1. (a) State one property that can be used to distinguish between a proton and a neutron.
charge / proton has a positive charge; neutron no charge //
Mass of a neutron is higher than that of proton.
(b) An ion of element Y has the formula:
4020Y2+
(i) Write the electron arrangement of the ion.
2:8:8 //
2,8,8
(ii) Identify the group and period in the Periodic Table to which the element belongs.
Group .....II // TWO // 2 √½ ..... (½ mark)
Period ... 4 // FOUR // IV √½ .... (½ mark)
2. (a) Complete Table 1 by writing the formula and naming the structure of the chlorides of the elements.
| Element | Sodium | Magnesium | Silicon | Phosphorus |
|---|---|---|---|---|
| Formula of chloride | NaCl | MgCl2 | SiCl4 | PCl3 / PCl5 |
| Name of the Structure of chloride | Giant ionic / Ionic | Giant ionic / Ionic | Simple molecular | Simple molecular |
(2 marks)
* (b) Select from Table 1 an acidic chloride and write the equation for its reaction with water.
...
(b) Select from Table 1 an acidic chloride and write the equation for its reaction with water.
with √ equation //
PCl3 (l) + 3H2O (l) → H3PO3 + 3HCl (aq)
PCl5 (s) + 4H2O (l) → H3PO4 (aq) + 5HCl (aq)
SiCl4 (l) + 2H2O (l) → SiO2 (s) + 4HCl (aq)
3. (a) Write a thermochemical equation for the formation of carbon(II) oxide. (1 mark)
C(s) + ½O2(g) → CO(g) ΔH = -110 kJ/mol
Determine the enthalpy change of:
(i) formation of carbon(II) oxide (1 mark)
(-110-0) = -110 kJ/mol
(ii) combustion of carbon(II) oxide (1 mark)
-400-(-110)
-400+110 = -290 kJ/mol
4. (a) Give a reason why painting or galvanising iron sheets protects them from rusting. (1 mark)
Creates a coating on the surface of iron preventing penetration of air & moisture
(b) Explain the advantage of galvanising over painting of iron sheets. (2 marks)
In galvanizing, zinc acts as a sacrificial metal since it is a more reactive metal than iron thus preventing rusting; In painting rusting will take place if scratched.
5. (a) The structure of compound A is:
Give its:
(i) name
Decane (1 mark)
(ii) empirical formula C5H11 (1 mark)
(b) Draw the structure of an alkanoic acid whose molecular formula is C5H10O2.
6. The following equilibrium exists in a closed system.
N2O4(g) \(\rightleftharpoons\) 2NO2(g); ΔH = +27.5 kJ
(Pale yellow)
State and explain two conditions under which the intensity of the brown colour of the equilibrium mixture can be increased.
Condition I
Reduction in pressure; Low pressure favours more gas volumes/molecules/production of NO2 is therefore favoured.
Increase in temperature; increase in temperature favors forward reaction for an endothermic reaction (formation of NO2).)
7. (a) Determine the oxidation numbers of:
(i) hydrogen in \(CaH_2\) (1 mark)
\(+2+2H=0\)
\(2H=-2\)
\(H=-1\)
accept -1
(ii) oxygen in \(OF_{2}\) (1 mark)
\(O+2(-1)=0\)
\(O-2=0\)
O=+2
(b) Write an ionic equation for the reaction between aqueous sodium hydrogen carbonate and ethanoic acid.(1 mark)
\(CH_3COOH_{(aq)} + HCO_3^-(aq) \longrightarrow CH_3COO^-(aq) + CO_2(g) + H_2O(l)\)
\(H^+ (aq) + HCO_3^- (aq) \longrightarrow CO_2(g) + H_2O(l)\)
8. The mass of one molecule of a hydrocarbon is \(9.33 \times 10^{-23}\) g.
(Avogadro's number = \(6.0 \times 10^{23}\) mol\(^{-1}\), C = 12.0; H = 1.0)
(a) Determine its:
(i) molecular mass
\(9.33 \times 10^{-23} \times 6.0 \times 10^{23} = 55.98\) g
\(\approx 56\) g
(ii) molecular formula
\(C_4H_8\)
\(C_4H_8\)
(b) Draw a structure of the hydrocarbon in 8(a).
9. (a) Water reacts with hydrogen ions:
(i) write the formula of the product formed (1/2 mark)
\(H_3O^+\)
(ii) Name the type of bond formed (1/2 mark)
Dative / Coordinate
(b) The melting point of iodine is higher than that of chlorine. Explain. (2 marks)
Iodine has a larger molecular size; has more no of Van der Waals forces.
10. A sample of ammonia gas can be prepared by heating a mixture of ammonium bromide and barium hydroxide.
(a) Write an equation for the reaction. (1 mark)
\(2NH_4Br_{(s)} + Ba(OH)_{2(s)} \longrightarrow BaBr_{2(aq)} + 2NH_{3(g)} + 2H_2O_{(l)}\)
(b) State why the gas cannot be dried using anhydrous calcium chloride. (1 mark)
\(NH_3\) reacts with \(CaCl_2\) forming a complex salt of \(CaCl_2 \cdot 2NH_3\).
(c) Name a suitable drying agent. (1 mark)
Calcium oxide / \(CaO\) / Quick lime
11. In an experiment to test for hardness of water from different boreholes, soap solution was added to 1000 cm\(^3\) of water and the volume of soap solution required for lather to start forming recorded.
The results are given in Table 2.
| Water sample (1000 cm\(^3\)) | Volume of soap solution added (cm\(^3\)) | |
|---|---|---|
| Before boiling | After boiling | |
| 1 | 25 | 3 |
| 2 | 12 | 8 |
| 3 | 10 | 10 |
| 4 | 3 | 3 |
| 5 | 25 | 24 |
(a) Select water samples that show:
(i) temporary hardness (1/2 mark)
Sample 1 / 2 / 5
(ii) no hardness (1/2 mark)
Sample 4
(iii) both temporary and permanent hardness (1/2 mark)
Sample 2 / 5
(b) Describe how water hardness can be removed using an ion exchange resin. (1½ marks)
Hard water is run into a column containing the ion exchange resin. \(Ca^{2+}\) & \(Mg^{2+}\) ions are exchanged for \(Na^+\) ions. Thus water coming out from the column is soft.
12. Products of electrolysis at the electrodes for aqueous solutions depend on three factors. Two of these factors are concentration of electrolyte and nature of electrode.
(a) State another factor that affects the products of electrolysis. (1 mark)
Position of the element / ion in the electrochemical series / reactivity series.
(b) Complete Table 3 to show products of electrolysis for dilute calcium chloride and concentrated calcium chloride at the anode and cathode.
Table 3
| Electrolyte | Anode | Cathode |
|---|---|---|
| Dilute calcium chloride | Oxygen | Hydrogen |
| Concentrated calcium chloride | Chlorine | Hydrogen |
Accept correct equations (2 marks)
13. (a) Carbon exhibits different boiling points. Explain. (1 mark)
It exists in different crystalline forms i.e (Diamond & graphite) in the same physical state hence different B.P // exhibits allotropy
(b) It takes 44 seconds for nitrogen(IV) oxide gas to effuse through an opening. Calculate how long it will take for an equal volume of chlorine gas to effuse through the same opening (N = 14.0; O = 16.0; Cl = 35.5).
(2 marks)
(b) It takes 44 seconds for nitrogen(IV) oxide gas to effuse through an opening. Calculate how long it will take for an equal volume of chlorine gas to effuse through the same opening (N = 14.0; O = 16.0; Cl = 35.5).
(2 marks)
\(\frac{T_{Cl_2}}{T_{NO_2}} = \sqrt{\frac{R.M.M_{Cl_2}}{R.M.M_{NO_2}}}\)
\(\frac{44}{x} = \sqrt{\frac{71}{46}}\)
RMM 46 = √2
RMM 71 = √2
\(\frac{x}{44} = \sqrt{\frac{71}{46}}\)
x = 54.66 secs
14. (a) Give an example of a natural polymer made of:
(i) cellulose material (½ mark)
Cotton / Cotton wool / Wood / Sisal / Paper
(ii) a hydrocarbon (½ mark)
Rubber / Latex
(b) Part of the structure of perspex is :
(i)Draw the structure of monomer of perspex
\(CH_2=CH-COOCH_3\)
or
methyl propenoate (methy acrylate)
(ii) Give two properties of perspex that make it suitable for use in making lenses. (1 mark)
Transparent:
Strong/Hard:
Refractive index total does not change.
15. Two allotropes of carbon are graphite and diamond.
(a) Explain why the density of diamond is higher than that of graphite. (1 mark)
Diamond uses all its four atoms to form a more compact tetrahedral structures; graphite uses 3 atoms to form a hexagonal structure. Graphite has hexagonal layers held by weak Van der Waals forces leaving larger gaps between layers hence less dense.
(b) Give one use of each of the allotropes and relate the use to properties of the allotrope.
I. Graphite
Use: Lubricant / Pencil leads / Electrode (½ mark)
Property: Soft & slippery / Conductor of electricity (½ mark)
II. Diamond
Use: Drilling bits (½ mark)
Use: Jewellery (½ mark)
Property
Hard and abrasive
Shiny lustre when polished
Use the graph to determine the:
(i) half-life of the radioactive isotope (1 mark)
85-90 hours
(ii) Rate of decay at time 150 hours
\( \frac{3.0 - 0.5}{80 - 240} = -0.016 \text{ g/hr} \quad (*\text{Ignore the sign}) \)
(0.014 - 0.018)g/hr (1 mark)
(b) The half-life of two radioactive isotopes A and B are 8 days and 5.2 years respectively. Given that both of them emit beta radiation, explain why A would be more suitable in the treatment of a disease. (1 mark)
A has a shorter half-life than B; it will clear from the body faster, hence does not expose the patient to radiation for long.
The formula of a hydrated salt of manganese is \( \text{MnSO}_4 \cdot X\text{H}_2\text{O} \). Given that the salt contains 24.7% manganese, determine the value of \( X \). (\(\text{Mn} = 55.0\); \(\text{S} = 32.0\); \(\text{O} = 16.0\); \(\text{H} = 1.0\)) (3 marks)
\( \frac{55}{\text{RFM}} \times 100 = 24.7 \)
\(\text{RFM} = 222.9\) ✓✓
\(\text{RFM of MnSO}_4 = 151\) ✓✓
\( 151 + 18X = 222.9 \)
\( 18X = 222.9 - 151 \)
\( X = \frac{71.9}{18} = 3.98 \)
\(X \approx 4\) ✓✓
Mass percentages:
\(\text{Mn} = 55 = 24.7\%\) ✓✓
\(\text{S} = 32 = 14.4\%\) ✓✓
\(\text{O} = 16 = 28.8\%\) ✓✓
\(\text{H}_2\text{O} = 18 = 32.2\%\) ✓✓
Final verification using mass ratio:
\( \frac{24.7}{55}, \quad \frac{14.4}{32}, \quad \frac{28.8}{16}, \quad \frac{32.2}{18} \quad \\ \text{(Checked and confirmed)} \)
Answer: \(X = 4\) ✓✓
Alternative:
RFM of MnSO4
55 + 32 + (16 × 4) = 151 ✓
55g = 24.7%
151g = ?
\(\frac{151 \times 24.7}{55}\)
= 67.8% ✓
(100 - 67.8)% = 32.2% ✓
| 67.8 | 32.2 ✓ |
| 151 | 18 ✓ |
| 0.449 | 1.789 ✓ |
1 : 4
\( x = 4 \) ✓
Describe the correct procedure of heating a liquid in a test tube using a Bunsen burner. (3 marks)
Hold using a test tube holder/tongs ✓, keep it slanting ✓ with the mouth facing away ✓ from you ✓. Heat ✓ while rotating ✓.
19. The melting and boiling points of naphthalene are 80°C and 218°C, respectively. A sample of naphthalene was cooled from 250°C to 25°C. On the axes provided, sketch and label the cooling curve that would be obtained. (3 marks)
20. Draw a labelled diagram of a setup that can be used to prepare a dry sample of chlorine gas using potassium manganate(VII) and concentrated hydrochloric acid. (3 marks)
Table 4 gives the boiling points of three liqids
| Liquid | Boiling point (°C) |
|---|---|
| Hexane | 68.7 |
| Butanol | 99.5 |
| Water | 100.0 |
Describe how the following mixtures can be separated:
(a) Hexane and butanol (1½ marks)
Fractional Distillation: Put the mixture in a fractionating column; heat gently. Hexane will distill off first at 68.7°C, leaving behind butanol.
(b) Hexane and water (1½ marks)
Funnel separation / Separating funnel: These are immiscible liquids. Hexane floats on water. Drain water from the bottom.
Accept Decantation.
22. Complete Table 5 by writing the observations made when aqueous ammonia and aqueous sodium sulphate are added to solutions containing calcium, aluminium, and iron(II) ions.
| Ions present | Aqueous ammonia | Aqueous sodium sulphate |
|---|---|---|
| \(\text{Ca}^{2+}\) | No white precipitate | White precipitate |
| \(\text{Al}^{3+}\) | White precipitate, soluble in excess | No white precipitate |
| \(\text{Fe}^{2+}\) | Green precipitate, insoluble in excess | No green precipitate |
23 (a). Iron is extracted from haematite ore. If the ore contains oxides of silicon and aluminium, explain how these impurities are removed. (2 marks)
\(\text{CaO}\) is added to the ore and heated, forming \(\text{CaSiO}_3\) and \(\text{Ca}_3\text{Al}_2\text{O}_6\), which are removed as slag.
Accept correct equations.
23 (b). The extraction process of iron produces waste gases. State how these waste gases can be used to lower the operational cost of the extraction process. (1 mark)
These gases are at high temperatures; the heat can be recycled to preheat the incoming air. \( \text{CO}_2 \) is recycled to form \( \text{CO} \) as a reducing agent.
24. When chlorine is bubbled into a sample of water, the solution smells strongly of chlorine. If aqueous sodium hydroxide is added to the solution, the smell of chlorine disappears.
The following equation shows the reaction that occurs:
\[ \text{Cl}_2(g) + \text{H}_2O(l) \rightleftharpoons \text{HCl}(aq) + \text{HOCl}(aq) \]
With reference to the equation for the reaction, explain why:
(a) Solution smells strongly of chlorine (1 mark)
Chlorine partially dissolves in water; the smell is due to the presence of chlorine molecules. Some chlorine molecules do not react with water, and equilibrium shifts to the left.
(b) Addition of sodium hydroxide removes the smell (2 marks)
\(\text{NaOH}\) reacts with both \(\text{HCl}\) and \(\text{HOCl}\), shifting equilibrium to the right. As a result, chlorine molecules are consumed.
Figure 3 shows how nitric (V) acid can be obtained
(a) Identify the chamber in which a catalyst is used. (1 mark)
Answer: Chamber I.
(b) Name substance Z. (1 mark)
Answer: Nitrogen (II) oxide \(\text{NO}\).
(c) Write an equation for the reaction that takes place in Chamber III. (1 mark)
Balanced chemical equation:
\[ 4NO_2(g) + 2H_2O(l) + O_2(g) \rightarrow 4HNO_3(aq) \]
26. The formula of the complex ion formed when aqueous zinc sulfate reacts with aqueous sodium hydroxide is given as:
\[ [Zn(OH)_4]^x \]
Explain how the value of \( x \) is determined. (2 marks)
Step 1: Determine the oxidation number of Zn:
\[ \text{Oxidation number of Zn} = +2 \]
Step 2: Determine the oxidation number of hydroxide ions (\(OH^-\)):
\[ \text{Oxidation number of } OH^- = -1 \]
Step 3: Apply the oxidation number rule:
\[ +2 + 4(-1) = x \]
Step 4: Solve for \( x \):
\[ +2 - 4 = x \]
\[ x = -2 \]
27. Copper can be obtained from copper(II) oxide using carbon(II) oxide or coke.
(a) Name another reagent that can be used to obtain copper from copper(II) oxide. (1 mark)
Hydrogen gas, Ammonia, Methane, Natural gas
(b) The equation for the reaction with carbon(II) oxide is:
\[ CuO(s) + CO(g) \rightarrow Cu(s) + CO_2(g) \]
Calculate the maximum mass of copper that would be obtained using 200 dm³ of carbon(II) oxide.
Given data:
Step 1: Determine the number of moles of CO used:
\[ \text{Moles of } CO = \frac{\text{Volume of } CO}{\text{Molar volume}} = \frac{200}{24.0} = 8.33 \text{ moles} \]
Step 2: From the balanced equation:
\[ 1 \text{ mole of } CO \text{ produces } 1 \text{ mole of Cu} \]
Therefore, 8.33 moles of CO will produce 8.33 moles of Cu.
Step 3: Calculate the mass of Cu produced:
\[ \text{Mass of Cu} = \text{Moles of Cu} \times \text{Atomic mass of Cu} \]
\[ = 8.33 \times 63.5 \]
\[ = 529.9 \text{ g} \]
Final Answer: The maximum mass of copper obtained is 529.9 g.
I. Moles of CO occupies 24 dm³:
For 200 dm³:
\[ \text{Moles of CO} = \frac{200 \times 1}{24} = 8.333 \text{ moles} \]
From the equation:
\[ \text{Ratio of CO : CuO} = 1:1 \]
So, the moles of Cu produced = 8.333 moles.
Mass of Cu:
\[ \text{Mass of Cu} = \text{Moles} \times \text{Atomic Mass of Cu} \]
\[ = 8.333 \times 63.5 \]
\[ = 529.9 \text{ g} \]
II. 24 dm³ reduces 63.5 g of Cu.
For 200 dm³:
\[ \frac{200 \times 63.5}{24} = 529.2 \text{ g} \]
Final Answer: The maximum mass of copper obtained is 529.2 g.