All 20 questions from the Post-UTME Screening (Post-UTME) Chemistry 2017 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.
What is quantity of iron is deposited when a current of 5A is passed through a solution containing Fe3+ ions for 19.3 minutes? [Fe=56, F=96,500C mol-1]
A. 0.56gB. 1.12gCorrect C. 2.24gD. 3.36gExplanation Q=It=5x(19.3x60)=5790C. Fe3+ + 3e -> Fe, so moles=Q/(3F)=5790/(3x96500)=0.02mol. Mass=0.02x56=1.12g.
Cu2+(aq) + 2e -> Cu(s), E = +0.34V; Zn2+(aq) + 2e -> Zn(s), E = -0.76V. From the equations above, it can be concluded that
A. Cu can easily reduce ZnB. Zn can easily oxidize CuC. Cu can easily oxidize ZnD. Zn can easily reduce any metalCorrect Explanation Zinc, being more reactive (more negative electrode potential) than most common metals, readily acts as a reducing agent towards other metal ions.
During the electrolysis of dilute sodium chloride solution, the product at the anode is
A. sodiumB. hydrogenC. oxygenCorrect D. chlorineExplanation In dilute NaCl solution, hydroxide ions are preferentially discharged at the anode, producing oxygen gas.
Palmwine exposed to the atmosphere for sometime becomes sour due to
A. oxidationCorrect B. reductionC. hydrolysisD. hydrationExplanation Exposure to air allows bacteria to oxidise the ethanol in palmwine to ethanoic (acetic) acid, making it sour.
2HI + H2O2 -> 2H2O + I2. In the equation above, the oxidation number of I has changed from
A. +1 to 0B. 0 to +1C. -1 to 0Correct D. 0 to -1Explanation Iodine in HI has oxidation number -1; in I2 it is 0. The oxidation number changes from -1 to 0.
A chemical substance reaction attains equilibrium when
A. delta H < 0B. delta G < 0C. delta G = 0Correct D. delta H = 0Explanation Chemical equilibrium is characterised by the Gibbs free energy change (delta G) being zero.
Substance / delta H kJ mol-1: CH4 -74.8, O2 0, C -393.5, H2O -286. Using the table above, calculate the delta H for the following reaction at 25C: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g).
A. -890.7kJCorrect B. -819.0kJC. +819.0kJD. +890.7kJExplanation delta H = [(-393.5)+2(-286)] - [(-74.8)+2(0)] = -965.5 - (-74.8) = -890.7 kJ.
N2(g) + O2(g) <=> 2NO(g), delta H + 92kJ. The equilibrium represented by the above equation can be shifted to right by
A. decreasing the temperatureB. using a catalystC. increasing the pressureD. using excess nitrogenCorrect Explanation Adding excess N2 (a reactant) shifts the equilibrium to the right (Le Chatelier's principle); the reaction is endothermic, so cooling would not help, and pressure has no effect since moles of gas are equal on both sides.
N2O4(g) <=> 2NO2(g). A high yield of nitrogen (IV) oxide can be obtained the reaction above by
A. increasing the temperatureCorrect B. lowering the temperatureC. increasing the pressureD. lowering the pressureExplanation The dissociation of N2O4 to NO2 is endothermic, so raising the temperature shifts equilibrium towards greater NO2 (nitrogen(IV) oxide) yield.
CuSO4(aq) + H2S(g) -> CuS(s) + H2SO4(aq). In the reaction above, hydrogen sulphide acts as
A. a weak acidB. a reducing agentC. an oxidizing agentD. a precipitating agentCorrect Explanation H2S provides sulphide ions that precipitate copper as insoluble CuS, acting as a precipitating agent.
The cleansing action of soap in hard water is not satisfactory because soap
A. is a salt of fatty acidsB. is not a detergentC. forms insoluble saltsCorrect D. forms soluble saltExplanation In hard water, soap reacts with calcium and magnesium ions to form insoluble scum (salts), reducing its cleansing effectiveness.
If the solubility of KHCO3 is 0.40 mol/dm3 at room temperature, calculate the mass of KHCO3 in 100cm3 of the solution at that temperature.
A. 100.0gB. 40.0gC. 10.0gD. 4.0gCorrect Explanation Moles = 0.40 x 0.100 = 0.040mol. Molar mass KHCO3 = 100g/mol. Mass = 0.040 x 100 = 4.0g.
An acid salt can be produced using
A. tetraoxsulphate (VI) acidCorrect B. ethanoic acidC. trioxonitrate (V) acidD. hydrochloric acidExplanation Acid salts form from polybasic (e.g. dibasic) acids like tetraoxosulphate(VI) acid, which can be only partially neutralised.
Which of the following is deliquescent?
A. CuSO4B. NaOHCorrect C. CuCl2D. Na2SO4Explanation Sodium hydroxide (NaOH) is a classic deliquescent substance, absorbing moisture from the air until it dissolves.
One of the products formed when chlorine is bubbled slowly through a solution of potassium iodide is
A. potassiumB. solid iodineCorrect C. potassium hydroxide solutionD. hydrogen iodide vapourExplanation Chlorine displaces iodine from potassium iodide (Cl2 + 2KI -> 2KCl + I2), producing iodine, which appears as a solid/dark precipitate.
Which of the following is produced by each of NaNO3, Pb(NO3)2 and AgNO3 when decomposed by heat?
A. OxygenCorrect B. Nitrogen (IV) oxideC. Nitrogen (II) oxideD. NitrogenExplanation All nitrates release oxygen gas as a common product upon thermal decomposition, alongside other products depending on the metal.
In the extraction of iron, limestone is fed into the furnace to provide
A. carbon (IV) oxide which reacts with hot cokeB. carbon (II) oxide which reduces iron (III) oxideC. both carbon (IV) and carbon (II) oxides for reduction processes in the furnaceD. calcium oxide which reacts with silicon (IV) oxideCorrect Explanation Limestone decomposes in the furnace to form calcium oxide, which reacts with silica impurities to form slag, removing them from the molten iron.
Electrolytic method is usually employed in the extraction of
A. aluminumCorrect B. copperC. zincD. ironExplanation Aluminium, being highly reactive, is extracted by electrolysis of its molten oxide (in cryolite).
The compound of sodium used in the manufacture of glass is
A. Na2SO4 10H2OB. NaHCO3C. Na2CO3 10H2OCorrect D. NaOhExplanation Sodium carbonate (washing soda, Na2CO3.10H2O) is used as a raw material in glass manufacture.
Which of the following is typical of transition metals?
A. They form colourless complexesB. They exhibit variable valencesCorrect C. They have low melting pointsD. They do not act as catalystsExplanation Transition metals characteristically exhibit variable (multiple) oxidation states/valences.
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