49 questions. Try each one, then open the answer to check yourself and read the explanation.
35 cm3 of hydrogen was sparked with 12 cm3 of oxygen at 110°C and 760 mm Hg to produce steam. What percentage of the total volume of gas left after the reaction is hydrogen?
Answer: B (31%)
2H2 + O2 → 2H2O: 12 cm3 of O2 uses 24 cm3 of H2 and makes 24 cm3 of steam (a gas at 110°C). 11 cm3 of H2 is left in a total of 35 cm3, so 11/35 × 100 ≈ 31%.
2.85 g of an oxide of copper gave 2.52 g of copper on reduction and 1.90 g of another oxide gave 1.52 g of copper on reduction. The data above illustrates the law of
Answer: D (multiple proportions)
Two different oxides of the same two elements are compared. Per gram of oxygen the copper masses are 2.52/0.33 ≈ 7.6 g and 1.52/0.38 = 4.0 g, a simple ratio of about 2 : 1, which is the law of multiple proportions.
Use the graph below to answer this question. A sample, X, solid at room temperature, was melted, heated to a temperature of 358 K and allowed to cool as shown in OPQR. The section PQ indicates that X is
Answer: D (a pure compound)
PQ is flat: the temperature stays constant while the liquid freezes. A sharp, fixed freezing point is a sign of a pure substance; a mixture freezes over a range of temperatures.
Use the graph below to answer this question. A sample, X, solid at room temperature, was melted, heated to a temperature of 358 K and allowed to cool as shown in OPQR. The section OP suggests that X is in the
Answer: A (liquid state)
X was melted and heated to 358 K, so along OP the liquid is simply cooling down until it starts to freeze at P.
An element, X, forms a volatile hydride XH3 with a vapour density of 17.0. The relative atomic mass of X is [H = 1]
Answer: B (31.0)
Relative molecular mass = 2 × vapour density = 34. Taking away 3 for the three hydrogens leaves 31 for X.
A mixture of 0.20 mole of Ar, 0.20 mole of N2 and 0.30 mole of He exerts a total pressure of 2.1 atm. The partial pressure of He in the mixture is
Answer: A (0.90 atm)
Total moles = 0.70. Mole fraction of He = 0.30/0.70, so its partial pressure = (0.30/0.70) × 2.1 = 0.90 atm.
If 30 cm3 of oxygen diffuses through a porous plug in 7s, how long will it take 60 cm3 of chlorine to diffuse through the same plug? [O = 16, Cl = 35.5]
Answer: C (21s)
By Graham's law, time is proportional to volume × √(molar mass). t = 7 × (60/30) × √(71/32) = 14 × 1.49 ≈ 21 s.
The temperature of a body decreases when drops of liquid placed on it evaporate because
Answer: C (the heat of vaporization is drawn from the body causing it to cool)
Evaporating molecules need energy (latent heat of vaporization). They take this heat from the body they are on, so the body cools.
The electron configurations of two elements with similar chemical properties are represented by
Answer: C (1s22s22p63s1 and 1s22s1)
Elements with similar properties are in the same group, so they have the same number of outer electrons. Both configurations in C end in s1 (sodium and lithium, Group 1).
In the periodic table, what is the property that decreases along the period and increases down the group?
Answer: D (Atomic radius.)
Across a period the rising nuclear charge pulls the electrons closer, so atoms get smaller; down a group new shells are added, so atoms get bigger.
Two elements, P and Q with atomic numbers 11 and 8 respectively, combine chemically to form the compound PxQy. The respective values of x and y are
Answer: C (2 and 1)
P (2,8,1) is sodium, which forms Na+; Q (2,6) is oxygen, which forms O2−. Two Na+ balance one O2−, giving P2Q (Na2O).
Oxygen is a mixture of two isotopes 168O and 188O with relative abundances of 90% and 10% respectively. The relative atomic mass of oxygen is
Answer: B (16.2)
Relative atomic mass = (90 × 16 + 10 × 18)/100 = (1440 + 180)/100 = 16.2.
200 cm3 of air was passed over heated copper in a syringe several times to produce copper (II) oxide. When cooled, the final volume of air recorded was 158 cm3. Estimate the percentage of oxygen in the air
Answer: C (21%)
The copper removes the oxygen, so oxygen = 200 − 158 = 42 cm3. 42/200 × 100 = 21%.
Which of the following gases is the most dangerous pollutant?
Answer: D (Carbon (II) oxide.)
Carbon(II) oxide is colourless and odourless, so it is not noticed, and it binds to haemoglobin far more strongly than oxygen, quickly causing suffocation and death.
A major process involved in the softening of hard water is the
Answer: A (conversion of a soluble calcium salt to its trioxocarbonate (IV))
Hardness is caused by dissolved calcium salts. Softening (by boiling or adding washing soda) changes these soluble salts into insoluble calcium trioxocarbonate(IV), which settles out and removes Ca2+ from the water.
On recrystallization, 20 g of magnesium tetraoxosulphate (IV) forms 41 g of magnesium tetraoxosulphate (IV) crystals, MgSO4.yH2O. The value of y is [Mg = 24, S = 32, O = 16, H = 1]
Answer: D (7)
Water taken up = 41 − 20 = 21 g = 21/18 ≈ 1.17 mol. MgSO4 = 20/120 ≈ 0.167 mol. y = 1.17/0.167 = 7.
A saturated solution of AgCl was found to have a concentration of 1.30 x 10−5 mol dm−3. The solubility product of AgCl therefore is
Answer: C (1.69 x 10−10 mol2 dm−6)
AgCl gives one Ag+ and one Cl−, so Ksp = (1.30 × 10−5)2 = 1.69 × 10−10 mol2 dm−6.
The hydroxyl ion concentration, [OH−], in a solution of sodium hydroxide of pH 10.0 is
Answer: C (10−4 mol dm−3)
pOH = 14 − 10 = 4, so [OH−] = 10−4 mol dm−3.
Which of the aqueous solutions with the pH values below will liberate hydrogen when it reacts with magnesium metal?
Answer: D (3.0)
Magnesium gives off hydrogen with acids. pH 3.0 is the only clearly acidic solution with plenty of H+ ions.
Given that 15.00 cm3 of H2SO4 was required to completely neutralize 25.00 cm3 of 0.125 mol dm−3 NaOH, calculate the molar concentration of the acid solution:
Answer: C (0.104 mol dm−3)
H2SO4 + 2NaOH → Na2SO4 + 2H2O. Moles NaOH = 0.025 × 0.125 = 0.003125, so acid = 0.0015625 mol in 0.015 dm3 = 0.104 mol dm−3.
When platinum electrodes are used during the electrolysis of copper (II) tetraoxosulphate (VI) solution, the solution gets progressively
Answer: A (acidic)
Cu2+ is removed at the cathode and OH− is discharged at the anode as oxygen. H+ and SO42− are left behind, so the solution becomes tetraoxosulphate(VI) acid.
How many faradays of electricity are required to deposit 0.20 mole of nickel, if 0.10 faraday of electricity deposited 2.98 g of nickel during electrolysis of its aqueous solution? [Ni = 58.7, 1F = 96 500 C mol−1]
Answer: C (0.40)
2.98 g of Ni is 2.98/58.7 ≈ 0.05 mol, and it needed 0.10 F (Ni2+ takes 2 F per mole). 0.20 mol therefore needs 0.20 × 2 = 0.40 F.
What is the oxidation number of Z in K3ZCl6?
Answer: B (+3)
The compound is neutral: 3(+1) + Z + 6(−1) = 0, so Z = +3.
2H2S(g) + SO2(g) + H2O(l) → 3S(s) + 3H2O(l) ......(i)
3CuO(s) + 2NH3(g) → 3Cu(s) + 3H2O(l) + N2(g) ......(ii)
In the equations above, the oxidizing agent (i) and the reducing agent in (ii) respectively are
Answer: C (SO2 and NH3)
In (i) sulphur in SO2 goes from +4 to 0, so SO2 is reduced and is the oxidizing agent. In (ii) nitrogen in NH3 goes from −3 to 0 (N2), so NH3 is oxidized and is the reducing agent.
2SO2(g) + O2(g) ⇌ 2SO3(g). In the reaction above, the standard heats of formation of SO2(g) and SO3(g) are −297 kJ mol−1 and −396 kJ mol−1 respectively. The heat change of the reaction is
Answer: B (−198 kJ mol−1)
ΔH = ΣΔHf(products) − ΣΔHf(reactants) = 2(−396) − 2(−297) = −792 + 594 = −198 kJ mol−1 (O2 is zero).
½N2(g) + ½O2(g) → NO(g) ΔH° 89 kJ mol−1. If the entropy change for the reaction above at 25°C is 11.8 J mol−1, calculate the change in free energy, ΔG°, for the reaction at 25°C.
Answer: B (85.48 kJ)
ΔG = ΔH − TΔS. Change ΔS to kJ: 11.8 J = 0.0118 kJ. ΔG = 89 − 298 × 0.0118 = 89 − 3.52 = 85.48 kJ.
If the rate law obtained for a given reaction is rate = k [X]n [Y]m, what is the overall order of the reaction?
Answer: C (n + m)
The overall order is the sum of the powers of the concentrations in the rate law: n + m.
One method of driving the position of equilibrium of an endothermic reaction forward is to
Answer: A (increase temperature at constant pressure)
An endothermic forward reaction absorbs heat. By Le Chatelier's principle, raising the temperature favours the reaction that takes in heat, so the equilibrium moves forward.
Oxidation of concentrated hydrochloric acid with manganese (IV) oxide liberates a gas used in the
Answer: D (sterilization of water)
MnO2 oxidizes HCl to chlorine gas. Chlorine kills germs, so it is used to sterilize drinking water.
mE + nF ⇌ pG + qH. In the equation above, the equilibrium constant is given by
Answer: C ([G]p[H]q/[E]m[F]n)
K is the product of the product concentrations divided by the product of the reactant concentrations, each raised to the power of its coefficient in the balanced equation.
A compound that will NOT produce oxygen on heating is
Answer: A (potassium dioxonitrate (III))
Lead(IV) oxide, KClO3 and KNO3 all give off oxygen when heated. Potassium dioxonitrate(III), KNO2, is the product of heating KNO3 and does not decompose further to give oxygen.
Coal gas is made up of carbon (II) oxide, hydrogen and
Answer: D (methane)
Coal gas, made by heating coal without air, is mainly hydrogen, methane and carbon(II) oxide.
In the diagram above the gas Y could be
Answer: A (hydrogen chloride)
The set-up is a fountain experiment: water is forced up into the flask because the gas dissolves very quickly in it. Hydrogen chloride is extremely soluble in water; the other gases are not.
The reaction that occurs during the laboratory test for the presence of tetraoxosulphate (VI) ion is
Answer: A (SO42−(aq) + Ba2+(aq) → BaSO4 (dil HNO3))
The test for SO42− is to add a barium salt in dilute acid: a white precipitate of barium tetraoxosulphate(VI) forms.
The removal of rust from iron by treatment with tetraoxosulphate (VI) acid is based on the
Answer: B (reaction of acid with base)
Rust is hydrated iron(III) oxide, a basic oxide. It reacts with the acid to form a salt and water, which is a neutralization (acid + base) reaction.
Which of the following additives could improve the quality of steel?
Answer: D (Chromium and nickel)
Alloying steel with chromium and nickel makes stainless steel, which is hard and resists rusting. Sulphur and phosphorus make steel brittle.
Sodium hydroxide is prepared commercially from sodium chloride solution by
Answer: A (electrolysis using mercury as cathode)
In the mercury cathode cell, brine is electrolysed; sodium forms an amalgam with the mercury, which then reacts with water to give sodium hydroxide.
A sample of a substance containing only C and H burns in excess O2 to yield 4.4 g of CO2 and 2.7 g of H2O. The empirical formula of the substance is [C = 12, O = 16, H = 1]
Answer: A (CH3)
4.4 g CO2 = 0.1 mol, so 0.1 mol C. 2.7 g H2O = 0.15 mol, so 0.3 mol H. C : H = 0.1 : 0.3 = 1 : 3, giving CH3.
An undesirable paraffin in the petroleum industry which is particularly prone to knocking is
Answer: B (n-heptane)
Straight-chain n-heptane knocks badly and is given an octane number of 0; iso-octane is the standard for good anti-knock fuel (100).
The IUPAC nomenclature of the organic compound with the above structural formula is
CH3—CH(CH3)—CH(CH2CH3)—CH2—CH(CH3)—CH3
Answer: A (3-ethyl-2, 5 - dimethylhexane)
The longest chain has six carbons (hexane). Numbering from the end that gives the lowest locants puts methyl groups on C2 and C5 and an ethyl group on C3: 3-ethyl-2,5-dimethylhexane.
The reaction of an alkanol with an alkanoic acid in the presence of concentrated H2SO4 will produce an
Answer: B (alkanoate)
An alkanol and an alkanoic acid react (esterification) to give an ester, which is named as an alkyl alkanoate.
The final product of the reaction of ethyne with hydrogen iodine is
Answer: A (CH3 — CHI2)
Ethyne adds two molecules of HI. The first gives CH2=CHI; by Markovnikov's rule the second iodine goes on the carbon that already has one, giving CH3CHI2.
CH3CH2—C(CH3)2—CH3
How many more isomers of the compound above can be obtained?
Answer: B (4)
The compound is 2,2-dimethylbutane, C6H14. Hexane has five structural isomers in all, so there are four others.
Synthetic detergents are preferred to soap for laundry using hard water because
Answer: B (the calcium salts of detergents are water soluble)
Soap reacts with Ca2+ and Mg2+ in hard water to form insoluble scum. The calcium salts of detergents stay dissolved, so no scum forms and no detergent is wasted.
The synthetic rubber obtained by the polymerization of chlorobutadiene in the presence of sodium is called
Answer: D (neoprene.)
Neoprene is the polymer of chlorobutadiene (chloroprene). Teflon comes from tetrafluoroethene and polythene from ethene.
25 cm3 of 0.02 M KOH neutralized 0.03 g of a monobasic organic acid having the general formula CnH2n + 1COOH. The molecular formula of the acid is [C = 12, H = 1, O = 16].
Answer: C (CH3COOH)
Moles of KOH = 0.025 × 0.02 = 0.0005 = moles of acid. Molar mass = 0.03/0.0005 = 60 g mol−1. 14n + 46 = 60 gives n = 1: CH3COOH.
When Fehling's solution is added to two isomeric carbonyl compounds X and Y with the molecular formula C5H10O, compound X gives a red precipitate while Y does not react. It can be inferred that X is
Answer: B (CH3CH2CH2CH2CHO)
Fehling's solution gives a red precipitate (Cu2O) with aldehydes but not with ketones. Only B has the –CHO group (pentanal); the others are ketones.
C6H5CH3
The compound above contains
Answer: D (sp3 and sp2 hybridized carbon atoms.)
The six ring carbons of the benzene ring are sp2 hybridized, while the CH3 carbon has four single bonds and is sp3.
CH3CH2CH(CH3)CHO
The compound above is the product of the oxidation of
Answer: B (2-methylbutan - 1 - ol)
The product is an aldehyde, 2-methylbutanal. Aldehydes come from oxidizing primary alkanols with the same carbon skeleton: 2-methylbutan-1-ol.