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post-utme2015chemistry
Q1. An elements used in production of matches is
Anitrogen
Baluminum
Ccopper
Dsulphur
Correct Answer: D
Sulphur is used in the production of matches as a key component of the match head, where it acts as a fuel that ignites easily when the match is struck.
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post-utme2015chemistry
Q2. P₁V₁ = P₂V₂ supports?
ACharles's law
BBoyle's law
CGraham's law
DAvogadro's law
Correct Answer: B
P₁V₁ = P₂V₂ is the mathematical expression of Boyle's Law, which states that at constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume.
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post-utme2015chemistry
Q3. Alloys are best prepared by
Aelectroplating
BArc-welding
Creducing and mixture of their metallic
Dcooking a molten of the metals
Correct Answer: D
Alloys are best prepared by melting (fusing) the component metals together in the molten state and allowing the mixture to cool and solidify. This ensures homogeneous mixing of the metals at the atomic level, producing a uniform alloy.
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post-utme2015chemistry
Q4. The filter in the cigarette reduce the nicotine construct by
Aburning
Badsorption
Cevaporation
DAbsorption
Correct Answer: B
The filter in cigarettes reduces nicotine content by adsorption. The filter's dense network of cellulose acetate fibers provides a large surface area onto which nicotine molecules and other smoke particles adhere (stick to the surface), preventing them from reaching the smoker's lungs.
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post-utme2015chemistry
Q5. Atomicity of ozone is
A1
B2
C3
D4
Correct Answer: C
The atomicity of ozone is 3 because each ozone molecule contains three oxygen atoms (O₃).
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post-utme2015chemistry
Q6. Which of the following gives a precipitation when treated with NaOH solution?
AAlCl₃
BNH₄Cl
CNa₂CO₃
DCH₃COONa
Correct Answer: A
AlCl₃ reacts with NaOH to form aluminium hydroxide, Al(OH)₃, a white gelatinous precipitate: AlCl₃ + 3NaOH → Al(OH)₃↓ + 3NaCl. NH₄Cl produces ammonia gas (not a precipitate), while Na₂CO₃ and CH₃COONa are fully soluble in NaOH.
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post-utme2015chemistry
Q7. In countries where the temperature fall below 273 K, salt is always spilled on the road in order to
Aincrease the boiling point
Bincrease the density of the ice
Cmake the ice impure
Dreduce the freezing point of ice
Correct Answer: D
Salt (NaCl) is spread on icy roads to reduce the freezing point of water (freezing point depression). Dissolving salt in water lowers its freezing point below 0°C (273 K), causing ice to melt even at sub-zero temperatures and preventing refreezing.
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post-utme2015chemistry
Q8. When sugar is dissolved in tea, the reaction is accomplished by?
Apositive change
Bnegative entropy change
Cno entropy change
Da minimum entropy change
Correct Answer: A
Dissolving sugar in tea is a spontaneous mixing process. The sugar molecules disperse throughout the liquid, increasing the disorder of the system. This represents a positive entropy change (ΔS > 0), as entropy increases when a solute dissolves in a solvent.
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post-utme2015chemistry
Q9. In the production of soap, concentrated sodium chloride solution is added to
Aincrease the solubility of soap
Bdecrease the solubility of the soap
Csaponify the soap
Demulsify the soap
Correct Answer: B
Concentrated NaCl solution is added during soap production to decrease the solubility of soap (salting out). The high ionic strength of NaCl reduces the solubility of soap in water, causing it to precipitate out and separate from the glycerol and excess reactants.
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post-utme2015chemistry
Q10. A major source of oxide of nitrogen is from the burning of
Acoal
Bwood
Cfuel
Dchlorofluorocarbons
Correct Answer: C
The burning of fuel (particularly fossil fuels in vehicle engines and power plants) at high temperatures causes atmospheric N₂ and O₂ to combine, producing oxides of nitrogen (NOₓ) such as NO and NO₂. These are major contributors to air pollution and acid rain.
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post-utme2015chemistry
Q11. Cu₂S(s) + O₂(g) = 2Cu(s) + SO₂(g).
In the above reaction, the oxidation number of Cu changes from ___ to ___?
A+1 to 0
B0 to +2
C+2 to +1
D0 to +1
Correct Answer: A
In Cu₂S, copper has oxidation number +1 (since S is -2 and there are 2 Cu atoms: 2x + (-2) = 0, x = +1). In the product Cu(s), copper is in its elemental state with oxidation number 0. Therefore Cu changes from +1 to 0 (reduction).