IB Chemistry Practice Test 18

Correct: (C)

Isotopes are atoms of the same element (same proton number) that differ in the number of neutrons, and so in mass number.

Correct: (D)

Ar = (0.75 × 35) + (0.25 × 37) = 26.25 + 9.25 = 35.5.

Correct: (A)

A Ca atom has 20 electrons; losing 2 electrons to form Ca^2+ leaves 18.

Correct: (B)

Cl^- has 18 electrons, filling the 3p subshell: [Ne] 3s^2 3p^6, the argon configuration.

Correct: (B)

Moles: C = 85.7/12.0 = 7.14; H = 14.3/1.0 = 14.3. Ratio 1 : 2, so the empirical formula is CH2 (C2H4 is a molecular formula).

Correct: (A)

V = 0.25 × 22.7 = 5.68, approximately 5.7 dm^3.

Correct: (D)

Moles of acid = 0.0250 × 2.0 = 0.050 mol; concentration = 0.050/0.250 = 0.20 mol dm^-3.

Correct: (C)

Percentage yield = (6.0/8.0) × 100 = 75%.

Correct: (A)

Moles of HCl = 0.0200 × 0.100 = 2.00 × 10^-3; the reaction is 1 : 1, so concentration of NaOH = 2.00 × 10^-3/0.0250 = 0.080 mol dm^-3.

Correct: (B)

Each formula unit of CaCl2 contains two chloride ions, so 0.50 × 2 = 1.0 mol of Cl^-.

Correct: (B)

An oxygen molecule has a double bond (O=O); H2, HCl and F2 all have single bonds.

Correct: (C)

Oxygen has two bonding pairs and two lone pairs; the lone pairs push the bonds together, giving a bent shape with an angle of about 104.5°.

Correct: (D)

Sulfur has five electron domains (four bonds and one lone pair) in a trigonal bipyramidal arrangement; the lone pair gives a seesaw molecular shape.

Correct: (A)

H2 is non-polar with no permanent dipole and no hydrogen bonding, so only weak London dispersion forces act between its molecules.

Correct: (C)

Each carbon in graphite forms three bonds, leaving one electron per atom delocalised within the layer, free to carry charge.

Correct: (D)

Silicon has a giant covalent structure like diamond, needing a lot of energy to break; the others are simple molecular substances.

Correct: (A)

In the solid the ions are locked in the lattice; on melting they become mobile and can carry charge to the electrodes.

Correct: (B)

The C=O bonds are polar, but the linear shape places them in opposite directions, so the dipoles cancel exactly.

Correct: (D)

Each carbon has three electron domains in a trigonal planar arrangement, corresponding to sp^2 hybridisation.

Correct: (C)

A double bond is made of one sigma bond (end-on overlap) and one pi bond (sideways overlap of p orbitals).

Correct: (A)

Down Group 1 the valence electron is further from the nucleus and more shielded, so it is lost more easily and reactivity increases; melting point, ionisation energy fall and radius increases.

Correct: (B)

The halogens are molecular; larger molecules have stronger London dispersion forces, so melting points increase down the group, while electronegativity, oxidising power and ionisation energy decrease.

Correct: (C)

Oxides become more acidic across Period 3: Na2O and MgO are basic, Al2O3 is amphoteric, and SO3 is acidic.

Correct: (D)

With three oxygen atoms at -2 each: 2x + 3(-2) = 0, so x = +3.

Correct: (D)

The ammonia molecules donate lone pairs to the central Cu^2+ ion, so NH3 is the ligand.

Correct: (A)

The -OH group bonded to a carbon chain is the hydroxyl group, characteristic of alcohols.

Correct: (B)

Alkanes fit the general formula CnH2n+2; C5H12 gives 2(5) + 2 = 12 hydrogens, so it is an alkane.

Correct: (C)

The chain has four carbons and the double bond starts at carbon 1: but-1-ene.

Correct: (D)

An element in its standard state is the reference point, so its ΔHf° is defined as zero.

Correct: (A)

Heat absorbed by the process = 100.0 × 4.18 × 5.00 = 2090 J; per mole = 2090/0.100 = 20 900 J = +20.9 kJ mol^-1 (positive, because the temperature fell: the dissolving is endothermic).

Correct: (B)

Adding the equations gives 2 C + 2 O2 → 2 CO2 with ΔH = -222 + (-566) = -788 kJ; halving gives -394 kJ mol^-1.

Correct: (C)

Bonds broken: 4(C-H) + 2(O=O) = 1652 + 996 = 2648 kJ. Bonds formed: 2(C=O) + 4(O-H) = 1486 + 1852 = 3338 kJ. ΔH = 2648 – 3338 = -690 kJ mol^-1.

Correct: (A)

The reaction forms 2 mol of NH3, so ΔH = 2 × (-46) = -92 kJ; the elements are in their standard states and contribute zero.

Correct: (B)

A gas is produced from a solid, greatly increasing the disorder (and the number of particles), so ΔS is positive; the other processes all decrease disorder.

Correct: (C)

ΔG = ΔH – TΔS; with ΔH negative and -TΔS also negative, ΔG is negative at every temperature.

Correct: (D)

ΔG = ΔH – TΔS = -200 – (500 × -0.100) = -200 + 50 = -150 kJ mol^-1.

Correct: (B)

More particles in the same volume collide more often, so successful collisions happen more frequently; particle energy, speed and activation energy are unchanged.

Correct: (C)

The rate depends on [A] squared (second order in A) and is independent of [B] (zero order), so rate = k[A]^2.

Correct: (D)

For rate = k[A], k = rate/[A] = (mol dm^-3 s^-1)/(mol dm^-3) = s^-1.

Correct: (A)

The total area (total number of molecules) stays the same, but the distribution spreads to higher energies and the peak lowers; the activation energy does not change.

Correct: (C)

Kc = [B]/[A] = 0.80/0.20 = 4.0.

Correct: (D)

The forward reaction is exothermic, so raising the temperature shifts the equilibrium to the left and Kc decreases (both reaction rates in fact increase).

Correct: (A)

A catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached sooner with no change in Kc or composition.

Correct: (B)

Q > Kc means there are too many products relative to equilibrium, so the net reaction proceeds in reverse until Q = Kc.

Correct: (A)

Rate is directly proportional to [A], which is the definition of first-order behaviour.

Correct: (B)

From ln k = -(Ea/R)(1/T) + ln A, the gradient is -Ea/R; the intercept is ln A.

Correct: (C)

Let x = [Cl2] at equilibrium: Kc = x^2/(1.00 – x) = 0.50 gives x^2 + 0.50x – 0.50 = 0, whose positive root is x = 0.50 mol dm^-3.

Correct: (D)

There are equal numbers of moles of gas on both sides (2 and 2), so a pressure change does not favour either side and Kc is unchanged.

Correct: (C)

NaOH fully dissociates, so [OH^-] = 1.0 × 10^-3 and pOH = 3.0; pH = 14.0 – 3.0 = 11.0.

Correct: (D)

A conjugate acid is formed by adding one H+; adding H+ to NH3 gives NH4^+.

Correct: (A)

pH = pKa + log([base]/[acid]); with equal concentrations the log term is log(1) = 0, so pH = pKa = 4.76.

Correct: (B)

At equivalence the solution contains NH4Cl; the NH4^+ ion is a weak acid, so the solution is acidic with pH less than 7.

Correct: (D)

With potassium at +1 and four oxygens at -2 each: +1 + x + 4(-2) = 0, so x = +7.

Correct: (A)

At the anode, bromide ions are oxidised to bromine (2 Br^- → Br2 + 2 e^-); lead forms at the cathode.

Correct: (B)

The strongest reducing agent is the metal most easily oxidised, which has the most negative E° value: zinc.

Correct: (C)

Ultraviolet light causes homolytic fission of Cl2 into two chlorine radicals, which then start the free radical substitution chain.

Correct: (B)

Esterification gives an ester named from the alcohol (ethyl) and the acid (ethanoate): ethyl ethanoate, plus water.

Correct: (C)

HBr adds across the double bond (electrophilic addition), giving bromoethane.

Correct: (D)

HCO3^- can donate a proton (forming CO3^2-) and accept a proton (forming H2CO3), so it is amphiprotic.

Correct: (A)

Cu^2+ gains electrons (it is reduced to Cu), so it is the oxidising agent; Zn is the reducing agent.