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100+ Free TASC Chemistry Level 4 Practice Questions
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Master TASC Chemistry Level 4 (CHM415115) with 100 aligned practice questions spanning thermochemistry, chemical equilibrium, redox electrochemistry, organic synthesis, and analytical spectroscopy. These practice questions are an English-language multiple-choice study aid for revising course knowledge and are not an official TASC paper or a simulation of the written external examination format.
Sample TASC Chemistry Level 4 Practice Questions
Try these sample questions to test your TASC Chemistry Level 4 exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1What is the ground-state electron configuration of the iron(II) ion, Fe²⁺ (atomic number Z = 26)?
A.[Ar] 3d⁶
B.[Ar] 4s² 3d⁴
C.[Ar] 4s¹ 3d⁵
D.[Ar] 3d⁵ 4s¹
Explanation: When transition metals form cations, electrons are lost first from the highest principal energy level (4s orbital) before the 3d subshell. Neutral Fe is [Ar] 4s² 3d⁶, so losing two electrons yields [Ar] 3d⁶.
2Across Period 3 of the Periodic Table (from Na to Ar), first ionization energy generally increases. Which statement best explains this trend?
A.Nuclear charge increases while shielding by inner shell electrons remains relatively constant, increasing electrostatic attraction on valence electrons.
B.Atomic radius increases across the period, bringing outer electrons closer to the nucleus.
C.The number of inner electron shells increases, increasing electron-electron repulsion.
D.Electronegativity decreases across the period, making it easier to gain electrons.
Explanation: As nuclear charge (number of protons) increases from Na to Ar across Period 3, extra electrons enter the same valence shell (n = 3), resulting in minimal change in core shielding. The higher effective nuclear charge exerts a stronger electrostatic pull on valence electrons, raising first ionization energy.
3Why does electronegativity decrease descending Group 17 (the halogens) from fluorine to iodine?
A.Atomic radius increases and core electron shielding increases, diminishing the nuclear attraction for bonding electrons.
B.Nuclear charge decreases down the group, reducing electron attraction.
C.Valence electrons occupy subshells of lower principal energy level n.
D.Halogens gain extra core shells without any change in atomic radius.
Explanation: Descending Group 17, each successive element possesses an additional electron shell, increasing atomic radius and core electron shielding. Consequently, the nucleus exerts a weaker electrostatic attraction on shared bonding pairs of electrons, causing electronegativity to decrease.
4Which dominant intermolecular force accounts for the surprisingly high boiling point of water (100 °C) compared to hydrogen sulfide (-60 °C)?
A.Hydrogen bonding between highly polar O-H bonds.
B.Permanent dipole-dipole forces between sulfur atoms.
C.Dispersion forces resulting from large electron clouds.
D.Covalent network bonding throughout liquid water.
Explanation: Oxygen is highly electronegative and small, creating strongly polarized O-H bonds. Hydrogen bonding forms between the lone pairs on oxygen and the electropositive hydrogen atoms of adjacent H₂O molecules, requiring substantial thermal energy to overcome compared to weaker dipole-dipole forces in H₂S.
5In a potential energy profile diagram for an exothermic reaction, how do the enthalpy of products and sign of ΔH compare to the reactants?
A.Products have lower enthalpy than reactants, and ΔH is negative.
B.Products have higher enthalpy than reactants, and ΔH is positive.
C.Products have lower enthalpy than reactants, and ΔH is positive.
D.Products have higher enthalpy than reactants, and ΔH is negative.
Explanation: Exothermic reactions release heat energy to the surroundings. Thus, the enthalpy of products (H_products) is lower than that of reactants (H_reactants), giving ΔH = H_products - H_reactants < 0 (negative).
6A 100.0 g sample of water is heated from 20.0 °C to 45.0 °C by burning ethanol. Given the specific heat capacity of water is 4.18 J g⁻¹ K⁻¹, calculate the heat energy q absorbed by the water.
A.10.45 kJ
B.18.81 kJ
C.8.36 kJ
D.104.5 kJ
Explanation: Using q = m c ΔT: q = 100.0 g × 4.18 J g⁻¹ K⁻¹ × (45.0 - 20.0 K) = 100.0 × 4.18 × 25.0 = 10450 J = 10.45 kJ.
7Given the thermochemical equations: (1) C(s) + O₂(g) → CO₂(g) ΔH₁ = -393.5 kJ mol⁻¹ (2) CO(g) + ½ O₂(g) → CO₂(g) ΔH₂ = -283.0 kJ mol⁻¹ Calculate ΔH for the partial oxidation: C(s) + ½ O₂(g) → CO(g).
A.-110.5 kJ mol⁻¹
B.-676.5 kJ mol⁻¹
C.+110.5 kJ mol⁻¹
D.+676.5 kJ mol⁻¹
Explanation: According to Hess's Law, reverse equation (2): CO₂(g) → CO(g) + ½ O₂(g) (ΔH = +283.0 kJ mol⁻¹). Adding equation (1) yields: C(s) + O₂(g) + CO₂(g) → CO₂(g) + CO(g) + ½ O₂(g), simplifying to C(s) + ½ O₂(g) → CO(g). Total ΔH = -393.5 + 283.0 = -110.5 kJ mol⁻¹.
8Using mean bond enthalpies: H-H = 436 kJ mol⁻¹, Cl-Cl = 242 kJ mol⁻¹, H-Cl = 431 kJ mol⁻¹, calculate ΔH for: H₂(g) + Cl₂(g) → 2 HCl(g).
A.-184 kJ mol⁻¹
B.+184 kJ mol⁻¹
C.-247 kJ mol⁻¹
D.+247 kJ mol⁻¹
Explanation: ΔH = Σ(bond energies broken) - Σ(bond energies formed). Energy broken = H-H + Cl-Cl = 436 + 242 = 678 kJ mol⁻¹. Energy formed = 2 × H-Cl = 2 × 431 = 862 kJ mol⁻¹. ΔH = 678 - 862 = -184 kJ mol⁻¹.
9Which chemical equation correctly represents the standard molar enthalpy of formation (ΔH°f) of liquid ethanol, C₂H₅OH(l)?
A.2 C(s, graphite) + 3 H₂(g) + ½ O₂(g) → C₂H₅OH(l)
B.2 C(g) + 6 H(g) + O(g) → C₂H₅OH(l)
C.C₂H₄(g) + H₂O(l) → C₂H₅OH(l)
D.4 C(s, graphite) + 6 H₂(g) + O₂(g) → 2 C₂H₅OH(l)
Explanation: Standard enthalpy of formation (ΔH°f) is defined as the enthalpy change when EXACTLY 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions (25 °C, 100 kPa). Graphite, H₂(g), and O₂(g) are standard states.
10When 0.0200 mol of methanol (CH₃OH) is completely burned in excess oxygen, 14.5 kJ of heat is released. What is the standard molar enthalpy of combustion (ΔH°c) of methanol in kJ mol⁻¹?
A.-725 kJ mol⁻¹
B.+725 kJ mol⁻¹
C.-0.290 kJ mol⁻¹
D.-14.5 kJ mol⁻¹
Explanation: ΔH°c = -q / n. ΔH°c = -14.5 kJ / 0.0200 mol = -725 kJ mol⁻¹. Combustion is exothermic, so ΔH is negative.
About the TASC Chemistry Level 4 Practice Questions
Verified exam format metadata for TASC Chemistry Level 4 (CHM415115) External Assessment is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.