48 questions. Try each one, then open the answer to check yourself and read the explanation.
Question 1
Separation of Mixtures
In the experiment shown in the figure, solid ammonium chloride is heated at the closed end of a tube. Ammonium chloride crystals deposit on the walls of the tube as a result of
A.evaporation
B.recrystallization
C.sublimation
D.fractional precipitation.
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Answer: C (sublimation)
Explanation
Ammonium chloride changes straight from solid to vapour on heating and back to solid on the cooler part of the tube, without melting. That is sublimation.
2.25 g of a sample of an oxide of copper on reduction gave 2.0 g of copper. 2.50 g of another oxide of copper on reduction also gave 2.0 g of copper. These results are in accordance with the law of
A.constant composition
B.conservation of matter
C.multiple proportions
D.definite proportions.
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Answer: C (multiple proportions)
Explanation
The same 2.0 g of copper combines with 0.25 g and 0.50 g of oxygen in the two oxides. These masses are in the simple ratio 1 : 2, which is the law of multiple proportions.
One mole of propane is mixed with five moles of oxygen. The mixture is ignited and the propane burns completely. What is the volume of the products at s.t.p? [G.M.V = 22.4 dm3 mol−1]
A.112.0 dm3
B.67.2 dm3
C.56.0 dm3
D.44.8 dm3.
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Answer: B (67.2 dm3)
Explanation
C3H8 + 5O2 → 3CO2 + 4H2O. At s.t.p. the water is liquid, so only 3 mol of CO2 is gas: 3 × 22.4 = 67.2 dm3.
9.0 dm3 of a gas at s.t.p was subjected by means of a movable piston to two times the original pressure with the temperature being now kept at 364 K. What is the volume of the gas in dm3 at this pressure?
An increase in temperature causes an increase in the pressure of a gas because there is an increase in the
A.average velocity of the molecules
B.number of collisions between the molecules
C.density of the molecules
D.free mean path between each molecule and the other.
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Answer: A (average velocity of the molecules)
Explanation
Heating gives the molecules more kinetic energy, so they move faster. Faster molecules hit the container walls more often and harder, which raises the pressure.
The forces holding naphthalene crystals together can be overcome when naphthalene is heated to a temperature of 354 K resulting in the crystals melting. These forces are known as
A.coulombic
B.ionic
C.covalent
D.Van der Waals
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Answer: D (Van der Waals)
Explanation
Naphthalene is a molecular solid. Its molecules are held together by weak van der Waals forces, which is why it melts at a low temperature.
Which of the following physical properties decreases across the Periodic Table?
A.Ionization potential
B.Electron affinity
C.Electronegativity
D.Atomic radius
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Answer: D (Atomic radius)
Explanation
Across a period the nuclear charge increases while electrons go into the same shell, so they are pulled closer and the atomic radius decreases. The other three properties increase.
1.34 g of hydrated sodium tetraoxosulphate (VI) was heated to give an anhydrous salt weighing 0.71 g. The formula of the hydrated salt is [Na = 23, S = 32, O = 16, H = 1]
A.Na2SO4.7H2O
B.Na2SO4.3H2O
C.Na2SO4.2H2O
D.Na2SO4.H2O
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Answer: A (Na2SO4.7H2O)
Explanation
Anhydrous Na2SO4: 0.71/142 = 0.005 mol. Water lost: 0.63 g = 0.035 mol. Ratio 0.035/0.005 = 7, so Na2SO4.7H2O.
Which of the following acts both as a reducing and an oxidizing agent?
A.H2
B.SO2
C.H2S
D.C
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Answer: B (SO2)
Explanation
Sulphur in SO2 is +4, in the middle of its range. It can be oxidised to +6 (acting as a reducing agent) or reduced to 0, for example by H2S (acting as an oxidising agent).
The mass of a substance, M, liberated at an electrode during electrolysis is proportional to the quantity of electricity, Q passing through the electrolyte. This is represented graphically by (see the graphs A–D in the figure)
A.Graph A
B.Graph B
C.Graph C
D.Graph D
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Answer: A (Graph A)
Explanation
Direct proportion (Faraday's first law) gives a straight line through the origin: M = 0 when Q = 0. Graph A is the only such line.
A mixture of starch solution and potassium iodide was placed in a test tube. On adding dilute tetraoxosulphate (VI) acid and then K2Cr2O7 solutions, a blue-black colour was produced. In this reaction, the
A.iodide ion is oxidized
B.tetraoxosulphate (VI) acid acts as an oxidizing agent
C.starch has been oxidized
D.K2Cr2O7 is oxidized.
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Answer: A (iodide ion is oxidized)
Explanation
Acidified dichromate oxidises iodide ions to iodine, which turns starch blue-black. The dichromate itself is reduced.
The diagram shows the reaction path of an exothermic reaction. The heat of reaction is represented by
A.I
B.II
C.III
D.IV.
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Answer: D (IV.)
Explanation
The heat of reaction is the energy difference between the reactants and the products. On the diagram that gap is labelled IV; the other labels measure from the peak (activation energy) or to the axis.
Which of the following will produce the greatest increase in the rate of the chemical reaction represented by the equation Na2S2O3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + SO2(g) + S(s)?
A.A decrease in temperature and an increase in the concentration of the reactants
B.An increase in the temperature and a decrease in the concentration of the reactants
C.An increase in the temperature and an increase in the concentration of the reactants.
D.A decrease in the temperature and a decrease in the concentration of the reactants.
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Answer: C (An increase in the temperature and an increase in the concentration of the reactants.)
Explanation
Both a higher temperature and a higher concentration speed up the reaction, so raising both gives the greatest increase in rate.
Which of the following compounds has the highest boiling point?
A.CH3CH2CH2CH2OH
B.CH3CH2CH2CHO
C.CH3CH2CH2CH3
D.CH3CH2OCH2CH3.
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Answer: A (CH3CH2CH2CH2OH)
Explanation
Butan-1-ol molecules form hydrogen bonds with each other through their –OH groups, so it boils much higher than the aldehyde, alkane or ether of similar size.