Page 62 - Spotlight A+ Form 4 & 5 Chemistry KSSM

Answer Chemistry
(iii) Dative bond. The nitrogen atom contributes a lone (b)
pair of electrons to share with the hydrogen ion, H . +
Acidified
8. (a) (i) Add water. potassium
(ii) Ethanoic acid is a weak acid that ionises partially in Reactant Iron(II) sulphate dichromate(VI)
water. Thus, the concentration of H in ethanoic acid solution
+
is low.
(iii) Number of moles of NaOH = 0.5 × 20 = 0.01 mol (i) Role Reducing agent Oxidising agent
1 000
Number of moles of H SO = 0.01 = 0.005 mol (ii) Transfer Donates electron. Accept/receive
2+
2 4 2 of electron Iron(II) ion, Fe electron.
2–
Molarity of H SO = 0.005 = 0.2 mol dm –3 donates electron Cr O ion
7
2
2 4 0.025 to produce accepts electrons
3+
©PAN ASIA PUBLICATIONS
(b) (i) K: Copper(II) carbonate iron(III) ion, Fe . to produce
L: Copper(II) oxide chromate ion,
3+
M: Carbon dioxide Cr .
W: Copper(II) sulphate (iii) Colour Pale green to Orange to green
3
(ii) Add 2 cm of hydrochloric acid, HCl and follow by change brown/
2 cm of barium chloride, BaCl solution. A white yellowish-brown
3
precipitate is formed. (c)
Section B Matter Cell P Cell Q
9. (a) (i) Change in the volume of hydrogen gas per second
// The volume of hydrogen gas released per second. (i) The energy Chemical Electrical energy
(ii) Temperature; Concentration of hydrogen ion changes. energy to to chemical
(b) Zn + 2HCl → ZnCl + H 2 electrical energy energy
2
(c) Number of moles of HCl= 50 × 0.5 = 0.025 mol (ii) The electron Electrode A to Electrode C to
1 000 flow electrode B electrode
Number of moles H = 0.025 = 0.0125 mol through D through the
2 2 the external external
Volume of H = 0.0125 × 24 dm = 0.3 dm 3 circuit. circuit.
3
2
= 300 cm 3 (iii) The product Magnesium ion, Oxygen, O
3
(d) Volume of hydrogen gas (cm ) formed at Mg 2+ 2
anode
Experiment III ANSWER FORM 5
(iv) Half Cu (aq) + 2e – Cu (aq) + 2e –
2+
2+
Experiment II equation at → Cu(s) → Cu(s)
cathode
Experiment I (v) The colour Blue to The blue
change of the colourless solution remains
Time (s) solution unchanged.
(e) Experiments I and II:
The rate of reaction in Experiment II is higher. The Section C
temperature of HCL acid in Experiment II is higher. H + 11. (a) (i) C H
2n + 2
n
ions in Experiment II has more kinetic energy and move (ii) Prercentage of carbon in Q = 72 × 100%
+
faster. The frequency of collision between H ions and 86
zinc atoms in Experiment II is higher. The frequency of = 83.72%
effective collision in Experiment II is higher. Prercentage of carbon in R = 72 × 100%
(f) Experiments II and III: 84
= 85.71%
The rate of reaction in Experiment III is higher. The Prercentage of carbon by mass in R is higher.
concentration of H ions in Experiment III is higher (b) (i)
+
because H SO is a diprotic acid while HCl is a monoprotic
4
2
+
acid. The number of H ions per unit volume solution Compound Homologues series
in Experiment III is higher. The frequency of collision
between H ions and zinc atoms in Experiment III is V Alcohol
+
higher. The frequency of effective collision in Experiment W Carboxylic acid
III is higher.
10. (a) Metal Q: Copper X Alkene
Oxidation half-equation: Y Ester
Cu(s) → Cu (aq) + 2e –
2+
Reduction half-equation: (ii) Functional group V: Hydroxyl group
–
Ag (aq) + e → Ag(s) Functional group X: Carbon-carbon double bond
+
577