Chemistry Paper 2 Questions and Answers - Momaliche Pre Mock Exams 2023

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INSTRUCTIONS TO CANDIDATES

Answer all the questions in the spaces provided after each question.
Mathematical tables and non-programmable electronic calculators may be used.
ALL working must be clearly shown where necessary.

The grid below forms part of the periodic table. Study it and answer the questions that follow. The letters do not represent the actual symbols of the elements
1. Write the general name given to the element P belong. (1mark)
2. An element N has an atomic number of 15. Write down its electronic arrangement and hence fix it in its right position on the grid above. (1mark)Electronic arrangement
3. Compare the size of the atom of R and that of its ion. Explain your answer. (2mks)
4. Give the formula of the compound formed between (1mark)
  1. P and W …………………………………………………………………………
  2. T and Y …………………………………………………………………………..
5. Compare the melting points of element Q and S. Explain (2Mks)
6. State the least reactive element in the grid. Give a reason for your answer (1mk)
7. Give two advantages that element S has over element Q in making electric cables (2mks)
8. Draw (a) dot (.) and cross (x) diagram to represent the bonding in compound formed between T and Y (2 marks)
1. Study the table below and answer the questions that follow
  Bond type bond energy kJmol^-1
  C-C 346
  C = C 610
  C-H 413
  C-Br 280
  Br-Br 193
  1. Calculate the enthalpy change for the following reaction (3 marks) C₂H_4(g) + Br_2(g) C₂H₄Br_2(g)
  2. Name the type of reaction that took place in (a) above. (1mark)
2. Butane C₄H₁₀ cannot be prepared directly from its elements but its standard heat of formation () can be obtained indirectly.
  The following heats of combustion are given.
  (Carbon) = -393kJ/mol
  (Hydrogen) = -286kJ/mol
  (Butane) =-2877kJ/mol
  1. Draw an energy cycle diagram linking the heat of formation of butane with its heat of combustion and the heat of combustion of its constituents elements. (2mk)
  2. Calculate the heat of formation of butane (C₄H₁₀) (2mks)
3. Given that the lattice enthalpy of potassium chloride is +690kJ/mol and hydration enthalpies of K⁺ and Cl^- are -322kJ and -364kJ respectively. Calculate the enthalpy of solution of potassium chloride. (3 mks)
The diagram below represents the Haber process for the manufacture of ammonia. Study it and answer the questions that follow.
1. Name any two impurities removed by the purifier. (1mark)
2. The catalyst used in the process is finely divided iron. Why iron is finely divided? (1mk)
3. In the Haber process the conversion of nitrogen and hydrogen into ammonia is only 10%.
  The remaining unreacted gases are recycled. What is the advantage of this? (1mk)
4. A part from iron catalyst and pressure of 500 atmospheres, name any other condition required for this process. (1mk)
5. Give any two uses of ammonia (1mk)
6. In the manufacture of nitric (v) acid from ammonia and air, ammonia is catalytically oxidized to nitrogen (ii) oxide
  1. Name the catalyst used in this reaction. (1mk)
  2. Write a balanced chemical equation for the reaction between ammonia and air. (1mk)
  3. State one environmental problem likely to be faced in an area where nitric (v) acid manufacturing plant is located. (1mark)
7. 1. In the preparation of chlorine gas in a school laboratory, either manganese (IV) oxide or potassium manganate(VII) may be used on concentrated hydrochloric acid. State one advantage of potassium manganate (VII) over manganese (IV) oxide in this reaction.(1mark)
  2. State and explain what would be observed when dry litmus papers are dipped in a gas jar of chlorine.(1mark)
  3. Freshly prepared chlorine water bleaches but chlorine water exposed to sunlight for sometime does not bleach. Explain.(2marks)
  4. When preparing hydrogen chloride gas from sodium chloride and sulphuric (VI) acid, two conditions are necessary. State the conditions.(1mark)
A label on the bottle containing Sulphuric (IV) acid has the following information
- Density = 1.836 g/cm3
- Percentage purity = 98%
- Relative formula mass = 98
  1. Calculate:
    1. The concentration of the acid (3 mks)
    2. The volume of concentrated sulphuric (IV) acid that should be diluted to produce 2 litres of 2 M Sulphiric (IV) acid (2 mks)
  2. A solution of sodium hydroxide was found to contain 12.4g/dm³ of sodium hydroxide. 25cm³ of this solution reacted with 15cm³ of a solution of sulphuric (VI) acid. (Na=23.0, H=1.0, S=32.0, O=16.0)
    1. Find the molarity of the sodium hydroxide solution. (1 mark)
    2. Calculate the number of moles of sodium hydroxide solution used. (1 mark)
    3. Calculate the number of moles of the acid used. (1 mark)
    4. Determine the concentration of the sulphuric (VI) acid solution in g/dm³. (3marks)

Define a saturated solution. (1 mark)

The table below represent the solubilities of sodium nitrate and Sulphur (IV) oxide at different temperatures.

Temp(°C)	10	18	26	34	42
Solubility of Sodium Nitrate (g/100g of water)	20	29	40	53	68
Solubility of Sulphur (IV) Oxide (g/100g of water)	78	55	45	40	36

On the grid provided below, plot a graph of solubilities of sodium nitrate and Sulphur (IV) oxide against temperature. (4 marks)
Using the graph;

Determine the solubility of Sulphur (IV) oxide at 16^oC. (1 mark)
The concentration, in moles per litre, of sodium nitrate at 16 ^oC. (assume density of solution is 1 g/cm³) (Na=23, 0=16, N=14). (3 marks)
Mass of crystals formed when a solution of sodium hydroxide is cooled from 40^oC to 26^oC.
(2 marks)
What is the relationship between solubility of sodium nitrate and temperature? (1 mark)

Give one advantage of hard water. (1 mark)

The diagram below shows the apparatus for the preparation of gas A and investigate on its properties . Study it and answer the questions that follow.
1. 1. Name gas A. (1 mark)
  2. suggest property of gas A under investigation (1mark)
  3. Write chemical equations for the reactions in the;
    1. Boiling tube I (1 mark)
    2. Combustion tube II (1 mark)
2. 1. State and explain the observation made in
    1. Tube I. (1 mark)
    2. Combustion tube II (1 mark)
3. 1. What is the use of hydrated copper (II) sulphate in the experiment? (1 mark)
  2. Name one other substance that comes out of tube III. (1 mark)
  3. Name liquid W. (1 mark)
  4. What is the role of sodium chloride in the ice (freezing mixture) (1 mark)
Study the condensed formulae below and answer the questions that follow
1. 1. CH₃CH (CH₃) CH₂CHCH₂
  2. CH₃CHCH(CH₃)CH₃
    1. Draw the structural formula of each of the compounds I and II. (2mks)
    2. Give the systematic name of each of the compounds represented by the formulae above
      (2mks)
    3. To which homologous series does the compound represented by I belong (1/2 mk)
2. The flow chart below shows some reactions starting with a long chain alkane. Study it and answer the questions that follows.
  1. Name substance. (1_1/2mks)
    1. A___________________________
    2. B___________________________
    3. C____________________________
  2. What is the name given to the process represented by
    1. Step I_______________________ ( ½ mk)
    2. Step III______________________ ( ½ mk)
    3. Step IV______________________ ( ½ mk)
    4. Step VI______________________ (½ mk)
  3. Write down the chemical equation represented by the reaction in step VI (1mk)

MARKING SCHEME

1. Alkali metals
2. 2.8.5
3. R has a bigger atomic size kthan its ions because R looses electrons hence an energy level lost therefore making the size of its ion to be small than the atom.
4. 1. P₂W
  2. TY₄
5. S has higher melting point than Q. This is because S has stronger metallic bond than. Q (3 electrons in outermost energy level requiring a lot of heat to break the bond in S than Q
6. M. It is stable doesn't need to react with other elements to gain stability.
7. 1. It is ductile
  2. Does not corrode
  3. Has higher electrical conductivity.
1. 1. Bond breaking
    C=C-1
    C-H=4
    Br-Br-1
    610 + 4(413) + 193
    = 2455
    Bond formation
    C-C=1× 346
    C-Br - 2× 280
    C-HF 4 × 413
    = j2558
    ΔH = 245J − 2558
    = − 103kJmol⁻¹
  2. Expthermic reaction/ Addition Reaction
2. 2. ΔH₁ = ΔH_F + ΔH₂
    ΔH_F = ΔH₁ − ΔH₂
    4(−393) + 5 (−286) −(−2877)
    − 30002 + 2877
    = − 125KJmol⁻¹
3. ΔH_Soln = ΔH_Latt + ΔH_hyd
  = 690 + 322 + −(364)
  = 690 − 686
  = +4kJmol⁻¹
1. - Carbon(IV)Oxide
  - Sulphur(IV) Oxide
  - Dust particles
    Any two ½mk each.
2. To increase Its surface Area
3. Reduces wastage
4. Temperature of 1450-500ºC (No range)
5. - As nitrogenous fertilizer
  - Softening of hard water
6. 1. Platinum - rhodium catalyst
  2. 4NH_3(g) + 5O_2(g) → 4NO + 6H₂O_(l)
  3. NO reacts nwith O₂ to form NO₂ which dissolve in moisture forming HNO₃ falls as acid rain which causes death of sea life.
7. 1. Heat is not required.
  2. There would be no change in both litmus mpapers. Dry chlorine doesn't have acidic property hence do not bleach.
  3. Freshly prepared chlorine water has chloric(I) acid therefore bleaches. But when exposed to sunlight Chloric(I) acid decomposes into hydrochloric acid and oxygen gas is released.
  4. 1. Heat
    2. The acid must be concentrated.
1. 1. Mass of acid in 1cm³ = 1.836g
    Mass of acid in 1000cm³
    1000 × 1.836 = 1836g
    1
    Mass of pure acid = ⁹⁸/₁₀₀ × 1836
    = 1799.28g
    M = Mass
    RMM
    = 1799.28
    98
    = 18.36M
  2. M₁V₁ = M₂V₂
    M₁ = 18.36
    V₁ = ?
    M₂ = 2M
    v₂ = 2L
    V₁ = M₂V₂
    M₁
    V₁ = 2 × 2
    18.36
    = 0.21786L
2. 1. g/L = Molarity × RFM
    M = 12.4
    40
    = 0.31M
  2. If 1000cm³ = 0.31
    25cm³ = ?
    25 × 0.31
    1000
    = 0.00775moles
  3. 2NaOH + H₂SO₄ → Na₂SO₄ + H₂O
    Mole ratio
    NaOH : H₂SO₄
    2 1
    0.00775 = 0.003875
    2
  4. If 15cm³ = 0.003875
    1000cm³ = ?
    1000 × 0.003875
    15
    = 0.25833M
    = 0.26M
    RFM of H₂SO₄ = 98
    g/L = M × R.F.M
    = 0.26 × 98
    = 25.32g/L
1. A solution that cannot dissolve anymore solute at a particular temperature
2. 1. 59g/100g ± 1 of water must be shown on graph
  2. At 16°C 26g/100g of water
    100cm³ = 26g
    1000cm³ = ?
    1000 × 26
    100
    = 260
    RFM = 23 + 14 + (16×3)
    = 85g
    M = ²⁶⁰/₈₅ = 3.0588M
  3. At 40°C = 62g/100gwater
    At 26°C = 40g/100g water
    Mass = 62 − 40
    = 22g
  4. As temperature increases, solubility of NaNO₃ increases
3. - Used in brewing
  - Provides Calcium - essential nutrient
4. HCl is a strong acid, CH₃COOH is a weak acid CH₃COOH partially dissociates while HCl dissociates fully since some energy is used in fully ionising of CH₃COOH
1. 1. Hydrogen
  2. Reducing property
  3. 1. Heat
      CuSO4.5H₂ CuSO4(s) + 5H₂O_(l)
    2. 3Fe(s) + H₂O_(l) → Fe₃O₄ + 4H₂O_(g)
2. 1. 1. Tube I
      - Blue solid turns white/ colourless liquid is formed at cooler part of test tube
      - CuSO₄.5H₂O losses water of crystallisation
    2. Combustion tube II
      - Black solid turns brown. Copper (II) oxide reduced by H_2(g) to copper metal.
3. 1. To provide steam
  2. Hydrogen
  3. Water
  4. Decrease freezing point of water
1. 2. 1. 4 - methylpent -1- ene
    2. 2 - methylbut - 2- ene
  3. Alkenes
2. 1. 1. A - 1, 2 - dichloropropane
    2. B - Propane
    3. C - Polythene/ polypropene
  2. 1. Step I - Cracking
    2. Step III - Hydrogenation
    3. Step IV - Dehydration
    4. Step VI - Combustion
  3. 2C₃H_6(g) + 9O_2(g) → 6CO_2(g) + 6H₂O_(l)