker and contents to stand overnight or even longer.
Do not disturb the contents of the beaker, it should contain precipitate and a colourless solution. 23 Carefully pour off as much of the liquid into another beaker.
Dispose of the liquid into the Zinc sulfate waste container provided, but keep the solid copper precipitate.
24 Carefully add 100 mL water to the precipitate. Allow the copper to settle again. 25 Repeat steps 23 and 24 once or twice more, as time permits, discarding waste into the liquid waste
container. 26 Use tweezers to scrape off copper deposits from any remaining zinc granules (if present).
Remove any zinc granules and discard into to the solid zinc waste container. 27 Weigh a piece of filter paper. Record the mass of the filter paper used in the results box on page 4 28 Use the filter paper to filter the copper precipitate.
Use a little distilled water to transfer all the copper pieces from the beaker onto the filter paper. 29 Place the filter paper and copper on a watch glass and dry in an oven overnight. 30 Weigh the dry filter paper and copper. Record the mass of filter paper and copper in the results box
on page 3. 31 Record all your observations on your flowchart.
Equations: For each step in the experiment a worded chemical equation is given. Use this information to write balanced symbol equations for each of the reactions taking place. Using state symbols as subscripts also indicate the state of the reactants and products in the reaction.
Part A: Copper + nitric acid o copper(II) nitrate + nitrogen dioxide + water
________________________________________________________________________________
Part B: Copper(II) nitrate + sodium hydroxide o copper(II) hydroxide + sodium nitrate
________________________________________________________________________________
Part C: Copper(II) hydroxide o copper(II) oxide + water
________________________________________________________________________________
Part D: Copper(II) oxide + sulfuric acid o copper(II) sulfate + water
________________________________________________________________________________
Part E: Copper(II) sulfate + zinc o copper + zinc sulfate
________________________________________________________________________________
Part A
Mass of Beaker (A) g
Mass of beaker and copper foil (B) g
Initial mass of copper (B-A) g
Part E
Mass of filter paper (C) g
Mass of filter paper and copper (D) g
Final Mass of copper (D-C) g
Questions
1 How does your final mass of copper compare with the starting mass of copper?
2 Calculate the percentage recovery of the copper.
% recovery = final mass of Cu x 100 initial mass of Cu
3 The percentage recovery of copper is unlikely to be 100%. Give reasons for possible losses or gains of the copper in your experiment.
4 In two of the Parts in this experiment, a solution was decanted. Explain the meaning of the term ‘decanted’.
5 Why was it necessary to repeat certain steps in Parts C and E of the procedure?
Theory Atoms are not created nor destroyed in a chemical reaction. A chemical reaction involves the rearrangement of atoms to form different substances. The total amount of each individual atom at the start and the end of a reaction should remain unchanged. In this experiment, you started with a sample of the element copper. You converted it into a number of different compounds of copper. In the final reaction, the compound of copper was converted back to the copper metal.
6 Complete the following table. For each of the copper-containing compounds formed, list the elements that combined to form the compound. The first one is done for you, as an example.
Elements present in each compound
Results & Observations
Copper metal Description of element:
______________________________________
_________________________________
Part E Part A Evidence for chemical change: Evidence for chemical change:
_____________________________________ _______________________________________
_____________________________________ _______________________________________
_____________________________________ _______________________________________
Copper(II) sulfate Copper(II) nitrate Description: Description of compound:
____________________________________ __________________________________
____________________________________ __________________________________
Part D Part B Evidence for chemical change: Evidence for chemical change:
__________________________________________ _______________________________________
__________________________________________ _______________________________________
__________________________________________ _______________________________________
Copper(II) oxide Copper(II) hydroxide Description of compound: Description of compound:
____________________________________ __________________________________
____________________________________ __________________________________
Part C Evidence for chemical change:
_____________________________________________
______________________________________
______________________________________
Conclusion: (This should (i) be related to the aim/s of the experiment, (ii) summarise the evidence for chemical change, (iii) discuss how this experiment relates to the law of conservation of mass)
Worksheet 1.1: Solutions Using nuclide symbol notation
© Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2008.
This page from the Chemistry: For use with the IB Diploma Programme SL Teacher’s Resource may be reproduced for classroom use.
No. Answer
1 a 98 b 154
2 2863Ni
3 a 94 protons, 147 neutrons and 93 electrons b 92 protons, 143 neutrons and 90 electrons
4 18
5 612C
6 51121Sb2+
7 a They have the same number of protons and electrons. b They react in the same way due to having the same electron configuration.
8 3888Sr and 39
88 Y have different chemical properties, due to their different numbers of valence electrons. 38
88Sr and 3886Sr are isotopes, and so are harder to separate as they have identical
chemical properties.
9 1123Na+ or 12
24 Mg2+ or 1327 Al3+
10 1632S2– and 17
35Cl–
Worksheet 1.2: Solutions Electron configurations
Copyright © Pearson Australia 2009 (a division of Pearson Australia Group Pty Ltd) ISBN 9780733993817
No. Answer
1 A shell is a major electron energy level within an atom, while an orbital is a region of space in which a maximum of two electrons may be found.
2 A subshell is a minor energy level. Like an orbital, it is also a region of space; however, a subshell is composed of one or more orbitals.
3 When writing the ground state electron configuration of sodium, 1s22s22p63s1, the last electron goes into an s orbital, therefore sodium belongs in the s-block.
4 10
5 Hund’s rule states that ‘Electrons will occupy orbitals singly until all orbitals of a subshell are half-full, then a second electron will be added to the orbitals.’
6 The relative energies of the subshells in a single energy level (in order of increasing energy) are as follows: s < p < d < f.
7 a 1s22s22p63s23p4 b 1s22s22p3 c 1s22s22p63s23p6 d 1s22s22p63s23p63d14s2
8 a 1s22s22p63s23p63d74s2 b 1s22s22p63s23p63d104s24p2 c 1s22s22p63s23p63d104s24p65s1 d 1s22s22p63s23p63d104s24p64d105s25p2
9 a 1s22s22p63s23p6 b 1s22s22p63s23p63d104s24p6 c 1s22s22p63s23p63d104s24p64d10 d 1s22s22p63s23p63d104s24p6
10 a 1s22s22p6. The nitride ion, N3–, has three more electrons than the nitrogen atom. Increased repulsion in the valence shell from the added electrons increases its radius.
b 1s22s22p6. The sodium ion, Na+, has one less electron than the sodium atom. It is smaller than the sodium atom because it has one less electron shell.
c The order of increasing atomic radius is Na+ < Ne < N3–. While all three have the same number of electrons, Na+ has 11 protons in its nucleus, while Ne has only 10. This means that the valence shell of Na+ will be held more tightly than that in Ne, resulting in a smaller radius. In N3–, there are only 7 protons in the nucleus and in conjunction with the increased repulsion in the valence shell of N3–, this produces the largest radius of the three species.
WKS001x019 © 2000 Cavalcade Publishing (http://www.cavalcadepublishing.com) All Rights Reserved
Balancing Chemical Equations – Answer Key
Balance the equations below:
1) 1 N2 + 3 H2 2 NH3
2) 2 KClO3 2 KCl + 3 O2
3) 2 NaCl + 1 F2 2 NaF + 1 Cl2
4) 2 H2 + 1 O2 2 H2O
5) 1 Pb(OH)2 + 2 HCl 2 H2O + 1 PbCl2
6) 2 AlBr3 + 3 K2SO4 6 KBr + 1 Al2(SO4)3
7) 1 CH4 + 2 O2 1 CO2 + 2 H2O
8) 1 C3H8 + 5 O2 3 CO2 + 4 H2O
9) 2 C8H18 + 25 O2 16 CO2 + 18 H2O
10) 1 FeCl3 + 3 NaOH 1 Fe(OH)3 + 3 NaCl
11) 4 P + 5 O2 2 P2O5
12) 2 Na + 2 H2O 2 NaOH + 1 H2
13) 2 Ag2O 4 Ag + 1 O2
14) 1 S8 + 12 O2 8 SO3
15) 6 CO2 + 6 H2O 1 C6H12O6 + 6 O2
16) 1 K + 1 MgBr 1 KBr + 1 Mg
17) 2 HCl + 1 CaCO3 1 CaCl2 + 1 H2O + 1 CO2
18) 1 HNO3 + 1 NaHCO3 1 NaNO3 + 1 H2O + 1 CO2
19) 2 H2O + 1 O2 2 H2O2
20) 2 NaBr + 1 CaF2 2 NaF + 1 CaBr2
21) 1 H2SO4 + 2 NaNO2 2 HNO2 + 1 Na2SO4
