QUESTIONS
- Study the electrode potentials for the half cells below and use them to answer the questions that follow. The letters do not represent the actual elements.
Eθ (volts)
A+(aq) + e- → A(s) -2.96
B+(aq) + e- → B(s) + 0.52
C+(aq) _ e- → 1⁄2C2 0.00
D2+ + 2e- → D(s) -0.44
1⁄2G2(g) +e- → G-(aq) +1.36- Identify the strongest oxidizing agent. Explain.
- Which of the two half cells would produce the highest potential difference when combined. (2 mks)
- Explain whether the reaction below can take place. D(s) + 2A+(aq) → D2+(aq) + 2A(s) (2 mks)
- Draw a well labelled diagram when combining A and B half cells.
- In an experiment to electroplate an iron watch with silver a circuit of 0.5A was passed fro 48 minutes. Calculate the amount of silver deposited on the watch. (IF -96,500C, Ag = 108) (3 mks)
- Study the scheme below and answer the questions that follow.
- Name
Compound P
Solution 'V'.
Substance R
Gas 'K' (3 mks) - Write the equation for the reaction in :
Step I
Step II
Step III - Name the type of reaction in step I and step II and give the conditions. (4 mks)
- A sample of polymer 'Q' is found to have a molecular mass of 4200. Determine the number of monomers in the polymer (C = 12, H =1) (2 mks)
- Describe an experiment to distinguish propane and propyne using burning. (2 mks)
- Name
-
- What is meant by the term enthalpy at formation. (1 mk)
- The enthalpies of combustion of carbon, hydrogen and ethanol are given below.
C(s) + O2(g) → CO2(g) ΔH = - 393kJ/mol
H2(g) + O2(g) → H2O(I) ΔH = -286 kJ/mol
If the enthalpy of combustion of ethanol is -1369kJ/mol. Draw the energy cycle diagram that links the enthalpy of formation of ethanol to enthalpies of combustion of carbon and hydrogen. (3 mks) - Study the information in the table below and answer the questions that follow.
BOND BOND ENERGY (kJ/Mol) C-H 414 C-CI 326 CI-CI 244 H-CI 431
CH3Cl(g) + Cl2(g) U.V. light → CH2Cl2(g) + HCl(g) - At standard temperature and pressure graphite changes to diamond as shown in the equation below.
C(graphic) → C(diamond) ΔH = +2.91kg/mol
Sketch a simple energy level diagram for the above change.
- 0.12M aqueous standard nitric (v) acid was titrated against 25cm3 of aqueous sodium hydroxide solutions in a conical flask. 11.5cm3 of acid was required to react completely with alkali.
- Name two apparatus that can be used in this experiment other than conical flask. (1 mk)
- Calculate the concentration of sodium hydroxide used in experiment.
- The table below shows the tests carried out on three portions of a compound and results obtained. Study it and answer the questions that follow.
Test Observations Addition of a few drops of aqueous ammonia to the first portion until excess White precipitate soluble in excess Addition of acidified barium nitrate to the second portion White precipitate formed Addition of few drops of lead (II) nitrate to the third portion White precipitate formed - Identify the cation and anion present in the compound.
- Write the ionic equations in first portion and third portion.
- Define the following terms.
- Solubility
- Saturated solution
- Fractional crystallization.
- The grid below represents port of the periodic table. Study it and answer the questions that follow. The letters do not represent the actual symbols of elements.
- Identify the most electronegative and electropositive element. Explain.
- Element Q reacted with steam at elevated temperatures to produce 150cm3 of a gas.
Determine the mass of Q which was reacted with steam
(M.G.V=24dm3) (R.A.M of Q = 27) - Explain the difference between the atomic radius of element J and it's ionic radius ( 2 mks)
- Write the formula of the most stable ion of element G when it ionizes. (1 mk)
- The ionization energies for elements A, B and L are 520Kj/mol, 500kJ/mol and 420Kj/mol Values. What does the valves indicate about their reactivity explain.(2 mks)
- Draw the atomic structure of a compound formed when element Q reacts with oxygen (Atomic number of oxygen = 8) (2 mks)
-
- The following reversible reaction represents the formation of methanol from hydrogen and carbon (II) Oxide.
CO(g) + 2H2(g) ⇌ CH3OH(g) ΔH- +91kJ/mol (4 mks)
What would be the effect on equilibrium and the yield of methanol/when:- Increasing pressure
- Decreasing temperature
- Using a catalyst.
- Adding ethanoic acid to the equilibrium in presence of few drops of concentrated sulphuric (vi) acid and warming
- An experiment was carried out using a given mass of magnesium ribbon and ImHel the results are as shown below.
Time (sec) 10 20 30 40 50 60 80 100 Volume of H2(g)(cm3) 49 90 117 136 147 150 150 150 - Plot a graph of volume of hydrogen gas produced against time. (3 mks)
- On the same axis sketch a graph that would be obtained if Im CH3COOH was used instead of IM HCI. (1 mk)
- Calculate the rate of reaction of magnesium with hydrochloric acid at 50 seconds and 60 seconds. Explain. (3 mks)
- State two factors that can affect the above rate of reaction a part from the one investigated above. (2 mks)
- The following reversible reaction represents the formation of methanol from hydrogen and carbon (II) Oxide.
- The diagram below is a flow chart for the extraction of copper. Study it and answer the question that follow.
- Give the name and formula of the major ore from which copper is extracted. (1 mk)
- Give the name of the process carried out in step II and III (1 mk)
- Write equation for the reactions taking place in step III and IV. (2 mks)
- Iron (ii) oxide is an impurity during extraction of copper. Write the equation for the reaction to show how it is removed. (1 mk)
-
- Draw a well labelled diagram of show blister copper is purified to form pure copper, (3 mks)
- Write the ionic equation for the anode and cathode reaction during purification of copper. (2 mks)
- When copper is exposed to the atmosphere for a long period of time it forms a green coating,
Name the green coating and write a balanced chemical equation for the reaction which leads to formation of green coating. (2 mks)
MARKING SCHEME
-
- G2 - the most positive E value
- G and A
- Eθ = Ee - Eo =-2.96 - (-0.44)
= - 2.54
the reaction will not take place because the calculated E value is negative -
- Q = It
0.5 x 48 x 60 = 1440C
Ag+(aq) + e- → Ag(s)
108 → 96,500C
1440C
108 x 1440 = 1.6116g
96,500
-
- compound p - propene/ prop-1-ene / C3H6
solution v - Sodium ethanate / CH3CH2COONa
substance r - Sodium propoxide / CH2CH2CH2ONa
gas K - Hydrogen/H2(g) - Step I - CH3CH2CH2OH + CH3COOH ⇌ CH3COOCH2CH2CH3 + H2O
Step II - CH3CH2CH2OH → CH3CH3COOH + H2O
Step III - 2CH3CH3CH3OH+2Na → 2CH3CH3CH2ON(g) + H2(g) -
- Estorification H2SO4/conc H2SO4/warming
- Oxidation H+/Acidifying / heat
- 4200 = 100monomers
42 - propane burns with blue unsooty flame
propyne burns with yellow sooty flame
- compound p - propene/ prop-1-ene / C3H6
-
- The heat change when a substance/compound is formed from its consistent elements in standard state
-
-
-
-
- pipete
- 0.12 x 11.5
1000
= 0.0013moles
HNOs : NaOH
1:1
0.0012moles of NaOH → 25cm3
? → 100cm3
= 0.0552m -
- Al3+ and SO42-
- Pb2+(aq) + SOO42-(aq) → PbSO4(s)
Al3+(aq) + 3OH-(aq) → Al(OH)3(s)
Al(OH)3(s) + OH-(aq) → [Al(OH)4]-(aq) -
- maximum amount of asolute that can dissolve in 100cm3 of water at specific temps
- A solution which cannot dissolve any more solute at a given temperature
- A process of separating salts using different solubilities at a given temperature
-
- M - Smallest atomic radius hence strongest nuclear attraction. L - Largest atomic radius
- 2Q(s) + 3H2O(g) → O3 + H2(g)
moles of H2(g) = 150cm3 = 0.00635moles
24000 - mole ratio
2:3
0.00625 = 0.0042moles
0.0042 x 27 = 0.11225g - J - Has a larger ionic radius compared to atomic radius due to electron repulsion when it forms
- G3-
- Reactivity increases down the
The larger the atomic radius the lower the ionization -
-
-
- Yield increases - equilibrium shifts to the left
Forward reaction formed. Large volume of reactants compared to product - Yield decreases equilibrium shifts to the right. backward reaction formed, reaction is endothermic
- No effect on equilibrium
- Equilibrium shiftf to the right yield increase as the concentration of methanol reduces
- Yield increases - equilibrium shifts to the left
-
-
- At 50s - determining the gradient
At 60s - Determining the gradient - Size of particles
Temperature
-
-
-
- Copper pyrites CuFes2
- II - froth floatation
III - Roasting - 2CuFeS2 + 4O → CuS(s) + 2FeO(s) + 3SO2(g)
2CuO(s) + Cu2S(s) → 6Cu(s) + SO2(g) - SIO2(s) + FeO(s) → FeSIO3(I)
-
-
- Cu(s) → Cu2+(aq) + 2e- - Anode
Cu2+(aq) + 2e- → Cu(s) - Cathode
-
- Basic copper (II) carbonate
2Cu(s) + O2(g) + H2O(I) + CO2(g) → CuCO3 . Cu(OH)2(s) (green coating)
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