- Give the name and formula of;
- A complex cation containing a transition metal (1mark)
- A complex anion containing a transition metal (1mark)
- The diagram below shows a set up of apparatus used to prepare oxygen gas and pass it over burning candle. The experiment was allowed to run for several minutes.
- Identify liquid M. (1mark)
- Write an equation for the reaction that forms oxygen gas in the set up. (1mark)
- The pH of the solution in flask II was found to be less than 7. Explain. (1mark)
- During heating of a hydrated copper (II) sulphate crystals, the following readings were got.
Mass of evaporating dish = 300g
Mass of evaporating dish + hydrated salt = 305g
Mass of evaporating dish + dehydrated salt = 303.2g
Calculate the empirical formula of hydrated copper (II) sulphate. (Cu = 63.5, S=32, O=16, H=1) (3marks) -
- Identify the following cleansing agents. (1 mark)
- State one disadvantage of using the cleansing agent in (a) (ii) above. (1mark)
- Identify the following cleansing agents. (1 mark)
- Study the energy level diagram below and answer the questions that follow.
- Identify the enthalpy changes represented by
ΔH1 ... ½ mark
ΔH2 ... ½ mark
ΔH4 ... ½ mark - Given that
ΔH1 = + 434 KJ/Mol
ΔH2 = + 371 KJ/Mol
ΔH3 = + 483 KJ/Mol
ΔH4 = - 781 KJ /Mol
Calculate the enthalpy change (Δ H) for the reaction. (1½ marks)
Na(s) + ½ Cl2(g) -> NaCl(s)
- Identify the enthalpy changes represented by
- Giving a reason in each case, identify an acid and a base in the equation.
H3O+(aq) + NH3(aq) → NH4+(aq) + H2O
Acid ........½ mark
Reason ..........½ mark
Base ... ½ mark
Reason ... ½ mark - Study the reaction scheme below and answer the questions that follow.
- What property of the white precipitate R is demonstrated by steps I and II. (1 mark)
- If the metal ion in solution W is divalent suggest its identity. (1mark)
- Write an ionic equation for the reaction taking place in step I. (1 mark)
- The diagram below shows a set up that was used to electrolyse aqueous magnesium chloride.
- On the diagram above, using arrows, show the direction of flow of electrons. (1mark)
- Identify the syringe in which oxygen gas would be collected. (1mark)
- State and explain the change of magnesium chloride concentration at the end of the experiment. (1 ½ marks)
- Study flow chat below and answer the questions that follow.
- Write a chemical equation to show how solid R is formed. (1mark)
- Write observation made in process (II). (1mark)
- The table below shows the standard reduction potentials for four half–cells. Study it and answer the questions that follow (letter are not the actual symbols for the elements)
(Volts)
F2(aq) + 2e → 2F-(aq) +0.54
G2+(aq) + 2e → G(s) -0.44
H2+(aq) + 2e → H(s) +0.34
2J+(aq) + 2e → J2(g) 0.00- Identify the strongest reducing agent. (1mark)
- Write the equation for the reaction which takes place when solid G is added to a solution containing H2+ ions. (1 mark)
- Calculate the E0 value for the reaction in (ii) above. (1mark)
-
- State the Charles’ law. (1mark)
- A certain mass of gas occupies 146dm3 at 291K and 98.31kPa. What will be its temperature if its volume is reduced to 133dm3 at 101.325 kPa? (2 marks )
- Below is the structure of a monomer used in polymerization.
- Draw the structure of part of the polymer involving 3 units. (1mark)
- What type of polymerisation takes place? (1mark)
- Give one advantage of artificial fibres over natural ones. (1mark)
- 20.0cm3 of NaOH solution containing 8.0gdm-3 were required for complete neutralization of 0.118g of a dibasic acid. Calculate the Relative Molecular Mass (R.M.M) of the acid. (Na=23, O=16, H=1) (3marks)
- The table below shows the results obtained when a current of 2 amperes is passed through copper II sulphate solution for 15 minutes.
Initial mass of cathode 1.0g Final mass on cathode 1.60g Change in mass at the cathode 0.60g
Calculate the quantity of electricity required to deposit one mole of copper (Cu= 63.5) (2marks) - The following equation shows a reversible reaction.
H2(g) + Br2(g) ⇌ 2HBr(g) ΔH = -74.4kJ
reddish brown colourless- State and explain the observation that can be made when:-
- Temperature is increased. (1½marks)
- Pressure is reduced. (1½marks)
- State and explain the observation that can be made when:-
- You are provided with:
A clean metallic spatula
Distilled water
Lead (II) nitrate solution
Source of heat
A rack of test – tubes
A white solid suspected to be sodium chloride
Required
Draft a procedure you would use to enable you test and confirm that the white solid is a chloride compound. (3marks)
Test Observation Inference - Sulphur is one of the elements with varying oxidation States
- What is oxidation state? (1mark)
- Determine the oxidation state of sulphur in SO32- (1mark)
- Give the electron pattern of sulphur in SO32- (1mark)
- When fuels burn in the internal combustion engine at high temperature, one of the products formed is nitrogen (II) oxide.
- Write the equation for the formation of nitrogen (II) oxide. (1mark)
- Give a reason why nitrogen (II) oxide is not formed at room temperature. (1mark)
- Describe how formation of nitrogen (II) oxide in the internal combustion engine leads to gaseous pollution? (1mark)
- Use part of the periodic table grid below to answer questions that follow. (Letters do not represent actual symbols of the elements)
- Which is the most reactive non-metal? Explain. (1 mark)
- Name the bond type formed when element A and B react. Explain. (1 mark)
- Giving a reason compare the atomic radius of element D and E. (1mark)
- Use the standard enthalpies of combustion given below to calculate the enthalpy of formation of ethane.
C(s) + O2(g) → CO2(g) ΔHC = -394Kjmol-1
H2(g) + O2(g) → H2O(l) ΔHC = -286Kjmol-1
C2H6(g) + 5O2(g) → 2CO2(g) + 3H2O(l) ΔHc = -1300Kjmol-1 (2 ½ marks) - Describe how a solid sample of Lead (II) Sulphate would be prepared using the following reagents. Dilute Sulphuric (VI) acid, Nitric (V) acid, solid lead (II) Carbonate. (3marks)
- Study the flow chart below and answer the questions that follow
- Identify
- Solid X (½ mark)
- Process W (½ mark)
- Write an equation for the reaction in step II. (1mark)
- Explain why Copper is suitable in making soldering equipment. (1mark)
- Identify
- The table below gives the rate of decay for a sample of radioactive element P.
Mass of P(g) Number of days 48
18
60
90
180- Determine its half-life (1mark)
- Complete the following nuclear equation (1mark)
146C → ___ + 0-1e
- Complete this set up below for the preparation and collection of dry Hydrogen Sulphate. (3marks)
- Using dots (.) and crosses (x) to represent elections draw diagram to represent.
- NH4+ (1½ mark)
- Mg3N2 (1½ mark)
- A sample of carbon (II) oxide gas was passed through hot iron (III) oxide in a combustion tube P.
- State the observation made in Tube P. (1 mark)
- Write a chemical equation for the reaction that took place in Tube P. (1 mark)
- A student performed three experiments on the reaction of magnesium with dilute hydrochloric acid and collected the hydrogen gas evolved for each experiment.
- Experiment I: Reacted 2g of magnesium ribbon with 1M hydrochloric acid.
- Experiment II: Reacted 2g of magnesium powder with 1M hydrochloric acid
- Experiment III: Reacted 2g of magnesium ribbon with 0.5M hydrochloric acid
On the grid below sketch the graphs for each of the three experiments performed. (3marks)
- In the manufacture of sodium carbonate by solvay process, ammoniated brine trickles down the carbonator while carbon (IV) oxide rises up the same tower.
- What is ammoniated brine? (1/2 mark)
- What is the main source of carbon (IV) oxide in the above process? (1/2 mark)
- Write two equations for the reactions in the carbonator (2 marks)
- The table below shows the solubility of a salt at various temperatures.
Temperature 0C Solubility (g/100g water) 0 36 40 30 80 25 100 22 120 20
Define the term Fractional Crystallization. (1 mark)- A saturated solution of the salt at 400C was heated to 1000C. State and explain the observation made. (1mark)
- Calculate the mass of salt formed when a saturated solution of the salt at 00C is place in a water bath maintained at 1000C. (1 mark)
Marking Scheme
-
- [Zn(NH3)4]2+√ 1 / Tetraamine zinc (II) ions
- [Zn(OH)4]2-√ 1 /Zincate ion
-
- water//H2O
- 2Na2O2(aq) + 2H2O(l) 4NaOH(aq) + O2(g)
- Carbon (IV) oxide formed from burning candle dissolved in the water to form a weak carbonic acid (pH less than 7).
- Mass of hydrated salt = 305 -300 =3g
Mass of anhydrous salt = 303.2 -300=3.2g
Mass of water of crystallization= 5-3.2 =1.8g
CuSO4 : H2O
3.2 1.8
159.5 18 Formula: CuSO4. 5H2O
0.02 0.1
1 5 -
-
- Soapy
- Soapless
- Is non-biodegradable thus environmental pollutant
-
-
- ΔH1 Atomization energy of Na
ΔH2 Ionization energy for Na
ΔH4 Lattice energy - ΔH = ΔH1 + ΔH2 + ΔH3 + ΔH4
= 434 +371 + 483 + -781
= 507 kJ/mole
- ΔH1 Atomization energy of Na
- Acid: H3O+ Reason: it donates a proton (H+) to NH3 to form NH4+
Base: NH3 Reason: accepts the donated H+ -
- It is amphoteric√ 1
- Lead ion√ 1 // Pb2+
- Pb(OH)2 (aq) + 4OH-(aq) → [Pb(OH)4]2-(aq)√ 1
-
- from left right
- Syringe 1
-
- Concentration increases
- In electrolysis of dilute MgCl2, Hydrogen and Oxygen are discharged (H2O) //equivalent to discharge of water molecules
-
- Na2CO3(s) + Cu(NO3)2(aq) → CuCO3(s) + 2NaNO3(aq) √1
- Blue precipitate formed √ ½ which dissolve to give deep blue solution. √ ½
-
- G √1
- G(s) + H2+(aq) → G2+(aq) + H(s) √1
- Ered –Eox
=+0.34 –(-0.44) √1
= 0.78V√1
-
- Volume of a fixed mass of an gas is directly proportional to its absolute temperature provided pressure is kept constant
-
-
- Addition√1
-
- Cheaper √1
- Not easily attacked by chemicals
- Do not Grease
- Are lighter
- Dry faster.
-
- molarity of NaOH = 8/40
=0.2M√ 1
Moles of NaOH
= 0.2 x 20
1000
= 0.004moles√ 1
Moles of acid =0.002moles
RMM =0.118/0.002
=56√ 1 - Equation: Cu2+(aq) + 2e → Cu(s)
1 mole (63.5g) Cu is discharged by 2 x 96500 C
0.6g will be discharged by
0.6 X 2 X 96500
63.5
=1823.622 C -
- Forward reaction is exothermic, thus increase in temperature shifts equilibrium to the left (backward reaction favoured) increasing the intensity of the red-brown colour of Br2 gas.
- The ratio of volumes of reactants to products is 1:1 therefore, decrease in pressure will have no effect on the equilibrium state of the reaction
-
Test Observation Inference Put less than half spatula end-full of the white solid in a test tube. Add about 5ml of distilled water and shake thoroughly. White solid dissolves to form a colourless solution -Soluble substance.
-coloured ions absent; Cu2+, Fe2+, and Fe3+To a small portion of solution formed above, add 3 drops of lead
(II) nitrate solution. Warm the mixture and allow it to coolWhite precipitate on addition of lead (II) nitrate which dissolves on warming and reappears on cooling Cl- confirmed present -
- A number showing a charge a species has in a compound or valency of an element or ion bearing the charge.
- S +3(-2) = -2
S – 6 =-2
S= +4 - S+4 has electron pattern 2.82
-
- N2(g) + O2(g) → 2NO (g)√ 1
- a lot of heat is required to break the triple bonds between the nitrogen atoms√01
- NO is readily oxidized to form nitrogen(IV)oxide which is responsible for acid rain√01
-
- C ; it is the most electronegative//it easily gain electrons// has smallest atomic radius
- Covalent bond. They react by sharing valence electrons
- E has smaller atomic radius than D. Across the period, nuclear force of attraction increases leading a stronger pull on the energy levels reducing overall size.
-
- React lead (II) carbonate in excess √ ½with dilute nitric acid until effervescence stops, √ ½ filter √ ½out the unreacted carbonate. To the filtrate add dilute sulphuric acid√ ½ to precipitate out lead sulphate, filter the mixture to obtain residue√ ½,wash with distilled water and dry√ ½ between filter papers
-
-
- Solid X: Iron silicate// FeSiO3
- Process W: Electrolysis
- 2CuFS2(s) + 4O2(g) → Cu2S (s) + 2FeO(s) + 3SO3(g)
- Copper has very high thermal conductivity
-
-
- Half life 48 → 24 → 12 → 6
3 half lives 180 days
1 half life ?
180 X1
3
=60 days - 146C → 147N + 0-1e
- Half life 48 → 24 → 12 → 6
-
-
-
- The red brown iron(III) oxide turns to a grey solid (Fe)
- Fe2O3(s) + 3CO(g)→ 2Fe(s) + 3CO2(g)
-
-
- Sodium chloride saturated with Ammonia√ ½
- Heating limestone/calcium carbonate √ ½
-
- NH3(aq) + CO3(aq) + H2O(l) → NH4HCO3(aq) √1
- NH4HCO3(aq) + NaCl(aq) → NH4Cl(aq) + NaHCO3(s) √1
-
- Separation of salts in a mixture due to difference in their solubilities at different temperatures
- Some crystals of the salt were deposited. The solublity of this salt decreased with increase in temperature
- Mass deposited 36 -20 = 16g
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