All Practice Exams
100+ Free TASC Physical Sciences Level 3 Practice Questions
TASC Physical Sciences Level 3 (Course Code: PSC315118) practice questions are available now; exam metadata is being verified.
✓ No registration✓ No credit card✓ No hidden fees✓ Start practicing immediately
100+ Questions
100% Free
Loading practice questions...
Same family resources
Explore More Tasmanian Assessment, Standards and Certification (TASC) Level 3
Continue into nearby exams from the same family. Each card keeps practice questions, study guides, flashcards, videos, and articles in one place.
2026 Statistics
Key Facts: TASC Physical Sciences Level 3 Exam
TASC Physical Sciences Level 3 (PSC315118) is Tasmania's Year 11/12 physical science course integrating chemistry and physics fundamentals. This 100-question practice bank features detailed worked calculations across stoichiometry, thermochemistry, kinematics, wave optics, DC circuits, and scientific error analysis. 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 Physical Sciences Level 3 Practice Questions
Try these sample questions to test your TASC Physical Sciences Level 3 exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1Radium-226 (\(^{226}_{88}\text{Ra}\)) undergoes alpha decay to form Radon (\(\text{Rn}\)). What are the mass number and atomic number of the resulting Radon isotope?
A.Mass number = 222, Atomic number = 86
B.Mass number = 224, Atomic number = 86
C.Mass number = 222, Atomic number = 87
D.Mass number = 226, Atomic number = 86
Explanation: An alpha particle is a helium nucleus (\(^4_2\text{He}\)), containing 2 protons and 2 neutrons. During alpha decay, the parent nucleus loses 4 in mass number and 2 in atomic number: \(^{226}_{88}\text{Ra} \rightarrow ^{222}_{86}\text{Rn} + ^4_2\text{He}\). Therefore, Radon-222 has mass number 222 and atomic number 86.
2Cobalt-60 has a half-life of 5.27 years. If a radioactive source originally contains 80.0 g of Cobalt-60, how much Cobalt-60 remains after 15.81 years?
A.10.0 g
B.20.0 g
C.40.0 g
D.5.0 g
Explanation: Calculate the number of half-lives elapsed: \(n = \frac{t}{t_{1/2}} = \frac{15.81}{5.27} = 3.0\) half-lives. Using the decay formula \(N(t) = N_0 \left(\frac{1}{2}\right)^n\), remaining mass = \(80.0 \times \left(\frac{1}{2}\right)^3 = 80.0 \times 0.125 = 10.0\text{ g}\).
3Carbon-14 (\(^{14}_6\text{C}\)) undergoes beta-minus (\(\beta^-\)) decay. Which equation correctly describes this nuclear reaction?
A.\(^{14}_6\text{C} \rightarrow ^{14}_7\text{N} + ^0_{-1}\text{e} + \bar{\nu}_e\)
B.\(^{14}_6\text{C} \rightarrow ^{14}_5\text{B} + ^0_{+1}\text{e} + \nu_e\)
C.\(^{14}_6\text{C} \rightarrow ^{10}_4\text{Be} + ^4_2\text{He}\)
D.\(^{14}_6\text{C} \rightarrow ^{13}_6\text{C} + ^1_0\text{n}\)
Explanation: In beta-minus decay, a neutron transforms into a proton, emitting an electron (beta particle, \(^0_{-1}\text{e}\)) and an electron antineutrino (\(\bar{\nu}_e\)). Atomic number increases from 6 to 7 (forming Nitrogen-14), while mass number stays 14: \(^{14}_6\text{C} \rightarrow ^{14}_7\text{N} + ^0_{-1}\text{e} + \bar{\nu}_e\).
4Naturally occurring chlorine consists of two isotopes: \(^{35}\text{Cl}\) (isotopic mass 34.97 u, relative abundance 75.78%) and \(^{37}\text{Cl}\) (isotopic mass 36.97 u, relative abundance 24.22%). What is the relative atomic mass of chlorine?
A.35.45
B.35.97
C.36.00
D.35.00
Explanation: Relative atomic mass \(A_r = \sum (\text{isotopic mass} \times \text{fractional abundance})\). \(A_r = (34.97 \times 0.7578) + (36.97 \times 0.2422) = 26.500 + 8.954 = 35.454 \approx 35.45\).
5According to Valence Shell Electron Pair Repulsion (VSEPR) theory, what is the molecular geometry of a water molecule (\(\text{H}_2\text{O}\))?
A.Bent (V-shaped)
B.Linear
C.Trigonal planar
D.Tetrahedral
Explanation: The central oxygen atom in \(\text{H}_2\text{O}\) has 4 electron pairs (2 bonding pairs with hydrogen and 2 lone pairs). While electron pair geometry is tetrahedral, lone pair repulsion distorts the molecular shape to a bent (V-shaped) geometry with a bond angle of approximately 104.5°.
6Why is carbon dioxide (\(\text{CO}_2\)) a nonpolar molecule despite containing polar C=O bonds?
A.The linear geometry causes the two opposing bond dipoles to cancel each other out.
B.Carbon and oxygen have identical electronegativity values.
C.The molecule forms hydrogen bonds that neutralize bond dipoles.
D.Oxygen atoms donate lone pairs to form nonpolar coordinate covalent bonds.
Explanation: Carbon dioxide has a linear molecular geometry (O=C=O). The two polar C=O bonds have equal dipole moments pointing in exactly opposite directions (180° apart), resulting in a net dipole moment of zero.
7Which type of intermolecular force accounts for the relatively high boiling point of ethanol (\(\text{C}_2\text{H}_5\text{OH}\)) compared to dimethyl ether (\(\text{CH}_3\text{OCH}_3\)) of identical molar mass?
A.Hydrogen bonding
B.Dispersion forces only
C.Dipole-induced dipole forces
D.Covalent network bonding
Explanation: Ethanol possesses a polar hydroxyl group (-OH) with a hydrogen atom directly bonded to oxygen, enabling strong intermolecular hydrogen bonding. Dimethyl ether lacks O-H bonds and can only form weaker dipole-dipole and dispersion forces.
8What is the mass of 0.250 moles of calcium carbonate (\(\text{CaCO}_3\))? (Molar mass of \(\text{CaCO}_3 = 100.09\text{ g/mol}\))
A.25.0 g
B.400 g
C.50.0 g
D.100 g
Explanation: Using the mole-mass formula \(m = n \times M\): \(m = 0.250\text{ mol} \times 100.09\text{ g/mol} = 25.02\text{ g}\), which rounds to 25.0 g.
9A chemist dissolves 11.7 g of sodium chloride (\(\text{NaCl}\)) in distilled water to prepare 500 mL of solution. What is the molar concentration of the solution? (Molar mass of \(\text{NaCl} = 58.44\text{ g/mol}\))
A.0.400 M
B.0.200 M
C.0.0234 M
D.0.800 M
Explanation: First calculate moles of NaCl: \(n = \frac{m}{M} = \frac{11.7}{58.44} = 0.2002\text{ mol}\). Convert volume to liters: \(V = 500\text{ mL} = 0.500\text{ L}\). Molar concentration \(c = \frac{n}{V} = \frac{0.2002}{0.500} = 0.4004\text{ M} \approx 0.400\text{ M}\).
10What volume of a 2.00 M stock \(\text{HCl}\) solution is required to prepare 250 mL of a 0.400 M \(\text{HCl}\) solution?
A.50.0 mL
B.100 mL
C.20.0 mL
D.125 mL
Explanation: Use the dilution equation \(c_1 V_1 = c_2 V_2\): \(2.00\text{ M} \times V_1 = 0.400\text{ M} \times 250\text{ mL}\). Solving for \(V_1\): \(V_1 = \frac{0.400 \times 250}{2.00} = \frac{100}{2.00} = 50.0\text{ mL}\).
About the TASC Physical Sciences Level 3 Practice Questions
Verified exam format metadata for TASC Physical Sciences Level 3 (Course Code: PSC315118) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.