IB Chemistry Practice Test 15

Correct: (B)

n = c × V = 0.200 × (250/1000) = 0.0500 mol.

Correct: (C)

Number of molecules = 0.500 × 6.02 × 10^23 = 3.01 × 10^23. Each molecule has 5 atoms, so atoms = 5 × 3.01 × 10^23 = 1.51 × 10^24.

Correct: (D)

Fe is [Ar] 4s^2 3d^6. The 4s electrons are removed first, then one 3d electron, giving [Ar] 3d^5.

Correct: (A)

V = nRT/p = (0.500 × 8.31 × 273) / 100 000 = 0.0113 m^3 = 11.3 dm^3.

Correct: (C)

The number of molecules is proportional to the amount in mol, and 2.0 mol is the largest amount.

Correct: (B)

Mole ratio: C = 40.0/12.01 = 3.33, H = 6.7/1.01 = 6.6, O = 53.3/16.00 = 3.33. Ratio 1 : 2 : 1, so the empirical formula is CH2O.

Correct: (D)

n = 5.85 / 58.5 = 0.100 mol. c = 0.100 / 0.500 = 0.200 mol dm^-3.

Correct: (A)

Protons = 37 – 20 = 17. The neutral atom has 17 electrons, and the X^- ion gains one more, giving 18 electrons.

Correct: (C)

n(Mg) = 2.43 / 24.3 = 0.100 mol, so n(H2) = 0.100 mol. V = 0.100 × 22.7 = 2.27 dm^3.

Correct: (B)

Mass of N per formula unit = 2 × 14.01 = 28.02. Percentage = (28.02 / 80.06) × 100 = 35.0%.

Correct: (A)

NH4Cl contains ionic bonds between NH4^+ and Cl^- ions, and covalent bonds between the N and H atoms within NH4^+.

Correct: (D)

Sulfur has 5 electron domains (4 bonding pairs and 1 lone pair). The lone pair occupies an equatorial position, giving a seesaw shape.

Correct: (B)

CH3Cl is tetrahedral with different atoms attached to carbon, so the bond dipoles do not cancel. CCl4, BF3 and CO2 are symmetrical and nonpolar.

Correct: (C)

Hydrogen is bonded to the highly electronegative fluorine atom, so molecules are held together by hydrogen bonds.

Correct: (A)

In diamond every carbon atom is bonded tetrahedrally to four others, so all carbons are sp^3. Graphite, graphene and C60 contain sp^2 carbons.

Correct: (D)

Carbon forms two double bonds (O=C=O), so 2 × 4 = 8 electrons are shared around the carbon atom.

Correct: (B)

MgO contains small Mg^2+ and O^2- ions. The combination of double charges and small ionic radii gives the strongest attraction and the most exothermic lattice enthalpy.

Correct: (C)

NH3 has 3 bonding pairs and 1 lone pair. The lone pair repels the bonding pairs more strongly, reducing the angle from 109.5° to 107°.

Correct: (D)

In the solid the ions are locked in the lattice and cannot move. When molten, the ions are free to move to the electrodes and carry charge.

Correct: (A)

NH3 molecules form hydrogen bonds, which need far more energy to break than the weak dispersion forces in the group 14 hydrides, despite its low molar mass.

Correct: (C)

Ionization energy generally increases across a period as the nuclear charge increases and the atomic radius decreases. Argon, at the end of period 3, has the highest value.

Correct: (A)

Transition elements form coloured ions, have variable oxidation states, and often act as catalysts; their d subshells are partially filled in at least one common ion.

Correct: (D)

The -COO- linkage between two carbon groups is the ester functional group.

Correct: (B)

Alkenes contain one C=C double bond and have the general formula CnH2n (alkanes are CnH2n+2 and alkynes CnH2n-2).

Correct: (B)

Atomic radius decreases across period 2 because the increasing nuclear charge pulls the electrons closer. Fluorine is furthest right (before the noble gas) and is smallest.

Correct: (D)

Aluminium oxide reacts with both acids and bases, so it is amphoteric. Na2O is basic, while SO3 and CO2 are acidic.

Correct: (A)

Oxidizing strength decreases down group 17 as atoms get larger and attract electrons less strongly, so F2 is the strongest oxidizing agent.

Correct: (C)

The -NH2 group attached to a carbon chain is the amine functional group.

Correct: (D)

q = mcΔT = 50.0 × 4.18 × 12.0 = 2508 J = 2.51 kJ released. ΔH = -2.51 / 0.0200 = -125 kJ mol^-1 (negative because combustion is exothermic).

Correct: (B)

By Hess's law, subtract the second equation from the first: ΔH = -393 – (-283) = -110 kJ mol^-1.

Correct: (A)

Bonds broken: 436 + 242 = 678 kJ. Bonds formed: 2 × 431 = 862 kJ. ΔH = 678 – 862 = -184 kJ mol^-1.

Correct: (C)

Melting converts the ordered solid into a less ordered liquid, so the entropy increases. The other processes all decrease disorder.

Correct: (B)

ΔG = ΔH – TΔS = -100 – (298 × -0.100) = -100 + 29.8 = -70.2 kJ mol^-1.

Correct: (D)

The standard enthalpy of formation is for 1 mol of a compound formed from its elements in their standard states: C(s) + 2 H2(g) → CH4(g).

Correct: (C)

Heat released = 0.0100 × 1367 = 13.67 kJ = 13 670 J. ΔT = q/(mc) = 13 670 / (200 × 4.18) = 16.3°C.

Correct: (A)

ΔG = ΔH – TΔS. With ΔH > 0 and ΔS > 0, ΔG becomes negative only when TΔS is large enough, i.e. at high temperatures.

Correct: (B)

Doubling [X] doubles the rate, so first order in X. Doubling [Y] multiplies the rate by 4 = 2^2, so second order in Y.

Correct: (D)

k = rate / [A]^2 = (mol dm^-3 s^-1) / (mol dm^-3)^2 = dm^3 mol^-1 s^-1.

Correct: (C)

At higher temperature the distribution spreads out: the peak moves right (higher energy) and becomes lower/flatter, keeping the same total area, and a larger fraction of molecules exceeds the activation energy.

Correct: (A)

For a first-order reaction t1/2 = ln 2 / k = 0.693 / 0.693 = 1.00 min.

Correct: (D)

Products are raised to their stoichiometric coefficients: Kc = [NO2]^2 / [N2O4]. (The coefficient 2 becomes a power, not a multiplier.)

Correct: (B)

Raising the temperature shifts an exothermic equilibrium in the endothermic (reverse) direction, reducing the yield. Higher pressure favours the side with fewer gas moles (products); a catalyst does not affect the yield; removing NH3 shifts the equilibrium forward.

Correct: (A)

Kc = [HI]^2 / ([H2][I2]) = (0.80)^2 / (0.10 × 0.10) = 0.64 / 0.010 = 64.

Correct: (C)

A catalyst lowers the activation energy of both the forward and reverse reactions equally, so equilibrium is reached faster but its position and Kc are unchanged.

Correct: (C)

Average rate = change in concentration / time = (0.100 – 0.060) / 40 = 0.040 / 40 = 1.0 × 10^-3 mol dm^-3 s^-1.

Correct: (A)

Gradient = -Ea/R, so Ea = 6000 × 8.31 = 49 860 J mol^-1 ≈ 50 kJ mol^-1.

Correct: (D)

If Q > K there is too much product relative to equilibrium, so the net reaction moves to the left, towards the reactants, until Q = K.

Correct: (B)

At equilibrium [PCl3] = [Cl2] = 0.20 and [PCl5] = 1.00 – 0.20 = 0.80 mol dm^-3. Kc = (0.20 × 0.20) / 0.80 = 0.050.

Correct: (A)

HCl dissociates completely, so [H^+] = 0.010 mol dm^-3 and pH = -log(0.010) = 2.00.

Correct: (C)

Ka = 10^-pKa = 10^-4.74 = 1.8 × 10^-5.

Correct: (D)

pH = pKa + log([salt]/[acid]) = 4.74 + log(0.100/0.100) = 4.74 + 0 = 4.74.

Correct: (B)

Let the oxidation number of S be x. x + 4(-2) = -2, so x = +6.

Correct: (C)

Cu^2+ + 2e^- → Cu, so 2 mol of electrons are needed per mol of Cu. Q = 2 × 0.0100 × 96 500 = 1930 C.

Correct: (A)

The strongest oxidizing agent has the most positive standard electrode potential, which is F2 at +2.87 V.

Correct: (D)

The Br2 adds across the C=C double bond with no atoms lost, which is an addition reaction.

Correct: (B)

The hydroxide ion donates a lone pair of electrons to the carbon atom, so it is the nucleophile. (Br^- is the leaving group.)

Correct: (A)

A conjugate acid is formed by adding one H^+ to the base: HCO3^- + H^+ → H2CO3.

Correct: (D)

[OH^-] = 0.0500, so pOH = -log(0.0500) = 1.30, and pH = 14.00 – 1.30 = 12.70.

Correct: (C)

[H^+] = sqrt(Ka × c) = sqrt(1.8 × 10^-5 × 0.100) = 1.34 × 10^-3 mol dm^-3, so pH = -log(1.34 × 10^-3) = 2.87.

Correct: (B)

The oxidation number of Sn increases from +2 to +4, so Sn^2+ is oxidized; the species that is oxidized is the reducing agent. (Fe^3+ is the oxidizing agent.)