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100+ Free New Jersey NJSLA Science Grade 11 Practice Questions

Prepare for the New Jersey Student Learning Assessments — Science Grade 11 exam with instant access — no signup required.

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Key Facts: New Jersey NJSLA Science Grade 11 Exam

NJDOE

State Agency

New Jersey Department of Education, Office of Assessments

Cambium Assessment, Inc.

Test Provider

NJDOE, NJSLA Science FAQ

240 min (4 units)

Science Time

NJSLA-Science Units Testing Times

Fixed-form (not adaptive)

Format

NJDOE, NJSLA Science FAQ

English only

Language

NJDOE, NJSLA Science FAQ

100 practice questions

Practice Bank Size

OpenExamPrep

Prepare for the NJSLA Science Grade 11 test with 100 practice questions covering chemistry, physics, biology, and Earth and space science, with detailed explanations.

Sample New Jersey NJSLA Science Grade 11 Practice Questions

Try these sample questions to test your New Jersey NJSLA Science Grade 11 exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A student compares the first ionization energies of sodium (Na) and magnesium (Mg), both in period 3. Magnesium has a higher first ionization energy than sodium. Which statement best explains this trend across the period?
A.As protons are added across a period, the nuclear charge increases while valence electrons stay in the same principal energy level, so they are held more tightly.
B.Valence electrons move into a higher principal energy level as you move left to right, placing them farther from the nucleus.
C.The number of occupied electron shells decreases from sodium to magnesium, so electrons are closer to the nucleus.
D.Electronegativity decreases from sodium to magnesium, weakening the attraction between the nucleus and valence electrons.
Explanation: Across a period, protons are added to the nucleus while valence electrons enter the same principal energy level. Shielding from inner electrons stays nearly constant, so the effective nuclear charge rises and valence electrons are pulled more strongly, increasing ionization energy from Na to Mg.
2A laboratory procedure calls for 0.250 mol of sodium chloride. Using Avogadro's number, 6.02 × 10^23 particles per mole, how many formula units of NaCl should be measured?
A.1.51 × 10^23 formula units
B.6.02 × 10^23 formula units
C.3.01 × 10^23 formula units
D.1.50 × 10^24 formula units
Explanation: The number of particles equals moles multiplied by Avogadro's number: 0.250 × 6.02 × 10^23 = 1.505 × 10^23, which rounds to 1.51 × 10^23 formula units. Avogadro's number converts a mole amount into a count of particles.
3Methane burns in oxygen by the reaction CH4 + 2 O2 → CO2 + 2 H2O. When 4 molecules of CH4 react with 6 molecules of O2, which reactant limits the amount of CO2 produced, and how many CO2 molecules form?
A.Oxygen is limiting; 3 molecules of CO2 form.
B.Methane is limiting; 4 molecules of CO2 form.
C.Oxygen is limiting; 6 molecules of CO2 form.
D.Both are consumed completely; 5 molecules of CO2 form.
Explanation: The equation shows 1 CH4 needs 2 O2. With 4 CH4, 8 O2 would be required, but only 6 O2 are available, so oxygen limits the reaction. Six O2 molecules can support 6 ÷ 2 = 3 CH4 molecules, producing 3 CO2. The limiting reactant sets the maximum product.
4A chemist compares two reactions. Reaction A releases 340 kJ per mole of product and Reaction B absorbs 340 kJ per mole. Which statement correctly relates bond breaking and bond forming to the energy change?
A.In Reaction A the bonds formed release more energy than is required to break bonds in the reactants; in Reaction B the bonds broken require more energy than the bonds formed release.
B.Both reactions release energy because bonds are always formed during a chemical change.
C.In Reaction B the bonds formed release more energy than the bonds broken require, making the change exothermic.
D.The energy change depends only on the number of atoms, not on the bond energies of reactants and products.
Explanation: The enthalpy change equals the energy absorbed to break bonds in reactants minus the energy released when bonds form in products. If forming bonds releases more than breaking absorbs, the reaction is exothermic (A); if breaking requires more than forming releases, it is endothermic (B).
5A power plant uses uranium-235 in a reactor. A U-235 nucleus absorbs a neutron and splits into two smaller nuclei, releasing several neutrons and a large amount of energy. Which process is occurring, and what allows the reaction to sustain itself?
A.Nuclear fission; the neutrons released can split additional U-235 nuclei, forming a chain reaction that is controlled by neutron-absorbing rods.
B.Nuclear fusion; the neutrons combine with U-235 to form heavier elements, releasing energy as mass is converted to energy.
C.Nuclear fission; the nuclei released by the split are absorbed by the control rods and converted directly into electrical energy.
D.Nuclear fusion; the splitting releases neutrons that fuse with each other to form new uranium nuclei, sustaining the reaction.
Explanation: Fission is the splitting of a heavy nucleus into smaller nuclei, releasing neutrons and energy. In a reactor, those neutrons strike other U-235 nuclei, sustaining a chain reaction that control rods moderate by absorbing excess neutrons.
6A student notes that fluorine (F) has a much smaller atomic radius than lithium (Li), even though both are in period 2. Which explanation best accounts for this difference?
A.Fluorine has more protons, so its valence electrons experience a greater effective nuclear charge and are pulled closer to the nucleus.
B.Fluorine has more electron shells than lithium, pushing its valence electrons closer to the nucleus.
C.Lithium has fewer protons, so its electrons are held more tightly than fluorine's electrons.
D.Atomic radius increases from left to right across a period because additional electrons expand the electron cloud.
Explanation: Across a period, protons are added while the valence electrons remain in the same shell. The greater nuclear charge in fluorine pulls the same-shell electrons more tightly, shrinking the atomic radius compared with lithium.
7Iron rusts when it reacts with oxygen and water: 4 Fe + 3 O2 → 2 Fe2O3. A sample containing 8.0 g of iron rusts completely. Which conservation principle lets a student predict the total mass of the iron oxide product, and what additional quantity is needed?
A.Conservation of mass; the mass of oxygen that reacted must be added to the iron mass.
B.Conservation of atoms; only the number of iron atoms is needed because oxygen atoms vanish in the product.
C.Conservation of charge; the charges on the iron ions must be balanced with the oxygen ions.
D.Conservation of energy; the heat released by the reaction must be subtracted from the iron mass.
Explanation: In a chemical reaction, atoms and mass are conserved: the total mass of Fe2O3 equals the mass of iron plus the mass of oxygen that combined with it. The student must know how much oxygen reacted to predict the product mass.
8Carbon-14 decays to nitrogen-14 with a half-life of about 5,730 years. A fossil contains 1/8 of its original carbon-14. Approximately how many half-lives have passed, and how old is the fossil?
A.3 half-lives; about 17,190 years old.
B.8 half-lives; about 45,840 years old.
C.1/8 of a half-life; about 716 years old.
D.2 half-lives; about 11,460 years old.
Explanation: Each half-life halves the remaining carbon-14. Reaching 1/8 requires three halvings: 1 → 1/2 → 1/4 → 1/8, so 3 half-lives have passed. At 5,730 years each, the age is about 17,190 years.
9Diamond and graphite are both made of carbon atoms, yet diamond is extremely hard and graphite is soft and slippery. Which explanation accounts for this difference in properties?
A.Diamond has a three-dimensional covalent network in which every carbon is bonded to four neighbors, while graphite has layered sheets held together by weak interlayer forces that slide past one another.
B.Diamond contains ionic bonds between carbon ions, while graphite contains metallic bonds that allow layers to slide.
C.Graphite's carbon atoms are bonded to four neighbors in a rigid network, while diamond's carbons are bonded to only three.
D.The two forms have different numbers of carbon atoms per molecule, making diamond harder than graphite.
Explanation: Both are pure carbon, but their atoms are arranged differently. Diamond's 3-D covalent network bonds each carbon to four others, producing rigidity; graphite's sheets are joined by strong covalent bonds within a layer but only weak London forces between layers, so the sheets slide and graphite feels soft.
10How many moles of carbon atoms are present in 36.0 g of carbon? Use 12.0 g/mol as the molar mass of carbon.
A.3.00 mol
B.0.333 mol
C.12.0 mol
D.432 mol
Explanation: Moles equal mass divided by molar mass: 36.0 g ÷ 12.0 g/mol = 3.00 mol of carbon atoms. The molar mass converts a measured mass into a mole amount.

About the New Jersey NJSLA Science Grade 11 Exam

The New Jersey Student Learning Assessments — Science (NJSLA-S) Grade 11 assessment measures proficiency on the 2014 New Jersey Student Learning Standards for Science (three-dimensional: Disciplinary Core Ideas, Science and Engineering Practices, and Crosscutting Concepts). It covers matter and interactions, forces, energy, waves, organisms and ecosystems, heredity and evolution, Earth's place and systems, Earth and human activity, and engineering design at the high-school level. It is administered online through the New Jersey Assessments Portal (Cambium Assessment, Inc.) each spring in four 60-minute fixed-form units, in English only. This 100-question practice bank uses four-option MCQs as an English-language study adaptation.

Assessment

Four 60-minute fixed-form units (240 minutes total) with selected-response, multi-select, technology-enhanced items, and phenomena-based item clusters aligned to the three-dimensional NJSLS-Science (DCI, SEP, CCC). Delivered online through the New Jersey Assessments Portal (Cambium Assessment, Inc.). A scientific calculator is embedded. This local bank contains 100 English four-option practice MCQs.

Time Limit

Science 240 minutes (four 60-minute units)

Passing Score

No pass/fail mark. Four performance levels are reported (Level 1 through Level 4).

Exam Fee

No fee to students (New Jersey Department of Education (NJDOE))

New Jersey NJSLA Science Grade 11 Exam Content Outline

NJSLS-Science HS-PS1

Matter and Its Interactions

Periodic trends, reaction outcomes, bond energies, conservation of mass, molecular structure.

NJSLS-Science HS-PS2 and HS-PS3

Forces and Energy

Newton's laws, electromagnetic forces, energy conservation, and energy conversion devices.

NJSLS-Science HS-PS4

Waves

Wave relationships, electromagnetic wave/particle models, and information technology.

NJSLS-Science HS-LS1 and HS-LS2

Organisms and Ecosystems

DNA/proteins, cellular division, cycling of matter, carrying capacity, and biodiversity.

NJSLS-Science HS-LS3 and HS-LS4

Heredity and Evolution

Genetic variation, common ancestry, and the four factors of evolution.

NJSLS-Science HS-ESS1 and HS-ESS2

Earth's Place and Systems

Stellar life cycles, Big Bang, plate tectonics, carbon cycling, and coevolution.

NJSLS-Science HS-ESS3 and HS-ETS1

Earth and Human Activity / Engineering

Resources, hazards, climate, and engineering solutions to global challenges.

How to Pass the New Jersey NJSLA Science Grade 11 Exam

What You Need to Know

  • Passing score: No pass/fail mark. Four performance levels are reported (Level 1 through Level 4).
  • Assessment: Four 60-minute fixed-form units (240 minutes total) with selected-response, multi-select, technology-enhanced items, and phenomena-based item clusters aligned to the three-dimensional NJSLS-Science (DCI, SEP, CCC). Delivered online through the New Jersey Assessments Portal (Cambium Assessment, Inc.). A scientific calculator is embedded. This local bank contains 100 English four-option practice MCQs.
  • Time limit: Science 240 minutes (four 60-minute units)
  • Exam fee: No fee to students

Keys to Passing

  • Work through all 100 available questions
  • Review every answer and explanation
  • Track weak areas and revisit them
  • Use our AI tutor for tough concepts

New Jersey NJSLA Science Grade 11 Study Tips from Top Performers

1Predict reaction outcomes from outermost electrons before balancing equations.
2Track energy across interactions; energy is conserved, only its form changes.
3Tie a phenotype back to its genotype and the selection pressure acting on it.
4Use Earth's internal heat and surface processes together to explain a landform.
5For climate items, separate the cause (emissions) from the modeled effect (forecast).

Frequently Asked Questions

What is the NJSLA Science Grade 11 test?

It is the New Jersey Student Learning Assessments — Science assessment for Grade 11, administered online through the New Jersey Assessments Portal (Cambium Assessment, Inc.) each spring in four 60-minute fixed-form units (240 minutes total) with selected-response, multi-select, technology-enhanced items, and phenomena-based item clusters.

Is the NJSLA Science test adaptive?

No. NJSLA Science is a fixed-form (non-adaptive) assessment. Only the NJSLA-Adaptive ELA and Mathematics assessments are computer-adaptive.

Is a calculator allowed on the NJSLA Science Grade 11 test?

Yes. Students have access to a scientific calculator embedded in the testing platform.

What standards does the NJSLA Science Grade 11 test cover?

It covers the 2014 New Jersey Student Learning Standards for Science, organized three-dimensionally across Disciplinary Core Ideas (Matter, Forces, Energy, Waves, Organisms/Ecosystems, Heredity/Evolution, Earth's Systems, Earth and Human Activity), Science and Engineering Practices, and Crosscutting Concepts.

Is this practice bank affiliated with NJDOE?

No. This is an independent study resource aligned to the public NJSLS-Science. It uses four-option MCQs as a study adaptation; it is not an official NJSLA-Science product or phenomena-cluster format simulation.