(b) 1 mol of CO (6 1023 (b) C H Empirical formula is MO2. 2
molecules) is produced together 82.76 % (ii) 2H2 + MO2 → M + 2H2O 2
with 3 mol of H2. Element 100 – 82.76 (e) No. Magnesium is more reactive
= 17.24 % than hydrogen. Thus, hydrogen
Hence, the number of CO Percentage gas cannot reduce magnesium
molecules produced together with mass
60 dm3 of hydrogen is Number of —8—2—1—.27——6 1——7—1.2——4 oxide. 2
= 3——6—0——2d—4m——d3—m—H—32——H—2 6 1023 moles
= 5 1023 molecules = 6.9 = 17.24 3 (a) Cation 1
6 (a) 3Zn(s) + 2P(s) → Zn3P2 Simplest
(b) 2 mol of P react with excess zinc ratio —66——..99— 1——67—..29——4 (b) Pb(NO3)2 1
to produce 1 mol of Zn3P2.
2 31 g of P react with excess =1 = 2.5 (c) (i) Pb(NO3)2(aq) + 2KI(aq) →
zinc to produce 257 g of Zn3P2. =132 = 2.5 3 2 PbI2(s) + 2KNO3(aq)
Therefore, 51.4 kg of Zn3P2 is =2 =5 Yellow 2
produced from
25——51—7.4——kk—gg——ZZ—nn—3—3PP—2—2 2 31 kg of P (ii) 1 mol of lead nitrate reacts
= 12.4 kg of phosphorus
with 2 mol of potassium
iodide to form 1 mol of
The empirical formula is C2H5. 2 lead(II) iodide and 2 mol of
(c) 1.2 dm3 of gas has a mass of 2.9 g.
potassium nitrate. 1
Hence, 1 mol of gas (24 dm3)
has a mass of —12—.42— 2.9 g = 58 g (iii) Lead(II) iodide 1
The relative molecular mass of X (iv) 2 mol of KI produce 1 mol
of PbI2(461 g).
Hence 0.04 mol of KI will
Self Assess 3.7 is 58. 2 produce —0—.0—4 461 g
1 (a) HBr
(b) Zn(OH)2 (d) Let the molecular formula be 2
(c) KNO3
(d) Na2SO4 (C2H5)n. = 9.22 g of PbI2 2
(e) AgNO3 (24 + 5)n = 58
F (f) MgCI2 4 (a) (i) 2Mg + O2 → 2MgO 2
O 2 (a) 1 mol of copper(II) oxide reacts
R n = —52—98— = 2 (ii) 2 mol of Mg produce 2 mol
M with 1 mol of sulphuric acid to
4 form 1 mol of copper(II) sulphate of MgO.
Hence the molecular formula is 1 mol of Mg (24 g) produce
C4H10. 2 1 mol of MgO (40 g).
(e) 2C4H10 + 13O2 → 8CO2 + 10H2O
Hence 3 g of Mg produce
2
—3— 40 g
and 1 mol of water. (f) (i) 1 mol of C4H10 (58 g)
24
(b) 2 mol of magnesium nitrate produces 5 mol of water = 5 g of MgO 2
decomposes to form 2 mol of (5 18 g). (iii) Used as an antacid medicine
magnesium oxide, 4 mol of 11.6 g of C4H10 produce 1
nitrogen dioxide and 1 mol of 1—5—1—8.6— 5 18 g
oxygen gas when heated. (iv) MgO + 2HNO3 →
Mg(NO3)2 + H2O 2
(b) (i) X : Copper(II) oxide
= 18 g of water 1
(c) 1 mol of hydrogen gas react with Gas P : Carbon dioxide 2
CHAPTER 3 1 mol of lead(II) oxide to form (ii) 1 mol of C4H10 (58 g) (ii ) CuCO3 hea→t CuO + CO2 1
1 mol of lead metal and 1 mol produces 4 mol of CO2
of water. (4 6 1023 molecules). (iii) (a) 1 mol of CuCO3 (124 g)
(d) 1 mol of calcium carbonate reacts produces 1 mol of CuO
with 2 mol of hydrochloric acid to 11.6 g of C4H10 produce
form 1 mol of calcium chloride, —15—1—8.—6 4 6 1023 (80 g).
1 mol of carbon dioxide and 1
mol of water. = 4.8 1023 molecules 1 6.2 g of CuCO3 produce
—16—2.2—4 80 g
SPM Exam Practice 3 2 (a) The air in the combustion tube
Multiple-choice Questions = 4 g of CuO 1
1 C 2 A 3 B 4 D 5 D must be displaced before lighting
6 C 7 B 8 A 9 A 10 C (b) 1 mol CuCO3 (124 g)
11 B 12 D 13 D 14 B 15 A the hydrogen gas. 1 produces 1 mol of CO2
16 A 17 C 18 C 19 B 20 A (24 dm3).
21 D 22 C 23 D 24 D 25 A (b) The heating, cooling and weighing
is repeated until a constant mass 6.2 g of CuCO3 produce
—6—.2— 24 dm3
is obtained. 1
124
(c) (i) Zinc and hydrochloric acid 1
(ii) Zn + 2HCI → ZnCI2 + H2 1 = 1.2 dm3 of CO2 1
(iii) Anhydrous calcium chloride 5 (a) 2 NaHCO3 heat →
1
2 6 B 27 B 28 D 29 C 30 B (d) (i)
3 1 C 32 D 33 C 34 B 35 A Element M O Na2CO3 +CO2 + H2O 2
36 C 37 A 38 C 39 D 40 B Mass 5.5 g 3.2 g (b) The gas bubbles is passed into
limewater. If CO2 is present, the
limewater turns cloudy. 2
Structured Questions
Number of moles 5.5 = 0.1 3.2 = 0.2 (c) (i) 2 mol of NaHCO3 (168 g)
1 (a) A hydrocarbon is a compound ——5—5— ——1—6— form 1 mol of CO2 (24 dm3)
at room temperature.
that contains the elements carbon Simplest ratio 1 2 Hence, 8.4 g of NaHCO3
and hydrogen only. 1
Answers 544
form —18—6.4—8 24 dm3 through the tube to cool (iii) The magnesium ribbon is
coiled and then placed in the
the lead and to prevent O2 crucible. The crucible, lid and
= 1.2 dm3 of CO2 3 from air mixing with the the magn esium ribbon is
(ii) 2 mol of NaHCO3 (168 g) weighed again and its weight
hot lead. recorded.
form 1 mol of Na2CO3
(106 g). • The tube with the lead (iv) The crucible is heated until
the magnesium ribbon starts
Hence, 8.4 g of NaHCO3 metal formed is weighed to burn.
form —18—6.4—8 106 and its mass recorded. (v) The lid is lifted and closed
occasionally during the
• The heating, cooling and burning of magnesium.
= 5.3 g of Na2CO3 3 weighing is repeated until (vi) After combustion, the crucible
(d) (i) NaHCO3 + HNO3 → and its contents is cooled
a constant mass is and weighed again. Its weight
NaNO3 + CO2 + H2O 2 is recorded.
(ii) 1 mol of NaHCO3 (84 g) obtained. 4
( vii) The heating, cooling and
produces 1 mol of NaNO3 (ii) Tabulation of data weighing is repeated until a
(85 g). constant weight is obtained.
Mass of combustion tube + x gram
Hence, 17 g of NaNO3 is empty porcelain dish 5
produced from (e) Tabulation of data
Mass of combustion tube + y gram
—8—15—7g—g—o—fN—Na—aN—NO—O—3 384 g porcelain dish + lead oxide Mass of empty crucible + lid
Mass of combustion tube + z gram Mass of crucible + lid +
= 16.8 g of NaHCO3 3 porcelain dish + lead metal magnesium (before heating)
(iii) As a food preservative 2
(e) (i) Sodium hydroxide 1 3 Mass of crucible + lid +
(iii) Calculation magnesium oxide (after
(ii) NaOH + HNO3 → heating) F
NaNO3 + H2O 2 Mass of lead formed = O
(z – x) gram 3 R
Essay Question Mass of oxygen that M
combines with lead = (y – z) 4
1 (a) (i) The empirical formula of gram
a compound shows the
simplest ratio of the atoms of Pb O
the elements that combine Mass (z – x) gram (y – z) gram
to form the compound. 2
(ii) The molecular formula of a Number z—x y—z
of moles —2——0—7— ——1—6——
compound shows the actual
numbers of atoms of the Simplest 4 Periodic Table of Elements
ratio
elements that combine to a b Self Assess 4.1
1 Valence electrons are electrons in the
form the compound. 2
outermost shell of an atom.
For example, the molecular Empirical formula is PbaOb. 4 (a) One valence electron
(b) Two valence electrons
formula of glucose is C6H12O6 Experiment (c) Five valence electrons
and its empirical formula is (d) Seven valence electrons
2
CH2O. 1 (a) Statement of the problem CHAPTER 3 & 4
(b) By reducing a sample of lead
What is the empirical formula of
oxide with hydrogen gas. 2
magnesium oxide? 3
(c) No. Magnesium is more reactive
(b) Variables
than hydrogen. Thus, hydrogen gas
Manipulated variable Mass of Element WX Y Z
cannot reduce magnesium oxide. 3
magnesium ribbon used. Electron 2.8.5 2.1 2.8.8.3 2.8.18.
(d) (i) Procedure
Responding variable Mass of arrangement 32.18.7
• A combustion tube with
magnesium oxide formed.
an empty porcelain dish Group 15 1 13 17
Constant variable Excess supply
is weighed and its mass
of oxygen. 3 Period 32 4 6
recorded.
(c) List of substances and apparatus
• A spatula of lead oxide
Materials Magnesium ribbon 3 Arsenic has a proton number of 33;
is put into the porcelain it has 33 electrons. The electron
about 20 cm long, oxygen arrangement of arsenic is 2.8.18.5.
dish. The combustion Arsenic belongs to Group 15 of the
Apparatus Sandpaper, crucible Periodic Table because it has five
tube and its contents is valence electrons.
with lid, tripod stand, Bunsen Arsenic belongs to Period 4 of the
weighed again. The mass Periodic Table because it has four
burner, clay pipe triangle, electron shells filled with electrons.
is recorded. Nitrogen (2.5) and phosphorus
electronic balance and tongs. 3 (2.8.5) have the same chemical
• Dry hydrogen gas is properties as arsenic because each
(d) Procedure element has five valence electrons.
passed through the tube
(i) An empty crucible with its
to expel all the air before
lid is weighed and its weight
the excess hydrogen is lit
recorded.
and the lead oxide heated.
(ii) A 20 cm length of
• After the lead oxide is
reduced to lead, a stream magnesium ribbon is
of hydrogen gas is passed polished with sandpaper.
545 Answers
4 (a) Element AB C D E FG H JK • Halogens have 7 valence electrons.
During a chemical reaction all halogens
Proton number 2 9 13 19 18 16 7 20 17 6
will accept one electron so that it can
Electron 2 2.7 2.8.3 2.8.8.1 2.8.8 2.8.6 2.5 2.8.8.2 2.8.7 2.4 attain a stable octet electron arrangement.
arrangement
—1 CI2 + e— → CI—
Group number 18 17 13 1 18 16 15 2 17 14
2
Period number 1 2 3 4 3 32 4 32 2.8.7 2.8.8
—1 Br2 + e— → Br—
(b) B and J have the same chemical properties because each element has seven 2
valence electrons.
2.8.18.7 2.8.18.8
Self Assess 4.2 2 Francium, caesium, rubidium, —12 I2 + e— → I—
1 Element P has a duplet (2) electron potassium, sodium, lithium. 2.8.18.18.7 2.8.18.18.8
arrangement and element S (2.8.8) 3 (a) Decreases (c) Increases • As we go down the group, the atomic
has an octet electron arrangement. (b) Increases (d) Increases size increases. The force of attraction
These elements do not need to 4 The burning sodium metal reacts with between the nucleus and electrons
accept, donate or share electrons with the brown bromine gas to produce become weaker.
other elements. Therefore, they exist white sodium bromide salt. • The elements down the group has a
as monatomic gases. 2Na(s) + Br2(g) → 2NaBr(s) lower tendency to attract an electron to
2 Argon: To fill electric light bulbs so as
Self Assess 4.4 form a halide ion.
to increase their lifespan. 1 Chlorine dissolves in water to form
Neon: Fill advertising bulbs. It gives a • Therefore, the elements down the
red light. hyd rochloric acid and hypochlorous(I)
3 The melting points and boiling points acid. group is less reactive.
increase as we go down the group
F because the sizes of the atoms CI2 + H2O → HCI + HOCI 5 Compound Uses
O increase. As the sizes of the atoms
R increase, the van der Waals forces of These acids ionise to give hydrogen Silver bromide To make photo
M ions, H+ resulting in the solution (AgBr) graphic films
becoming acidic.
Potassium iodide Added to
attraction become stronger. The hypochlorous acid (HOCI) formed (KI) iodised salt to
4 4 The electronic configuration of neon is
has a bleaching property. prevent goitre
2.8. Neon has attained an octet in its 2 (a) Test both solutions with starch Silver chloride To make photo
electron arrangement. It does not need solution in separate test tubes. (AgCI) chromic glass
to share, donate or accept electrons from Iodine solution will form a dark
other elements to form a compound. blue precipitate with starch Dichlorodiphenyl- An insecticide
solution. Bromine solution does trichloroethane, used to kill
Self Assess 4.3 not show any reaction. CCI3CH(C6H4CI)2 mosquito larvae
1 (a) Pentium has a silvery surface; it is (b) See experiment 4.6 (Reaction of
Tin(II) fluoride, Added to
soft and can be cut by a penknife halogen with iron wool) SnF2 fluoridated
and it is a conductor of electricity. 3 (a) Yes, it shows similar chemical toothpaste
Pentium reacts with oxygen to
CHAPTER 4 form pentium oxide. This oxide properties. The electron Self Assess 4.5
dissolves in water, forming arrangement of iodine is 1 (a) Decreases
pentium hydroxide solution. 2.8.18.18.7 and the electron (b) Increases
arrangement of chlorine is 2.8.7. (c) The oxides change from basic to
4Pn(s) + O2(g) → 2Pn2O(s) Both elements have 7 valence
Pn2O(s) + H2O(I) → 2PnOH(aq) electrons. During a chemical amphoteric to acidic.
reaction, both iodine and chlorine
will accept one electron to attain 2 (a) The electron arrangement of X is
Pentium metal reacts with cold
an octet electron arrangement. 2.1 and Y is 2.7.
water, forming pentium hydroxide
(b) (i) Higher (ii) Higher Both elements have two electron
solution and hydrogen gas. shells. Thus, they belong to
4 (c) c((aiii))n IN2b2e(FaseI()e(asx+qp))l+a2+iNn3eNaIdO2a(OHsb)Iy((aa→aqqt))o2m+→FeicHI32s(Osiz)(eI)s. Period 2 of the Periodic Table.
2Pn(s) + 2H2O(l) → This (b) Y is more electronegative. Y has
2PnOH(aq) + H2(g) 7 valence electrons. Thus, it has
Element Proton Electron a tendency to accept an electron
Pentium burns in chlorine gas, number arrangement during a reaction to form a negative
forming pentium chloride salt. ion. It is more electronegative.
Chlorine 17 2.8.7 Element X has one valence
2Pn(s) + CI2(g) → 2PnCI(s) Bromine 35 2.8.18.7 electron. It will donate the valence
Iodine 53 2.8.18.18.7 electron during a reaction to form
(b) Pentium is more reactive than a positive ion. It is electropositive.
sodium because the atomic
size of Pn is larger than Na. The
force of attraction between the
nucleus and the valence electron
in pentium atom is weaker and 3 Na2O MgO AI2O3 SiO2 P4O10 SO2 or SO3 CI2O7
thus the electron is more easily
transferred to other elements Basic Basic Amphoteric Acidic Acidic Acidic Acidic
during a chemical reaction.
Answers 546
4 (a) The melting point of silicon is (c) (i) 2Rb(s) + 2H2O(I) → of attraction of the nucleus of
higher than that of sulphur. 2RbOH(aq) + H2(g)
atom R towards an electron is
(b) Silicon is a weak conductor of (ii) 2Rb(s) + CI2(g) → 2RbCI(s)
electricity but iron is a good 2 stronger than atom T. Thus, R
conductor of electricity.
is more electronegative. 2
Self Assess 4.6
1 To each solution, add sodium (d) Each Group 1 element has one 5 (a) Good conductors of electricity and
hydroxide reagent. The solution valence electron. This electron will high melting points 2
containing Cu2+ ion will form a blue
precipitate. be released in a chemical reaction. (b) Blue 1
Cu2+(aq) + 2OH—(aq) → Cu(OH)2(s) The atomic radius of rubidium is (c) (i) Iron powder
blue precipitate
larger and the valence electron is (ii) Nickel powder 2
The solution containing the blue food
colouring does not form any blue further from the nucleus compared (d) They have more than one oxidation
precipitate.
2 The alloy is strong, light and can to potassium. The electrostatic force number in their compounds.
withstand high temperatures caused
by combustion of the jet fuel. of attraction between the nucleus They form complex ions. 2
SPM Exam Pracitce 4 and the valence electron is weaker. (e) Aqueous ammonia solution or
Multiple-choice Questions
1 D 2 A 3 D 4 C 5 D Hence, the valence electron of sodium hydroxide 1
6 B 7 A 8 C 9 B 10 D
11 A 12 C 13 C 14 D 15 A rubidium is more easily released 6 (a) Na, Mg or AI 1
16 B 17 A 18 B 19 D 20 C
21 A 22 D 23 A 24 B 25 A compared to potassium. 2 (b) The atomic radius becomes smaller.
26 B 27 C 28 D 29 B 30 C
31 D 32 D 33 A 34 D 35 B (e) (i) RbNO3 (ii) Rb2SO4 2 The charge of the nucleus increases
3 6 C 37 C 38 D 39 A 40 B
(f) White 1 by one unit from one element to
Structured Questions
1 (a) Liquid 1 3 (a) E, R, C, G, or H 2 the next element across the period
(b) Br2(I) + H2O(I) → (b) (i) E+ (ii) Q— 2 but the number of filled electron
HBr(aq) + HOBr(aq)
(c) G 1 shells is the same. Thus, the force
hydrobromic hypobromous(I)
acid acid 1 (d) B. It is used to fill airships or of attraction between the nucleus
(c) meteorological balloons. 2 and the valence electrons
(e) Q 1 becomes stronger. 3 F
O
(f) R is more reactive. (c) Argon (Ar). Argon has attained an R
M
Both E and R donate one valence octet in its electron arrangement. It 4
electron each during chemical does not need to donate, receive
reactions. The atomic radius of R is or share electrons with other
larger than the atomic radius of E. elements. 2
The valence electron in R is further (d) Na2O, MgO, AI2O3, SiO2,
P4O10, SO2, CI2O7 3
from the nucleus and the force (e) (i) Na2O
(ii) Na2O(s) + H2O(I) →
of attraction is weaker. Thus, the
2NaOH(aq)
element R can donate its valence
e lectron more easily. 2 sodium hydroxide
4 (a) R and T or P and U 1 2
(b) (i) Group 1 2 (f) (i) Sulfur dioxide
(ii) Period 4 (ii) SO2(g) + H2O(I) → H2SO3(aq)
sulphurous acid
(Note that the electron
2
arrangement of U is 2.8.8.1) CHAPTER 4
(c) P is used in the hydrogenation of (g) Amphoteric oxide 1
oil to produce margarine. (h) Silicon is used as a semiconductor.
(P is hydrogen) 1 1
(d)
Essay Questions
1 (a) — Each Group 1 element has
one valence electron. 1
2 2 — During chemical reactions, a
(e) (i) U
(d) S even 1 Group 1 element will donate its
(e) Bromine is more reactive. (ii) 2U(s) + 2H2O(I) → valence electron to attain the
2UOH(aq) + H2(g) 2
Each halogen atom has seven stable octet in its electron
(f) U+ 1
valence electrons. During reactions, arrangement. 1
a halogen will accept one electron (g) (i) Electronegativity is a — The reactivity of Group 1
to attain the stable octet in its measurement of the elements depends on the
electron arrangement. A bromine tendency of an element to tendency of the elements to
atom has a smaller atomic radius attract electrons towards the donate their valence electrons.
than an iodine atom. Thus, nucleus. 1 1
bromine has a greater tendency to — When descending Group 1 from
accept an electron than iodine. (ii) R is more electronegative. 1 lithium to potassium, the atomic
3 T is below R in the same size becomes larger. 1
2 (a) It is a soft metal and is a group (Group 17). Atom R — The force of attraction between
conductor of electricity. 2 has less filled electron shells the protons in the nucleus and
(b) Store in paraffin oil. 1 than T. The atomic radius of R the valence electron becomes
is less than that of T. The force weaker. 1
547 Answers
— The elements lower in Group 1 (ii) Each element in Period 3 has 3 (a) Helium: used to fill airships.
three filled electron shells. 1
loses its valence electron more Neon: used to fill advertising
The proton number increases
easily. 1 bulbs.
by one unit from one element
— Therefore, reactivity increases Argon: used to fill electric
to the next across Period 3. 1
down Group 1. 1 bulbs. 5
As the number of protons in
(b) Physical properties (b) Helium has attained a duplet in
the nucleus increases, the
Rubidium is a soft metal, it is a good its electron arrangement. The
electrostastic force of attraction
conductor of electricity and heat. 3 other noble gases have attained
between the nucleus and the
Chemical properties the octet in their electron
valence electrons become
(i) Rubidium reacts with cold arrangement. 2
stronger. 2
water to form an alkaline Therefore, they do not donate,
The valence electrons are
solution. accept or share electrons with
attracted closer to the nucleus,
2Rb(s) + 2H2O(I) → other elements, thus they exist
2RbOH(aq) + H2(g) 2 causing the atomic radius to
(ii) Rubidium reacts with chlorine as monatoms. 3
to form rubidium chloride salt. decrease. 1
(c) (i) — A piece of sodium is
(b) — Transition elements have
heated in a gas jar spoon
variable oxidation numbers
2Rb(s) + CI2(g) → 2RbCI(s) until it starts to burn. 1
2 in their compounds. 1
— The ignited sodium in
For example, iron forms two
the gas jar spoon is
(iii) Rubidium burns in the air to oxidation states:
form rubidium oxide. This Fe2+ and Fe3+ 1 placed in a gas jar of
oxide dissolves in water to — Transition elements form chlorine gas. 1
form rubidium hydroxide coloured compounds in — The sodium burns brightly
solution. 1 aqueous solutions. 1 with a yellowish flame. A
white solid is obtained. 2
F 4Rb(s) + O2(g) → 2Rb2O(s) For example, aqueous Fe2+
O 1
R Rb2O(s) + H2O(I) → solution is green while an 2Na(s) + CI2(g) →
M 2RbOH(aq) 1 2NaCI(s) 1
4 (c) The number of protons in sodium aqueous Fe3+ solution is
atom is 23 – 12 = 11.
The number of protons in chlorine brown. 1 (ii) — Iron wool inside a
atom is 35 – 18 = 17.
The number of electrons is equal — Transition elements or their combustion tube is heated
to the number of protons. 1
compounds have catalytic strongly. 1
properties. 1
For example, iron powder is — Chlorine gas is passed
used as a catalyst in the Haber through the heated iron
process to manufacture wool. 1
ammonia. 1 — The iron wool glows
Therefore, the electron arrangement — Transition elements can
of Na is 2.8.1 and the electron forming brown iron(III)
form complex ions. 1
arrangement of CI is 2.8.7. 1 chloride. 2
For example, copper(II) ions
As both atoms have three filled 2Fe(s) + 3CI2(g) →
form the dark blue tetraammine 2FeCI3(s) 1
electron shells, they belong to
Period 3 of the Periodic Table. 1 copper(II), Cu(NH3)42+ ion. 1
CHAPTER 4 2 (a) (i) — Each halogen atom has 4 (a) (Period 3 is used as an example)
seven valence electrons. 1 Sodium Magnesium Aluminium Silicon Phosphorus Sulphur Chlorine Argon
— During a chemical reaction,
Metal Metal Metal Semi-metal Non-metal Non-metal Non-metal Non-metal
a halogen atom will accept
one electron so that it can 5
attain a stable octet in its (b) The electronegativity increases Procedure
1 A little sodium oxide powder is
electron arrangement. 1 from sodium to argon in Period 3. 1 added to two separate test tubes.
2 5 cm3 of nitric acid and 5 cm3
— The reactivity of Group 17 Across Period 3, the number of of sodium hydroxide are added
separately to the two different
elements depend on their protons increases by one for each test tubes.
3 The contents in each test tube
tendency to accept an element. 1 are heated slowly while being
stirred with glass rods.
electron. 1 The force of attraction of the 4 The solubility of sodium oxide in
the two solutions is recorded.
— The atomic size of Group nucleus increases as the number 5 The experiment is repeated by
replacing the sodium oxide with
17 elements increase of protons increases. 1 magnesium oxide, aluminium
down the group as the The atomic radius decreases
number of filled electron across Period 3. 1
shell increases. 1 This causes an increase in the
— The force of attraction force of attraction between the
between the nucleus nucleus and electrons. 1
and the valence electron (c) PN4aO2O10,, MSOg2O, ,CAI2IO2O7 3, Si O2, 2
becomes weaker when
descending the group. 2
— The lower elements in the Experiment: Reaction of oxide, silicon(IV) oxide
group has a lower tendency oxides of Period 3 with 2 mol and phosphorus(V) oxide. 3
to accept an electron to dm—3 nitric acid and 2 mol dm—3
form an ion. 1 sodium hydroxide solutions
Answers 548
Results (b) Atoms P and Q will take part in
chemical bonding. This is because
Oxides Solubility in NaOH Solubility in HNO3 Inference they have less than 8 electrons in
Na2O Not soluble the outermost shell.
MgO Dissolves in nitric acid Sodium oxide is a base
Not soluble 3 Ionic bond and covalent bond.
AI2O3 forming a colourless because it reacts with
SiO2 Dissolves in sodium Self Assess 5.2
P4O10 hydroxide forming a solution an acid 1 (a) Ca2+
colourless solution
Dissolves in sodium Dissolves in nitric acid Magnesium oxide is a (b) P3—
hydroxide forming a (c) S2—
colourless solution forming a colourless base because it reacts (d) K+
Dissolves in sodium (e) N3—
hydroxide forming a solution with an acid 2 (a) P + e— → P–
colourless solution
Dissolves in nitric acid Aluminium oxide is ampho 2.7 2.8
forming a colourless teric because it reacts Q → Q2+ + 2e—
2.8.8.2 2.8.8
solution with both acid and alkali
(b)
Not soluble Silicon(IV) oxide is acidic
because it reacts with an
alkali
Not soluble Phosphorus(V) oxide is
acidic because it reacts
with an alkali
Thus, we can conclude that the oxides change from basic to amphoteric and then
to acidic. 5
Experiments (i) The way to manipulate a variable: Pass F
1 (a) (i) Manipulated variable: the different halogen gases over the O
heated iron wool R
Types of halogens M
4
(ii) Responding variable: (ii) What to observe in the responding 3 (a) E+, F2+, G2—, H —
(b) (i) E2G
The glow of iron wool variable: Brightness of flame or glow (ii) EH
(iii) FG
(iii) Constant variable: (iii) The way to maintain the constant variable: (iv) FH2
Quantity of iron wool Use the same amount of iron wool 6
(b) The higher the position of a 2 The lithium metal is then Self Assess 5.3
dropped into a basin of water
halogen is in Group 17, the more carefully using a pair of tongs. 1 (a) H (b) H H
reactive it is with heated iron 3 The observation is recorded.
4 The solution formed in the
wool. 1
basin is tested with a piece of
(c) Chlorine, bromine, iodine 1 blue litmus paper. (c) H (d)
5 The experiment is repeated
(d) The reactivity of the halogens in using small pieces of sodium 2 (a) (i) VZ4 CHAPTER 4 & 5
and potassium metal. (ii) VY2 (ii)
Group 17 decreases down the (e) Tabulation of data (iii) XZ3
group. 1 (b) (i)
(e) Least reactive, it will only glow
faintly with heated iron wool. 1
2 (a) Problem statement What is the
trend of the reactivity of Group Metal Observation
1 metals when descending the Lithium
Sodium (iii)
group from lithium to potassium? Potassium
(b) Hypothesis The reactivity of 15
Group 1 metals increases down
the group.
(c) (i) Covalent compound
(c) Materials Small pieces of lithium,
(ii) Covalent compound
sodium and potassium metals, (iii) Covalent compound
basin filled with water, filter paper 5 Chemical Bonding
and blue litmus paper. Self Assess 5.1 3 (a) (b)
1 Neon: 2.8
Apparatus Penknife, basin and
Argon: 2.8.8
tongs. Both are stable because each of them
has 8 valence electrons corresponding
(d) Procedure to the stable octet electron arrangement.
2 (a) Atom R is chemically inert. This is
1 A piece of lithium metal is
because atom R has 8 electrons
removed from the bottle. A in the outermost shell.
small piece of the metal is cut
(c)
using a penknife. A piece of
filter paper is used to absorb
the paraffin oil from the metal.
549 Answers
Self Assess 5.4 2 (a) P: 2.8.1, Q: 2.6 2 (ii) Element Q has a low
1 (a) Aluminium oxide, copper(II) (b) P is in Group 1, Q is in Group 16 melting point because the
chloride/silicon dioxide/ 2 intermolecular forces of
ammonium nitrate attraction between molecules
(c) (i) P → P+ + e– (van der Waals forces) are
(b) Ammonium nitrate, copper(II) (ii) Q + 2e– → Q2– 2
chloride (d) (i) P2Q 1 weak. 1
(c) Ethanol (d) Silicon dioxide (ii) Ionic bond
(e) Ethanol, tetrachloromethane (iii) 1 Essay Questions
(f) Aluminium oxide 1 1 (a)
2 (a) R and S (c) Q and S Element C Na Cl
No. of protons 6 11 17
(b) Q and S
3 Carbon dioxide is a covalent
compound consisting of simple (e) Low melting point and boiling No. of neutrons 6 12 18
covalent molecules with weak
intermolecular forces of attraction point/does not conduct Electron arrangement 2.4 2.8.1 2.8.7
electricity in any state. 1
between the molecules. Little energy 3 (a) (i) Covalent bond
is required to overcome these weak 1 No. of valence 417
intermolecular forces of attraction. (ii) Ionic bond 1 electrons
Hence, it exists as a gas. Magnesium
chloride is an ionic compound with (b) The weak intermolecular forces 4
strong ionic bonds holding the ions (b) • One carbon atom forms 4
together. High energy is required to require little energy to overcome
separate the strong bonds. It has a covalent bonds with 4 chlorine
high melting point and hence it exists it. 2 atoms to form a CCl4 molecule.
as a solid. • One carbon atom shares
Magnesium chloride can conduct (c) Any ionic compound, example: 4 valence electrons with 4
electricity in the molten state. The ionic chlorine atoms while each
bonds between ions are overcome in sodium chloride, magnesium chlorine atom shares 1 valence
the molten state. The ions are free to electron with the carbon atom.
move to conduct electricity. chloride 1 • The sharing of 4 pairs of
electrons enables carbon and
F (d) R 1 chlorine to achieve the stable
O octet electron arrangement as
R (e) There are freely moving ions
M
4 in the liquid state. The ions are
not able to move freely in the
solid state because they are held
together by the strong
electrostatic force of attraction. 2
(f) Particles in R are atoms. Particles in the noble gases.
• The electron arrangement of
SPM Exam Practice 5 in S are molecules.
Multiple-choice Questions R: any metal, examples: copper/ CCl4 is as follows:
aluminium
1 B 2 C 3 B 4 B 5 D
S: macromolecules , examples:
6 C 7 A 8 C 9 C 10 B
diamond/silicon dioxide. 2
11 B 12 D 13 D 14 A 15 C
16 D 17 B 18 B 19 A 20 C 4 (a) 19 1
21 C 22 D 23 A 24 C 25 A (b) (i) Covalent bond 1
CHAPTER 5 26 A 27 D 28 B 29 A 30 B (ii) WY4 1 • A sodium atom donates 1
31 D 32 D 33 A 34 C 35 B (iii) 88 (12 + 4 3 19) 1 valence electron to form a
36 D sodium ion with the stable
(c) W and X. They have the same octet electron arrangement as
in the noble gases:
Structured Questions number of protons but different Na → Na+ + e–
2.8.1 2.8
1 (a) (i) Ionic bond 1 number of neutrons. 2
(ii) • A chlorine atom accepts 1
(d) (i) Y and Z 1 electron to form a chloride ion
with the stable octet electron
(ii) They have the same number arrangement as in the noble
gases:
of valence electrons. 1 Cl + e– → Cl–
2.8.7 2.8.8
(e) ((iii)) FHeigZh3 melting point and 1
boiling • The sodium ions and the chloride
ions are attracted by the strong
2 point/soluble in water/conducts electrostatic force of attraction.
(iii) High melting point and boiling electricity in liquid state or in • The electron arrangement of
sodium chloride is as follows:
point/soluble in water/conducts aqueous solution. 1
electricity in liquid state or in 5 ((ba)) PPQis4 in Group 14, Q is in Group 1
17
aqueous solution. 1
2
(b) (i) 2.8 1
(c) Covalent bond 1
(ii)
(d) It has low melting point and
boiling point, it is insoluble in
2 water but soluble in organic
(c) No, because U has a stable octet
solvents and it cannot conduct
in the electronic configuration. U
does not gain, lose or share electricity in any state. 3
(e) (i)
electrons with other elements. 2
(d) 2V + 2H2O → 2VOH + H2 1 2 8
Answers 550
(c) — An ionic compound has high — Two nitrogen atoms share 3 pairs
boiling and melting points but of electrons between them. 1
CCl4 NaCl a covalent compound has low
melting and boiling points. — Diagram showing the electron
Low melting and High melting and arrangement of a nitrogen
boiling points boiling points 1+1 molecule:
Insoluble in water Soluble in water — An ionic compound is soluble
in water but insoluble in
Soluble in organic Insoluble in organic organic solvents whereas a
1
solvents solvents covalent compound is soluble
in organic solvents but (b) An ionic compound consists of
Does not conduct Conducts electricity
electricity in any in the liquid and insoluble in water. 1 + 1 ions that carry charges. 1
state aqueous state
— An ionic compound conducts • In the liquid state, heat energy
electricity in the liquid state or has overcome the ionic bond;
4 in an aqueous solution but a the ions are free to move. 1
2 (a) (i) — Two elements that form
covalent compound does not • In aqueous solution, water
an ionic compound are conduct electricity in any state. molecules separate the ions so
W and Z. 1 1+1 that the ions are free to move.
— The electron arrangement 3 (a) — A single covalent bond is 1
of atom W is 2.8.1 and formed from the sharing of • A covalent compound consists
the electron arrangement one pair of electrons. 1 of molecules. 1
of atom Z is 2.8.7. 1 + 1 — Example: a chlorine molecule • There are no freely moving ions
— To achieve the stable octet consists of a single covalent and therefore it does not
electron arrangement as in bond formed between 2 conduct electricity. 1
the noble gas: 1 chlorine atoms. 1 Experiments
1 (a)
— An atom W donates one — A chlorine atom has 7 valence F
O
electron to form a W+ ion. electrons and requires 1 more R
M
1 electron to achieve the stable Chemical Physical Observation when 4
compound state shaken with
— An atom Z accepts one octet electron arrangement. 1
Water Acetone
electron to form a Z– ion. — Two chlorine atoms share 1
1 pair of electrons. 1
— The positive ions and — Diagram showing the electron Potassium Solid Forms a Does not
chloride colourless dissolve in
negative ions formed are arrangement of a chlorine solution acetone
attracted by the strong molecule:
electrostatic force of
attraction. 1 3
1 (b) X is soluble in water but is
— A double covalent bond is insoluble in acetone.
formed from the sharing of 2 Y is insoluble in water but is
pairs of electrons. 1 soluble in acetone.
1 — For example, an oxygen Z is soluble in both water and
(ii) — Two elements that form a molecule consists of a double acetone. 3 CHAPTER 5
covalent compound are Y covalent bond formed (c) (i) X and potassium chloride
and Z. 1 between 2 oxygen atoms. 1 (ii) Y and Z 3
— The electron arrangement — An oxygen atom has 6 valence (d) X is potassium nitrate because
of atom Y is 2.4 and the electrons and requires 2 potassium nitrate is a solid ionic
electron arrangement of electrons to achieve the stable compound, soluble in water and
atom Z is 2.8.7. 1 octet electron arrangement. 1 insoluble in acetone. 3
— One atom of Y contributes — Two oxygen atoms share 2 pairs (e) Z is ethanol because ethanol is a
4 electrons to be shared of electrons between them. 1 liquid and is soluble in both water
with 4 atoms of Z. 1 — Diagram showing the electron and acetone. 3
— Four pairs of shared arrangement of an oxygen 2 (a) Problem statement Do ethanol
electrons form 4 single molecule: and aqueous sodium nitrate
covalent bonds. 1 solution conduct electricity?
(b) Hypothesis Ethanol does not
conduct electricity whereas
1 aqueous sodium nitrate solution
— A triple covalent bond is formed conducts electricity.
from the sharing of 3 pairs of (c) Variables
electrons. 1 Manipulated variable Types of
— For example, a nitrogen molecule compounds.
consists of a triple covalent bond Responding variable Electrical
between 2 nitrogen atoms. 1 conductivity.
2 — A nitrogen atom has 5 valence Constant variable Physical state
(b) Three differences between an
electrons and requires 3 of compound.
ionic compound and a covalent
compound: electrons to achieve the stable (d) Substances Ethanol and
octet electron arrangement. 1 aqueous sodium nitrate solution.
551 Answers
Apparatus Beaker, measuring are not free to move and hence cannot conduct electricity. In the molten state, heat
cylinder, graphite rods, batteries, energy has overcome the ionic bonds. The ions are freely moving and hence able to
light bulb, switch and connecting conduct electricity. Ions produced Ions that move to
wires. Self Assess 6.2 Zn2+, Cl– anode cathode
(e) Procedure 1
(i) 30 cm3 of ethanol is measured CI— Zn2+
Name of compound
by a measuring cylinder and
is poured into a beaker. (a) Zinc chloride
(ii) Two graphite rods are
immersed in ethanol and (b) Magnesium oxide Mg2+, O2— O2— Mg2+
the circuit is completed by
connecting to batteries, light 2 Half-equation at Products formed at
bulb and a switch. Name of
(iii) The switch is turned on and compound anode cathode anode cathode
the bulb is checked if it lights (a) Calcium 2O2— → O2 + 4e— Ca2+ + 2e— → Ca Oxygen gas Calcium metal
up. oxide
(iv) The experiment is repeated
using aqueous sodium nitrate (b) Aluminium 2I— → I2 + 2e— AI3+ + 3e— → AI Iodine vapour Aluminium
solution to replace ethanol. iodide metal
(f) Tabulation of data
Chemical 3 (a) (i) Shiny grey metal is deposited (ii) Greenish-yellow gas is evolved
compound
Observation (b) (i) Lead metal (ii) Chlorine gas
(c) (i) Pb2+ + 2e— → Pb (ii) 2CI— → CI2 + 2e—
F Ethanol Bulb does not light Self Assess 6.3 Ions present Ions discharged Ions discharged
O up when circuit is 1 at the cathode at the anode
R completed.
Aqueous solution
M Aqueous sodium Bulb lights up when (a) Aqueous nitric acid H+, NO3—, OH— H+ OH—
nitrate solution circuit is completed. (b) Silver nitrate Ag+, NO3—, H+, OH— Ag+ OH—
(c) Very dilute Cu2+, H+, CI—, OH— Cu2+ OH—
4 17
copper(II) chloride
CHAPTER 5 & 6 6 Electrochemistry 2 (a) Cell I: copper(II) ions, hydrogen (c) (i) Ag+ + e— → Ag
ions, sulphate ions, hydroxide ions (ii) Ag → Ag+ + e—
Self Assess 6.1 Cell II: silver ions, hydrogen ions, (d) No change. The concentration
1 (a) (i) An electrolyte is a chemical nitrate ions, hydroxide ions
of silver ions remains constant
substance that conducts (b) (i) Colourless gas bubbles are because the rate of formation
electricity in the molten state formed at electrode P. A of silver ions at the anode is the
or in aqueous solution and brown deposit is formed at same as the rate of discharge of
undergoes chemical changes. electrode Q. silver ions at the cathode.
Examples: molten lead(II) (e) The iron key is polished with a
chloride, aqueous potassium (ii) The blue colour becomes piece of sandpaper. Electrolysis
iodide solution (molten or paler. The concentration of is carried out using a low current
aqueous solution of any copper(II) ions decreases. over a longer period of time.
other ionic compounds).
(ii) A non-electrolyte is a chemical (iii) Electrode R becomes thinner. Self Assess 6.5
compound that does not Electrode S becomes thicker. 1 (a)
conduct electricity in any state.
Examples: naphthalene, glucose (iv) Electrode P: (b) Iron metal is the negative
(or any covalent compounds). 4OH— → 2H2O + O2 + 4e— terminal
(b) A conductor can conduct electric Electrode R: Ag → Ag+ + e—
current. An electrolyte can (c) At the negative terminal:
conduct electric current as well as (c) (i) Electrode Q: Positions of ions Fe → Fe2+ + 2e—
be decomposed by electricity. in the electrochemical series. At the positive terminal:
2 Electrolytes: molten zinc oxide, aqueous
zinc chloride solution, aqueous ethanoic (ii) Electrode R: Types of Cu2+ + 2e— → Cu
acid solution and molten sodium chloride. electrodes used. (d) The iron metal corrodes while the
Non-electrolytes: molten sulphur, solid
zinc oxide, molten zinc metal, solid Self Assess 6.4 copper metal thickens.
zinc metal, acetone and aqueous 1 Three uses of electrolysis:
glucose solution.
3 Solid magnesium chloride consists of (a) To extract reactive metals from
positive ions and negative ions held their ores
together by strong ionic bonds. The ions
(b) To purify impure metals
(c) To electroplate a metal object
with another metal
2 (a) Silver
(b) The iron key is deposited with a
layer of shiny silver metal. The
silver electrode becomes thinner.
Answers 552
2 Type of cell Advantages Disadvantages 4 (a) Cu2+, SO42–, H+, OH– 1
(b) Electrons flow from the zinc plate
Dry cells • Cheap • Cannot last long to the copper plate through the
• No spillage occurs • Cannot be recharged
Alkaline • Produces moderately stable • Zinc metal will dissolve and external circuit. 1
cells
current and voltage the electrolyte leaks, corroding (c) (i) The cathode increases in
• Can be carried around easily electrical instruments
• Can be made into different sizes • Current and voltage produced is low mass. 1
• Can last longer than a dry cell • Not rechargeable (ii) Cu2+ + 2e– → Cu 1
• Produces a higher and more • More expensive than a dry cell
• If leakage occurs, the electrolyte is (iii) The colour intensity remains
stable voltage
more corrosive the same. 1
(iv) Oxygen gas 1
(d) (i) The zinc plate becomes
thinner. 1
3 Lead-acid accumulator (c) At the anode: brown gas is (ii) Zn → Zn2+ + 2e– 1
At the negative terminal: Lead metal
releases electrons. Pb → Pb2+ + 2e— evolved. At the cathode: grey (iii) Zn + Cu2+ → Zn2+ + Cu 1
Electrons that are released flow to the
positive terminal through the external metal is deposited. 2 (iv) The cell voltage increases. 1
circuit, producing electric current.
At the positive terminal: Electrons are (d) At the cathode: Pb2+ + 2e— → Pb 5 (a) Positive terminal is copper
received by lead(II) oxide. At the anode: 2Br– → Br2 + 2e—
PbO2 + 4H+ + 2e— → Pb2+ + 2H2O 2 metal.
Self Assess 6.6 Negative terminal is lead metal. 2
1 (a) X, Z, Y, W
(e) Zinc metal will form at the (b) (i) Pb → Pb2+ + 2e— 1
(b) (i) W becomes the negative
terminal. cathode and chlorine gas will (ii) Cu2+ + 2e— → Cu 1
(ii) 0.7 volt [(1.2 – 0.9) + 0.4] form at the anode. 2 (c) Pb + Cu2+ → Pb2+ + Cu 1
(c) (i) Y will displace metal Z and Y
2 (a) H+, Cl–, OH– 1 (d) To complete the circuit for the ions
ions are produced.
(ii) No change will occur. (b) Carbon/graphite/platinum 1 to flow. 1
(iii) W will displace metal X and
(c) (i) Gas X is oxygen gas, gas Y is (e) Higher than 0.57 V. 1 F
W ions are produced. O
2 (a) No change occur. hydrogen gas. 2 6 (a) Electrode L and electrode S. 2 R
M
(b) Displacement reaction occurs: (ii) 4OH—→ 2anHd2OO+H–Oi2on+s 4e— 1 ((cb)) C(ui)2 +C, SuO→42–,CHu2++, OH– 2 4
copper will displace silver from (d) (i) H+ ions are + 2e–
silver nitrate. Copper metal
dissolves, silver metal deposits and discharged at the cathode (ii) Cu2+ + 2e– → Cu 2
a blue colour solution is formed.
Cu + 2Ag+ → 2Ag + Cu2+ and anode respectively in (d) (i) Copper metal
(c) Displacement reaction occurs: electrolysis. The effect is (ii) Oxygen gas 2
magnesium will displace silver
from silver nitrate. Magnesium equal to the removal of (e) Intensity of blue colour in beaker I
metal dissolves and silver metal
deposit is formed. water which causes the is constant. Intensity of blue colour
Mg + 2Ag+ → 2Ag + Mg2+
concentration of hydrochloric in beaker II decreases. 2
acid to increase. 2 (f) (i) Rheostat.
(ii) Chlorine gas is collected at (ii) To control and to ensure that
the anode. This is because the magnitude of electric
the concentration of Cl– ions current remains constant. 2
is higher than OH– ions, Essay Questions
1 (a) Electrolysis is a process involving
hence Cl– ions are selectively
the decomposition of an ionic
discharged at the anode. 2 compound (electrolyte) in the CHAPTER 6
molten form or in aqueous
3 (iii) 2Cl— →cheCml2ic+al 2e— 1 solution by an electric current. 2
(a) In cell A, energy is (b) Magnesium chloride is an ionic
compound. In the solid state, the
converted into electrical energy. In ions are held by strong electrostatic
force of attraction. Hence, the
cell B, electrical energy is converted ions are not free to move and
subsequently cannot conduct
SPM Exam Practice 6 into chemical energy. 2 electric current. When it melts, the
Multiple-choice Questions ions are free to move and
(b) (i) Mg → Mg2+ + 2e— 1 can conduct electric current. 4
1 C 2 C 3 B 4 C 5 B (c) The arrangement of apparatus for
6 D 7 A 8 D 9 A 10 C (ii) Cu2+ + 2e— → Cu 1 the experiment to be carried out is
11 C 12 C 13 D 14 B 15 D shown in the diagram below.
16 D 17 D 18 A 19 A 20 D (c) (i) Carbon electrode Q 1
21 A 22 C 23 B 24 A 25 A (ii) Q is connected to the
26 C 27 D 28 D 29 C 30 A
3 1 C 32 C 33 C 34 B 35 B negative terminal (magnesium
3 6 C 37 B 38 C 39 C 40 C
electrode) in cell A. 1
(iii) Electrons flow from the
magnesium electrode
through the external circuit to
Structured Questions electrode Q. 1
1 (a) Naphthalene does not contain (d) (i) Copper metal,
freely moving ions. 1 Cu2+ + 2e— → Cu 2
(ii) Oxygen gas,
(b) In lead(II) bromide solid, the ions
4OH— → 2H2O + O2 + 4e— 2
are held in a lattice by strong (e) Electrons flow from the copper
bonds and are not free to move. electrode to the silver electrode. 1
In the molten form, the ions are (f) Electrode P will dissolve forming
free to move when the bonds copper(II) ions. Copper metal
holding them are overcome. 2 will deposit on electrode Q. 1
553 Answers
Magnesium chloride is heated Anode: 4OH— → • P releases electrons to form P
until it melts and the circuit is 2H2O + O2 + 4e— 1 ions, P2+.
completed to allow the flow of If concentrated copper(II) P → P2+ + 2e— 2
electric current. Shiny grey solids solution is used, chloride ions • Q ions receive electrons to
will deposit at the cathode and are discharged at the anode and form metal Q.
greenish-yellow gas will evolve chlorine gas is produced. 1 Q2+ + 2e— → Q 2
at the anode. In terms of the This is because in concentrated • Electrons flow from metal P to
ionic theory, molten magnesium copper(II) chloride solution, metal Q. 1
chloride consists of freely moving the concentration of chloride • Deflection of the voltmeter
magnesium ions, Mg2+ and ions is higher than that of 1 needle indicates the flow of
chloride ions, Cl—. hydroxide ions. Hence, chloride an electric current. 1
MgCl2 → Mg2+ + 2Cl— ions are discharged. (c) Displacement reaction will occur.
Magnesium ions, Mg2+ move to Anode: 2Cl— → Cl2 + 2e — 1 P will displace Q from Q(NO3)2
the cathode to accept electrons (b) Electroplating of an iron spoon solution. Metal P will dissolve to
and form magnesium metal. with copper in electrolysis: form P2+ ions. Metal Q will be
Mg2+ + 2e– → Mg In the experiment, copper(II) deposited.
sulphate solution is used as the P + Q2+ → P2+ + Q 4
Chloride ions move to the anode electrolyte. The iron spoon is 1
and donate electrons to form Experiments 3
chlorine atoms. Two atoms of used as the cathode and 1 1 (a) (i) Anode: oxygen gas
chlorine combine to form a copper metal is used as the anode.
chlorine molecule. Copper anode will dissolve and 1 Cathode: hydrogen gas
2Cl— → Cl2 + 2e— 14 copper metal will be deposited on (ii) Positions of ions in the
2 (a) (i) In the electrolysis of aqueous the surface of the iron spoon. 1 electrochemical series.
F copper(II) chloride solution, (b) (i) Iodine
O the ions present are Cu2+
R ions and Cl— ions from CuCl2, (ii) Add a little starch solution, a
H+ ions and OH– ions from dark blue colour will be
water. If carbon is used as
M formed.
(iii) Concentration of ions in the
the anode and cathode, 1
4 hydroxide ions, OH— ions are 2 electrolyte 3
(c) (i) Anode: 2I— → I2 + 2e—
discharged at the anode and Anode: Cu → Cu2+ + 2e— 1
oxygen gas is released. 1 Cathode: Cu2+ + 2e– → Cu 1 Cathode: 2H+ + 2e— → H2
Copper is deposited at the 3 (a) An electrolytic cell is a circuit (ii) 4OH— → O2 + 2H2O + 4e— 3
(d) (i) Copper metal
cathode.
arrangement consisting of electrodes, (ii) Bromine gas
Anode: 4OH— →
connected by wires to batteries, (iii) Copper(II) bromide 3
2H2O + O2 + 4e—
Cathode: Cu2+ + 2e— → Cu immersed in an electrolyte. 1 2 (a) Aim of experiment
The passage of an electric current To investigate the effect of the
1 produces a chemical reaction. 1
If copper metal is used as concentration of an electrolyte on
CHAPTER 6 the cathode and anode, A voltaic cell is a circuit the types of products formed in
copper anode dissolves to arrangement that consists of electrolysis. 2
form copper(II) ions. 1 different metals as electrodes (b) Hypothesis Ions with higher
Anode: Cu → Cu2+ + 2e— immersed in an electrolyte. A 1 concentration will be selectively
chemical reaction takes place discharged during electrolysis. 3
Cathode: Cu2+ + 2e— → Cu 1 and chemical energy is converted
When different substances (c) Variables
are used as the anode, into electrical energy. 1 • Manipulated variable
different products will be (b) Concentration of aqueous
produced at the anode, copper(II) chloride solutions 1
even though copper metal is • Responding variable
deposited at the cathode in Products of electrolysis 1
both cases. 1 • Constant variables Types of
(ii) In the electrolysis of copper(II) electrodes and electrolytes
chloride aqueous solution using used 1
2 (d) Substances 1.0 mol dm–3
carbon as the anode and
cathode, the ions present are • 30 cm3 of P(NO3)2 solution copper(II) chloride solution and
Cu2+ ion and Cl— ion from and 30 cm3 of Q(NO3)2 1 0.001 mol dm–3 copper(II)
solution are placed into chloride solution.
CuCl2, H+ ions and OH– ions
from water. 1 different beakers. 1 Apparatus Batteries, carbon
electrodes, rheostat, ammeter,
If dilute copper(II) chloride • Metal P is dipped in P(NO3)2 connecting wires with crocodile
solution and metal Q is dipped
solution is used, hydroxide
ions are discharged at the in Q(NO3)2 solution. 1 clips, switch and test tubes. 3
anode producing oxygen gas. • A salt bridge is used to complete (e) Procedure
1
the circuit by allowing the ions 1 1.0 mol dm–3 aqueous
to flow. 1 copper(II) chloride solution is
Answers 554
placed in an electrolytic cell 5 Hydrochloric acid dissociates to H+ (c) Number of moles of NaOH in
with carbon electrodes. ions in aqueous solution. H+ ions react 100 cm3 of 0.2 mol dm—3 NaOH
2 Two test tubes filled with with magnesium to produce hydrogen
copper(II) chloride solutions gas: 2HCl + Mg → MgCl2 + H2 = —0—.2—1—0—0—01—0—0 = 0.02
are inverted on top of the Hydrogen chloride dissolved in Number of moles of NaOH in
anode and cathode. methylbenzene exists as molecules 20 cm3 of 2.0 mol dm—3 NaOH
3 The switch is turned on and and is not acidic. H+ ions are not
electrolysis is carried out. produced to react with magnesium. = —2.—10—0—0—02—0 = 0.04
4 Observations at the anode Total number of moles of NaOH
and cathode are recorded. Self Assess 7.2 = 0.02 + 0.04 = 0.06
5 The experiment is repeated Total volume = 100 + 20
using 0.001 mol dm–3 1 (a) (i) E (ii) A = 120 cm3
copper(II) chloride solution. 3 Molarity of new solution
(f) Tabulation of data (b) (i) E (ii) A (iii) D/F (iv) C = (—1—2—0—0/—.0—16—0—0—0)
= 0.5 mol dm—3
(c) (i) B (ii) E (iii) C (iv) A
Self Assess 7.4
2 A strong acid is a chemical substance 1 (a) MgCl2, H2
that dissociates completely into (b) ZnSO4, 2H2O
(c) CH3COOH
Chemical Observation Observation hydrogen ions in water. An example (d) Ca(OH)2, H2SO4
substance at the anode at the cathode 2 (a) 2HNO3 + Mg(OH)2 →
is hydrochloric acid, which undergoes
Mg(NO3)2 + 2H2O
Copper(II) Greenish- Brown deposit complete dissociation in water to form (b) From the equation, 1 mol of
is formed
chloride yellow gas is hydrogen ions, H+ and chloride ions, Cl—: Mg(OH)2 reacts with 2 mol of
HNO3. Hence, 0.001 mol of
solution of 1.0 evolved HCl → H+ + Cl— Mg(OH)2 will react with 0.002
A weak acid is a chemical substance mol of HNO3.
mol dm–3 0.002 mol = —1M—0—0V—0 = —M1—0—0—10—0
that dissociates partially into hydrogen Concentration of HNO3,
M = 0.002 —1—01—00—0
Copper(II) Colourless Brown deposit ions in water. An example is ethanoic = 0.2 mol dm—3
chloride gas is evolved is formed acid, which undergoes partial 3 —MM——BAVV——AB = —12—
∴ —11—.0.—25———25—V.—A0 = —21—
solution of dissociation in water to form hydrogen V olume of HCl, VA = 2 25.0 —11—.2.—05 F
O
0.001 mol dm–3 ions, H+ and ethanoate ions, CH3COO—: = 40 cm3 R
CH3COOH CH3COO— + H+ 4 H2SO4 + 2NH3 → (NH4)2SO4 M
3 pH value of the strong acid is lower —MM—BA—VV—AB— = —12— 4
∴ M A = —M——B—V—A— —VB ——12 = —2—1.0—5—.0——2—52—.0
7 Acids and Bases than that of the weak acid of the
= 1.67 mol dm—3
same concentration.
Self Assess 7.1 3 (a) Propanoic acid, ethanoic acid,
1 (a) (i) An acid is a chemical methanoic acid
compound that produces (b) pH of methanoic acid < 3, pH of
hydrogen ions when it propanoic acid > 3
dissolves in water. Self Assess 7.3
1 (a) C oncentration = —V—oM—lua—ms—s—e
(ii) A base is a chemical
∴46.4 g dm—3 = —2——.m0——ad—sms— 3
compound that neutralises an
Mass of KOH = 2.0 46.4 g
acid to produce salt and water. = 92.8 g
(b) Relative molecular mass of KOH
(iii) An alkali is a chemical = 39 + 16 + 1 = 56
Number of moles = —1—M—0——0—V0
compound that produces Number of moles of KOH CHAPTER 6 & 7
= 2.0 —11—00—000— = 0.2
hydroxide ions when it Mass of KOH = 0.2 56 = 11.2 g
2 ( a) Concentration = —V—oM—lua—ms—se
dissolves in water.
= —————2—.1—2——g— ——— = —2—.—1—2—g—
(b) An acid changes moist blue litmus
(500/1000) dm3 0.5 dm3
paper to red. An alkali changes
= 4.24 g dm—3
moist red litmus paper to blue. (b) Relative molecular mass of Na2CO3
2 Uses Acids or bases = (23 2) + 12 + (16 3)
= 106
To make antacid Mg(OH)2 Molarity = 4—1.—026—4 = 0.04 mol dm—3 SPM Exam Practice 7
To make fertiliser H2SO4, HNO3, 3 (a) Relative molecular mass of NaOH Multiple-choice Questions
NH3, NaOH = 23 + 16 + 1 = 40 1 A 2 D 3 C 4 B 5 C
To make soap NaOH 6 D 7 B 8 A 9 C 10 B
To neutralise M olarity = —84—.0—0 g dm—3 11 C 12 D 13 D 14 D 15 D
acidity in soil CaO, Ca(OH)2 = 0.2 mol dm—3 16 B 17 A 18 A 19 C 20 C
21 B 22 B 23 D 24 D 25 D
3 Q = MgCO3 R = HNO3 (b) M1V1 = M2V2 26 B 27 C 28 D 29 B 30 B
X = HCl Y = KOH Z = NH3 M final (100 + 100) = 0.2 100 31 C 32 A 33 B 34 A 35 B
M final = 0.1 mol dm—3 36 C 37 B 38 C 39 C 40 B
4 (a) H2SO4 + MgO → MgSO4 + H2O Structured Questions
(b) 6HNO3 + 2Al → 2Al(NO3)3 + 3H2
(c) 2HCl + CaCO3 → 1 (a) To prevent the solution from
CaCl2 + CO2+ H2O being sucked up. 1
(d) CH3COOH + NaOH → (b) (i) No noticeable change in 1
CH3COONa + H2O beaker A. Effervescence
(e) KOH + NH4Cl → NH3 + KCl + H2O occurs in beaker B. 1
555 Answers
(ii) Hydrogen chloride in (e) Concentration of H2A Essay Questions
methylbenzene does not have = —02—.5———22—65—.1.—0 = 0.24 mol dm—3 2 1 (a) Neutralisation is a reaction
any acidic properties. In 1 between an acid and a base to
(f)
beaker B, hydrogen chloride produce salt and water only. 1
in water produces H+ ions For example, sulphuric acid reacts
which react with magnesium with sodium hydroxide to 1
to produce hydrogen gas. 1 produce the salt, sodium sulphate
(c) (i) Hydrogen chloride molecules and water only. 1
1 H2SO4 + 2NaOH → Na2SO4 + 2H2O
1
(ii) Hydrogen ions and chloride
ions 1 (b) Sodium hydroxide is a strong
(d) (i) Carbon dioxide gas. Deliver the alkali that undergoes complete 1
gas produced into limewater. dissociation in aqueous solution. 1
Limewater will turn cloudy. 1 2 Ammonia is a weak alkali that 1
(ii) In the presence of water, 4 (a) (i) H+ ion. 1 undergoes partial dissociation 1
hydrogen chloride dissociates (ii) Mg + 2H+ → Mg2+ + H2 1 only. The concentration of hydroxide
to H+ ions that will react with (b) (i) Experiment I. 1 ion in sodium hydroxide is thus
sodium carbonate solution to (ii) Sulphuric acid is a strong acid higher than that in ammonia. 1
produce carbon dioxide gas. 1 whereas ethanoic acid is a Hence the pH value of sodium
(iii) 2H+(aq) + CO32—(aq) → weak acid. Sulphuric 1 hydroxide is higher than that of
CO2(g) + H2O(l) 1
acid undergoes complete ammonia. 1
(iv) Hydrogen chloride can only
dissociation to produce more (c) Molar mass of KOH
act as an acid in the
hydrogen ions whereas 1 = 39 + 16 + 1 = 56 g mol–1 1
presence of water. 1
F ethanoic acid undergoes 250 cm3 of 1.0 mol dm—3 KOH
O 2 (a) (i) A standard solution is a
R partial ionisation in the contains 1—.—01—0—0—02—5—0 mol KOH
M solution with a known
4 presence of water. 1
concentration. 1
(c) pH value of sulphuric acid is lower which is 0.25 56
(ii) Molarity of the solution gives
than that of ethanoic acid. 1 = 14.0 g of KOH. 1
the number of moles of
The concentration of hydrogen • Weigh exactly 14.0 g of KOH
sodium hydroxide dissolved
ions is higher in sulphuric acid accurately in a weighing bottle. 1
in 1 dm3 of the solution. 1
than in ethanoic acid. 1 • Dissolve 14.0 g of KOH in a little
(b) Number of moles of NaOH
(d) pH value will be higher. 1 water in a beaker and transfer 1
= —0—.—5—1—03—0——01—0 ——0 = 0.05
1 Addition of water will dilute the the contents into a 250 cm3
Mass of NaOH = 0.05 3 40 sulphuric acid. This will lower the volumetric flask. 1
= 2 g 1 hydrogen ion concentration and • Rinse the beaker with distilled
(c) (i) A measuring cylinder is hence increase the pH value. 1 water and transfer all the contents
unsuitable because it cannot 5 (a) pH value is 7 (see graph). 2 into the volumetric flask. 1
measure the volume of water • Distilled water is added to
CHAPTER 7 accurately. 1 the volumetric flask until the
(ii) A volumetric flask 1 graduation mark. The solution
(d) Parameter I: mass of sodium produced is 1.0 mol dm—3
hydroxide 1 potassium hydroxide. 1
Parameter II: volume of the • To prepare a 0.1 mol dm—3
potassium hydroxide solution,
solution 1
25.0 cm3 of 1.0 mol dm—3 1
(e) Water is added carefully (drop by potassium hydroxide solution is
transferred to a 250 cm3 1
drop) so that the level of the solution (b) Volume of nitric acid is 20.0 cm3
volumetric flask using a 25.0 cm3
does not exceed the graduation (see graph). 2 pipette.
mark of the volumetric flask. 1 (c) HNO3 + NaOH → • Distilled water is added to the
volumetric flask until the 1
The volumetric flask is stoppered NaNO3 + H2O 1 graduation mark.
2 (a) Chemical formula of magnesium
to prevent the evaporation of water (d) From yellow to orange 1 hydroxide is Mg(OH)2. 1
Magnesium hydroxide is a base 1
which can change the concentration (e) Concentration of HNO3
= —0—.5—2—0—.0—2—5—.0 = 0.625 mol dm—3 2
of the solution prepared. 1
3 (a) 1 mol of acid reacts with 2 mol of
OH— ions (or 1 mol of acid (f)
produces 2 mol of H+ ions) 1 which can neutralise the excess
(b) H2A(aq) + 2KOH(aq) → hydrochloric acid in the stomach. 1
K2A(aq) + 2H2O(l) 1
(c) From pink colour to colourless. 1 Another chemical found in antacid
(d) (i) 26.05 cm3, 26.15 cm3, 26.10
is aluminium hydroxide. 1
cm3 1
(b) (i) Solution X is an acid whereas
(i i) (—2—6—.0—5——+——26—.—13—5—+——2—6—.1—0—)—c—m—3
solution Y is an alkali. 1
• Solution X is sour in
= 26.10 cm3 1 2 taste. 1
Answers 556
• It can react with a base to ions, H+ which is responsible (d)
produce salt and water. 1 for its acidic properties. 1
• It can react with a reactive • CH3COOH(aq)
H+(aq) + CH3COO—(aq) 1
metal to produce a salt • H+ ions change blue litmus
and hydrogen gas. 1 to red and react with reactive
• It can react with a metal metals or carbonates or bases.
carbonate to produce a 1
salt, carbon dioxide gas (b) (i) Test with magnesium ribbon.
and water. 1 • About 3 cm of magnesium
• Solution Y is bitter in taste ribbon is added to about
and it feels soapy. 1 5 cm3 of acidic solution
• It can react with an acid to in a test tube. 1
produce salt and water. 1 • The gas evolved is tested
• When it is heated with by placing a glowing
ammonium salts, ammonia wooden splint near 1 3
gas is produced. 1 the mouth of the test tube. (e) The pH value of a 0.020 mol dm—3
• It forms a metal hydroxide If a ‘pop’ sound is nitric acid is 1.69 (accept 1.68-
when it is added to an produced, indicating the 1.70). 3
aqueous salt solution. 1 presence of hydrogen gas, 2 (a) Problem statement How to
An example of solution X is the solution in the test 1 determine the volume of sulphuric
hydrochloric acid. An example tube is proven to be acidic. acid required to neutralise 25.0 cm3
of solution Y is sodium • Mg + 2H+ → Mg2+ + H2 1 of 0.5 mol dm—3 potassium
(ii) Tests with sodium carbonate
hydroxide solution. 1 hydroxide solution?
(ii) When solution X and solution • A little sodium carbonate (b) Variables F
O
Y are mixed together, powder is added to about Manipulated variable Volume of R
M
neutralisation reaction takes 1 5 cm3 of acidic solution in potassium hydroxide solution used 4
place, salt and water is a test tube. 1 Responding variable Volume of
produced. The solution 1 • The gas evolved is tested sulphuric acid required
produced has a pH value of 7. by passing the gas into lime- Constant variables
1 water. If the limewater 1 Concentrations of sulphuric acid
The ionic equation is: turns milky, indicating the and potassium hydroxide used
H+ + OH— → H2O 1 presence of carbon dioxide (c) Statement of hypothesis
(iii) M1V1 = M2V2 gas, the solution in the test The volume of sulphuric acid
M final (80 + 20) 1 tube is proven to be acidic. required in neutralisation depends
= 0.1 20 1 on the volume of potassium
M final = 0.02 mol dm—3 1 • CO32— + 2H+ → CO2 + H2O hydroxide used.
3 (a) • Pure and dry acid, for 1
(d) Apparatus 25 cm3 pipette,
example, glacial ethanoic acid (iii) Test with copper(II) oxide
pipette filler, 50 cm3 burette,
does not show any acidic • A little copper(II) oxide
retort stand and clamp, conical CHAPTER 7
property. 1 powder is added to about
flasks, filter funnel, dropper and
• Glacial ethanoic acid exists as 5 cm3 of acidic solution in
white tile.
molecules. 1 a test tube and the mixture
Materials Sulphuric acid, 0.5
• Hydrogen ions are not present is heated. 1
mol dm—3 potassium hydroxide
• If the black copper(II)
to show any acidic properties. solution and methyl orange.
oxide powder dissolves and
1 (e) Procedure
a blue solution is formed,
• Glacial ethanoic acid does not 1 A 25 cm3 pipette is cleaned
the solution in the test tube
change blue litmus to red. with distilled water and
is proven to be acidic. 2
It also does not react with rinsed with a little potassium
• CuO + 2H+→ Cu2++ H2O 1
reactive metals or carbonates hydroxide solution.
or bases. 1 Experiments 2 Using a pipette, 25 cm3 of 0.5
1 (a) Manipulated variable
• When water is added to mol dm—3 potassium hydroxide
Concentration of nitric acid solutions.
glacial ethanoic acid, aqueous Responding variable pH values is transferred to a clean conical
ethanoic acid produced shows flask. Three drops of methyl
acidic properties. 1 Constant variable Type of acid orange indicator are added to
• In the presence of water, used 3 the alkali and the colour of the
aqueous ethanoic acid (b) The higher the concentration of H+ solution is noted.
dissociates to form hydrogen ions, the lower the pH value. 3 3 A 50 cm3 burette is cleaned
(c) Concentration of H+ ions 0.100 0.060 0.040 0.025 0.015 0.010 with distilled water and is rinsed
(mol dm—3)
with a little sulphuric acid.
4 The burette is then filled with
pH value 1.0 1.2 1.4 1.6 1.8 2.0 sulphuric acid and is clamped
3 to a retort stand. The initial
burette reading is recorded.
557 Answers
5 The conical flask containing 25 Hence, —00—..33 mol of Pb2+ ions 5 Add aqueous ammonia a little at a
cm3 of potassium hydroxide time until in excess to the salt solution.
reacts with —00—..63 mol of B r – ions.
is placed below the burette. If a white precipitate that dissolves in
1 mol of Pb2+ ions reacts with excess aqueous ammonia is formed,
A piece of white tile is placed 2 mol of Br — ions. the salt is zinc nitrate. If no precipitate
below the conical flask for Pb2+ + 2Br — → PbBr2 is formed, the salt is calcium nitrate.
5 Number of moles of Ba2+ ions Add potassium iodide solution to the
clearer observation of the colour = 0.2 —1—05—0—0 = 0.001 salt solution. If a yellow precipitate is
formed, the salt is lead(II) nitrate.
change.
6 Sulphuric acid is added
slowly from the burette to the
potassium hydroxide solution in Number of moles of CrO42— ions 6 Inference
the conical flask while swirling = 0.1 1—01—00—0 = 0.001 Test
the flask gently.
7 Titration is stopped when methyl 0.001 mol of Ba2+ ions react with
1 Zn2+, Al3+ or Pb2+ ions may be present.
0.001 mol of CrO42— ions.
orange changes from yellow Hence, 1 mol of Ba2+ ions react with
colour to orange colour. The 1 mol of CrO42— ions. 2 Zn2+ ions may be present.
Ba2+ + CrO42— → BaCrO4
final burette reading is recorded. 3 NO2 gas and O2 gas are evolved.
8 Steps 1 to 7 are repeated 6 N um ber of moles of Ag+ ions = 0—.—51—0—0—01—0 NO3— ions may be present.
= 0.005 The residue may be ZnO, Zn2+ ion may
until accurate titration values be present.
are obtained, that is, until the
difference in the volumes of Number of moles of CO32— ions
= —0—1.5—0—0—0—5 = 0.0025
sulphuric acid used in two 4 SO42— ions are not present.
experiments is less than 0.005 mol Ag+ react with 0.0025 mol
0.10 cm3. of CO32—. Salt Z contains Zn2+ ions and NO3— ions.
(f) Tabulation of data Hence, 2 mol of Ag+ react with 1 mol
of CO32— . SPM Exam Practice 8
F Rough Accurate Multiple-choice Questions
O 123 2Ag + + CO32— → Ag2CO3 1 C 2 B 3 B 4 B 5 D
R 6 C 7 C 8 B 9 C 10 C
M Final burette 7 (a) 3MgO + 2H3PO4 → 11 D 12 D 13 B 14 B 15 D
4 reading (cm3) Mg3(PO4)2 + 3H2O 16 C 17 C 18 A 19 C 20 D
2 1 C 22 A 23 A 24 A 25 A
Initial burette (b) 3 mol of MgO produce 1 mol of 26 C 27 D 28 C 29 B 30 D
reading (cm3) Mg3(PO4)2. Hence, 3 1.2 = 3.6
mol of MgO is used to produce Structured Questions
Volume of 1.2 mol of Mg3(PO4)2. 1 (a) (i) Nitric acid 1
sulphuric acid
used (cm3) Self Assess 8.2 (ii) PbO + 2HNO3 →
Pb(NO3)2 + H2O 1
17 1 (a) PbCl2, ZnCO3, CuCO3
(b) PbCl2, ZnSO4, ZnCO3, Na2CO3, (b) (i) Gas B is nitrogen dioxide. 1
Al2(SO4)3, Mg(NO3)2 Solid D is lead(II) oxide. 1
8 Salts (c) CuCl2, Cu(NO3)2 (ii) Gas C will rekindle a glowing
CHAPTER 7 & 8 (d) ZnSO4, Al2(SO4)3 wooden splint. 1
Self Assess 8.1 (e) ZnSO4, Al2(SO4)3 (c) (i) Lead(II) iodide. Yellow 1
1 (a) Neutralisation reaction between (f) Al2(SO4)3, Mg(NO3)2 (ii) Pb2+ + 2I– → PbI2 2
(a) Iron(III) carbonate
NaOH and H2SO4. 2 (d) (i) Lead(II) hydroxide 1
(b) Neutralisation reaction between
(b) Fe2(CO3)3 + 6HNO3 → (ii) The white precipitate formed
NH3 and H2SO4. 2Fe(NO3)3 + 3CO2 + 3H2O
(c) Reaction between excess Al dissolves in excess aqueous
(c) Y decomposes to iron(III)
powder/ Al2O3/ Al2(CO3)3 and sodium hydroxide to produce
H2SO4.
oxide and carbon dioxide gas is a colourless solution. 1
(d) Reaction between excess Pb evolved. (e) Lead(II) carbonate or lead metal
powder / PbO/ PbCO3 and HNO3. (d) A blood red colour solution is 1
(e) Reaction between excess Zn
formed. 2 (a) MgO + 2HNO3 →
Mg(NO3)2 + H2O 1
(f) pPorewcidpeitra/tiZonnOb/etZwneCeOn3 and HCl. (e) Fe3+ + 3OH— → Fe(OH)3
Pb(NO3)2 3 P contains chloride ions, Q contains (b) Number of moles of HNO3
(g) PanredciNpiata2StiOon4/bKe2tSwOe4e/nHA2gSNOO4.3 and sulphate ions and R contains = 2——.0————5—0 = 0.1 1
carbonate ions. 1000
NaCl /KCl/ HCl. 4 Add barium chloride solution to each 0.1 mol of HNO3 will produce
0.05 mol of Mg(NO3)2. 1
2 Magnesium powder (or magnesium type of the acid. If a white precipitate Mass of the salt, Mg(NO3)2
produced
carbonate or magnesium oxide) and is formed, sulphuric acid is present.
nitric acid. Add silver nitrate solution to each type = 0.05 {24 + 2(14+16 3)}
Mg + 2HNO3 → Mg(NO3)2 + H2 of the acid. If a white precipitate is = 7.4 g 1
3 (a) Mole ratio of CAla32++: :OSHO—42=— = 1: 1 formed, hydrochloric acid is present. (c) (i) A white precipitate is formed,
(b) Mole ratio of 1 :3
If there is no precipitate formed with 1
insoluble in excess aqueous
4 (a) 2Ag+ + CrO42— → Ag2CrO4 either barium chloride solution or
(b) 0.3 mol of Pb2+ ions react with silver nitrate solution, nitric acid is sodium hydroxide. 1
0.6 mol of Br— ions. present.
Answers 558
(ii) Magnesium nitrate 1 (ii) To prepare lead(II) chloride, • The reaction that occurs:
(d) (i) Gas E is nitrogen dioxide. the reactants are lead(II) nitrate • NPThbaeC2COpOr3e3+c+ip2PiNtabat(NeNOfOo3r3 m)2e→d is lead(1II)
Solid C is magnesium oxide. solution and sodium chloride /
2 potassium chloride solution. 3 carbonate salt.
(ii) Gas D will rekindle a glowing (b) • The unsaturated salt solution is • Filter the mixture to obtain the
wooden splint. 1 evaporated until it is saturated. precipitate. 1
3 (a) (i) Carbon dioxide 1 1 • The residue is rinsed with
(ii) Carbonate ion, CO32– 1 • This is tested by placing a drop distilled water and dried
(b) Zinc ion, Zn2+ 1
of the solution on a piece of between filter papers. 1
(c) ZnCO3 → ZnO + CO2 1 glass plate. If crystals are formed, (c) • Add ammonia solution to
(d) Yellow when hot, white when
then the solution is saturated. 1 2 cm3 of solution X in a test tube, a
cold. 1 • The saturated salt solution is little at a time until the ammonia
(e) (i) ZnO + 2HCl → H2O + ZnCl2 cooled to allow crystallisation to solution is in excess. 1
occur. 1 • Repeat the test on solution Y
1 • The salt crystals formed are followed by solution Z to replace
(ii) Residue B dissolves in HCl then filtered. 1 solution X. 1
effervescence occurs 1 • If a white precipitate that does not
(f) (i) Zinc(II) hydroxide 1 dissolve in excess ammonia is
(ii) Zn2+ + 2OH– → Zn(OH)2 1 formed, the solution is
(g) A white precipitate is formed
magnesium sulphate. 2
which is soluble in excess sodium
• If a white precipitate that dissolves
hydroxide solution. 1
in excess ammonia is formed,
4 (a) ZnO + 2HCl → ZnCl2 + H2O 1 the solution is zinc sulphate or
(b) Hydrochloric acid is heated in
diagram 1 , label 1 zinc nitrate. 2 F
O
a beaker. Add zinc oxide to (c) (i) The anion present is nitrate • Add a little barium nitrate solution to R
M
hydrochloric acid a little at a time ion. 1 2 cm3 of the solutions in separate 4
until a little of solid zinc oxide is Chemical test: Add a little test tubes. 1
in excess. Filter the mixture and 1 dilute sulphuric acid and • If the solution produces a white
the filtrate is salt solution P. 1 iron(II) sulphate solution precipitate that dissolves in excess
(c) (i) Sodium carbonate solution 1 to about 5 cm3 of salt X ammonia and also a white
(ii) Double decomposition 1 solution in a test tube, 1 precipitate with barium nitrate
(iii) Zn2+ + CO32— → ZnCO3 1 followed by concentrated solution, the solution is zinc
(d) (i) Number of moles of HCl
sulphuric acid along the sides sulphate. 1
= 2.0 1——03—00——0 = 0.06 1 of the test tube. 1 • If the solution produces a white
A brown ring will be formed, precipitate that dissolves in 1
ZnO + 2HCl → thus confirming the presence excess ammonia but does not
ZnCl2 + H2O
of nitrate ions. 1 produce any precipitate with
(ii) The 3 cations present are barium nitrate solution, the
0.06 mol of HCl will produce Al3+ ions, Pb2+ ions and Zn2+ solution is zinc nitrate. 1
0.03 mol of salt P. 1 ions. 3 Experiment CHAPTER 8
(ii) ZnCl2 + Na2CO3 → Add potassium iodide 1 (a) 50.00 cm3 burette 3
ZnCO3 + 2NaCl
0.03 mol P will produce solution to a little salt X (b) Manipulated variable Volume of
solution in a test tube. 1 potassium chromate(VI) solution
0.03 mol of ZnCO3 1 The formation of yellow Responding variable Height of
Mass of ZnCO3 produced precipitate will confirm the precipitate
= 0.03 125 = 3.75 g 1
presence of Pb2+ ions. 1 Constant variable Volumes of
2 (a) A salt is an ionic compound
(e) Heat zinc carbonate strongly, zinc barium nitrate solution and size of
formed by replacing the hydrogen
carbonate will decompose to zinc test tubes 3
ion of an acid by a metal ion or
oxide. 1 (c)
ammonium ion. 2
5 (a) Filtration 1 (b) • Add lead metal powder a little
(b) Ammonia gas, ammonium ion 2 at a time to 30 cm3 of dilute
(c) Barium sulphate, sulphate ion 2 nitric acid, 1
(d) Zinc ion 1 • until the lead metal powder is in
(e) Metal Q is zinc, gas T is hydrogen excess, with a little remaining
gas. 2 undissolved. 1
• The reaction that occurs:
Essay Questions Pb + 2HNO3 → H2 + Pb(NO3)2
1 (a) (i)
1
Soluble salt Insoluble salt • Filter off the excess undissolved 3
(d) (i) 5.0 cm3
lead metal powder, keep the (ii) Number of moles of Ba2+ ion
= —0—.—51—0—0——05—.—0— = 0.0025 3
Sodium sulphate Lead(II) chloride filtrate. 1
Magnesium nitrate
• 30 cm3 of sodium carbonate
3 solution is added to the filtrate.
1
559 Answers
(e) Fmrooml otfhBeaf2o+rimonuslawBilal CprrOec4,ip0it.a0t0e25 Self Assess 9.3 • Reduce, reuse and recycle
1 (a) An alloy is a mixture of two or • Produce biodegradable polymers
with 0.0025 mol of CrO42– ions.
Molarity of CrO42– ion more elements with a fixed Self Assess 9.5
= 0.0025 mol —1—0—50—0— dm–3 composition of which the major 1 (a) Silicon dioxide or silica
component is a metal. (b) Sand
= 0.5 mol dm–3 3 (b) (i) Carbon (c) Fused glass, soda lime glass,
(ii) Chromium and nickel
(f) All the barium ions have been (c) (i) borosilicate glass and lead glass
2 (a) Silicon dioxide
precipitated. The chromate ions (b) Similarities: Hard, brittle,
are in excess. 3 inert to chemicals, withstand
compression, poor conductors of
9 Manufactured Substances in heat and insulators of electricity.
Industry Differences:
Self Assess 9.1 Glass Ceramics
1 (a) (i) Ammonium sulphate,
(ii)
2NH3 + H2SO4 → (NH4)2SO4 • Transparent • Opaque
(ii) Potassium sulphate,
• Becomes soft • Withstand heating
(b) BHa2SSOO2K442KS++OOHB24 a+H+(2O2OHHH2S)2OO2 4→→ when heated
To make fertilisers, detergents • Usually porous
• Impermeable unless glazed
2 and (c) • To make lenses
white paint pigment • To make laboratory apparatus
3 (a) To absorb water • To make light bulbs
F (d) Stainless steel does not rust, whereas • To make mirrors and window
O (b) (i) Carbon iron rusts.
R panes
4 (ii) C6H12O6 → 6C + 6H2O 2 (a) Tin, lead and antimony
(a) Contact process
M (b) Harder, stronger and more shiny • To make cooking utensils
(b) Gas X is sulphur dioxide, gas Y when polished (d) In making cement, in making bricks
is sulphur trioxide and liquid Z is (c) To make decorative ornaments 3
3 (a) Copper alloys are stronger, harder Differences
4 oleum.
(c) Step 1 : S + O2 → SO2 and more resistant to tarnish than
(d) TSetempp2era: t2uSreOo2 f 4+5O02°C – 5502S°OC,3 pure copper. Soda lime Borosilicate
glass glass
pressure of one atmosphere and (b) Bronze and brass (a) Composition Sodium Boron oxide
vanadium(V) oxide as the (c) Pure copper metal contains atoms oxide,
catalyst. of the same size arranged in a calcium oxide
(e) A lot of heat will be evolved regular and organised manner. Pure
when gas Y dissolves in water, copper is soft and ductile because (b) Property Does not Withstand
this vaporises the sulphuric acid the orderly arrangement of atoms withstand heating
produced as acid mist. enables the layers of atoms to slide heating
CHAPTER 8 & 9 over each other easily when an
Self Assess 9.2 external force is applied. There are Self Assess 9.6
1 (a) Nitrogen gas and hydrogen gas empty spaces in the structures of 1 (a) Composite material is a structural
pure copper due to imperfections.
(b) Iron powder Pure copper is malleable because material made by combining two
groups of metal atoms may glide or more materials with different
(c) Temperature of 450 °C – 550 °C; into new positions in these empty properties into a complex mixture.
spaces when pressed. (b)
pressure of 200 – 500
atmospheres
((de)) NTo2 m+a3kHe2fertiliser2sNaHnd3 nitric acid Self Assess 9.4 Composite Uses
1 (a) A polymer is a large molecule materials
2 (a) Gas A is nitrogen gas, gas B is
made up of many smaller and
ammonia, acid C is nitric acid identical repeating units, joined Reinforced Construction of
together by covalents bonds. concrete framework of high rise
(b) Haber process (b) Cotton and silk buildings
(c) Nylon and Terylene
(c) Air 2
(d) White fumes are formed. Superconductor To make stronger and
lighter electromagnets
((ef)) NReHla3ti+veHmNoOle3c→ulaNr mH4aNsOs 3of
NH4NO3 Ethene Polythene Fibre optic Transmit data in
= 14 + 4(1) + 14 + 3(16) telecommunications
= 80 Chloroethene Polychloroethene or PVC Fibreglass To make boats, fishing
rods and containers
1 mmooll-aotfoNmHs4oNfOn3itcroognesnis.ts of Propene Polypropene
2
Percentage of nitrogen in 1 mol of Styrene Polystyrene Photochromic To make light sensitive
glass optical lens and
N H4NO3 = —2—(81—04—) 100% 3 Nonbiodegradable and produces camera lens
= 35% poisonous gas when burnt
Answers 560
2 (a) Fibreglass is a composite material (e) To make fertilisers and to make (d) (i)
made by embedding glass fibre
filaments in plastic. Fibreglass is paint pigments/ detergent/ as an 1
lighter, stronger and tougher than (ii)
glass and is not brittle. It is more electrolyte in lead-acid
resilient and flexible than plastic
and is not flammable. accumulator 2
(b) Reinforced concrete is a composite 2 (a) Process P is the Contact process. 1
material made by adding a concrete
mixture of cement, water, sand and Process Q is the Haber process.
small stones into a frame of steel
bars or wire netting. 1
It is strong, hard, can withstand
compression and pressure. It (b) Temperature of 450 – 550 °C 1
does not rust like iron.
Pressure of 200 – 500
3 Lenses of spectacles made of
photochromic glass will darken when atmospheres. 1
the light intensity is high and becomes
clear when the light intensity is low. This Presence of iron powder as
eliminates the necessity for a separate pair catalyst 1
of sunglasses.
Photochromic glass is produced when a (c) (i) Ammonium sulphate 1
dispersion of silver chloride, AgCl or silver
bromide, AgBr is added to normal glass. (ii) H2SO4 + 2NH3 → 1
When exposed to ultraviolet light, AgCl (NH4)2 SO4 1
or AgBr decomposes to form silver and
halogen atoms. The fine silver deposited (d) 0.4 mol of NH3 is required to
in the glass is black and the glass is react with 0.2 mol of sulphuric
darkened.
acid. 1 (e) (i) Aluminium, copper,
Mass of ammonia = 0.4 3 17 magnesium 1
= 6.8 g 1
(ii) Stronger, harder and does
(e) Urea 1
not tarnish easily compared
3 (a) Temperature of 450 °C – 550 °C, 1
to aluminium. 1
pressure of 200 – 500 1
(iii) In the making of aircraft
atmospheres and in the presence body 1 F
6 (a) Polymerisation 1 O
of iron powder as a catalyst. 1 R
M
(b) (i) Ostwald process 1 (b) (i) Ethene 1 4
(ii) Platinum 1 (ii) To make plastic bag / 1
containers / toys
(c) Physical property: Pungent smell/
AgBr ultrav iole→t Ag + —21 Br2 H H
very soluble in water. 1
(c) (i) H – C = C – Cl
When the ultraviolet ray intensity Chemical property: Dissolves 1
decreases, silver atoms and bromine in water to produce an alkaline (ii) Polyvinyl chloride is lighter 1
gas recombine to form silver bromide. solution/forms white fumes with and does not rust. 1
hydrogen chloride gas. 1 (iii) It is not biodegradable. It 1
Ag + —21 Br2 → AgBr (d) Nitric acid 1 produces poisonous gas
SPM Exam Practice 9 NH3 + HNO3 → NH4NO3 1 when burnt. 1
Multiple-choice Questions (e) (i) Potassium hydroxide 1
1 B 2 D 3 B 4 B 5 C (d) Petroleum 1
6 B 7 D 8 D 9 C 10 C (i i) KKONOH3++HHN2OO3 → 1
11 D 12 A 13 A 14 B 15 D Essay Questions
16 D 17 D 18 D 19 A 20 A
21 D 22 C 23 B 24 D 25 B 4 (a) (i) Silicon dioxide 1 1 (a) A polymer is a big molecule CHAPTER 9
2 6 A 27 B 28 D 29 C 30 B
3 1 A 32 C 33 C 34 B 35 B (ii) Sand 1 formed by many small molecules
36 D 37 D
(iii) Boron oxide 1 (monomers) joined together by
(b) (i) Polypropene 1 covalent bonds. For example, in
(ii) Lighter / does not rust 1 polythene, many molecules of
ethene are joined by covalent
(c) (i) Magnesium metal 1 bonds to form the big molecule
Structured Questions (ii) Magnesium atoms disrupt called polythene. 4
1 (a) (i) Sulphur dioxide 1 the orderly arrangement of (b) (i) Three examples of natural
(ii) S + O2 → SO2 1 aluminium atoms. Hence 1 polymers are:
(b) (i) Sulphur trioxide 1
layers of aluminium atoms Natural rubber: to make car
(ii) Temperature of 450°C are prevented from sliding tyres
– 550°C, one atmosphere over each other easily. 1 Protein: as a class of food
pressure and in the presence (d) Z is reinforced concrete. 1 Carbohydrate: as a class of
of vanadium(V) oxide as a (e) (i) T is a composite material. 1 food 3
catalyst. 2 (ii) It is sensitive to light intensity (ii) Three examples of synthetic
(darkens when light intensity polymers are:
(c) (i) Oleum 1 is high, becomes clear when Polythene: to make plastic
(ii) Gas Z (or sulphur trioxide) is light intensity is low). 1 bags or bottles
dissolved in concentrated 5 (a) An alloy is a mixture of two or Polyvinyl chloride (PVC): to
sulphuric acid. 1 more elements in which the make water pipes
major component is a metal. 1
(d) A lot of heat will be evolved when Polystyrene: to make
sulphur trioxide gas is dissolved (b) (i) Zinc 1 packaging materials 3
in water. This will vaporise the (ii) Tin 1 (c) Sulphuric acid is manufactured by
sulphuric acid produced into fine the Contact process in industry.
1
droplets as acid mist. 1 (c) Copper atoms 1
561 Answers
• Sulphur is burned in air to (b) An example of an alloy is brass, a Experiments
produce sulphur dioxide gas. mixture of copper and zinc.
• SSu+lpOhu2 r→dioSOxid2 e gas is converte2d To compare the hardness of 1 (a) Brass : 3.0 mm
to sulphur trioxide gas by excess brass and copper, the following Copper : 4.0 mm 3
air in the presence of vanadium experiment is carried out. (b) Type of
Diameter of
Materials
Block depression (mm)
(V) oxide as a catalyst, Copper block, brass block, ball Brass 3.0
450 – 550 °C temperature and bearing, 1 kg weight, metre rule,
retort stand with clamp, cellophane Copper 4.0
one atmosphere pressure.
tape and thread.
2SO2 + O2 2SO3 3 Procedure 3
(c) Alloy is a mixture of two or
• Sulphur trioxide gas is dissolved 1 A metre rule is clamped to more elements with a fixed
in concentrated sulphuric acid to a retort stand and a piece of
produce oleum. copper block is placed on the composition in which the major
base of the retort stand. component is a metal. 3
SO3 + H2SO4 → H2S2O7 2 (d) The smaller the diameter of the
• Oleum is diluted with an equal 2 A steel ball bearing is placed on depression, the harder the metal.
the copper block and a piece of
volume of water to produce cellophane tape is used to hold 3
concentrated sulphuric acid. the ball bearing in place. (e) Brass is harder than copper /
H2S2O7 + H2O → 2H2SO4 2 3 A 1 kg weight is hung at a copper is softer than brass. 3
2 (a) A composite material is a height of 50 cm above the (f) Mass of weight used, height of
structural material formed by copper block. The weight is
combining two or more materials dropped onto the ball bearing weight from which it is dropped,
placed on the copper block. size of steel ball bearing. 3
with different properties into a 2 (a) Problem statement
complex mixture. 4 The diameter of the dent
• An example of composite made by the ball bearing is How to compare the rate of rusting
material is reinforced concrete measured to the nearest of an iron nail and a stainless steel
F made from concrete and steel. 0.5 mm. nail?
O (b) Variables
R • Concrete is hard but brittle, 5 The experiment is repeated
M with low tensile strength and three times using different • Manipulated variable Types
areas of the surface of the of nails (iron nail and stainless
can crack under the action of copper block. steel nail).
bending forces.
4 • Steel is strong with high tensile 6 The average diameter of the • Responding variable Rate of
dent is calculated. rusting.
strength but can corrode and is • Constant variable Size of nails,
expensive. 7 The whole experiment from
• Reinforced concrete is formed steps 1 to 6 is repeated duration for rusting and conditions
using a piece of brass block of experiment (temperature,
by adding a concrete mixture of in place of the copper block. supply of water and air).
cement, water, sand and small
stones into a frame of steel bars (c) Hypothesis
Stainless steel can resist rusting
or steel wire netting and allowed better than iron.
to set. The structural material
formed on setting combines (d) Materials
Iron nail, stainless steel nail,
the compressive strength of sandpaper, 5% jelly solution and
CHAPTER 9 concrete and the tensile strength
of steel. It can withstand very potassium hexacyanoferrate(III)
solution.
high pressures and can support (e) Procedure
heavy loads.
• Reinforced concrete is used in 1 Two test tubes are half-filled
with jelly solution.
the construction of framework 2 1 cm3 of potassium
for highways, bridges and high- To compare the hardness of an hexacyanoferrate(III) solution
rise building. 5 alloy with a pure metal is added to each test tube.
• Another example of composite Results
3 An iron nail and a stainless
material is fibreglass made from steel nail are polished with
plastic and glass. Metal Diameter of the dent (mm) sandpaper and placed in the
• Plastic is light, flexible and elastic block I II III Average two test tubes separately.
but is weak and inflammable. 4 The two test tubes are
• Glass is strong but is heavy, Copper allowed to stand for 2
brittle and non-flexible. Brass days after which they are
• Fibreglass is formed by embedding examined to see if any colour
glass filaments in polyester resin (a
type of plastic). The new material If the average diameter of the change has taken place.
(f) Tabulation of data
formed is strong, tough, resilient, dents made by the steel ball
flexible, impermeable to water and bearing on the copper block is
bigger than that on the brass block, Test tube Type of nail Observation
is not inflammable (does not catch the steel bearing has been pressed 1 Iron nail
fire easily). deeper onto the surface of copper 2 Stainless steel
• Fibreglass is used in the making metal than that on brass. This will
of boats, containers, pipes and nail
show that, brass, a type of alloy, is
fishing rods. 5
harder than pure copper. 10 17
Answers 562
Form 5 (d) From the graph, the total volume of From the graph:
H2 produced at 1 minute = 4 cm3. Volume of O2 = 62 cm3
1 Rate of Reaction Average rate of reaction
∴ The average rate of reaction = —1—6——4—2—4— = 0.43 cm3 s–1
Self Assess 1.1 between magnesium and or,
hydrochloric acid
1 (a) Reactions I and IV. Average rate of reaction
= —v——o——l—tu——i—mm————ee————t—o—a—f—k——eh———ny——d——(—r—mo———g—i—ne———nu——t——e(——cs——m)————3——) = ——2—6——.2—4— = 25.8 cm3 min–1
(b) Reactions II and III. = ——14— = 4 cm3 min–1 (b) Time taken for the overall
reaction
2 (a) • Place zinc in dilute sulphuric acid. The average rate of reaction = 4 minutes (from the table
between hydrochloric acid and given in the question) = 240 s
• Using a stopwatch, record the another metal = 12 cm3 min–1. Average rate of the overall
Thus, the average rate of reaction reaction
time taken for all the zinc to between hydrochloric acid and = —2——7—4—4——0 = 0.31 cm3 s–1
another metal is higher. (c) The average rate of reaction
dissolve. 5 Average rate of reaction between between 1 minute and 3 minutes
nickel and hydrochloric acid = —(——6—(——93————––—————13———)2—— ) = ——3—2——7— cm3 min–1
(b) Let the time taken for zinc to = —v——o———l—u——m————e—t—i—m——o——ef———h—t——ay——kd——e—r—o—n——g——(e——s—n—)———(——c——m————3——) = 18.5 cm3 min–1
= 1——6—2——0—0— = 0.5 cm3 s–1 (d) Time taken = 150 s
dissolve completely be t s. Average rate of reaction between zinc = 1——6—5——0—0— = 2.5 minutes
and hydrochloric acid Rate of reaction at 2.5 minutes
Mass of zinc dissolved = 2.0 g = —45———56— = 0.8 cm3 s–1 = gradient of the curve at 2.5
The reaction between zinc and minutes
Units of reaction rate are g s–1. hydrochloric acid is a faster reaction. = —3—7—.—86—————––————51———0.—5 = 13.3 cm3 min–1, or
Hence, zinc is a more reactive metal = 1——6—3——0—.—3 = 0.22 cm3 s–1
(c) Assumptions: than nickel.
6 (a) The reaction stops at 70 s. Self Assess 1.2
Sulphuric acid is in excess. Average rate = —t—o———t—a——l———v——o——t—l—iu—m——m——e——e————o———f———H——2 1 The rate of reaction of zinc powder with
= —37———06— = 0.51 cm3 s–1
All the zinc (2.0 g) dissolves (b) After 0.5 minutes, amount dilute sulphuric acid can be increased by
reacted = 0.0030 3 0.5 3 60 (a) increasing the temperature of
completely in sulphuric acid.
= 0.09 mol sulphuric acid,
3 Method 1: Measure the total volume Mass of Mg reacted = 0.09 3 24 (b) increasing the concentration of
of hydrogen given off at = 2.16 g sulphuric acid,
Mass of Mg unreacted = 4.0 – 2.16 (c) using a catalyst (for example,
regular time intervals.
= 1.84 g copper(II) sulphate solution).
Plot the graph of volume 7 (a) Time = 144 s = —1—6——4—0——4 = 2.4 min. 2 (a) At 6.6 seconds (the point when
of hydrogen released Since the volume of O2 gas the graph becomes flat)
collected at 2.4 minutes is not (b) The reaction stops because all
against time. given in the table, we need to F
plot a graph of volume of O2 the hydrochloric acid has reacted. O
Method 2: Measure the mass of evolved against time and use the (Comment: As stated in the R
graph to determine this value. question, excess calcium carbonate M
the conical flask and is used in this experiment). 5
(c)
its contents (mixture of
iron and sulphuric acid)
at regular time intervals.
Calculate the loss of mass
at regular intervals.
Plot the graph of loss of
mass (of the conical flask
and its contents) against
time.
(Comments: The loss of mass of the
conical flask and its contents is equal
to the mass of hydrogen released.)
4 (a) 5.6 minutes CHAPTER 1
(Comments: The time at which
the curve becomes completely
flat is the time at which the
reaction is complete).
(b) The slope of the graph at 1
minute is less steep than the
slope at 2 minutes. Hence, the
rate of reaction at 1 minute is
lower than the rate at 2 minutes.
(c) It is not a normal behaviour.
Normally, the curve is steepest at
the start because the reaction is
most vigorous at the start. But as
the reaction proceeds, it becomes
less vigorous and the graph
becomes less steep.
This unusual behaviour is due
to the fact that the magnesium
ribbon is not cleaned and is
therefore covered with a layer of
magnesium oxide. The acid must
first dissolve the magnesium
oxide layer before it can react
with the magnesium ribbon.
563 Answers
Comments: At a lower 4 (a) Hypothesis: The higher the (d) (i) A blue precipitate will be
temperature, the rate of reaction temperature of the reaction, the formed, insoluble in excess
is lower. Hence, the curve is faster the reaction, that is, the sodium hydroxide.
less steep. Since the reaction is higher the rate of release of (ii) Cu2+(aq) + 2OH–(aq) →
slower, it takes a longer time to carbon dioxide. Cu(OH)2(s)
complete. The total volume of Comment: The experimental (e) A catalyst remains chemically
CO2 collected at the end is the results in the table below confirm unchanged at the end of the
same because the same amounts the hypothesis. The decrease in reaction.
of CaCO3 and HCl are used. mass corresponds to the mass of Self Assess 1.3
(d) Constant variables: Concentration carbon dioxide released. 1 Only I and II will increase the rate of
and volume of hydrochloric acid, Experiment Temperature Decrease in reaction.
mass of calcium carbonate. I (°C) mass (g) Explanation:
3 (a) II 28 I: Effect of concentration on reaction
III 270.35 – 270.04
35 = 0.31 rate
271.42 –270.01 When the concentration of
40 = 1.41 sulphuric acid increases, the
268.20 –266.00 number of reacting particles per
= 2.20 unit volume also increases. As
a result, the particles are closer
F (b) Step 1: Calculate the total (b) Constant variable: Size and mass together and they collide more
O number of moles of oxygen of marble, concentration and frequently. Hence, the frequency
R released. volume of hydrochloric acid. of effective collisions increases and
M Experiment I: this causes the rate of reaction to
5 Number of moles of H2O2 used (c) There is a loss of mass due to increase.
= —1——M—0———0V———0 = ——1——.—0——1———0———0———10———0———0— the release of carbon dioxide gas II: Effect of temperature on reaction
= 0.1 into the atmosphere. rate
From the equation, The loss of mass = mass of When the temperature is increased,
1 o f.0Om2. ol of H2O2 produces —12 mol carbon dioxide released the particles gain energy and they
= 270.35 – 270.04 move faster and collide more
= 0.31 g frequently. At a higher temperature,
the number of reactant particles
(d) Different beakers are used for the having the activation energy
experiments. increases. As a result, there are
more effective collisions and the
Although the beakers are the same rate of reaction increases.
size, their masses are different. III: An increase in the volume of
sodium thiosulphate solution will
5 (a)
∴Total number of moles of O2 not increase the rate of reaction.
released
IV: The sodium hydroxide solution
CHAPTER 1 = —12 0.1
added will neutralise the sulphuric
acid. Hence, the concentration of
= 0.05 sulphuric acid decreases and this
Experiment II: will decrease the rate of reaction.
Number of moles of H2O2 used 2 (a) Number of moles of H2O2 used
= 1———0M———0—V——0 = —0———.—2—1———0———0———30———0———0 = 0———.—2—1——0——3—0———0——3——0— = 0.0060
= 0.06 2H2O2(aq) → 2H2O(l) + O2(g)
Total number of moles of O2 (b) (i) Reaction II is faster than
released Number of moles of O2 produced
= —12 0.06 reaction I because it is carried = —0———.—0——2—0——6———0 = 0.0030
= 0.03 out at a higher temperature.
Step 2: Predict which reaction The higher the temperature, Volume of O2 produced
has a higher rate of reaction. = 0.0030 3 24 000 = 72 cm3
(b) (i)
Since the concentration of the higher the rate of reaction.
hydrogen peroxide used in (ii) Reaction III is a faster reaction
Experiment I is higher than than reaction II because the
the concentration of hydrogen catalyst, copper(II) sulphate,
peroxide in Experiment II, the increases the rate of reaction.
rate of reaction in Experiment (c) This shows that the catalyst is
I is higher. Thus, the graph in the Cu2+ ion. Both the Na+ and
Experiment I is steeper than the SO42– ions are not the catalyst for
graph in Experiment II. the reaction.
Answers 564
(ii) • The graph is steep at the the activation energy (iii) Both reactions have stopped.
start of reaction because decreases. Consequently,
the rate of reaction is the the frequency of effective The reaction for experiment
highest. collisions decreases and
the rate of reaction also II was completed earlier than
• As the reaction proceeds, decreases.
the concentration of 4 Negative catalysts reduce the rate of experiment I.
hydrogen peroxide reaction by increasing the activation
decreases. As a result, energy of the reaction. In the (iv) The mass of zinc, the
the graph becomes less presence of a negative catalyst, the
steep because the rate of reaction occurs via a pathway with a concentration and volume of
reaction decreases. higher activation energy.
Ea = activation energy without a sulphuric acid used are the
• The graph becomes flat catalyst
when the reaction stops, Ea’ = activation energy with a negative same for both experiments.
that is, when all the catalyst
hydrogen peroxide has As a result, fewer particles have 4
decomposed. The total sufficient energy to overcome the
volume of oxygen gas higher activation energy (Ea’). (e) The smaller the particle size,
produced is 72 cm3. Hence, the frequency of effective
collisions decreases and this causes the larger the total surface area
(c) (i) It acts as a catalyst. the rate of reaction to decrease.
exposed for reaction and the
(ii) Manganese(IV) oxide lowers
the activation energy of the higher the rate of reaction. 1
reaction by providing an
alternative reaction route. (f)
Thus the effective rate of
collision between hydrogen 2 (a) (i) C3H8 + 5O2 → 3CO2 + 4H2O F
peroxide molecules increases 1 mol of propane needs O
and this causes the rate of 5 mol of O2 for complete R
decomposition of hydrogen reaction. M
peroxide to increase. 5
Rate of consumption of O2
3 (a) (i) The rate of reaction at the 1st = 0.20 3 5
minute is higher than the rate
of reaction at the 2nd minute. = 1.0 mol s–1 1
(ii) When the reaction continues, (ii) 1 mol of propane produces
the total surface area of 3 mol of CO2.
calcium carbonate decreases, Rate of production of CO2
and the concentration = 0.2 3 3 = 0.6 mol s–1 1
of hydrochloric acid also
decreases. As a result, the SPM Exam Practice 1 (b) (i) Mg(s) + 2H+(aq) →
effective collisions that Multiple-choice Questions
cause the reaction to occur 1 C 2 C 3 D 4 A 5 D Mg2+(aq) + H2(g) 1 CHAPTER 1
decrease. Thus, the reaction 6 A 7 C 8 C 9 C 10 D
rate decreases. 11 D 12 C 13 C 14 A 15 A (ii) The time, t1, is shorter
16 D 17 D 18 C 19 A 20 D because HCl is a strong acid
(b) (i) 21 D 22 A 23 B 24 B 25 C but CH3COOH is a weak
26 A 27 A 28 A 29 C 30 B acid. 1
31 A 32 C 33 A 34 D 35 A
36 D A weak acid dissociates
only partially in water but
Structured Questions a strong acid dissociates
completely in water. Hence
1 (a) The size (total surface area) of the concentration of H+ ions
in Experiment I is higher.
reactant. 1 1
(b) Temperature/concentration of The higher the concentration,
the faster the reaction, the
acid/mass of zinc (any two shorter the time needed
to dissolve magnesium
variables). 1 completely. 1
(c) (i) Hydrogen
(ii) The volume of hydrogen (c) (i) Curve X: Experiment II
can be measured easily. 2 Curve Y: Experiment III
(ii) When the experiment (d) (i) Experiment II. Curve Z: Experiment I 3
is carried out at a lower
temperature, (ii) The gradient of curve II is (ii) The amount (in moles) of
• the reactant particles move HCl used in Experiment II
slower, steeper than the gradient of (curve X) is half the amount
• the number of reactant (in moles) of HCl used in
particles that possess curve I. Experiment I (curve Z). 1
Alternative answer: It takes a
shorter time for reaction II to
complete.
565 Answers
3 (a) Mg(s) + 2HCl(aq) → (c) Rate of reaction ∝ ——t—i—m————e———1—t——a——k—e——n— (iii)
MgCl2(aq) + H2(g) The time taken for Experiment
III is longer than Experiment II Experiment Concentration Time (s)
1 (mol dm–3)
(b) Number of moles of magnesium
A 0.15 65
= —0—2.—14—2— = 0.005 because the reaction rate in C 0.10 85
From the equation, Experiment III is lower than in E 0.05 105
Experiment II. This is because
volume of hydrogen liberated hydrogen peroxide used in The higher the concentration
= 0.005 3 24 = 0.12 dm3 Experiment III has a lower of the acid, the shorter the
= 120 cm3 2 concentration after dilution with time taken for a fixed mass
(c) Maximum volume of hydrogen water. When the concentration of sulphur to be precipitated.
= 120 cm3. of a reactant decreases, the 1
CuSO4(aq) acts as a catalyst. number of particles per unit As rate of reaction ∝ —t—i—m—————e ,
A catalyst increases the rate of
volume decreases. Therefore, the
reaction but does not affect the frequency of effective collisions this shows that the higher
yield of the products. 2 decreases. This causes a lower the concentration of the acid,
(d) Volume of hydrogen produced in rate of reaction. 3 the higher the rate of
Experiment II = 120 cm3 (d) Increase the temperature of the reaction. 1
Rate of reaction = —13—2—20— 1 hydrogen peroxide solution. 1 (c) (i) Experiments B and C or
Fe2O3 Experiments D and E. 1
(e) 2H2O2(aq) → 2H2O(l)+O2(g)
= 3.75 cm3 s–1 (ii) Comparing Experiments B
(e) (i) Experiment III has the highest 1 and C, the acid concentration
(f) is fixed at 0.10 mol dm–3,
rate of reaction. It takes the but the temperature varies
from 30 °C to 20 °C.
F shortest time to complete 1 Comparing Experiments D
O the reaction. and E, the acid concentration
is fixed at 0.05 mol dm–3 but
R (ii)
M
5 the temperature varies from
30 °C to 20 °C. 1
(iii) Comparing Experiments B
and C or Experiments D and
E show that the higher the
temperature, the shorter the
time taken for the cross ‘X’ to
disappear from view.
2 1
S ince rate of reaction ∝ t—i—m—————e ,
2 5 (a) (i) Na2S2O3 + H2SO4 →
CHAPTER 1 S + Na2SO4 + SO2 + H2O this shows that the higher
4 (a) Average rate of reaction (ii) S2O32- + 2H+ → the temperature, the higher
= —3———0— S + SO2 + H2O 2
35 the rate of reaction. 1
= 0.857 cm3 s–1 (b) (i) Experiments A, C and E. 1
(ii) In these three experiments, (d) At a higher temperature, the H+
2 the temperature was kept and thiosulphate (S2O32–) ions
constant at 20 °C but the have greater kinetic energy and
1 acid concentration varies
(b) Rate of reaction ∝ ———————————————————— move faster. The frequency of
time taken
effective collisions between the
The time taken for experiment from 0.15 mol dm–3 to ions increases. Hence, the rate of
II is shorter than Experiment 0.05 mol dm–3. 1
reaction increases. 2
I. Hence, reaction rate in 6 (a)
Experiment II is higher than in
Experiment I. This shows that Experiment 1234
iron(III) oxide acts as a catalyst in
Volume of hydrogen peroxide (cm3) 40 40 40 80
Experiment II.
A catalyst provides an alternative Volume of water (cm3) 40 40 40 0
reaction pathway that has a lower Temperature (°C) 30 30 32 30
activation energy than the original
reaction. As a result, a larger Mass of MnO2 used (g) 1.0 0 1.0 1.0
number of reactant particles have
sufficient energy to overcome the • Comparing Experiments 1 and 2
lower activation energy. Thus the Constant variable: Concentration and temperature of hydrogen peroxide
frequency of effective collisions Manipulated variable: Mass of catalyst
increases and the rate of reaction Explanation
increases. 3 Decomposition of hydrogen peroxide does not occur because a catalyst is not used.
Answers 566
• Comparing Experiments 1 and 3 Essay Questions (ii) A rubber stopper with a
delivery tube is immediately
Constant variable: 1 (a) (i) Average rate of reaction inserted into the conical flask
and the stopwatch is started
Concentration of hydrogen 480 simultaneously.
= ————
peroxide and mass of catalyst (iii) The total volume of oxygen
100 evolved is determined from
Manipulated variable: the burette readings at half-
= 4.8 cm3 s–1 2 minute intervals.
Temperature of hydrogen peroxide
(ii) The gradient of graph I at (iv) The experiment is repeated
Explanation using 0.5 g of copper(II)
any time is steeper than that oxide instead of 0.5 g of
The total volume of oxygen lead(IV) oxide.
of graph II. Hence, the rate
released is the same for both
of reaction for Experiment
Experiments 1 and 3. Thus, the
I is greater than that of
concentration and volume of
Experiment II. 2
hydrogen peroxide are the same
This is because 1 mol of
for both Experiments 1 and 3.
H2SO4 dissociates in water to
The volume of oxygen released form 2 mol of H+ ions but 1 Tabulation of data
Experiment I: The decomposition
(Table 4) shows that the rate mol of HCl dissociates to form of hydrogen peroxide in the
presence of PbO2
of reaction in Experiment 3 only 1 mol of H+ ions. Thus
is higher than Experiment 1 the concentration of H+ ions in
because a higher temperature Experiment I is higher. 2
is used in Experiment 3. Time 0 —12 1 1—21 2 2—12 3 3—21
(minute)
• Comparing Experiments 1 and 4 At a higher concentration,
Constant variable: Mass of catalyst the frequency of effective Burette
reading
and temperature of reaction collisions between H+ ions (cm3) 50
Manipulated variable: and iron atoms is greater.
Concentration of hydrogen Hence the rate of reaction is F
O
peroxide higher. 2 Volume of 0 R
O2 (cm3) M
Explanation (b) 5
The total volume of oxygen
released in Experiment 4 is Experiment II: The
decomposition of hydrogen
twice the total volume of oxygen peroxide in the presence of CuO
released in Experiment 1. This
implies that the concentration Time 0 —21 1 1—21 2 2—21 3 3—21
(minute)
of hydrogen peroxide used
in Experiment 4 is twice the
concentration of hydrogen Burette
reading
peroxide in Experiment 1. (cm3) 50
The volume of oxygen released
(Table 4) shows that the rate Volume of
O2 (cm3)
of reaction in Experiment 4 0
is higher than Experiment 1 Experiment To investigate the
effectiveness of lead(IV) oxide
because a higher concentration and copper(II) oxide on the rate Based on the experimental results obtained, CHAPTER 1
of decomposition of hydrogen two graphs of total volume of oxygen against
of hydrogen peroxide is used peroxide. time for the decomposition of hydrogen
Hypothesis Lead(IV) oxide is peroxide are plotted on the same axes.
in Experiment 4. 5 a more effective catalyst than
copper(II) oxide.
(b) (i) The volume of oxygen collected Variables
Manipulated variable Type of
is less than 18 cm3 because of catalyst used.
Responding variable Volume of
the lower rate of reaction. oxygen liberated at particular time
intervals.
(ii) The total volume of oxygen Constant variables Volume and
concentration of hydrogen peroxide,
collected remains the same temperature of experiment and
mass of catalyst used.
(that is, 38 cm3) because a Procedure
(i) Using a measuring cylinder,
catalyst does not affect the
25 cm3 of 0.2 mol dm–3
yield of the product obtained hydrogen peroxide is measured
out and poured into a conical
when the reaction is flask. 0.5 g of lead(IV) oxide
is then added to the conical
complete. 2 flask.
(c) (i) This means that the mass
and chemical composition
of the catalyst are the same
before and after the reaction. Graph I shows the effect of PbO2 on the
decomposition of hydrogen peroxide while
(ii) The catalyst is weighed graph II shows the effect of CuO on the
decomposition of hydrogen peroxide.
before the experiment. After
Conclusion
the reaction, the reaction The gradient of graph I is steeper than
the gradient of graph II at any particular
mixture is filtered. The catalyst time. This shows that the rate of reaction
is collected and dried. It is
then weighed again. The
mass of the catalyst remains
the same after the reaction. 3
567 Answers
in Experiment I is higher than the rate of refrigerator. The souring of milk is Variables
reaction in Experiment II. Thus, lead(IV) a decomposition reaction caused Manipulated variable
oxide is more effective as a catalyst for the by bacteria. If the milk is kept at a Concentration of sodium
decomposition of hydrogen peroxide. The low temperature in the fridge, the thiosulphate solution.
Responding variable Time taken for
hypothesis is accepted. 12 rate of decomposition is reduced the cross 'X' to disappear from view.
Constant variables The
2 (a) The dissolving of iron in dilute and it will remain fresh for a total volume of solution, the
concentration of hydrochloric
hydrochloric acid is a chemical longer period of time. 4 acid and the temperature of the
experiment.
reaction. When iron dissolves in (d) Hydrogen peroxide decomposes Chemicals Sodium thiosulphate
solution (0.2 mol dm–3),
hydrochloric acid, the following slowly at room temperature to hydrochloric acid (1 mol dm–3)
and distilled water.
reaction occurs: give water and oxygen. Apparatus Measuring cylinders,
conical flask, stopwatch and white
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g) 2H2O2(aq) → 2H2O(l) + O2(g) paper marked with a cross ‘X’.
Procedure
Iron powder has a much larger The human blood contains (i) A white piece of paper
surface area than an equal enzymes. Enzymes are
mass of coarse iron filings. biological catalysts. The enzyme
With a larger surface area, the in the blood catalyses the
frequency of effective collisions decomposition of hydrogen
increases. Hence, even at room peroxide. Hence, adding a drop
temperature, the reaction rate of blood suddenly speeds up
of iron powder with dilute
the decomposition of hydrogen
hydrochloric acid is much greater with an ink cross ‘X’ on
peroxide to produce bubbles of
than the rate of coarse iron filings it, 0.2 mol dm–3 sodium
with acid. This explains why iron oxygen gas. Thus, a vigorous thiosulphate solution and
F powder dissolves readily in cold effervescence occurs. 3 1 mol dm–3 hydrochloric
O dilute hydrochloric acid at room acid are prepared.
R temperature. 3 (a) (i) The rate of reaction is (ii) Using a measuring
M In the case of coarse iron filings, defined as the amount cylinder, 50 cm3 of sodium
the rate of reaction is low at of a reactant used up or thiosulphate solution is
the amount of a product
room temperature. Hence, a measured out and poured
5 higher temperature is required obtained per unit time. 2 into a conical flask.
to speed up the reaction. When (iii) The conical flask is placed
a substance is heated, the (ii) The collision theory states on the white piece of paper
particles in the substance absorb that for a chemical reaction marked with ‘X’.
to occur, the reactant
energy. This causes the number particles must collide with (iv) Using a small measuring
of reactant particles having the each other. cylinder, 5 cm3 of 1 mol
activation energy to increase. As a At high pressures, the dm–3 hydrochloric acid is
result, the frequency of effective nitrogen and hydrogen measured out.
collisions increases, and the rate molecules are packed closer (v) Hydrochloric acid is quickly
of iron dissolving in dilute together. This means that poured into the sodium
CHAPTER 1 hydrochloric acid increases. 7 the number of gaseous thiosulphate solution and
(b) Coal is a solid fuel and thus molecules per unit volume is the mixture is shaken.
flammable. But big pieces of coal increased. (vi) The stopwatch is then
are safe because they do not
catch fire easily. In contrast, coal Consequently, the frequency started simultaneously.
dust is highly explosive because of effective collisions (vii) The cross ‘X’ is viewed
it is in the finely divided state. increases and the rate of
Explosions are caused by very reaction also increases. from the top of the conical
rapid chemical reactions. flask. The solution becomes
In the coal mine, the air is filled 5 cloudy because sulphur is
(b) precipitated.
(viii) The time is taken as soon as
with coal dust which forms an the cross disappears from view.
explosive mixture with air. It only (ix) Steps (ii) to (viii) are repeated
needs a spark to trigger off an using different volumes of
explosion because the reaction sodium thiosulphate solution
between finely divided coal and which are diluted with distilled
oxygen is a very fast reaction. 6 water as shown in the
(c) A reaction can be made to following table.
go slower by decreasing the Calculations
temperature. Conversely, a The concentration of sodium
reaction can be made to go faster thiosulphate solution after
by increasing the temperature. Hypothesis The rate of reaction mixing
Milk will turn sour very quickly between hydrochloric acid volume of
if it is exposed to the air at and sodium thiosulphate is = 0.2 ——N——a—2—S—2—5O—0—3—(—a——q—)— mol dm–3
room temperature, but it will proportional to the concentration
keep for several days in a of sodium thiosulphate used.
Answers 568
Experiment 12345 energy to overcome the
Volume of sodium thiosulphate (cm3) 50 40 30 20 10 lower activation energy
required for effective
Volume of water (cm3) 0 10 20 30 40 collisions. 1
Total volume of solution (50 cm3) 50 50 50 50 50 • Hence, the frequency
of effective collisions
Concentration of Na2S2O3(aq) after mixing 0.20 0.16 0.12 0.08 0.04 increases and the rate of
Time taken (s)
reaction increases. 2
T—i—m1——e (s–1) (b) (i) Combustion of charcoal
The rate of reaction between
charcoal and oxygen in the
air depends on the area of
Results Based on the graph obtained, contact between them. A
A gra ph of —t—im—1—e— against
concentration of sodium thiosulphate the rate of reaction between large block of charcoal will
solution is plotted.
sodium thiosulphate solution not catch fire easily and thus
Conclusion
R ate ∝ —ti—m——e—1—t—a—k—e—n— and hydrochloric acid is directly will burn slowly. If the block
proportional to the concentration of charcoal is cut into smaller
of sodium thiosulphate used. The pieces, the total surface area
hypothesis is accepted. 10 of the charcoal increases and
(c) 20 cm3 of ethanedioic acid is the rate of burning is higher.
added to 5 cm3 of potassium The larger the total surface
manganate(VII) solution. The area, that is, the smaller the
time taken for the purple colour size of the charcoal pieces,
of potassium manganate(VII) to the higher the rate of F
O
disappear (that is, for decolourisation combustion. 3 R
M
to occur) is recorded. The (ii) Cooking food in a pressure 5
experiment is repeated by adding 20 cooker
cm3 of ethanedioic acid to 5 cm3 of The pressure cooker is a
potassium manganate(VII) solution cooking utensil that allows
containing one or two drops of (i) the pressure inside it to
manganese(II) sulphate solution, become greater than the
(ii) sodium sulphate solution. atmospheric pressure.
All other conditions are kept Pressure cookers are used
unchanged. to speed up cooking. In the
pressure cooker, the high
Experiment H2C2O4 + KMnO4 H2C2O4 + KMnO4 H2C2O4 + KMnO4 pressure causes the water to
Time taken (s) t1 + +
boil at a temperature above
MnSO4(aq) Na2SO4
100 °C or the cooking oil to
t2 t3
boil above its boiling point.
Furthermore, an increase in CHAPTER 1
The results of the experiment show that t1 = t3 but t2 is very much less than t1 or pressure causes the number
t3. This proves that the manganese(II) ion, Mn2+ and not the sulphate ion, SO42–,
increases the rate of reaction. Thus, sulphate ion does not act as a catalyst. 3 of water vapour or cooking oil
molecules that are in contact
(ZnCl2 + H2) is lower with the food to increase.
than the energy content
4 (a) (i) Experiment I: Sulphuric acid Thus, food cooks faster in
Experiment II: Hydrochloric of the reactants (Zn + a pressure cooker than at
acid normal atmospheric pressure.
(ii) Zn(s) + H2SO4(aq) → 2HCl). Hence, the reaction Faster cooking preserves the
ZnSO4(aq) + H2(g)
is exothermic and the essential vitamins and other
(iii)
temperature of the reaction nutrients in the food. 3
increases. 2
• The energy profile diagram Experiments
also shows that the 1 (a) t1 = 52 s t4 = 36 s
activation energy for a t2 = 46 s t5 = 32 s
catalysed reaction is lower t3 = 38 s 3
than that of an uncatalysed (b) Manipulated variable The
reaction. 1 temperature of sodium
• A catalyst lowers the thiosulphate
activation energy by Responding variable Time taken
providing an alternative for the ‘X’ sign to disappear
Explanation reaction pathway for the Constant variable The
• The energy profile diagram
reaction. 1 concentrations and volumes of
shows that the energy
content of the products • As a result, more reacting both sodium thiosulphate solution
particles possess sufficient and dilute sulphuric acid 3
569 Answers
(c) (c) The volume of gas evolved 3 (a) C3H6 + Cl2 → C3H6Cl2
(b) C3H6 + H2O → C3H7OH
Temperature decreases as the reaction proceeds. (c) C3H6 + H2 → C3H8
(°C) (d) 2C3H6 + 9O2 → 6CO2 + 6H2O
25 30 35 40 45 Thus it can be concluded that (e) C3H6 + H2O + [O] → C3H6(OH)2
rate of reaction decreases as the
Time (s) 52 46 38 36 32 reaction proceeds. 3
—T—i—m1——e (s–1) 0.019 0.022 0.026 0.028 0.031 2 Carbon Compounds Self Assess 2.4
Self Assess 2.1 1 (a) 2-methylbut-2-ene
2 (b) 3,4-dimethylpent-1-ene
(d) (i) (c) 2,4-dimethylpent-2-ene
1 Organic Inorganic 2 C5H10 has five isomers.
compounds compounds
H H H H H
Rubber, sugar, Limestone, carbon | | | | |
vinegar, dioxide, calcium
H—C=C—C—C—C—H
| | |
polyvinylchloride, carbonate, sand, H H H
urea sodium chloride, pent-1-ene
ammonium H H H H H
sulphate | | | | |
H—C—C=C—C—C—H
| | |
2 (a) Hydrocarbons are organic
compounds that contain the H H H
elements carbon and hydrogen only.
pent-2-ene
F (b) Three sources of hydrocarbon are H
O |
R petroleum, natural gas and coal. H—C—H
M 3 Two main products formed are carbon
5 H H H
dioxide and water. | | |
Rubber is an organic compound
containing carbon and hydrogen. H—C=C—C—C—H
When rubber is burnt in excess air, | |
the carbon combines with oxygen to H H
form carbon dioxide and the hydrogen
2-methylbut-1-ene
4 combines with oxygen to form water. H
|
(ii) The rate of reaction —t—im—1—e—. Self Assess 2.2 H H H — C — H H
1 (a) Propane, C3H8 | | | |
The graph shows that the (b) Pentane, C5H12 HC=C C CH
| |
rate of reaction increases as (c) Hexane, C6H14
the temp erature of sodium 2 (a) C8H18 H H
thiosulphate solution (b) 2C8H18 + 25O2 → 18H2O + 16CO2 3-methylbut-1-ene
3 (a) The presence of ultraviolet light or
CHAPTER 1 & 2 increases. 3 H
(e) At 50 °C, —t—im—1——e = 0.0345 s–1 sunlight |
H—C—H
3 (b) Substitution reaction
Time = 29 s (c) C2H6 + Cl2 → C2H5Cl + HCl H H H
4 The general formula of a saturated | | |
(f) The higher the temperature,
H—C—C=C—C—H
the higher the rate of reaction. hydrocarbon is CnH2n+2. | |
Conversely, the lower the CnH2n+2 = 58 H H
temperature, the lower the rate 12n + 2n+2 = 58 2-methylbut-2-ene
of reaction. 3 n = —51—46— = 4 3 Cl H
| |
(g) Meat and fish decay (turn bad) H—C—C—H
| |
rapidly at room temperature. The X is C4H10, butane. Cl H
lower the temperature, the lower 1,1-dichloroethane
the rate of decay. Hence, meat Self Assess 2.3
1 (a) CnH2n, where n = 2, 3, 4…
and fish are kept in refrigerators to
prevent them from turning bad. 3 (b) (i) C3H6 and
2 (a) Rate of reaction is the volume (ii) C5H10
(iii) C7H14
of gas evolved per second. 2 (a) Both have low boiling points and H H
Rate of reaction | |
both are insoluble in water but
= ——v——o———l—u——m—————e————o——f———g———a——s————e——v———o——l—v——e———d————(——c——m—————3—)——— H—C—C—H
| |
time taken (s) 3 soluble in organic solvents. Cl Cl
(b) At 30 s (b) Both will undergo complete 1,2-dichloroethane
Burette reading = 28.10 cm3 combustion in excess oxygen 1,2-dichloroethane is formed when
At 40 s to produce carbon dioxide and ethene reacts with chlorine in addition
Burette reading = 24.70 cm3 3 water. reaction.
Answers 570
Self Assess 2.5 (b) The functional group of X is the temperature whereas oils have
carboxyl group, –COOH. lower melting points / exist as
1 (a) The hydroxyl (–OH) group The general formula is liquids at room temperature.
CnH2n+1 COOH, where n = 0, 1, 2 (a) Saturated fatty acids are long-
(b) CnH2n+1OH where n = 1, 2, 3… 2, 3… chained carboxylic acids that do
(c) C4H9OH has four isomers. not contain carbon-carbon double
(c) 2CH3COOH + CaCO3 → bonds. Unsaturated fatty acids are
H H H H (CH3COO)2Ca + CO2 + H2O long-chained carboxylic acids that
| | | | Effervescence occurs and a gas that contain carbon-carbon double
turns limewater cloudy is produced. bonds.
H—C—C—C—C—H (b) Shake 5 cm3 of each type of oil
| | | | 2 (a) Pentanoic acid, C4H9COOH with 1 cm3 of bromine dissolved
O H H H (b) C4H9COOH + NaOH → in trichloromethane (or acidified
| potassium manganate(VII)).
H C4H9COONa + H2O Oleic acid will decolourise the
2C4H9COOH + Mg → brown colour of bromine (or
butan-1-ol (C4H9COO)2Mg + H2 the purple colour of potassium
3 (a) Compound Q is propanoic acid, manganate(VII)) whereas there
H H H H C2H5COOH. is no noticeable change in stearic
| | | | Compound R is propyl propanoate, acid.
C2H5COOC3H7. 3 (a) Hydrogenation
H—C—C—C—C—H (b) Reflux the reagents in the (b) Heat with nickel as a catalyst
| | | | presence of concentrated (c) Palm oil has a lower melting point
H O H H sulphuric acid as a catalyst. than margarine.
| (c) Step I is oxidation.
H Step II is esterification.
butan-2-ol Self Assess 2.7
1 (a) Methyl methanoate, methanol
H Self Assess 2.9 Monomer F
| and methanoic acid 1 Natural polymer Amino acid O
H—C—H (b) Propyl ethanoate, propanol and Isoprene R
Protein Glucose M
H H ethanoic acid Glucose 5
| | (c) Methyl butanoate, methanol and Rubber
H—C—C—C—H butanoic acid Carbohydrate
| | | 2 (a) Methyl propanoate,
O H H Starch
| O
H i
CH3 — CH2 — C — O — CH3
2-methylpropan-1-ol
(b) Propyl ethanoate,
H O
| i
H—C—H CH3 — C — O — CH2 — CH2 — CH3
H H (c) Propyl methanoate, 2 (a) Acids such as methanoic acid or
| | O ethanoic acid; the liquid latex is
i changed to solid rubber.
H—C—C—C—H H — C — O — CH2 — CH2 — CH3
| | | (b) Sulphur of sulphur compounds;
H O H 3 (a) CH3OH + C3H7COOH → the rubber becomes harder,
| C3H7COOCH3 + H2O stronger, more elastic and more
H resistant towards oxidation and
(b) Reflux methanol and butanoic acid heat.
2-methylpropan-2-ol in the presence of concentrated CHAPTER 2
(d) C4H9OH + 2[O] → sulphuric acid as a catalyst. 3 The organic acids produced by
bacteria present in latex neutralise
C3H7COOH + H2O (c) Has a sweet fruity smell, insoluble the negative charges on the
2 (a) Compound P is ethanol, C2H5OH. in water. membranes of the latex colloidal
particles. Coagulation occurs when
Compound Q is ethanoic acid, Self Assess 2.8 the membranes of the latex particles
CH3COOH. 1 Similarity: break during collision.
Compound R and compound S (a) Both molecular structures have To prevent this, aqueous ammonia is
are carbon dioxide and water. added to neutralise the organic acids
Compound T is ethene, C2H4. the carboxylate group, –COO– as produced by the bacteria.
3 (a) Dehydration, W is C3H7OH. the functional group.
(b) Hydration, X is C3H6. (b) Both are insouble in water but
(c) Oxidation, Y is C3H7COOH. soluble in organic solvents.
(d) Combustion, Z is C4H9OH. Difference:
(a) Fats do not have carbon-carbon
Self Assess 2.6 double bond, C = C in their
1 (a) (CH2O)n structures whereas oils have.
(12 + 2 = 60 (b) Fats have higher melting
+ 16)n points / exist as solids at room
= 60
n = —63—00— = 2 SPM Exam Practice 2
Multiple-choice Questions
Molecular formula of X is C2H4O2 1 B 2 A 3 A 4 A 5 B
or CH3COOH. 6 D 7 B 8 A 9 D 10 B
Structural formula of X is 11 C 12 C 13 B 14 D 15 C
16 C 17 D 18 C 19 D 20 A
H O 21 B 22 C 23 B 24 A 25 A
| i 26 B 27 C 28 C 29 B 30 D
31 A 32 C 33 D 34 D 35 C
H—C—C—O—H
|
H
571 Answers
Structured Questions (e) (i) C2H4 + H2O → C2H5OH 1 (b) (i)
1 (a) Alkane, CnH2n+2, n = 1, 2, 3… (ii) C2H5OH + 2[O] → Element C H O
1+1 CH3COOH + H2O 1
(b) Propane, C3H8 1 (iii) Acidified potassium Mass in 40 6.7 53.3
Butane, C4H10 1 dichromate(VI) / potassium 100 g
H manganate(VII) 1 Mol —41—20— —6—1.—7 —5—13—6—.3—
|
(c) H — C — H 4 (a) (i) Fermentation 1
|
H (ii) H H = 3.3 = 6.7 = 3.3 1
| |
(d) (i) Gas 1
1 H — C — C — H Mol ratio 1 2 1 1
| |
(ii) Liquid 1 O H The empirical formula is
| CH2O. 1
(e) Hexane, C6H14 1 H 1
(CH2O)n = 60
(f) 2C2H6 + 7O2 → 4CO2 + 6H2O 1 (b) (i) Oxidation 1 (12 + 2 + 16)n = 60
2 (a) Alkene 1 (ii) C2H5OH + 2[O] → n = 2 1
CH3COOH + H2O 1 The molecular formula of
(b) Hydrogenation 1 carbon compound X is
(c) Add magnesium powder / C2H4O2. 1
H H H (ii) The structural formula of X is
| | | magnesium carbonate powder H O
(c) nH — C = C — C — H | i
| to liquid A and liquid B in two
H
propene separate test tubes.
Effervescence occurs / gas bubbles
H| C| H3 form in liquid B but no noticeable H—C—C—O—H
C—C |
F polymerisation change occurs in liquid A. 1 + 1
O → – H 1
R | | (d) (i) Ester 1
M (iii) The name is ethanoic acid.
5 H H n (ii) Ethyl ethanoate 1
1
polypropene (iii) As a solvent 1
1 (iv) The general formula is
(e) Hydration of ethene 1
CnH2n+1COOH, n = 0, 1, 2…
(d) (i) Heat propene vapour with 5 (a) Process I is coagulation. 1
steam over phosphoric(V) Process II is vulcanisation. (c) Structures
1+1
acid as a catalyst 1 + 1 Similarity:
(b) Ethanoic acid / methanoic acid
H H H Both margarine and palm oil are
| | | 1
esters / they have –COO– as the
(ii) H — C — C — C — H (c) Vulcanised rubber 1
| | | functional group 1
(d) Compound X is stronger /
O H H Differences:
| harder / more elastic / more Margarine is saturated / has
H covalent single bonds
1 resistant to heat / more resistant Palm oil is unsaturated / has 1
covalent double bonds 1
(e) (i) Purple colour changes to to oxidation than natural rubber Physical properties
colourless 1 (any 2) 1+1
CHAPTER 2 (ii) C3H6 + H2O + [O] → (e) Heat natural rubber with sulphur
C3H6(OH)2 1 1
(f) The quantity of soot produced
(f) (i) Aqueous ammonia 1 Similarity:
by the combustion of propene Both are insoluble in water /
(ii) The hydroxide ions of non-conductors of electricity
is more than that by propane. Differences:
aqueous ammonia neutralise Margarine is a solid / has a 1
This is because the percentage of higher melting point
the organic acids produced Palm oil is a liquid / has a 1
carbon by weight in propene is lower melting point 1
more than in propane. 1 + 1 by the bacteria in latex. 1 Palm oil can be converted to
3 (a) For complete reaction between
(g) To make tyres / shoe soles /
ethanol and ethanoic acid to
mattress / balloons 1
take place / to prevent the vapour
Essay Questions
H H H H
| | | | margarine by hydrogenation. 1
of ethanol and ethanoic acid to
1 (a) H — C — C = C — C — H Hydrogen gas and palm oil is
escape into the atmosphere. 1 | |
(b) As a catalyst (or as a dehydrating 1 passed over heated nickel
H H 1
agent) 1 at 200 °C 1
but-2-ene
(c) (i) Esterification 1 2 (a) To prepare ethanoic acid from
H
(ii) CH3COOH + C2H5OH → | ethanol
CH3COOC2H5 + H2O 1
H—C—H Materials Ethanol, concentrated
(d) (i) Methyl ethanoate 1 H H sulphuric acid, potassium
| |
O H—C=C—C—H dichromate(VI) 1
i |
Apparatus Boiling tubes, delivery
(ii) CH3 — C — O — CH3 1 H tube with stopper, test tube,
1 Bunsen burner, test tube holder,
(iii) Has a sweet / fragrant / 2-methylpropene 1 blue litmus paper 1
fruity smell 1
Answers 572
potassium dichromate + ethanoic acid 6 Equation (c) Reduced.
concentrated sulphuric acid + C2H5OH + CH3COOH → Loss of oxygen.
ethanol CH3COOC2H5 + H2O 1
(d) Oxidised.
Experiment Loss of hydrogen or gain of oxygen.
1 (a) Statement of the problem 2 (a) (i) Calcium
(ii) Chlorine
water How to differentiate between
(b) Oxidising agent: chlorine
hexane, a saturated hydrocarbon Reducing agent: calcium
and hexene, an unsaturated 3 (a)
heat hydrocarbon? 3 Compound Oxidation number
+6
Diagram + label (b) All the variables (i) CaCrO4 +3
1+1 (ii) HNO2 +5
Manipulated variable (iii) NaClO3 +3
(iv) Cr2(SO4)3 +3
Procedure Types of hydrocarbon, hexane (v) FeCl3.6H2O +6
1 A boiling tube is filled with (vi) K2MnO4
and hexene
about 5 cm3 of potassium Responding variable
dichromate(VI) solTutCioAn 2/1
Reactions towards addition
and 2 cm3 of concentrated reaction / bromine water /
sulphuric acid. About 3 cm3 acidified potassium manganate(VII)
of ethanol is added to the Constant variables
tube. 1 Quantity of hydrocarbon and (b)
2 A piece of tiny porcelain is quantity of reagent 3
put in to promote smooth (c) Hypothesis Ion Oxidation number
boiling. The boiling tube Hexene undergoes addition (i) SO42– +6
(ii) CuCl43– +1
is closed with a stopper reaction whereas hexane does (iii) ClO– +1 F
O
connected with a delivery not undergo addition reaction / R
M
tube. The end of the delivery Hexene decolourises the brown 5
tube is placed in a test tube colour of bromine water whereas
cooled by cold water in a hexane does not. 3 (iv) NO3– +5
beaker (or The content in the (d) List of materials and apparatus 4 (a) (i) Cu2SO4
(ii) MnCl2
boiling tube is heated and the Hexene, hexane, acidified potassium (iii) NiSO4
(b) (i) Chromium(III) chloride
distillate is collected in the test manganate(VII) solution, bromine
tube cooled by water). 1 water, test tubes. 3
3 Observation The potassium
(e) Procedure (ii) Cobalt(II) oxide
dichromate(VI) changes from
1 5 cm3 of hexene is put in a (iii) Iron(III) sulphate
orange to green 1
4 Test The distillate changes test tube. (c) (i) Fe: +3
blue litmus to red, indicating 2 1 cm3 of acidified potassium Al: +3
an acid is produced. 1 manganate(VII) solution is (ii) Iron(III) oxide
5 Equation C2H5OH + 2[O] → added to hexene. Aluminium oxide
CH3COOH + H2O 1
3 The mixture is then shaken (iii) Iron shows variable oxidation
6 The carboxylic acid produced
and any colour changed is states. CHAPTER 2 & 3
is ethanoic acid. 1 recorded. The Roman numeral III refers
(b) Ester prepared Ethyl ethanoate 1 4 The experiment is repeated to the oxidation number of
Alcohol used Ethanol 1 using hexane to replace iron in the compound.
Carboxylic acid used hexene. Aluminium has only one
Ethanoic acid 1 5 Steps 1 to 4 are repeated oxidation state, that is, +3.
Apparatus Beaker, boiling tube, using bromine water Hence, the Roman numeral
Bunsen burner, test tube holder 1 to replace the acidified figure, III, is not shown in
Procedure potassium manganate(VII) the names of aluminium
1 About 3 cm3 of ethanol and solution. 3 compounds.
5 (a) Magnesium is oxidised.
3 cm3 of ethanoic acid are (f) Tabulation of data (b) (i) Loss of electrons to form
put in a boiling tube. 1 Hydrocarbon Observation Mg2+ ion in the compound,
2 About 1 cm3 of concentrated Hexene MgO.
sulphuric acid is added to the Hexane Mg(s) → Mg2+(s) + 2e—
mixture. 1 3
3 A piece of tiny porcelain is put loss of electrons
in to promote smooth boiling. (ii) Increase in oxidation number
of magnesium from 0 in Mg
1 3 Oxidation and Reduction to +2 in MgO.
4 The contents of the boiling Self Assess 3.1 6 (a) Zinc powder dissolves.
1 (a) Oxidised. The brown solution of Fe3+(aq)
tube are heated to boiling
Oxidation number of Fe increases. changes to the green solution of
for a few minutes. 1 (b) Oxidised. Fe2+(aq).
5 The contents are then poured Oxidation number of Zn increases.
into a beaker half-full of water.
1
573 Answers
(b) Zn(s) → Zn2+(aq) + 2e– (a) reaTc>tiviPty>deQcr>eaRses→
Fe3+(aq) + e– → Fe2+(aq)
(c) Zinc is oxidised because it has donated electrons to form Zn2+ ions. (b) Metal Q does not react with the
oxide of metal P because metal
7 (a) (i), (ii) and (v) are redox reactions. Q is less reactive than metal P.
(b) Reaction (i) Reaction (ii) Reaction (v) 2 (a) Metal X is less reactive than
carbon but more reactive than
Cl2 + 2e– → 2Cl– Fe → Fe3+ + 3e– Fe3+ + e– → Fe2+ hydrogen. Thus, X is zinc.
2Br– → Br2 + 2e– O2 + 4e– → 2O2– Sn2+ → Sn4+
(c) Oxidising agent
Reducing agent
(decrease in oxidation number) (increase in oxidation number) reacCtiv>ityZdne>creHases→
(i) FeCl3 (Fe: +3 to +2) SnCl2 (Sn: +2 to +4)
(ii) Pb(NO3)2 (Pb: +2 to 0) Zn (Zn: 0 to +2) (b) (i) 2Al(s) + 3ZnO(s) →
(iii) Cl2 (Cl: 0 to –1) KBr (Br: –1 to 0) 3Zn(s) + Al2O3(s)
(iv) H2O2 (O: –1 to –2) KI (I: –1 to 0) (ii) Oxidising agent : ZnO
Reducing agent : Al
3 Method 1: Reduction of hot lead(II)
8 (a) Oxidising agent: potassium dichromate(VI) oxide by carbon in
Reducing agent: iron(II) sulphate powdered form
(b) At the negative terminal 2PbO(s) + C(s) → 2Pb(s) + CO2(g)
Fe2+ → Fe3+ + e–
Method 2: Reduction of lead(II) oxide
At the positive terminal by hydrogen gas
Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O
PbO(s) + H2(g) → Pb(s) + H2O(l)
F (c) At the negative terminal, oxidation process occurs. Fe2+ is oxidised to Fe3+. The
O oxidation number of iron increases from +2 to +3. Electrons are donated to the Method 3: Electrolysis of molten
R electrode. At the positive terminal, reduction process occurs. Cr2O72– is reduced lead(II) oxide
M to Cr3+. The oxidation number of chromium decreases from +6 to +3. Electrons Molten lead is produced at
are accepted from the electrode. the cathode
5 (d) At the negative electrode Pb2+(l) + 2e– → Pb(l) … reduction
The colour changes from green (Fe2+) to brown (Fe3+).
At the positive electrode Method 4: Displacement of lead from
The colour changes from orange (Cr2O72–) to green (Cr3+). its salt solution
Step 1: Prepare lead(II) nitrate solution
Self Assess 3.2 by dissolving lead(II) oxide in
1 (a) A brown substance is formed on the surface of iron. dilute nitric acid.
(b) Iron reacts with oxygen in the air and water to form rust, PbO(s) + 2HNO3(aq) →
Fe2O3. xH2O. Pb(NO3)2(aq) + H2O(l)
Step 2: Add zinc powder to lead(II)
2 (a) 4Fe(s) + 3O2(g) + 2H2O(l) → 2Fe2O3.H2O(s)
CHAPTER 3 (b) A layer of grease can prevent iron from having contact with air (oxygen) and nitrate produced
water.
Without oxygen and water, rusting of iron cannot occur. Zn(s) + Pb(NO3)2(aq) →
3 (a) Galvanised iron refers to iron that has been coated with a layer of zinc to protect Pb(s) + Zn(NO3)2(aq)
it from corrosion.
(b) Zinc is more electropositive than iron. Hence, zinc donates electrons more readily. Self Assess 3.4
1 (a) Electrolysis of concentrated
When galvanised iron is exposed to air and water, zinc will corrode first.
potassium iodide solution
Zn(s) → Zn2+(aq) + 2e– At the anode (carbon)
The electrons donated by zinc will move to iron. It is, therefore, difficult for
Anions present: I– and OH–
iron(II) ions to form. Oxidation of iodide ions to
iodine occurs because of its high
Hence, rusting of iron does not occur. concentration.
4 (a) (i) 2H+(aq) + 2e– → H2(g) Iodide ions are oxidised to form
(ii) Mg(s) → Mg2+(aq) + 2e– iodine and electrons are
released.
(b) Magnesium is more electropositive than zinc This means that magnesium
donates electrons more readily than zinc to form Mg2+ ions. Hence, magnesium
prevents the corrosion of zinc.
Self Assess 3.3 Observation Inference 2I–(aq) → I2(aq) + 2e–
1 Mixture At the cathode (carbon)
Cations present: K+ and H+
Metal P + oxide of metal Q Glows P is more reactive than Q Reduction of H+ ions to hydrogen
Metal Q + oxide of metal R Glows Q is more reactive than R occurs.
H+ ions accept electrons and are
Metal Q + oxide of metal T Does not glow T is more reactive than Q preferentially discharged to form
Metal P + oxide of metal T Does not glow T is more reactive than P hydrogen gas.
2H+(aq) + 2e– → H2(g)
Answers 574
(b) Electrolysis of copper(II) Structured Questions (ii) A mixture of hydrogen and
sulphate solution 1 (a) Fe(s) + Cu2+(aq) → air explodes easily when burnt.
At the anode (copper) Fe2+(aq) + Cu(s) 1 In order to avoid an explosion,
Anions present: SO42– and OH– ions (b) Iron is less electropositive than all the air in the apparatus
Both OH– and SO42– ions are not
discharged. magnesium. Hence there is must be removed before
The copper anode dissolves no reaction between iron and hydrogen gas is ignited. 1
because oxidation of copper to magnesium chloride solution. (iii) Melting point: 0 °C
Cu2+ ions occurs. Iron is more electropositive than Boiling point: 100 °C 1
Cu(s) → Cu2+(aq) + 2e– copper. Hence it displaces copper (iv) The oxide of X is iron(III)
At the cathode (copper) from copper(II) nitrate solution. oxide. 1
Cations present: Cu2+ and H+ 2 Reason: Iron(III) oxide is
Reduction of Cu2+ ions to Cu occurs. (c) (i) Copper metal 1 brown and can be reduced
Cu2+ ions accept electrons and (ii) A brown precipitate of copper to iron which is grey in colour.
are reduced to form copper. is formed. (v) The oxide of Y is zinc oxide.
Cu2+(aq) + 2e– → Cu(s) The solution changes from (vi) Y > H > X 1
2 (a) (i) The diagram shows that blue to green. 2 (vii) Hydrogen reduces the oxide
electrode X is an anode (d) (i) Cu2+ ion (ii) Fe 2 of X into metal X. 1
because it is connected to (e) Cu2+ oxidises Fe to Fe2+. Hence it This means that hydrogen
the positive terminal of the is an oxidising agent. Fe reduces is more reactive than the
battery. Cu2+ to Cu. Hence it is reducing metal X.
Anions present: OH– and agent. 2 Hydrogen does not react with
SO42– ions 2 (a) (i) From +2 to 0. 1 the oxide of Y.
Oxidation occurs. OH– ions
(ii) Ammonia is oxidised. The Hence, hydrogen is less
are oxidised to oxygen. oxidation number of nitrogen reactive than Y. 2 F
O
4OH–(aq) → increases from –3 to 0. 1 (b) (i) To produce oxygen for the R
M
O2(g) + 2H2O(l) + 4e– (b) (i) An oxidising agent is a combustion of magnesium, 5
Electrode Y is the cathode
substance that brings about carbon and zinc. 1
because it is connected to the the oxidation of another (ii)
negative terminal of the battery. substance but is itself Observation
Element
Cations present: H+ and Fe2+ reduced. 1
Flame/glow Residue (if any)
ions (ii) Oxidising agent
Reduction occurs. Hydrogen Potassium manganate(VII) Magnesium Very bright White powder
flame
ions are reduced to hydrogen solution 1
Substance reduced
gas. The manganate(VII) ion, Zinc The glow The residue is
2H+(aq) + 2e– → H2(g) MnO4– 1 spreads yellow when
Comment: Hydrogen
hot and white
is below iron in the (iii) Electrode Y 1
when cold
electrochemical series. (iv) Fe2+(aq) → Fe3+(aq) + e– 1
(v) Potassium nitrate solution 1
Hence, hydrogen ions are Carbon Bright flame No residue
preferentially discharged. (c) At electrode X: Green to brown 1 3 CHAPTER 3
(ii) The electrons are transferred At electrode Y: Purple to (iii) The powdered element
from electrode X to electrode colourless 1 (magnesium, carbon and
3 (a) (i)
Y through the external circuit. zinc) is first heated strongly.
(b) In electrolysis, electrical energy is When the powdered element
converted into chemical energy. has become very hot,
3 (a) Oxidising agent: NiO2 potassium manganate(VII) is
Reducing agent: Cd
then heated strongly.
(b) (i) Cd + 2OH– →
The oxygen released is
Cd(OH)2 + 2e–
(ii) NiO2 + 2H2O + 2e– → allowed to pass over the
Ni(OH)2 + 2OH– elements. 2
(c) From Cd electrode to NiO2
4 (a) The presence of water and
electrode.
oxygen. 1
Some pieces of zinc are put (b)
into the round-bottomed
SPM Exam Practice 3 flask. 3
Multiple-choice Questions By means of a thistle funnel,
1 B 2 D 3 C 4 A 5 C dilute sulphuric acid is added (c) The electrons flow to the edge of
6 C 7 D 8 B 9 C 10 B to the zinc.
11 B 12 A 13 A 14 B 15 C Hydrogen gas released is the water droplet. Oxygen accepts
16 D 17 C 18 A 19 C 20 A passed through a U-tube
21 B 22 B 23 A 24 B 25 D containing a drying agent
26 D 27 A 28 C 29 B 30 D such as anhydrous calcium
31 B 32 A 33 D 34 A 35 D chloride. 3
36 A 37 B 38 A 39 B 40 D
575 Answers
these electrons and is reduced to The coke burns in the hot air Oxide (O2–) ion has the same
hydroxide ions. and is oxidised to carbon stable electron arrangement as
O2(g) + 2H2O(l) + 4e– → 4OH– dioxide. neon (2.8). Chloride (Cl–) ion
(aq) C(s) + O2(g) → CO2(g) has the same stable electron
Fe2+ and OH– combine to form Carbon dioxide then ascends to arrangement as argon (2.8.8).
Fe(OH)2. the top part of the furnace and 10
Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s) is reduced by coke to carbon (c) Galvanising is the coating of iron
The iron(II) hydroxide produced is monoxide.
or steel with zinc for protection
oxidised by oxygen to form C(s) + CO2(g) → 2CO(g) against rusting.
Fe2O3. xH2O(rust). 3 Iron ore: Reduced to iron. Even if the zinc coating is
5 (a) Metal X is magnesium. This is a The iron ore is reduced by coke scratched, the iron below it does
displacement reaction. Copper and carbon monoxide to produce not rust.
and lead are less reactive than molten iron. Rusting occurs when iron donates
iron. They will not displace iron Fe2O3(s) + 3C(s) → electrons to form Fe2+ ions.
from its salt solution. Potassium 2Fe(l) + 3CO(g) Fe(s) → Fe2+(aq) + 2e–
will react vigorously with water Fe2O3(s) + 3CO(g) → In galvanised iron, zinc is
2Fe(l) + 3CO2(g) corroded and not iron. This
and hence cannot be used. 3 Molten iron flows to the bottom is because zinc is more
of the furnace and is collected. electropositive than iron. When
(b) (i) Magnesium: Changes from 0 Limestone: To remove impurities zinc corrodes, zinc ions are
Limestone decomposes to formed and electrons are
to +2. calcium oxide and carbon dioxide. released.
CaCO3(s) → CaO(s) + CO2(g)
Iron: Changes from +3 to +2. The calcium oxide produced reacts Zn(s) → Zn2+(aq) + 2e–
with the impurities such as sand
2 to form calcium silicate (slag). The electrons released will
(ii) Mg → Mg2+ + 2e– prevent iron from forming Fe2+
Fe3+ + e– → Fe2+ 2
F (c) Iron(III) chloride acts as the
O oxidising agent in this reaction. It
oxidises Mg to Mg2+ and is itself
R reduced to Fe2+. 2 CaO + SiO2 → CaSiO3 ions and thus prevent the rusting
M 6 (a) (i) Oxidation and reduction 1 sand slag
of iron. 4
5 (ii) In cell P, electrical energy is (silica) (calcium silicate) 2 (a) Experiment 1
The molten slag floats on top of
coverted to chemical energy. Apparatus
the molten iron. The two layers
In cell Q, chemical energy is Wire gauze, tripod stand and
are run off from separate tap
converted to electrical energy. Bunsen burner.
1 holes and allowed to solidify. 6 Materials
(b) Metals: reducing agents
Reducing agents are electron
(b) (i) 2Cl–(aq) +→2Cel–2(→g) + 2e– Magnesium powder, copper(II)
(ii) 2H+(aq) H2(g) 2 donors. Metals are reducing oxide and asbestos paper.
agents because they tend to Procedure
(c) (i) 2NaCl(aq) + 2H2O(l) → donate electrons to form stable 1 Magnesium powder is added
2NaOH(aq) + H2(g) + Cl2(g)
(ii) Cl– ions are oxidised to to copper(II) oxide and
metal ions (cations) with the
chlorine and H+ ions are mixed uniformly.
noble gas electron arrangement.
CHAPTER 3 reduced to hydrogen. 2 2 The mixture is placed on an
For example,
(d) (i) The copper plate is coated asbestos paper.
Na → Na+ + e–
with a brown deposit of 3 The asbestos paper with the
(2.8.1) (2.8)
copper metal. mixture is placed on a wire
(ii) The zinc plate becomes Mg → Mg2+ + 2e– gauze over a tripod stand.
thinner. 2 (2.8.2) (2.8) 4 The mixture of powdered
(e) (i) Zn(s) + Cu2+(aq) → 1 Al → Al3+ + 3e– magnesium and copper(II)
Zn2+(aq) + Cu(s) (2.8.3) (2.8) oxide is then heated strongly.
(ii) Zinc is oxidised to Zn2+ ions. Na+, Mg2+ and Al3+ ions have the
Cu2+ ions are reduced to same stable electron arrangement
copper. 2 as a neon atom (2.8).
Essay Questions Non-metals: oxidising agents Observation
1 (a) Iron can be extracted from Oxidising agents are acceptors A glow appears and spreads
of electrons. Non-metals are across the surface of the mixture.
haematite by reduction with oxidising agents because they Conclusion
carbon. Haematite is the iron ore accept electrons readily to form
that contains iron(III) oxide. stable anions (negative ions) with
Iron ore, coke (carbon) and the stable electron arrangement
limestone (calcium carbonate) of noble gases.
are added at the top of the blast For example,
furnace.
A blast of hot air is blown into O2 + 4e– → 2O2–
the furnace from the bottom. (2.6) (2.8)
Coke: It is oxidised to carbon Cl2 + 2e– → 2Cl– Magnesium is more reactive with
monoxide. (2.8.7) (2.8.8) oxygen than copper.
Answers 576
Discussion as iron is precipitated. When (ii)
1 The experiment shows that excess sodium hydroxide is Metals more reactive Metals less reactive
magnesium reacts with added to the reaction mixture, a than carbon than carbon
copper(II) oxide. white precipitate is formed. The
precipitate dissolves in excess Magnesium Copper
2 Magnesium reduces sodium hydroxide solution to Sodium Lead
copper(II) oxide to copper
and is itself oxidised to form a colourless solution. This 2
magnesium oxide. shows the presence of Zn2+ ions. (c) (i) The purpose is to produce
Mg(s) + CuO(s) → In this reaction, iron(II) sulphate oxygen gas from potassium
Cu(s) + MgO(s) acts as the oxidising agent. Iron(II) manganate(VII). 1
sulphate oxidises zinc to Zn2+ ions (ii)
Experiment 2 and is itself reduced to iron (Fe). Metal Observation
Apparatus Iron
Spatula and test tube. Zn(s) → Zn2+(aq) + 2e– • Moderately bright flame
Materials …oxidation • Bright glow
Fe2+(aq) + 2e– → Fe(s)
Magnesium powder and …reduction Lead • Faint flame
copper(II) sulphate solution. The overall redox reaction Magnesium • Moderately bright glow
Procedure between zinc and iron(II)
A spatula of magnesium powder sulphate can be represented by • Very bright flame
is added to copper(II) sulphate the equation: • Very bright glow
solution in a test tube. The FeSO4(aq) + Zn(s) → Fe(s) + 3
mixture is shaken gently. oxidising reducing ZnSO4(aq)
a gent agent (d) The more reactive the metal, the
The reaction of iron(II) sulphate
brighter the flame or the glow is
on the metal. 3 F
O
(e) R
M
as a reducing agent Magnesium Iron Lead Copper 5
When chlorine water is added
to iron(II) sulphate solution, the 3
green colour of iron(II) sulphate (f)
changes to a brown colour. Metal Metal
When sodium hydroxide solution
is added to the reaction mixture, (i) Manipulated (i) Use different
a brown precipitate is formed,
which is insoluble in excess variable metals for the
experiment
Observation sodium hydroxide. This shows (ii) Responding (ii) Observe the
1 The magnesium powder variable brightness of the
the presence of Fe3+(aq) ions. In flame or the glow
dissolves. on the metal
2 The blue colour of copper(II) this reaction, iron(II) sulphate acts
sulphate changes to colourless. as the reducing agent. It reduces
3 A reddish-brown precipitate is
chlorine to chloride ions and is (iii) Constant (iii) Use the same
produced. variable mass of potassium
Conclusion itself oxidised to Fe3+ ions. manganate(VII) CHAPTER 3
A more reactive metal will and the same
displace a less reactive Cl2(aq) + 2e– → 2Cl–(aq) mass of each
metal from its salt. Hence, … reduction metal
the experiment shows that
magnesium is more reactive than 2Fe2+(aq) → 2Fe3+(aq) + 2e–
copper.
Discussion … oxidation
The overall redox reaction 3
2 (a) Problem statement
between chlorine and iron(II) Is steel more resistant to rusting
sulphate can be represented by than iron?
(b) Hypothesis
the ionic equation:
2Fe2+(aq) + Colx2i(daisqi)ng→
reducing
1 The brown precipitate is Iron rusts more readily than steel.
agent agent
copper. (c) Manipulated variable
2Fe3+(aq) + 2Cl–(aq)
2 Magnesium displaces copper 10 Steel and iron nails
from copper(II) sulphate Responding variable
solution. Experiments Colour change in the gelatin
Mg(s) + CuSO4(aq) → 1 (a) Copper(II) oxide reacts vigorously solution
Cu(s) + MgSO4(aq) 10 Constant variables
(b) The reaction of iron(II) sulphate with carbon. Conditions of experiment
(b) (i) If carbon is more reactive
as an oxidising agent than the metal, a reaction will (d) Apparatus
When excess zinc powder is added occur and the mixture will Test tubes
to iron(II) sulphate solution, zinc glow. Materials
dissolves to form zinc sulphate. Conversely, if carbon is less Iron nail, steel nail, gelatin,
The green colour of iron(II) reactive than the metal, the sandpaper and potassium
sulphate slowly disappears mixture will not glow. 3 hexacyanoferrate(III).
577 Answers
(e) Procedure 4 (a) The formation of covalent bonds Ag+(aq) + Cl–(aq) → AgCl(s)
1 The iron nail and steel nail between nitrogen and oxygen to 0.02 mol of Ag+ ions on mixing
are cleaned with sandpaper. form NO2 molecule. with 0.02 mol of Cl– ions will
2 They are then placed in test (b) The breaking of covalent bonds produce 0.02 mol of AgCl.
tubes A and B respectively. in nitrogen molecules and oxygen Step 2: Calculate the heat
3 A solution of gelatin in hot molecules. released when 0.02 mol of AgCl
water is prepared. A few (c) Total energy content in nitrogen is precipitated
drops of potassium hexacyano and oxygen is less than the total The precipitation of 1.0 mol of
ferrate(III) are added to the hot energy content in NO2 molecule. AgCl releases 66 kJ of heat.
gelatin solution. The reaction between nitrogen Heat change for the precipitation
4 The mixture is stirred and and oxygen to form nitrogen of 0.02 mol of AgCl
poured into test tubes A and B. dioxide is endothermic, that is, = –0.02 66
5 The test tubes are set aside heat energy is absorbed when = –1.32 kJ
for three days and examined. the reaction occurs. 2 (a) M2+(aq) + CO32–(aq) → MCO3(s)
(b) Heat change (mcθ)
(f) Tabulation of results 5 (a) Number of moles of anhydrous
Intensity of Inference on copper(II) sulphate needed = (100 + 100) 3 4.2 3
blue colour rate of rusting (30.5 – 29.5)
Test tube = —1—66—6—.5— = 840 J = 0.84 kJ
= 0.25 (c) Number of moles of M2+
A (iron nail) 1 mol → 66 kJ = 1—M—0—0—V—0
? mol → 16.5 kJ = —0—.—5——3———1—0——0
B (steel nail)
Relative molecular mass of CuSO4
17 = 64 + 32 + (4 16)
= 160
1000
F 4 Thermochemistry Mass of CuSO4 needed = 0.05
O = 0.25 160 Number of moles of CO32–
R Self Assess 4.1 = 40 g = 1——M0—0—V—0 = 1——.0—1—0—3—0—01——0—0—
M 1 (a) Endothermic (b) Number of moles of nitrogen = 0.10
5 used Number of moles of MCO3
(b) Exothermic = —21—24— precipated = 0.05
(c) Endothermic = 0.5
Number of moles of oxygen used ∆H of precipitation
(d) Exothermic = —22—44— = —–—0——.8—4—
2 (a) The temperature of the solution 0.05
= 1.0
increases until the maximum
temperature is reached. The
temperature then decreases until
room temperature is reached. 0.5 mol of N2 will react with = –16.8 kJ mol–1
(b) 0.5 mol of O2 to form 1.0 mol (d)
of NO2.
Heat absorbed
CHAPTER 3 & 4
= —2—1.—0 3 66
= 33 kJ
∆H = +33 kJ
3 (a) Exothermic Self Assess 4.2 3 Experiment 1
(b) Number of moles of M2+ ions
1 (a) (i) 2AgNO3(aq) + CaCl2(aq) → = 0——.5—1——0—0—10——0—0
Characteristic Reaction in the 2AgCl(s) + Ca(NO3)2(aq) = 0.05
hot pack Number of moles of Cl– ions
(i) Temperature (ii) Ag+(aq) + Cl–(aq) → AgCl(s) = 1——.0—1——0—0—10——0—0
change Temperature = 0.1
increases (b) Step 1: Calculate the number of Number of moles of MCl2 formed
(ii) Energy content moles of AgCl precipitated = 0.05
of reactants Energy content of Total volume of solution = 200 cm3
and products products is higher Number of moles of AgNO3 used (a) Experiment 2
than energy 0.1 200 Number of moles of M2+ ions
(iii) Energy content of the = ———————————
involved in reactants 1 000 = 0——.5—1——0—0—20——0—0
bond breaking = 0.02 = 0.1
and bond Energy absorbed Number of moles of CaCl2 used
forming in bond breaking 0.1 100
is less than = ———————————
energy released 1 000
in bond forming = 0.01
Number of moles of Cl– ions
produced
= 2 0.01
= 0.02
Answers 578
Number of moles of Cl– ions 2 Step1: Calculate the number of moles Step 2: Calculate the number of F
= 1——.0—1——0—0—20——0—0 of copper displaced moles of iron displaced O
Number of moles of Fe2+ used R
= 0.2 Number of moles of copper displaced = —0—.—2—15—0—0——0—5—0— M
M2+(aq) + 2Cl–(aq) → MCl2(s) = —r—e—l—a—ti—v—e——ma—t—oa—sm—s—i—c—m———a—s—s = 0.0125 5
0.1 mol of M2+ ions react with = —2—6—.5—46— = 0.04
0.2 mol of Cl– ions to form 0.1 Step 2: Calculate the heat energy Mg(s) + Fe2+(aq) →
mol of MCl2. released
Total volume = 400 cm3 From the thermochemical equation Mg2+(aq) + Fe(s)
Rise in temperature is also 10 °C. given, the displacement of 1.0 mol
(Comment: The total number of copper releases 210 kJ of heat 0.0125 mol of Fe2+ reacts with
of moles of MCl2 increases two energy. excess magnesium to form
times (from 0.05 mol to 0.1 Heat change when 0.04 mol of 0.0125 mol of Fe.
mol), but the total volume of copper is formed Step 3: Calculate the heat
solution also increases 2 times. = –210 0.04 of displacement of iron by
Hence, the temperature remains = –8.4 kJ magnesium
unchanged). From steps (1) and (2),
(b) Experiment 3 3 Step 1: Calculate the number of displacement of 0.0125 mol
Number of moles of M2+ ions moles of copper displaced of Fe releases 2.52 kJ of heat
= —1—.—0—1—0—0——10—0—0— = 0.10 Number of moles of Cu2+ used energy.
Number of moles of Cl– ions = —0—.—5—10—0—0——0—5—0— = 0.025 Heat released for the formation of
= —2—.—0—1—0—0——10—0—0— = 0.2 Number of moles of iron used 1.0 mol of Fe
Number of moles of MCl2 formed = —r—e—l—a—ti—v—e——ma—t—oa—sm—s—i—c—m———a—s—s = 2.52 —0—.—0—11——2—5
= 0.10 = —05——.67— = 201.6 kJ
Total volume = 200 cm3 = 0.0125 That is, heat of displacement of
Rise in temperature = 2 10 °C iron by magnesium
Fe(s) + Cu2+(aq) → Fe2+(aq) +Cu(s) = –201.6 kJ mol–1
= 20 °C 0.0125 0.0125 (in excess) (d) Number of moles of iron
(Comment: Number of moles of displaced
MCl2 precipitated is increased two Number of moles of Cu displaced = —2—20—5—1—2.—6 = 1.25
times (from 0.05 mol to 0.10 = 0.0125
mol), but the total volume of Mg(s) + Fe2+(aq) →
solution does not change. Thus, Step 2: Calculate the heat energy
the temperature of the solution is released in the experiment Mg2+(aq) + Fe(s)
increased two times). The heat of displacement
= –150 kJ mol–1 Number of moles of Mg needed
Self Assess 4.3 Displacement of 0.0125 mol of Cu = 1.25
1 (a) (i) Magnesium powder, lead releases 0.0125 150 Relative atomic mass of
= 1.875 kJ of heat energy. magnesium = 24.3
nitrate solution. Step 3: Calculate the maximum Mass of magnesium needed
(ii) Magnesium is used in excess temperature reached = 1.25 24.3 = 30.4 g
∆H (in J) = mcθ
to ensure all the lead metal From step 2, Self Assess 4.4 CHAPTER 4
is displaced from lead 1.875 1000 = 50 4.2 θ
nitrate. ∆T = 8.9 °C 1 (a) ∆H represents the heat energy
(b) (i) Maximum temperature reached evolved when 1.0 mol of H2SO4
= 30 + 8.9 = 38.9 °C reacts with 2.0 mol of NaOH.
(ii) Measuring cylinder, electronic 4 (a) As the mass of magnesium
balance. That is, when 2 mol of H+ ions
added increases, more iron is
(c) (i) Maximum increase in displaced from iron(II) chloride react with 2 mol of OH– ions to
temperature, volume and solution. Hence more heat
concentration of lead nitrate energy is released and the rise in form 2 mol of water molecules,
solution. temperature increases.
(b) All the iron present in iron(II) 114 kJ of heat energy is released.
(ii) Mass of zinc added. chloride is displaced on the
addition of 0.3 g of magnesium. (b) Heat of neutralisation.
(c) Step 1: Calculate the heat
released in the experiment = –——1—21—4— = –57 kJ mol–1
Heat released (mcθ)
= 50 4.2 12 (c) Heat change (mcθ)
= 2520 J = (50 + 50) 3 4.2 3 6.5 J
= 2.52 kJ
= 2730 J
= 2.73 kJ
H+(aq) + OH–(aq) → H2O(l);
∆Hneut = –57 kJ mol–1
Number of moles of OH– ion
= H——e—a—∆—t H—c—hn—eau—tn——g—e
= —2—5.—77—3—
= 0.045
579 Answers
0.045 = ———M——V— Concentration Volume (cm3) The molecular formula of
1000 (mol dm–3) propane is CH3CH2CH3.
= ——————————————————————————————— H2(g) + —21 O2(g) → H2O(l);
= —M—1——03—0—0—5—0 1000
M = 0.9 ∆H = –286 kJ mol–1
(a) Reactions I and IV
Reason Same number of moles
Concentration of NaOH of water molecules produced C(s) + O2(g) → CO2(g);
from strong acid and strong base ∆H = –394 kJ mol–1
= 0.9 mol dm–3 reaction.
2 Number of moles of acid/alkali One molecule of butane contains
Reaction Number of moles of Number of moles of Number of moles of one CH2 group more than one
H2O formed molecule of propane.
H+ ions OH– ions Expected heat of combustion of
0.025 butane
I + NaOH 1——.01——03—0——05——0 = 0.05 —1—1—3—0—0—2—0—5— = 0.025 = heat of combustion of propane
HNO3 + heat of combustion of carbon
II 1——.01——03—0——05——0 = 0.05 1——.01——03—0——05——0 = 0.05 0.05 + heat of combustion of
HCl + NaOH hydrogen
III 1——.01——03—0——05——0 = 0.05 1——.01——03—0——02——5 = 0.025 0.025 = (–2220) + (–286) + (–394)
CH3COOH +
NH3 = –2900 kJ mol–1
2 (a)
IV 1——.01——03—0——05——0 3 2 = 0.10 1——.01——03—0——02——5 = 0.025 0.025 H H H H H H
H2SO4 + KOH
H – C – C – C – OH; H – C – C – C – H
F V 1——.01——03—0——02——5 = 0.025 —2—.—01—0—3—0—0—2—5— = 0.05 0.025 H H H
O CH3COOH + H OH H
R NaOH propan-1-ol
M propan-2-ol
(b) Reaction II the formation of 0.08 mol of water (b) The temperature rise is 18.2 °C
because propan-1-ol and propan-
5 Reason The largest number will increase the temperature by 2-ol are isomeric, that is, containing
of moles of water molecules — 0—0—.0—.1—8 5.5 = 4.4 °C the same number of carbon,
CHAPTER 4 produced from strong acid and 4 (a) Number of moles of NaOH hydrogen and oxygen atoms.
strong base reaction. The heat energy released
(c) Reaction III used depends on the molecular
Reason 0.025 mol of water = —1—.0—1—0—0——0—5—0— = 0.05 formula and not on the structural
produced from weak acid and 0.05 mol of OH– ions will react formula of the isomer.
weak base reaction. with 0.05 mol of H+ ions to form (c) Increase in temperature
3 Experiment 1 0.05 mol of H2O molecules. = —12—..—12—48— 18.2
Number of moles of CH3COOH Formation of 0.05 mol of water = 36.4 °C
= 1——.0—1——0—0—10——0—0— = 0.1 releases 2.65 kJ of heat. (d) The mass of propan-1-ol decreases
Number of moles of NaOH Heat of neutralisation by two times, that is, from 1.14 g
= —1—.—0—1—0—0——10—0—0— = 0.1 = –2.65 —0——.10—5— to 0.57 g, but the mass of water
CH3COOH(aq) + NaOH(aq) → = –53 kJ mol–1 also decreases by two times, that
CH3COONa(aq) + H2O(l) (b) For weak acid/strong alkali is, from 500 cm3 to 250 cm3.
0.1 mol of CH3COOH reacts with reaction, the heat of neutralisation The rise in temperature is the
0.1 mol of NaOH to form 0.1 mol of is less than –57 kJ mol–1. same, that is, 18.2 °C.
water. Hence, the acid, HA, is a weak (e) Step 1: Calculate the heat change
when 1.14 g of propan-1-ol is
Experiment 2 acid. burnt
Number of moles of CH3COOH (c) The calculation is based on the Heat change (mcθ)
= —0—.—8—————1—0——0 = 0.08 assumption that the apparatus
1000 is well insulated and there is no = –500 4.2 18.2 J
Number of moles of NaOH heat loss to the surroundings. = –38 220 J
= –38.2 kJ
= —0—.—8—1—0—0——10—0—0— = 0.08
Step 2: Calculate the number of
Self Assess 4.5 moles of propan-1-ol used
1 (a) Mass of pentane used. Relative molecular mass of
CH3COOH(aq) + NaOH(aq) → The volume of water heated by propan-1-ol
CH3COONa(aq) + H2O(l) pentane. = (3 12) + 8 + 16
0.08 mol of CH3COOH reacts with The initial temperature and the = 60
0.08 mol of NaOH to form 0.08 mol final temperature of water after Number of moles of propan-1-ol
of water. heating. used
The total volume of solution for the (b) The molecular formula of butane = —1—6.—10—4— = 0.019
two experiments are the same. Thus, is CH3CH2CH2CH3.
Answers 580
Step 3: Calculate the heat of Increase in temperature in (ii) Heat released = mcθ
combustion of propan-1-ol = 100 4.2 (61.5 – 30.0)
The combustion of 0.019 mol of Experiment II
propan-1-ol releases 38.2 kJ of heat.
Heat change when 1 mol of = 51.2 – 30.4 = 13 230 J
propan-1-ol is burnt
= –38.2 —0—.—01—1——9 = –2011 kJ = 20.8 °C. 2 = 13.23 kJ 1
(d) (i) The blue colour of copper(II) (iii) ∆H of combustion
sulphate becomes colourless. 13.23
= —————
A brown precipitate of copper
0.012
is produced. 2
= 1102 kJ mol–1 1
Step 4: Calculate the value of x (ii) The solution must be stirred
(d) The heat of combustion obtained
In the equation given, ∆H = –x kJ continuously to obtain a
from the experiment is less than
refers to 2 mol of propanol burnt. uniform temperature. 1
the theoretical value because
Hence, x = 2 2011 = 4022 (e) (i) Experiment I shows that
• of the loss of heat to the
3 Number of moles of heptane used the release of 1.0 kJ of
surroundings,
= —15—15——02—40— heat energy increases the
1 mol → 5 520 kJ • the combustion of ethanol
2 mol → 1 104 kJ temperature by 2.4 °C.
is incomplete (or, some of
Energy change in Experiment II
= 0.2 the ethanol escaped due to
= 1.0 ——22—0—.4.—8—
Relative molecular mass of heptane evaporation).
(C7H16)
= (7 12) + (16 1) = 100 = 8.67 kJ NaHCO3(aq) + HCl(aq) →
NaCl(aq) + H2O(l) + CO2(g)
Mass of heptane used = 0.2 100 (ii) Relative atomic mass of zinc
= 20.0 g
4 (a) Coal as fuel. = 65.4 3 (a) (i) Heat of neutralisation is the
Advantages: Cost per gram is cheap.
Can be stored easily and safely. Number of moles of zinc used heat energy released when
Disadvantages: Low fuel value. A
= —3—.—2—7 1 mol of H+ ions react with F
lot of residue is produced after 65.4 O
combustion. 1 mol of OH– ions to form R
(b) Natural gas as fuel = 0.05 M
Advantages: High fuel value. 1 mol of water. 1 5
Products of combustion do not Zn(s) + Cu2+(aq) →
pollute the air. (ii)
Disadvantages: Explodes easily if Zn2+(aq) + Cu(s)
not handled properly.
It is a fossil fuel which is non- From the equation, 1 mol of
renewable.
Zn displaces 1 mol of copper
metal.
0.05 mol of Zn displaces
0.05 mol of Cu.
(iii) Displacement of 0.05 mol of
Cu releases 8.67 kJ of heat.
Heat change in the
displacement of 1 mol of Cu
SPM Exam Practice 4 = –8.67 —0—.1—0—5 (b) (i) • The acid must be added
Multiple-choice Questions
1 B 2 C 3 A 4 A 5 D = –173.4 kJ quickly to the alkali to CHAPTER 4
6 A 7 D 8 C 9 D 10 A
11 C 12 A 13 D 14 A 15 A Heat of displacement of Cu prevent heat loss to the
16 A 17 B 18 D 19 B 20 C
21 D 22 D 23 B 24 C 25 C by Zn = –173.4 kJ mol–1. 3 surroundings.
26 C 27 C 28 C 29 A 30 C
31 C 32 D 33 A 34 A 35 A (f) The polystyrene container A • The mixture must be stirred
36 C 37 C
should be used. If polystyrene continuously to ensure
A is used, the surface area of that the temperature is
the solution exposed to the constant.
atmosphere is reduced. This will (ii) Number of moles of nitric
reduce the loss of heat to the acid = 1——0M—0—V—0
Structured Questions surroundings. 2
1 (a) The heat of displacement is the
2 (a) The heat of combustion of = 1——.01——03—0——05——0 = 0.05
heat evolved when 1 mol of
copper is displaced by zinc from ethanol is the heat released
copper(II) sulphate solution. 1
(b) The polystyrene beaker is an when 1 mol of ethanol is burnt Number of moles of ammonia
insulator of heat. The polystyrene
beaker is used to reduce the loss completely in excess oxygen. 1 = 1——0M—0—V—0 = 1——.01——03—0——05——0 = 0.05
of heat to the surroundings. 1
(c) The highest temperature reached (b) C2H5OH + 3O2 → 2CO2 + 3H2O
in Experiment 1 = 32.8 °C. 1
Increase in temperature in HNO3(aq) + NH4+(aq) + OH–(aq)
Experiment 1 (c) (i) Mass of of ethanol used ammonia solution
= 32.8 – 30.4 = 2.4 °C.
The highest temperature obtained = 150.38 – 149.83 = 0.55 g
in Experiment II = 51.2 °C
Relative molecular mass of → NH4NO3(aq) + H2O(l)
C2H5OH = 46 From the equation, the
Number of moles of ethanol number of moles of water
used
molecules formed = 0.05
= —0—.—5—5— = 0.012 (iii) Temperature change
46
2 = 34.2 – 28.2
= 6 °C
581 Answers
∆H = mcθ (iv) The mixture is stirred and the
= –(50 + 50) 3 highest temperature is recorded.
4.2 3 6 J Results
= –2520 J Initial temperature of sodium
= –2.52 kJ chloride solution = t1 °C
Initial temperature of lead(II)
Formation of 0.05 mol of nitrate solution = t2 °C
Maximum temperature of the
water molecules releases reaction mixture = t3 °C
Rise in temperature
2.52 kJ of heat. = t3 – (——t1——+2——t—2 ) = t °C
Assumptions
∴ Heat evolved in the Specific heat capacity for solution
= 4.2 J g–1 °C–1
formation of 1.0 mol of (b) (i) Heat change = mcθ Density of solution = 1.0 g cm–3
Calculation
water molecules = (25 + 25) 4.2 3 Heat change
= mcθ
= 2.52 3 —0—.—10—5— = 630 J = 0.63 kJ 1 = –(25 + 25) 4.2 t J
= –y J = —1—0—–—0y—0— kJ
(ii) Step 1: Calculate the number Number of moles of Pb2+ ions
used
= 50.4 kJ of moles of AgCl precipitated = —0—.—5—10—0—0——0—2——5
= 0.0125
∴ Heat of neutralisation Number of moles of AgNO3 Number of moles of Cl– ions used
between HNO3(aq) and used = 0——.5—1—0—0—0——0—2—5— = 0.0125
NH3(aq) = —0—.—51—0——0—0—2—5 = 0.0125
= –50.4 kJ mol–1
(c) In Experiment I, sulphuric acid Number of moles of NaCl
is a strong acid and potassium used
hydroxide is a strong alkali. Strong = —0—.—51—0——0—0—2—5— = 0.0125
acids and strong alkalis are fully
dissociated in aqueous solution. Ag+(aq) + Cl–(aq) → AgCl(s)
F In Experiment II, nitric acid is a
O strong acid but ammonia solution
R is a weak alkali. A weak alkali is Number of moles of AgCl
precipitated = 0.0125
M only partially dissociated in Step 2: Calculate the heat of
precipitation of silver chloride
aqueous solution. Part of the
5 heat energy released during Precipitation of 0.0125 mol of Pb2+(aq) + 2Cl–(aq) → PbCl2(s)
AgCl releases 0.63 kJ of heat. The equation shows that 2 mol
neutralisation is absorbed of Cl– ions react with 1 mol of Pb2+
Precipitation of 1.0 mol of ions to produce 1 mol of PbCl2.
to produce hydroxide ions. 0.0125 mol of Cl– ions reacts
Consequently, heat of AgCl releases with 0.00625 mol of Pb2+ ions to
neutralisation of nitric acid and form 0.00625 mol of PbCl2.
ammonia solution is less than 0 .63 —0——.0—1—1—2—5— Heat evolved on the precipitation
–57 kJ mol–1. of 0.00625 mol of PbCl2
(d) Both nitric acid and hydrochloric = 50.4 kJ of heat = —1—0—y—0—0— kJ
acid are strong acids. Heat evolved on the precipitation
When the volume of acid Heat of precipitation of AgCl of 1.0 mol of PbCl2
and alkali is increased by two = —1—0—y—0 —0— —0——.0——01—6——25— kJ
(from 50 cm3 to 100 cm3), the is –50.4 kJ mol–1. 2 Heat of precipitation of lead(II)
number of moles of water chloride
molecules produced is also (iii) Rise in temperature = 3.0 °C = —1—0–——0y— 0— —0—.—0—0—1—6—2—5— kJ mol–1
increased by two. This means that = —6—–—.2—y—5— kJ mol–1
CHAPTER 4 the heat released is increased The volume of the reaction
by two. However, the total Energy level diagram
volume of solution also increased mixture remains constant,
by two (from 100 cm3 to
200 cm3), hence the rise in that is, 50.0 cm3.
temperature remains the same,
that is, 6 °C. The concentrations of Ag+ ions
4 (a) (i) Exothermic reaction: and Cl– ions remain constant.
Experiment I. Hence, the temperature rise
remains constant, that is,
3.0 °C. 2
Endothermic reactions: Essay Questions
1 (a) Procedure
Using a measuring cylinder,
25 cm3 of 0.5 mol dm–3
sodium chloride solution is
measured into a plastic cup. The
initial temperature of sodium
Experiments II and III. 2 chloride solution is measured
(ii) X is sodium hydroxide. and recorded. Using another
measuring cylinder, 25 cm3 of
Y is ammonium sulphate. 0.5 mol dm–3 lead(II) nitrate
is measured out. The initial
Z is sodium hydrogen temperature of lead(II) nitrate is
measured and recorded. Lead(II)
carbonate. 2 nitrate solution is poured quickly
into the sodium chloride solution.
(iii) NaHCO3(aq) + HCl(aq) →
NaCl(aq) + H2O(l) + CO2(g)
1
12
Answers 582
(b) CH3OH(l) + —23 O2(g) → hydroxide solution. The mixture Sodium hydroxide and potassium
CO2(g) + 2H2O(l); is stirred and the highest hydroxide are strong alkalis that
∆H= –715 kJ mol–1 temperature is recorded. dissociate completely in aqueous
Results solution.
CH3CH2OH(l) + 3O2(g) → Initial temperature of sodium
2CO2(g) + 3H2O(l); hydroxide = t1 °C NaOH(aq) → Na+(aq) + OH–(aq)
Initial temperature of sulphuric KOH(aq) → K+(aq) + OH–(aq)
∆H = –1371 kJ mol–1 acid = t2 °C
Similarities in combustion Maximum temperature of reaction In the reaction between 1 mol
Both the alcohols burn in excess air mixture = t3 °C of hydrochloric acid and 1 mol
to form carbon dioxide and water. Average initial temperature of sodium hydroxide or 1 mol
Both the alcohols release a large = —21 (t1 + t2) °C of nitric acid and 1 mol of
amount of heat energy when Rise in temperature potassium hydroxide, the change
burnt in excess air. = t3– —12 (t1 + t2) °C = t °C that occurs is the combination of
Difference in heat of combustion Assumptions 1 mol of H+ ions with 1 mol of
The heat of combustion of Specific heat capacity of solution OH– ions to form 1 mol of water
ethanol is higher than the heat = 4.2 J g-1 °C–1 molecules.
of combustion of methanol. This Density of solution = 1.0 g cm–3
is because a molecule of ethanol Calculation H+(aq) + OH–(aq) → H2O(l)
contains one carbon atom and Total amount of heat energy
two hydrogen atoms (that is, released Heat of neutralisation is the heat
the –CH2 group) more than a = mcθ
molecule of methanol. = (100 + 100) 4.2 t J released when 1 mol of H+ ions
Hence, the combustion of 1 mol y
of ethanol produces 1 mol of CO2 = y J = —1—0—0——0 kJ react with 1 mol of OH– ions to
and 1 mol of H2O more than the
combustion of 1 mol of methanol. Number of moles of H2SO4 used form 1 mol of water.
= —0—.—51——0—0——10—0——0 = 0.05
Hence, the heats of neutralisation
for all reactions between a strong F
O
acid and a strong base are the R
M
same, that is, –57 kJ mol–1. 4 5
C(s) + O2(g) → CO2(g); (c) When N2 and H2 react, the bonds
∆H = –x kJ mol–1 between N2 and H2 molecules are
broken and new N – H bonds in
H2(g) + —12 O2(g) → H2O(l);
NH3 are formed. Bond breaking
absorbs heat and bond forming
∆H = –y kJ mol–1 Number of moles of NaOH used releases heat. In this reaction, the
= —1—.—01——0—0——10—0——0 = 0.1
Hence, the heat released on the heat released in bond forming is
combustion of 1 mol of ethanol greater than the heat absorbed in
increases by (1371 – 715) bond breaking. Hence the
= 656 kJ. H2SO4(aq) + 2NaOH(aq) → reaction is exothermic. 4
Na2SO4(aq) + 2H2O(l)
That is, (x + y) = 656 5 Experiments
1 (a) Initial temperature of the mixture
(c) (i) Relative molecular mass of From the equation, 1 mol of
propanol (C3H7OH) = 60 = 29 °C
Fuel value H2SO4 reacts with 2 mol of NaOH Highest temperature of the CHAPTER 4
to produce 2 mol of water. mixture = 41 °C
= ———H—e——a—t—o——f—c—o——m——b—u—s—t—i—o—n—— Increase in temperature = 12 °C
Relative molecular mass 0.05 mol of H2SO4 reacts with 3
0.1 mol of NaOH to form 0.1 mol (b)
= —2—0—1——0
60 of water.
Formation of 0.1 mol of water
= 33.5 kJ g–1 r eleases —1—0—y0——0—kJ of heat energy. Experiment I Experiment II
(ii) The factors are high fuel Heat energy released when 1 Experiment (HCOOH + (HCl +
value and no pollution mol of water is formed NaOH) NaOH)
problem (or low fuel price). = 1——0—y0— —0 —0—1.—1 kJ Initial
temperature
3 of the mixture
(°C)
2 (a) Experiment
• Using a measuring cylinder, 100 Heat of neutralisation for the
cm3 of 1.0 mol dm–3 sodium reaction between sulphuric acid
hydroxide solution is measured and sodium hydroxide Highest
temperature
into a plastic cup. The initial = 1——0–—0y—— 0 —0—1.—1 kJ mol–1 of the mixture
(°C)
temperature of the alkali is
determined and recorded. = —1—–—0y—0— kJ m ol–1
• Using another measuring 12 Increase in
temperature
cylinder, 100 cm3 of 0.5 mol (b) Hydrochloric acid and nitric acid (°C)
dm–3 sulphuric acid is measured are strong acids that dissociate
out and its initial temperature completely in aqueous solution. 3
determined. HCl(aq) → H+(aq) + Cl–(aq) (c) The polystyrene cup gets warmer
HNO3(aq) → H+(aq) + NO3–(aq) when reaction occurs.
• The sulphuric acid is poured
quickly into the sodium
583 Answers
The mercury level in the calorimeter, tripod stand, 5 Chemicals for Consumers
thermometer rises. measuring cylinder
The smell of methanoic acid (e) Procedure Self Assess 5.1
disappears. 3 1 (a) Soaps are sodium or potassium
(d) The heat of neutralisation salts of long chain fatty acids that
between methanoic acid and contain 12 to 18 carbon atoms
sodium hydroxide is lower per molecule.
than the heat of neutralisation (b) Animal fats or vegetable oils,
between hydrochloric concentrated sodium hydroxide or
acid and sodium hydroxide. 3 potassium hydroxide, water and
sodium chloride.
(e) Concentrations of acids and (c) Naturally occurring ester +
3NaOH
sodium hydroxide solution.
↓
Volume of acids and sodium 1 200 cm3 of water is measured 2C17H35COONa + C15H31COONa
hydroxide solution. out using a measuring soap
+
Polystyrene cup 3 cylinder and poured into a CH2OH
(f) In Experiment I, the reaction metal calorimeter. CHOH
2 The initial temperature of
is between a weak acid and a
water is read and recorded.
strong alkali. In Experiment II, the 3 The metal calorimeter is put
reaction is between a strong acid on a tripod stand.
4 A canister containing fuel of
and a strong alkali. Therefore, the
rise in temperature in Experiment
II is higher than 12 °C. 3 brand X is weighed and its
(g) (i) The heat of neutralisation is mass recorded. CH2OH
glycerol
F the heat released when 1.0 5 The canister with fuel of
mol of H+ ions react with 1.0 brand X is then placed below 2 (a) The detergent ion has two parts.
O the metal calorimeter and lit. (i) The ‘tail’ is a long
mol of OH– ions to produce
R 1.0 mol of water molecules. The flame is protected from hydrocarbon chain, C12H25,
M air movement by putting a that is insoluble in water but
3
screen around it as shown in
(ii) ∆H = mcθ dissolves readily in grease.
the diagram.
5 = –50 g 3 4.2 J g–1 °C–1 6 The water in the metal (ii) The ‘head’ is a negatively-
charged ion, –O – SO3–, that
3 12 °C calorimeter is stirred dissolves readily in water but
= – 2520 J continuously with a
Number of moles of methanoic thermometer. does not dissolve in grease.
acid used 7 The flame from the burning (b) When the detergent is dissolved
of fuel X is extinguished
CHAPTER 4 & 5 = —2——.0———3———2—5— = 0.05 when the temperature of in water, the detergent molecule
1000 water has gone up by 30 °C. dissociates into the detergent
Number of moles of NaOH ion (negatively-charged) and the
used 8 The canister containing fuel sodium ion (positively-charged).
= —2——.0———3———2—5— = 0.05 of brand X is weighed again C12H25 – O – SO3–Na+ →
1000 and its mass recorded. C12H25 – O – SO3– + Na+
CH3COOH + NaOH → detergent ion
CH3COONa + H2O 9 The experiment is repeated • When detergent ions come
Number of moles of water using canister of brand Y to
molecules formed = 0.05 replace canister of brand X. into contact with grease-stained
Heat of neutralisation cloth,
(f) Tabulation of data – the ‘head’ dissolves in water
Experimental results
= –———2—5—2——0 = – 50.4 kJ mol–1 because it is hydrophilic,
0.05 Type of fuel Brand X Brand Y – the ‘tail’ dissolves in the
2 (a) Aim
To compare the fuel values of Initial mass of canister greasy stain because it is
fuel X and Y. (g) hydrophobic.
(b) Variables • On agitation, the greasy stain is
Manipulated variable Type of fuel. Final mass of canister lifted from the cloth.
(g) • Finally, the stain is broken up to
Responding variable Heat of Mass of fuel used (g) form tiny oily drops, which are
combustion. Mass of water in metal then suspended in the solution.
Constant variable Volume of calorimeter (g) • When rinsed with water, the
water and the metal container. Initial temperature of cloth becomes clean, because
(c) Hypothesis water (°C) the oily droplets are removed
Fuel X has a higher heat of by water.
combustion than fuel Y. Final temperature of 3 Soap
(d) Substances Water, canisters X water (°C) Advantages
and Y Temperature rise (°C) (a) Effective in soft water
Apparatus 17 (b) Biodegradable, do not cause
Screen, thermometer, metal pollution
Answers 584
Disadvantage Eating food that are not fresh can (ii) The long hydrocarbon chain
Forms scum in hard water cause food poisoning and eating
Detergent food containing unapproved dyes (the alkyl group). 1
Advantages may affect the health.
(a) Effective in both soft water and (iii) ⊕ represents Na+ ions and
Self Assess 5.3 represents the anion
hard water 1 (a) Both aspirin and paracetamol
(b) Different types of detergent can part of the detergent ion. 2
are analgesics, that is, medicine
be synthesied for specific uses that relieves pains. However, (iv) When the detergent particles
Disadvantages aspirin is strongly acidic and can
(a) Non-biodegradable cause bleeding in the stomach. come into contact with
(b) Phosphates in detergents cause In contrast, paracetamol is neutral
in nature and does not cause grease, the negatively-
water pollution stomach bleeding.
(b) Add aspirin and paracetamol charged ‘heads’ of the
Self Assess 5.2 separately to two solutions of
1 (a) Food additives are chemical sodium carbonate. detergent ions dissolve in
For aspirin, effervescence occurs
substances added in small and the gas evolved turns lime- water and the hydrocarbon
quantities to food for specific water cloudy.
purposes, such as preserving food This is because aspirin contains ‘tails’ dissolve in the grease
or enhancing the flavour or smell –COOH group which is acidic.
of food. For paracetamol, no effervescence (Diagram 1 (a)). On agitation,
(b) Any method that prevents the is seen. This is because
growth of microorganisms or paracetamol does not contain the grease particles begin to
retards oxidation is an effective – COOH group.
preservation process. 2 Similarities separate from the cloth
Method 1: By adding preservative • Both penicillin and streptomycin are
By adding a preservative such as organic compounds. (Diagram 1 (b)). 3
sodium nitrate or benzoic acid, • Both penicillin and streptomycin
the growth of microorganisms are antibiotics. They are used to (d) (i) On further agitation, each
such as bacteria or mould is kill bacteria so that illnesses and
retarded. As a result, the food can infections caused by bacteria can be large drop of grease particle
be kept longer. cured.
Method 2: By adding antioxidant Differences is completely separated from
By adding an antioxidant such as Penicillin is used to treat infectious
ascorbic acid, the oxidation of fats diseases such as pneumonia and the cloth and broken into
and oils by atmospheric gonorrhaea. Streptomycin is used
oxygen can be prevented. In this to treat tuberculosis. small droplets and
way, the food does not go rancid. 3 (a) Aspirin is used as a an analgesic
2 (a) Acacia gum acts as a stabiliser (painkiller) to relieve pain. suspended in water. 1
to improve the texture of food (b) Aspirin can cause stomach bleeding.
so that the different ingredients (ii)
(for example, oil and water) can
blend (mix together) properly. –– + – + F
For example, acacia gum is O
added to ice cream so that the +– – –– R
oil and water ingredients can M
mix properly. In this way, tiny ice – – + – – 5
crystals do not form in the ice – –
cream. –
(b) Aspartame is an artificial
sweetener. It is used to enhance –– – –+
the sweetness of food.
– + ––
3 Advantages
Artificial colourings 2 (a) (i) Cleansing agent Y is a
• restore colours lost during food
processing, detergent.
• enhance natural colours of food, and
• make food look more interesting or (ii) Cleansing agent X is a
appetising.
Disadvantages soap. 2
• The food may not be fresh and the
dyes are used to make the food (b) Solution X produces little lather.
look fresh.
• The dyes used may not be the Solution Y produces a lot of
permitted dyes.
lather. 2
(c) The oily stain disappears and CHAPTER 5
lather is formed. 1
(d) Y is a more effective cleansing
agent because Y can remove oily
SPM Exam Practice 5 stains in both hard water and soft
Multiple-choice Questions
1 D 2 A 3 B 4 D 5 A water. In contrast, the cleansing
6 B 7 C 8 B 9 A 10 B
11 D 12 D 13 A 14 D 15 C agent X can only remove oily
16 B 17 C 18 C 19 C 20 B
2 1 B 22 A 23 C 24 C 25 C stains in soft water. 2
26 D 27 A 28 B 29 C 30 D
(e) (i) M anipulated variable Type
of cleansing agent and type
of water.
(ii) Responding variable
Disappearance of oily stains.
Structured Questions (iii) Constant variable Amount of
1 (a) Saponification 1 cleansing agent, amount of
(b) (i) Potassium palmitate 1 oil stains and volume of
(ii) Ester 1 water. 3
(iii) O 3 (a) (i) Sodium lauryl sulphate: To
act as cleansing agent
C – O– 1 Mint oil: To provide a fragrant
O flavour
Saccharin: To make the
(c) (i) – O – S – O– toothpaste taste sweet 3
(ii) A thickening agent is a food
O 1 additive to absorb water and
585 Answers
prevent food from becoming (c) (i) Aspirin dissolves slowly. 3 The mixture is boiled and
liquid. 1 Effervescence occurs as stirred continuously until the
(iii) Tooth decay is caused –COOH reacts with Na2CO3 layer of oil disappears.
by bacteria that converts and a colourless gas is
(B) Precipitation of soap
sugars to acids. Magnesium released. 2 1 100 cm3 of water is added
hydroxide is used to (ii) It is used as an analgesic to to the reaction mixture
neutralise the acid and thus relieve pain. 1 followed by three spatulas of
prevents tooth decay. 1 (d) Aspirin is an acid. It irritates sodium chloride.
(iv) Lauryl alcohol (dodecan-1-ol), the stomach to cause gastric 2 The mixture is boiled for 5
concentrated sulphuric acid problems. In severe cases, it can minutes with constant stirring
and sodium hydroxide. 2 cause bleeding of the stomach and then set aside to cool.
(v) CH3(CH2)10CH2OH + H2SO4 lining. 1 3 When cooled, a white
→ CH3(CH2)10CH2OSO3H +
H2O (e) Aspirin should be taken after food precipitate of soap is
CH3(CH2)10CH2OSO3H + NaOH
→ CH3(CH2)10CH2OSO3–Na+ to reduce irritation of the stomach produced.
+ H2O 2
wall. 1 4 The solution is filtered to
(b) (i) Wind in the stomach 1
(f) Paracetamol 1 obtain soap. The soap is then
(g) The formula of aspirin is C9H8O4. washed with distilled water
Relative molecular mass
and dried. 8
= (9 12) + (8 1) + (4 16) (b) Four types of additives in liquid
(ii) X 1 = 180 detergent are builder, whitening
(iii) Boil part X in water. Drink the Number of moles of aspirin agent, biological enzyme and
ginger water obtained. 1 = —01—.8—3—02— brightening agent.
= 0.00178
4 (a) • To prevent the growth of 1 mol of aspirin contains one Builder: sodium
–COOH group.
F microorganisms Let formula of aspirin be RCOOH. tripolyphosphate
O • To prevent the oxidation of fats 2RCOOH + Na2CO3 →
12RmCOolOoNf aas+pirHin2Orea+ctsCOw2ith —21 mol Function: It is used to soften hard
and oils in food
R • To enhance the taste of food water. In the presence of builders,
M 3 Ca2+ and Mg2+ ions are removed.
Whitening agent:
(b) Gelatin acts as a stabiliser/ sodium perborate
5 thickener. 1 Function: Whitening agents
(c) Analgesics are medicines that remove the colour of dirty stains
relief pain. of Na2CO3. (such as coffee or tea stains) by
Psychotherapeutic medicines Number of moles of Na2CO3
are medicines that change the needed the oxidation process.
When the stain is oxidised, the
mood and behaviour of the = —12 0.00178 colour of the stain disappears.
= 0.00089
patient. 2 0.00089 = —0——.10——50——0—0——V— Biological enzyme: amylase
V = 17.8 cm3 3
(d) Function: To break down protein
Name of Type of stains such as blood stains or
medicine medicine food stains
CHAPTER 5 Aspirin Analgesic Brightening agent:
Barbiturate Antidepressant Blancophor R
Essay Questions Function: It is used to make white
1 (a) Soaps can be prepared in the
Amphetamine Stimulant fabrics whiter and coloured
laboratory by boiling a mixture
Codeine Analgesic of coconut oil and concentrated fabrics brighter. 7
sodium hydroxide solution.
(c) Soaps act as good and effective
4 cleansing agents in soft water.
(e) (i) Can cause bleeding in the However, soaps are not effective
stomach in hard water. This is because
(ii) Can cause addiction 2 the Mg2+ and Ca2+ ions in hard
5 (a) C9H8O4 1 water react with soap ions to
(b) Aspirin molecule dissociates in
form scum. Soaps are also not
water to form H+ ion and the effective in acidic water because
following ion: 1 the H+ ions from the acid react
with soap ions to form a white
precipitate of long chain fatty
The following steps are taken in acid. The advantage of soaps is
the preparation of soap.
(A) Saponification of coconut that they do not cause pollution
oil because they are biodegradable.
1 10 cm3 of coconut oil is
They can be decomposed by
poured into a beaker.
2 50 cm3 of 5.0 mol dm–3 bacteria.
sodium hydroxide solution is Detergents are effective cleansing
added to the coconut oil.
agents in both hard and soft
586
water because they do not form
precipitates with hard or soft
Answers
water or acidic water. However, Antidepressants make a person • Aspartame (artificial sweetener)
detergents cause water pollution. is a good substitute for sugar
For example, the phosphates feel calm and more cheerful. for diabetic patients
in detergents act as nutrients
for water plants and algae and Example: Prozac Disadvantages
speed up their growth. When • Artificial food additives, when
the water plants and algae die, Antipsychotic medicine:
they are decomposed by bacteria consumed in large quantities,
which use up dissolved oxygen. Antipsychotic medicine are may harm our health
Consequently, fish and aquatic • Some food additives can
life will die because of lack of medicine used to reduce the cause allergy, cancer, brain
oxygen. 5 damage and hyperactivity 5
2 (a) (i) Medicine are substances symptoms of mental illness such
Experiment
used for treating illnesses, as fear, hallucination and hearing 1 (a) Aim of experiment
preventing illnesses or
reducing pain or suffering. of voices. To compare the effectiveness of
(ii) Traditional medicine is the cleansing action of a soap
medicine derived directly Example: Chloropromazin 8 and a detergent in soft water and
from plants (leaves, roots hard water
or barks of the plants) or (d) Wrong methods of taking (b) All the variables
animals without being Manipulative variable Type of
chemically processed. 3 medicine include cleansing agent and type of
water
(b) Two examples of traditional • not taking the dosage directed Responding variable Effect of
medicine derived from plants: the cleansing agent on the oily
Garlic and quinine. by the doctor. stain
Garlic Constant variables Amount of
— reduces high blood pressure If a diabetic patient takes a the cleansing agent and oil stain,
— prevents attack of common volume of water, soft water and
colds or influenza higher dosage of insulin than hard water used.
(c) Hypothesis
Quinine – for treating malaria 2 recommended, the patient (i) Both soap and detergent are
(c) Analgesics: Medicine to relieve may suffer or die from insulin effective in soft water.
pain. (ii) Soap is ineffective but
shock.
Examples: Aspirin, paracetamol detergent is effective in hard
and codeine. • not taking the medicine at the water.
Antibiotics: Medicine to destroy (d) Materials
or prevent the growth of proper time. Soap solution, detergent solution,
infectious microorganisms. 1 mol dm–3 magnesium sulphate
Examples: Penicillin and If aspirin or other acidic solution, four pieces of cloth of
streptomycin. the same size and stained with
Psychotherapeutic medicine: drugs are taken on an empty the same volume of oil. F
For treating mental or emotional Apparatus O
illness. stomach, the patient may Beaker, measuring cylinder and R
Types of psychotherapeutic glass rod. M
medicine: Stimulant, suffer from gastritis (irritation of (e) Procedure 5
antidepressant and antipsychotic 1 Four beakers are labelled as
medicine. stomach wall). A, B, C and D.
Stimulants: 2 Beakers A and C are filled
Stimulants are chemical • not completing the whole with the cleansing agent and
substances that can stimulate distilled water.
the activities of the brain and course of treatment. Beakers B and D are
the central nervous system and filled with the cleansing
hence increase alertness and If antibiotics given by the doctor agent, distilled water and
physical activity. magnesium sulphate.
Examples of stimulants are are not taken until the course 3 A piece of oil-stained cloth
caffeine and amphetamine. is placed in each of the
Stimulants stimulate the heart to is completed, the bacteria will beakers and the solution is
beat faster and make a person stirred.
more alert and less tired. develop resistance towards the 4 The observation of the
Antidepressants: effect of soap and detergent
Antidepressants are medicine that antibiotic and cause the on the oily stain for each
decrease the activities of the brain beaker is recorded.
and the central nervous system. antibiotic to be ineffective. 6
3 (a) Food additives are chemical
substances added in small
quantities to food for specific
purposes, such as preventing
food spoilage, or enhancing the CHAPTER 5
flavour, smell and colour of
food. 3
(b) • Azo compound is used as a
dye to give colour to food.
• Citric acid is used as an
antioxidant to prevent the
oxidation of fats and oils by
oxygen in the air. In this way,
food does not become rancid.
• Benzyl ethanoate is an ester
and has a fruity smell. It is
used as artificial flavour.
• Sodium benzoate is used as a
preservative to stop the growth
of microorganisms.
• Sucrose is used as an artificial
sweetener to give a sweet
taste to food. 12
(c) Advantages
• Reduce food spoilage
• Make food taste better,
smell better and look more
attractive
587 Answers
Beaker Chemicals used (c) Fine iron powder
(d) 450 °C
A 50 cm3 of soap solution + 50 cm3 distilled water (e) N2(g) + 3H2(g) → 2NH3(g)
(f) The ammonia gas is cooled to
B 50 cm3 of soap solution + 20 cm3 of MgSO4(aq) + 30 cm3 distilled water
C 50 cm3 of detergent solution + 50 cm3 distilled water liquefy it
(g) To make nitric acid and urea
D 50 cm3 of detergent solution + 20 cm3 of MgSO4(aq) + 30 cm3 distilled water 4 (a) 2Li(s) + 2H2O(l) →
(f) Tabulation of data
2LiOH(aq) + H2(g)
Experiment Solution Observation on oily stain Inference (b)
A Soap + soft water
B Soap + hard water volume of
C Detergent + soft water gas (cm3)
D Detergent + hard water
360 I
17 80 time (s)
SPM Model Test (c) (i) Rate decreases with time
Paper 1 (ii) The quantity of lithium
1 B 2 A 3 D 4 C 5 C 6 D 7 B 8 A 9 A 10 D decreases with time.
F 11 B 12 C 13 C 14 B 15 A 16 D 17 B 18 C 19 A 20 D ( d) Number of moles = —2—43—60—00—0—
2 1 D 22 B 23 A 24 C 25 C 26 A 27 B 28 A 29 D 30 B
= 0.015 mol
O 3 1 D 32 C 33 C 34 A 35 D 36 B 37 A 38 C 39 B 40 D (e)
R 41 B 42 D 43 C 44 A 45 A 46 C 47 B 48 D 49 B 50 B
M Paper 2 volume of gas (cm3)
4 1 (a) P = 2.1; Q = 2.4; R = 2.5; S = 2.6; T = 2.8.3; U = 2.8.7; V = 2.8.8.2 360 II I
& (b) III
5 Pair of elements Formula of compound Ionic or covalent Relative formula mass 180
Q and S QS2 Covalent 12 + 2(16) = 44
P and R P3R Ionic 3(7) + 14 = 35 time (s)
SPM Model Test T and U TU3 Ionic 27 + 3(35) = 132 5 (a) To prevent the KClO3 from mixing
R and U RU3 Covalent 14 + 3(35) = 119 with the metal powders
V and R V3R2 3(40) + 2(14) = 148
Q and U QU4 Ionic 12 + 4(35) = 152 ( b) 2KClO3 heat→ 2KCl + 3O2
T and S T2S3 Covalent 2(27) + 3(16) = 102
(c) R, P, S, Q
Ionic (d) (i) Zinc
(ii) Lead
Note: ZnO is yellow when hot
Note: Metals and non-metals form ionic compounds. and white when cold;
Non-metallic and non-metallic elements form covalent compounds. PbO is brown when hot
(c) H x o Rox xx xo H 1 mol of NaNO3 salt and yellow when cold
H = (23 + 14 + 48) g (e) (i) Copper
= 85 g (ii) Copper(II) oxide
2 (a) Pipette 0.05 mol of NaNO3 salt
= 0.05 x 85 g (iii) 2Cu + O2 → 2CuO
(b) From pink to colourless = 4.25 g Note: Copper(II) oxide is black
(iii) Used as food preservative
(c) (i) 24.00 cm3 (e) The mixture is heated in an (f) Carry out displacement reactions.
(ii) So that the salt obtained is evaporating dish to expel excess A more electropositive metal can
water so as to saturate the
not contaminated with the solution. When the saturated displace a less electropositive
solution is cooled, sodium nitrate
phenolphthalein indicator metal from its salt solution.
salt will form. 6 (a) (i) Antibiotic
(d) (i) NaOH(aq) + HNO3(aq) → 3 (a) (i) From fractional distillation of (ii) Psychotherapeutic medicine
NaNO3(aq) + H2O(l) (iii) Analgesic
liquid air (b) (i) To kill pathogenic bacteria
(ii) Number of moles of sodium (ii) From reaction of methane (ii) Treat psychotic patients
(iii) Relieve pain and fever
hydroxide neutralised, n with steam (c) (i) Tuberculosis
(b) 200 atmosphere (ii) If the dosage is not
= ——25——3——2– mol
1000 completed, the surviving
bacteria will continue to
= 0.05 mol
0.05 mol NaOH will produce
0.05 mol of NaNO3 salt.
Answers 588
multiply and become At anode solutions. 1
resistant to the antibiotic.
A higher dosage of the Silver anode ionises. Each silver The mixture which produces a
antibiotic is needed to kill off
the mutant bacteria. atom releases one electron to white precipitate contains
(d) (i) It is acidic and can cause
intestinal wall bleeding. form silver ion. 1 sulphate ions. 1
(ii) Paracetamol
Ag → Ag+ + e– 1 Barium ions react with sulphate
(e) It can cause dependency /
At cathode ions to form insoluble white
addiction.
The silver ions are then attracted barium sulphate salt according to
to the cathode. the equation:
At the cathode, the silver ion will Ba2+(aq) + SO42–(aq) →
BaSO4(s) 1
discharge by accepting one The othe two solutions either
electron and form silver atom. 1
Section B Ag+ + e– → Ag 1 contain nitrate or chloride ions.
7 (a) (i) Gas P = Oxygen ; The silver atoms will deposit on Each of the two remaining
Gas Q = Hydrogen ; the iron chain. solutions is poured into separate
Gas R = Chlorine ; 8 (a) 25 cm3 of nitric acid is poured test tubes.
Gas S = Hydrogen 3 into a 100 ml beaker. The Dilute nitric acid is added followed
(ii) Dilute sodium chloride solution is heated. 1 by silver nitrate solution. 1
solution Magnesium oxide powder is The mixture which produces a
The ions present in dilute added into the acid solution while white precipitate is the one which
NaCl solution are Na+, H+, stirring the mixture with a glass contains chloride ions. 1
Cl– and OH–. 1 rod until some magnesium oxide Silver ions react with chloride ions
The Cl– and OH– ions move remains undissolved. 2 to produce insoluble white silver
to the anode. The hydroxide The reaction taking place can chloride salt according to the
ions are selected for be represented by the chemical equation: F
O
discharge to produce oxygen equation: Ag+(aq) + Cl–(aq) → AgCl(s) 1 R
M
gas because it is lower in the 2HNO3(aq) + MgO(s) → To confirm the presence of nitrate 4
Mg(NO3)2(aq) + H2O(l) 1 &
electrochemical series. 1 The mixture is filtered to remove ions in the remaining solution, the 5
4OH– → 2H2O + O2 + 4e– 1 following procedure is performed:
the excess magnesium oxide
The Na+ and H+ ions move powder. 1 Dilute sulphuric acid is added to
to the cathode. 1 The filtrate contains the soluble the solution, followed by iron(II)
The H+ ions are selected magnesium nitrate salt. sulphate solution and then
for discharge to produce Now excess sodium carbonate concentrated sulphuric acid is
hydrogen gas because it is solution is poured into the filtrate added. 1
lower in the electrochemical and stirred. 1 A brown ring formed confirms the
series. 1 Double decomposition takes presence of nitrate ions.
2H+ + 2e– → H2 1 place and insoluble magnesium
Concentrated sodium
carbonate salt is formed. 1 Section C
chloride solution
The reaction can be represented 9 (a) (i) An exothermic reaction is one
The Cl– and OH– ions
by the chemical equation: that releases heat energy to SPM Model Test
move to the anode. The
Mg(NO3)2(aq) + Na2CO3(aq) → the surroundings. 1
chloride ions are selected for MgCO3(s) + 2NaNO3(aq) 1
The mixture is filtered to separate Thus the temperature of the
discharge to produce chlorine
surrounding increases. 1
gas because its concentration the insoluble magnesium
In the exothermic reaction,
is higher. 1 carbonate salt. 1
the energy content of
2Cl– → Cl2 + 2e– 1 The salt is rinsed with distilled
products is lower than the
The Na+ and H+ ions move water and then dried by pressing
energy content of the
to the cathode. 1 with dry filter paper. 1
reactants. 1
The H+ ions are selected (b) Each solution is poured into four
for discharge to produce different test tubes. energy reactants
hydrogen gas because it is Dilute nitric acid is added into
lower in the electrochemical each solution. 1
series. The solution which releases gas ∆H = –x kJ mol-1
products
2H+ + 2e– → H2 1 which turns limewater cloudy,
(b) The iron chain is connected to contains carbonate ions. 1
the negative terminal of the cell. Carbonate ions react with
1 hydrogen ions from acid to
A piece of silver metal is connected release carbon dioxide gas (ii) An endothermic reaction
to the positive terminal of the according to the equation: is one that absorbs heat
cell. 1 2H+ + CO32– → CO2 + H2O 1 energy from the
The other three solutions do not
The electrolyte used is aqueous surroundings. 1
silver nitrate solution. show any reaction. Thus the temperature of the
The switch is closed and current Now barium nitrate solution is surrounding decreases. 1
is passed through the solution. 1 added to the three remaining
589 Answers
In the endothermic reaction, the energy content of products is higher than HHH H H H
the energy content of the reactants. 1 H C C C C C HH C C
energy products
H HHH H H
pentane
∆H = +y kJ mol-1
reactants
HHH H H H H CH3 H H CH
H C C C C C HH C C C C HH C C
(b) (Lowest) ∆H2, ∆H1, ∆H3 (Highest) H H H2 H H HHH H H CH
The reactivity series of metals is shown below:
Mg 2-methylbutane
∆ H2 ZAnl HHH H H H H CH3 H H CH3 H
Fe
∆ H1 PSnb ∆H3 H C C C C C H H C C C C H H C C C H
Cu HHH H H HHH H H CH3 H
Hg 2,2-dimethylpropane
Ag 1
3
F The further apart the two metals, the higher the heat of displacement. 1 (b) When a mixture of methane
O The distance between the metals Mg/Ag > Fe/Cu > Al/Zn. 1 and chlorine gas is exposed
R Thus the value of ∆H2 < ∆H1 < ∆H3. 1 to ultraviolet light, substitution
M (c)
reaction will take place. The
thermometer hydrogen atoms in methane will
thermometer be substituted by the chlorine
4 atoms. 2
& CH4 + Cl2 → CH3Cl + HCl
5 methane chloromethane
polystyrene
cup
AgCl CH3Cl + Cl2 → CH2Cl2 + HCl
chloromethane dichloromethane
NaCI(aq) AgNO3(aq)
Procedure CH2Cl2 + Cl2 → CHCl3 + HCl
dichloromethane trichloromethane
(i) The initial temperatures of 25.0 cm3 of 0.5 mol dm–3 silver nitrate solution
and 25.0 cm3 of 0.5 mol dm–3 of sodium chloride were recorded. 1 CHCl3 + Cl2 → CCl4 + HCl
trichloromethane tetrachloromethane
SPM Model Test (ii) The solutions are then mixed in a polystyrene cup and stirred. 1
(iii) The highest temperature attained by the mixture was recorded. 1 4
Tabulation
Initial temperature of AgNO3 solution = a °C
Initial temperature of NaCl solution = b °C (c) (i) X = CnH2n
Highest temperature of mixture = c °C Y = CnH2n + 2 2
2 (ii) X and Y solutions are poured
CAvaelcruaglaetioinnitial temperature of reactants is ——a—+——b— = t °C into two different test tubes.
Bromine water is added into
2 each solution.
Rise in temperature of mixture = (c – t) = T °C 1 The mixture is shaken. 3
1
Heat released in the experiment = 50 3 4.2 3 T joule Observation
1
= 210T joule Solution Observation
= 0.21T kJ X Brown colour of bromine
water is decolourised
Number of moles of silver chloride formed is —0—.—5—3——2—5— = 0.0125 mol
1000
The heat released when 1 mol of AgCl is formed is —0—.—21—T— kJ Y Brown colour of bromine
0.0125 water is not decolourised
= 16.8T kJ
(iii) Hydrogenation
Heat of precipitation of silver chloride is, ∆H = –16.8T kJ mol–1 C6H12 + H2 → C6H14 2
2
10 (a) Isomerism is a phenomena whereby there exists organic molecules with the
same molecular formula with different structural formulae. 2
For example, pentane has three isomers as shown:
Answers 590
Paper 3 Responding variable
1 (a) & (b) Removal of greasy spots
Controlled variable
Time (min) 0 —21 1 1—21 2 2—21 3 3—21 4 Mass of soap and detergent
79.0 82.0 85.0
Temperature 73.0 75.0 77.0 77.0 77.0 77.0 added
(°C) 6 (d) Apparatus
Two 500 ml beakers, glass rod,
(Note : All temperature readings must be in 1 decimal place)
(c) temperature (ºC) fabric stained with five spots of
grease, weighing balance and
90 measuring cylinder
Chemicals
80 5 g of soap and 5 g of detergent
powder, magnesium sulphate salt
70 and distilled water
(e) Procedure
0 1 2 3 4 time (min) 3 • Two portions of 400 ml of
(d) To ensure even heating. 3 distilled water were poured into
two beakers labelled A and B.
(e) The temperature remains constant. • Two spatulas of MgSO4 salt F
were added into each beaker O
Heat supplied by the Bunsen burner is used to overcome the force of attraction and stirred until it dissolved. R
(Dissolution of MgSO4 in water M
between the molecules. 3 produces hard water) 4
• 5 g of soap was added into &
(f) (i) 77 °C 3 beaker A and 5 g of detergent 5
powder was added into beaker
(ii) The substance from the laboratory is contaminated / impure. 3 B.
• Two pieces of fabric stained
2 (a) Mass of oxides added to the acid (c) Magnesium oxide is basic, with five greasy spots are
silicon(IV) oxide is acidic and dropped into each beaker and
and alkali solutions aluminium oxide is the mixtures were stirred with a
glass rod for 10 minutes.
Concentrations and volumes of • The fabrics were removed and
observation recorded.
the nitric and sodium hydroxide amphoteric. 3 (f) Tabulation
solutions 3 (d) (i) Copper(II) oxide
(b) Experiment I : Magnesium oxide (ii) Phosphorus pentoxide
(iii) Tin(II) oxide / lead(II) oxide
is soluble in nitric Beaker A B
Observation (Soap (Detergent
acid solution 3 added) added)
but insoluble in 3 (a) Problem statement Greasy Greasy
spots spots
sodium hydroxide Is the cleansing action of soap remain on removed
fabric from fabric
solution. and detergent affected by hard SPM Model Test
17
Experiment II : Silicon(IV) oxide water?
is soluble in (b) Hypothesis
sodium hydroxide Cleansing power of soap is
solution but lower in hard water whereas the
insoluble in nitric cleansing power of detergent is
acid solution. unaffected in hard water.
Experiment III : Aluminium oxide (c) Variables
is soluble in both Manipulated variable
nitric acid and Addition of soap and detergent to
sodium hydroxide hard water
solutions. 3
591 Answers
Glossary
Form 4 Relative atomic mass Chapter 6: Electrochemistry
Chapter 1: Introduction to Chemistry The number of times an atom of an Anion
Hypothesis element is heavier than one-twelfth of the Negatively-charged ion
A statement that relates a manipulated mass of a carbon-12 atom. Anode
variable with a responding variable. Relative molecular mass An electrode connected to the positive
Constant variable The number of times a molecule of a terminal of a battery or power supply in
The factor that is kept constant throughout compound is heavier than one-twelfth of an electrolytic cell.
an experiment. the mass of a carbon-12 atom. Cathode
Manipulated variable An electrode connected to the negative
The factor that is changed during the Chapter 4: Periodic Table of Elements terminal of a battery or power supply in
experiment. Amphoteric oxide an electrolytic cell.
Responding variable An oxide which react with both acid and Cation
The variable that responds to the change base to form salt and water. Positively-charged ion.
due to the manipulated variable. Catalyst Electrochemical series
Variable A chemical which alter the rate of a An arrangement of metals based on the
A factor that affects the result of an reaction. tendency of each atom to donate electrons.
experiment. Electronegativity Electrolysis
A measure of the tendency of an element A process in which an electrolyte is
Chapter 2: The Structure of the Atom to gain electron and form a negative ion. decomposed by the passage of an electric
Compound Electropositivity current.
A compound is made up of two or more A measure of the tendency of an element Electrolyte
elements bonded together. to lose electron and form a positive ion. A chemical compound which conducts
Element Group electricity in the molten state or in
A substance that cannot be split into The horizontal row of elements in the aqueous solution and undergoes chemical
simpler substances by a chemical reaction. Periodic Table. changes.
Ion Inert Electrolytic cell
A charged particle. The positively-charged Unreactive A cell that consists of two electrodes
ion is called cation whereas the negatively- Period connected to a battery and immersed in
charged ion is called anion. The vertical column of elements in the an electrolyte.
Isotopes Periodic Table. Voltaic cell
Isotopes are atoms having the same Metalloids A cell that produces electrical energy from
number of protons but different number Elements which show metallic and non- chemical energy.
of neutrons. metallic properties.
Nucleon number Chapter 7: Acids and Bases
The total number of protons and neutrons Chapter 5: Chemical Bonds Acid
in the nucleus of an atom. Covalent bond A chemical compound that ionises in
A chemical bond formed from the sharing water to produce hydrogen ions, H+.
Chapter 3: Chemical Formulae and of electrons between two atoms. Alkali
Equations Duplet electron arrangement A chemical compound that ionises in
Chemical formulae A stable electron arrangement in which water to produce hydroxide ions, OH–.
A concise way of writing the exact number the only electron shell is occupied with Base
of atoms of elements that make up a two electrons. A chemical compound that can neutralise
compound. Ionic bond an acid to produce salt and water.
Empirical formula A chemical bond formed by the transfer Concentration
A chemical formula which gives the of electrons from a metal atom to a non- The number of grams or moles of solute
composition of elements in a molecule in metal atom. that is present in 1 dm3 of solution.
their lowest relative proportions. Octet electron arrangement Dilution
Mole A stable electron arrangement in which A process of diluting a concentrated
A mole is the amount of substance which the outermost electron shell is occupied solution by adding a solvent such as water
contains the same number of particles with eight electrons. to obtain a dilute solution.
as there are in 12 grams of carbon-12 Volatile compound End point
isotope. A compound with low boiling point that The point in a titration process when an
Molecular formula can be changed to the vapour state easily. acid-base indicator changes colour to
A formula which gives the actual number Volatility indicate the end of titration.
of atoms of each element present in a The ability of a compound to be changed Molarity
compound. to the vapour state. The number of moles of solute that is
present in 1 dm3 of solution.
Glossary 592
Neutralisation a certain composition in which the major Alkene
A reaction in which an acid reacts with a component is a metal. A family of unsaturated hydrocarbons that
base to produce salt and water only. Catalyst have the general formula, CnH2n.
pH scale A substance that changes the rate of a Amino acids
A set of numbers which measures the chemical reaction. Organic compounds containing the amino
degree of acidity or alkalinity of a solution. Composite material group ((––CNOHO2)Ha)n. d the carboxylic acid
pH value A structural material formed by combining group
A value that measures the concentration two or more materials with different Carbohydrates
of hydrogen ions, H+ in a solution. physical properties, producing a complex Organic compounds that have the
Standard solution mixture with more superior properties. empirical formula, CH2O, and the general
A solution of known concentration. Monomer Cfoarmrbuolxay, lCicn(aHc2iOd )n.
Strong acid A small molecule that make up the An organic acid that has the general formula,
A chemical that dissociates completely to repeating unit of a polymer. DCneHh2yn+d1rCaOtiOonH, where n = 0, 1, 2, 3…
form hydrogen ions, H+ in water. Polymer A chemical reaction in which a water
Strong alkali A large molecule made up of many molecule is removed from the organic
A chemical that dissociates completely to smaller and identical repeating units joined compound.
form hydroxide ions, OH– in water. together by covalent bonds. Diol
Weak acid Polymerisation A saturated alcohol with two hydroxyl
A chemical that dissociates partially to A chemical process in which monomers (–OH) groups on adjacent carbon atoms.
form hydrogen ions, H+ in water. are joined together to form big molecules Esterification
Weak alkali known as polymers.
A chemical that dissociates partially to
form hydroxide ions, OH– in water. The reaction between a carboxylic acid
Titration and an alcohol to produce ester and water.
A quantitative analysis that involves the Form 5 Ester
gradual addition of a chemical solution
from a burette to another chemical Chapter 1: Rate of Reaction An organic compound produced from the
solution of known quantity. reaction between a carboxylic acid and an
Chapter 8: Salts alcohol.
Crystallisation Activation energy Fermentation
A process in which salt crystals are formed The minimum energy that the reactant The chemical process in which
from saturated salt solution. molecules must possess at the time of microorganisms such as yeast act on
Double decomposition collision in order for a chemical reaction carbohydrates to produce ethanol and
A reaction in which two aqueous solutions to occur. carbon dioxide.
react to produce an insoluble salt as a Catalyst Functional group
precipitate and another soluble salt. A substance than changes the rate of a An atom or a group of atoms that
Ionic equation chemical reaction but is itself chemically determines the characteristic properties of
An equation that shows the ions that take unchanged at the end of the reaction. an organic compound.
part in a chemical reaction. Contact process Glycerol
Precipitate The industrial process of making sulphuric An alcohol that contains three hydroxyl
An insoluble solid produced from a acid from sulphur dioxide and oxygen. (–OH) groups per molecule.
solution during a chemical reaction. Effective collision Halogenation
Precipitation The collision that is successful in The addition reaction between an alkene
A reaction used in the preparation of an producing a chemical reaction. and a halogen (chlorine or bromine).
insoluble salt. Haber process Homologous series
Qualitative analysis of salts The industrial process of making ammonia A family of organic compounds that have
A technique used to determine the from nitrogen and hydrogen. the same functional group and similar
cations and anions of a salt. Inhibitor chemical properties.
Recrystallisation A negative catalyst that decreases the rate Hydration
A process used to purify a salt by the of a chemical reaction. The reaction in which a water molecule is
repeat crystallisation process. Rate of reaction added across the carbon-carbon double
Salt Change in amount or concentration of a bond.
A compound formed when the hydrogen reactant or a product per unit time. Hydrocarbons
ion of an acid is replaced by a metal ion
or ammonium ion, NH4+. Chapter 2: Carbon Compounds Organic compounds that contain carbon and
Chapter 9: Manufactured Substances in
Industry Addition reaction hydrogen only.
Acid rain Hydrogenation
Rain that has a pH value between 2.5 A reaction in which an unsaturated The addition of a hydrogen molecule
to 5.0 when acidic gases dissolve in rain organic compound combines with another
water. element of compound to form a single across a carbon-carbon double bond.
Alloy Isomers
A mixture of two or more elements with new compound which is saturated. Organic compounds which have the same
Alcohol
An organic compound that contains the molecular formula but different structural
formulae.
hydroxyl (–OH) group and have the Oils and Fats
general formula, CnH2n+1OH.
Alkane Naturally occurring esters produced from
glycerol and carboxylic acids with long
A family of saturated hydrocarbons that hydrocarbon chains.
have the general formula, CnH2n+2.
593 Glossary
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