49 questions. Try each one, then open the answer to check yourself and read the explanation.
Chromatography is used to separate components of mixtures which differ in their rates of
Answer: B (migration)
In chromatography the solvent carries each component along the paper or column at its own speed. Components separate because they migrate at different rates.
Which of the following is an example of a chemical change?
Answer: B (Rusting of iron)
Rusting makes a new substance (hydrated iron(III) oxide) that cannot easily be turned back into iron. The other three are physical changes.
The number of hydrogen ions in 4.9 g of tetraoxosulphate (VI) acid is [S = 32, O = 16, H = 1, NA = 6.02 x 1023]
Answer: B (6.02 x 1022)
H2SO4 has M = 98, so 4.9 g is 0.05 mol. Each molecule gives 2 H+, so 0.10 mol of H+ = 0.10 × 6.02 × 1023 = 6.02 × 1022 ions.
What volume of oxygen will remain after reacting 8 cm3 of hydrogen with 20 cm3 of oxygen?
Answer: D (16 cm3)
2H2 + O2 → 2H2O, so 8 cm3 of hydrogen uses only 4 cm3 of oxygen. Oxygen left = 20 − 4 = 16 cm3.
A gas sample with an initial volume of 3.25 dm3 is heated and allowed to expand to 9.75 dm3 at constant pressure. What is the ratio of the final absolute temperature to the initial absolute temperature?
Answer: A (3:1)
By Charles' law, V/T is constant at fixed pressure, so T2/T1 = V2/V1 = 9.75/3.25 = 3, a ratio of 3:1.
Two cylinders A and B each contains 30 cm3 of oxygen and nitrogen respectively at the same temperature and pressure. If there are 5.0 moles of nitrogen, then the mass of oxygen is
Answer: D (160.0 g)
Equal volumes at the same temperature and pressure hold equal numbers of moles (Avogadro's law), so there are 5.0 mol of O2. Mass = 5.0 × 32 = 160.0 g.
A liquid begins to boil when
Answer: C (its vapour pressure equals the atmospheric pressure)
Boiling starts when the vapour pressure of the liquid becomes equal to the pressure of the air above it. Bubbles of vapour can then form throughout the liquid.
Use the section of the periodic table above. Which of the letters indicate an alkali metal and a noble gas respectively?
Answer: D (G and L)
G has atomic number 3 (lithium), an alkali metal in group 1. L has atomic number 2 (helium), a noble gas.
Use the section of the periodic table above. Which letter represents a non-metal that is a solid at room temperature?
Answer: A (T)
T has atomic number 16 (sulphur), a solid non-metal. J (8) is oxygen, a gas, while R (13, aluminium) and X (4, beryllium) are metals.
In the oil drop experiment, Millikan determined the
Answer: C (charge of the electron)
Millikan balanced charged oil drops between charged plates and found that every charge was a multiple of one small value, the charge of the electron.
The stability of ionic solids is generally due to the
Answer: B (crystal lattice forces)
In an ionic solid, each ion is held by strong electrostatic attractions to many oppositely charged neighbours in the crystal lattice. This lattice energy makes ionic solids stable.
Which of the following statements is FALSE about isotopes of the same element? They have the
Answer: D (same atomic number but different number of electrons)
Isotopes differ only in the number of neutrons. Neutral isotopes of an element have the same number of protons and therefore the same number of electrons, so D is false.
Helium is often used in observation balloons because it is
Answer: B (light and non-combustible)
Helium is much less dense than air, so it gives lift. Being a noble gas, it does not burn, which makes it safer than hydrogen.
When plastics and packaging materials made from chloroethene are burnt in the open, the mixture of gases released into the atmosphere is most likely to contain
Answer: C (hydrogen chloride)
PVC (poly(chloroethene)) contains chlorine. When it burns, the chlorine combines with hydrogen to give hydrogen chloride gas.
Deliquescent substances are also
Answer: C (hygroscopic)
A deliquescent substance absorbs so much moisture from the air that it dissolves. Since it absorbs water from the air, it is also hygroscopic.
The difference between colloides and suspensions is brought out clearly by the fact that while colloids
Answer: D (do not settle out on standing, suspensions do)
Colloid particles are small enough to stay dispersed indefinitely, but the larger particles of a suspension settle out when left to stand.
In general, an increase in temperature increases the solubility of a solute in water because
Answer: D (most solutes dissolve with absorption of heat)
For most solids, dissolving absorbs heat (it is endothermic). By Le Chatelier's principle, raising the temperature favours the heat-absorbing process, so more solute dissolves.
Neutralization involves a reaction between H3O+ and
Answer: B (OH−)
Neutralization is acid + base, which at the ionic level is H3O+ + OH− → 2H2O.
Which of the following solutions will have a pH < 7?
Answer: D (NH4Cl(aq))
NH4Cl is the salt of a strong acid and a weak base. It hydrolyses in water to give an acidic solution, while Na2CO3 is alkaline and the other two are neutral.
What is the pH of a 2.50 x 10-5 M solution of sodium hydroxide?
Answer: C (9.4)
[OH−] = 2.50 × 10-5, so pOH = −log(2.50 × 10-5) = 4.6. pH = 14 − 4.6 = 9.4.
The graph above shows the pH changes for the titration of a
Answer: B (weak acid versus strong base)
The pH starts at about 3 (a weak acid) and the sharp jump runs from about 6 to 11, ending high (a strong base). This shape is typical of a weak acid titrated with a strong base.
In the process of silver-plating a metal M, the metal M is the
Answer: D (cathode and a direct current is used)
Silver ions are positive, so they are deposited on the negative electrode, the cathode. The current must be direct (one-way) so that M stays the cathode throughout.
How many moles of copper would be deposited by passing 3F of electricity through a solution of copper (II) tetraoxosulphate (VI)? [F = 96 500 C mol-1]
Answer: C (1.5)
Cu2+ + 2e− → Cu, so 2 F deposit 1 mol of copper. 3 F deposit 3/2 = 1.5 mol.
2Cl−(aq) → Cl2(g) + 2e−. The above half-cell reaction occurring at the anode during the electrolysis of dilute ZnCl2 solution is
Answer: B (oxidation)
The chloride ions lose electrons. Loss of electrons is oxidation, which always happens at the anode.
Which of the following is a redox reaction?
Answer: B (2FeBr2(aq) + Br2(aq) → 2FeBr3(aq))
In B, iron goes from +2 to +3 (oxidized) and bromine from 0 to −1 (reduced). No oxidation numbers change in the other three reactions.
Cr2O72−(aq) + 14H+(aq) + 6I−(aq) → 2Cr3+(aq) + 3I2(g) + 7H2O(l). The change in the oxidation number of oxygen in the equation above is
Answer: A (0)
Oxygen is −2 in Cr2O72− and still −2 in H2O, so its oxidation number does not change. Only chromium and iodine change.
If an equilibrium reaction has ΔH < O, the reaction will proceed favourably in the forward reaction at
Answer: A (low temperatures)
ΔH < 0 means the forward reaction is exothermic. Lowering the temperature favours the heat-releasing direction, so the forward reaction is favoured at low temperatures.
Which of the following processes leads to increase in entropy?
Answer: C (Boiling a sample of water)
Entropy measures disorder. Boiling turns an ordered liquid into a gas whose molecules move freely, a large increase in disorder. Condensing and cooling reduce disorder.
Which of the following equilibria is shifted to the right as a result of an increase in pressure?
Answer: B (2NO2(g) ⇌ N2O4(g))
Higher pressure favours the side with fewer gas molecules. In B, 2 molecules of NO2 form 1 of N2O4, so the equilibrium shifts right.
The arrangement above can be used for the collection of
Answer: C (nitrogen)
The gas is collected over water, which only works for a gas that is nearly insoluble in water. Nitrogen is insoluble, but SO2, NH3 and HCl dissolve readily.
Use the diagram above. The activation energy of the uncatalysed reaction is
Answer: A (x)
Activation energy is the height of the energy barrier above the reactants. For the uncatalysed path (I, the higher curve) this is the distance x from the N2 + 3H2 level to the top. y is the enthalpy change.
Use the diagram above. It can be deduced that the rate of the reaction
Answer: B (for path II is higher than path I)
Path II has a lower energy barrier (it is the catalysed path). A lower activation energy means more molecules have enough energy to react, so the reaction is faster.
In the industrial production of hydrogen from natural gas, carbon (IV) oxide produced along with the hydrogen is removed by
Answer: A (washing under pressure)
CO2 is much more soluble in water under pressure than hydrogen is. Washing the gas with water under pressure dissolves the CO2 cheaply on a large scale.
Sulphur exists in six forms in the solid state. This property is known as
Answer: B (allotropy)
Allotropy is the existence of an element in two or more different forms in the same physical state, like rhombic and monoclinic sulphur.
A gas that will turn orange potassium heptaoxodichromate (VI) solution to clear green is
Answer: C (sulphur (IV) oxide)
SO2 reduces orange dichromate(VI) to green chromium(III), giving a clear green solution. H2S also reduces it but leaves a cloudy yellow deposit of sulphur.
Which of the following ions will give a white precipitate with aqueous NaOH and soluble in excess of the base?
Answer: C (Zn2+)
Zn2+ forms white Zn(OH)2, which is amphoteric and dissolves in excess NaOH. Ca(OH)2 and Mg(OH)2 stay undissolved, and Cu(OH)2 is blue.
In the extraction of iron in the blast furnace, limestone is used to
Answer: D (remove impurities)
Limestone decomposes to CaO, which combines with the sandy impurities (SiO2) to form molten slag (CaSiO3) that is run off.
Which of the following compounds will impart a brick-red colour to a non-luminous bunsen flame?
Answer: C (CaCl2)
Calcium compounds give a brick-red flame. Sodium gives golden yellow, lithium crimson, and magnesium no characteristic colour.
Group 1 A metals are not found free in nature because they
Answer: D (are very reactive)
Alkali metals react quickly with oxygen and water, so in nature they are always found combined in compounds, never as free metals.
CH3COOH + CH3CH2OH → X + Y (conc. H2SO4)
X and Y in the reaction above are respectively
Answer: C (CH3COOCH2CH3 and H2O)
An acid and an alcohol react (esterification) to form an ester and water. Ethanoic acid and ethanol give ethyl ethanoate, CH3COOCH2CH3, and H2O.
CHCl3 + Cl2 → HCl + CCl4. The reaction above is an example of
Answer: B (a substitution reaction)
A chlorine atom takes the place of the last hydrogen atom in CHCl3. Replacing one atom by another is substitution.
CH3—CH(CH3)—CH=CH—CH3
The IUPAC nomenclature for the compound above is
Answer: D (4-methylpent-2-ene)
The longest chain has 5 carbons with a double bond. Numbering from the end nearest the double bond puts it at carbon 2 and the methyl group at carbon 4: 4-methylpent-2-ene.
Which of the following pairs has compounds that are isomers?
Answer: A (Propanal and propanone)
Isomers have the same molecular formula but different structures. Propanal (CH3CH2CHO) and propanone (CH3COCH3) are both C3H6O.
Aromatic and aliphatic hydrocarbons can be distinguished from each other by the
Answer: C (action of heat)
Aromatic hydrocarbons such as benzene have a high carbon content and burn with a very smoky, sooty flame, while aliphatic ones burn more cleanly. Bromine does not help, since neither benzene nor alkanes decolourise it readily.
The role of sodium chloride in the preparation of soap is to
Answer: B (separate the soap from glycerol)
Adding salt ('salting out') makes the soap less soluble, so it floats out as a solid. The glycerol stays dissolved in the salty water.
CH3CH2=CH2—CHO
The functional group represented in the compound above is
Answer: B (alkanal)
The –CHO group (a C=O with an H on the end carbon) is the aldehyde group, found in alkanals.
CxHy + 4O2 → 3CO2 + 2H2O. The hydrocarbon, CxHy, in the reaction above is
Answer: C (propyne)
3CO2 means 3 carbon atoms and 2H2O means 4 hydrogen atoms, so the hydrocarbon is C3H4, propyne.
An example of a secondary amine is
Answer: B (di-butylamine)
A secondary amine has two alkyl groups on the nitrogen (R2NH). Dibutylamine has two butyl groups; methylamine is primary and trimethylamine is tertiary.
The relatively high boiling points of alkanols are due to
Answer: D (hydrogen bonding)
The –OH groups of neighbouring alkanol molecules form hydrogen bonds, which need extra energy to break, so alkanols boil higher than alkanes of similar mass.