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Grade 11 · Science · curriculum collection

Grade 11 Science, NGSS domains

8 units for Grade 11 science, organised by NGSS disciplinary core idea domain: Physical Sciences, Life Sciences. Each unit has a short lesson, a practice worksheet and a review worksheet, both with answer keys.

8 units16 worksheets192 questions with answers
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Answer-checked and approved on 29 October 2026

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Aligned by StudyBrix at domain level. Not an official publication of, or endorsed by, the standards' authors. Framework: Next Generation Science Standards.

Units in teaching order

01

PS Physical Sciences

Motion and energy

Calculations with velocity, acceleration, force, work, energy and power.

Short lesson: Motion, forces and energy

Study guide for Grades 7 to 11: 20 key ideas, including distance, displacement and speed; 3 practice questions.

Zero net force means zero acceleration, not necessarily zero velocity. An object can move steadily in equilibrium.

  • Distance. The total length of the path travelled by an object.
  • Displacement. The directed change from initial position to final position.
  • Speed. The rate at which distance is travelled.
  • Velocity. The rate of change of position, including direction.

Read the full study guide

Worksheets

A practice worksheet and a review worksheet, 12 and 12 questions.

Worksheets (PDF)Answer keys (PDF)
Practice worksheet: 12 questions

  1. Velocity changes from 15 m/s to 26 m/s in 11 s. Find average acceleration in m/s².

  2. A perpendicular force of 558 N acts over 18 m². Find the pressure in Pa.

  3. A device transfers 261 J in 9 s. Find average power in W.

  4. A spring has constant 220 N/m and extension 14 cm within its linear range. Find the restoring-force magnitude in N.

  5. A 18 kg object has speed 14 m/s. Find its classical kinetic energy in J.

  6. A device receives 2300 J and delivers 1840 J of useful energy. Find its efficiency as a percentage.

  7. A mass of 13 kg has velocity -21 m/s. Find its signed momentum in kg·m/s.

  8. An object of mass 33 kg accelerates at 7 m/s². Find the net force in N.

  9. A 11 kg sample has specific heat 400 J/(kg·°C). Its temperature rises 23°C with no phase change. Find the energy gained in J.

  10. A constant 58 N force acts in the same direction as a 21 m displacement. Find work in J.

  11. Lift a 7 kg mass vertically by 11 m. Using g = 10 m/s², find the increase in gravitational potential energy in J.

  12. Motion, forces and energy: which term fits this description? A contact force perpendicular to the contact surface.

    1. Gravitational potential energy
    2. Scalar
    3. Normal force
    4. Inertia
Answer key
  1. 1 m/s² (26 − 15)/11 = 1
  2. 31 Pa 558 ÷ 18 = 31
  3. 29 W 261 ÷ 9 = 29
  4. 30.8 N 220 × 0.14 = 30.8
  5. 1764 J ½ × 18 × 14² = 1764
  6. 80 % 1840/2300 × 100 = 80%
  7. -273 kg·m/s 13 × -21 = -273
  8. 231 N 33 × 7 = 231 N
  9. 101200 J 11 × 400 × 23 = 101200
  10. 1218 J 58 × 21 = 1218
  11. 770 J 7 × 10 × 11 = 770
  12. Normal force Normal force: A contact force perpendicular to the contact surface.
Review worksheet: 12 questions

  1. Velocity changes from 1 m/s to -35 m/s in 12 s. Find average acceleration in m/s².

  2. A 2 kg object has speed 2 m/s. Find its classical kinetic energy in J.

  3. A 8 kg sample has specific heat 200 J/(kg·°C). Its temperature rises 3°C with no phase change. Find the energy gained in J.

  4. A spring has constant 340 N/m and extension 16 cm within its linear range. Find the restoring-force magnitude in N.

  5. A mass of 44 kg has velocity -5 m/s. Find its signed momentum in kg·m/s.

  6. A device receives 6700 J and delivers 2010 J of useful energy. Find its efficiency as a percentage.

  7. An object of mass 20 kg accelerates at 13 m/s². Find the net force in N.

  8. A constant 69 N force acts in the same direction as a 36 m displacement. Find work in J.

  9. A perpendicular force of 980 N acts over 14 m². Find the pressure in Pa.

  10. Lift a 38 kg mass vertically by 11 m. Using g = 10 m/s², find the increase in gravitational potential energy in J.

  11. A device transfers 432 J in 9 s. Find average power in W.

  12. Motion, forces and energy: which term fits this description? A physical quantity described by magnitude and direction.

    1. Vector
    2. Acceleration
    3. Tension
    4. Impulse
Answer key
  1. -3 m/s² (-35 − 1)/12 = -3
  2. 4 J ½ × 2 × 2² = 4
  3. 4800 J 8 × 200 × 3 = 4800
  4. 54.4 N 340 × 0.16 = 54.4
  5. -220 kg·m/s 44 × -5 = -220
  6. 30 % 2010/6700 × 100 = 30%
  7. 260 N 20 × 13 = 260 N
  8. 2484 J 69 × 36 = 2484
  9. 70 Pa 980 ÷ 14 = 70
  10. 4180 J 38 × 10 × 11 = 4180
  11. 48 W 432 ÷ 9 = 48
  12. Vector Vector: A physical quantity described by magnitude and direction.

Related ready-made worksheets: Grade 11 Motion & energy worksheet

02

PS Physical Sciences

Electricity & circuits

Charge, current, voltage, resistance and electrical power.

Short lesson: Charge transferred

Study guide for Grades 8 to 12: 2 worked examples and 2 practice questions. This relation uses constant current in amperes and time in seconds.

Q = It

This relation uses constant current in amperes and time in seconds.

Worked example

A constant current of 6 A flows for 88 s. How much charge passes in C?

6 × 88 = 528

Answer: 528 C

Read the full study guide

Worksheets

A practice worksheet and a review worksheet, 12 and 12 questions.

Worksheets (PDF)Answer keys (PDF)
Practice worksheet: 12 questions

  1. A constant current of 1.4 A flows for 96 s. Find transferred charge. Round to 3 decimal places if needed. Answer in C.

  2. A theoretical 32 V ideal source supplies parallel ohmic resistors 36 Ω and 27 Ω. Find total source current. Round to 3 decimal places if needed. Answer in A.

  3. A constant current of 4 A flows for 74 s. How much charge passes in C?

  4. An ideal ohmic resistor of 14 Ω carries 9 A. Find the voltage across it in V.

  5. An ohmic resistor of 2 Ω has potential difference 44 V. Find current. This is a paper calculation, not a construction instruction. Round to 3 decimal places if needed. Answer in A.

  6. Resistors of 10 Ω and 42 Ω are connected in parallel. Find the equivalent resistance in Ω as an exact fraction or integer.

  7. In a theoretical circuit, 34 Ω and 39 Ω are in series across an ideal 12 V source. Find their total electrical power. Round to 3 decimal places if needed. Answer in W.

  8. A battery supplies 2.2 A to two parallel branches. One branch carries 1.9 A. What current flows in the other branch? (Assume ideal wires and an ideal battery.)

  9. A model device uses constant power 375 W for 3 h. Find the energy in kilowatt-hours. Round to 3 decimal places if needed. Answer in kWh.

  10. Three ideal resistors of 87 Ω, 48 Ω and 23 Ω are in series. Find total resistance in Ω.

  11. A steady DC device operates at 49 V and 15 A. Find power in W.

  12. A 12 V ideal source is across a 3 Ω resistor. Find current in that branch.

Answer key
  1. 134.4 C Q = It 1.4 × 96 = 134.4
  2. 2.074 A I = V/R₁ + V/R₂ 32/36 + 32/27 = 2.074
  3. 296 C 4 × 74 = 296
  4. 126 V 9 × 14 = 126
  5. 22 A I = V/R 44 / 2 = 22
  6. 105/13 Ω 10 × 42 / (10 + 42) = 105/13
  7. 1.973 W P = V²/(R₁ + R₂) 12² / (34 + 39) = 1.973
  8. 0.3 A 2.2 − 1.9 = 0.3
  9. 1.125 kWh E(kWh) = P(W) × t(h)/1000 375 × 3 / 1000 = 1.125
  10. 158 Ω 87 + 48 + 23 = 158
  11. 735 W 49 × 15 = 735
  12. 4 A I = V/R = 12/3 = 4 A.
Review worksheet: 12 questions

  1. An ohmic resistor of 45 Ω has potential difference 12 V. Find current. This is a paper calculation, not a construction instruction. Round to 3 decimal places if needed. Answer in A.

  2. An ideal ohmic resistor of 25 Ω carries 6 A. Find the voltage across it in V.

  3. A constant current of 1.5 A flows for 104 s. Find transferred charge. Round to 3 decimal places if needed. Answer in C.

  4. In a theoretical circuit, 32 Ω and 8 Ω are in series across an ideal 20 V source. Find their total electrical power. Round to 3 decimal places if needed. Answer in W.

  5. A constant current of 6 A flows for 33 s. How much charge passes in C?

  6. A theoretical 26 V ideal source supplies parallel ohmic resistors 32 Ω and 17 Ω. Find total source current. Round to 3 decimal places if needed. Answer in A.

  7. Resistors of 44 Ω and 49 Ω are connected in parallel. Find the equivalent resistance in Ω as an exact fraction or integer.

  8. Three ideal resistors of 86 Ω, 11 Ω and 86 Ω are in series. Find total resistance in Ω.

  9. A model device uses constant power 2000 W for 13 h. Find the energy in kilowatt-hours. Round to 3 decimal places if needed. Answer in kWh.

  10. Two branches are connected in parallel to a battery. The branch currents are 0.6 A and 0.8 A. What current flows through the battery? (Assume ideal wires and an ideal battery.)

  11. A steady DC device operates at 50 V and 13 A. Find power in W.

  12. Waves and electric circuits: which term fits this description? A material that strongly resists charge flow under ordinary conditions.

    1. Insulator
    2. Electromagnetic wave
    3. Series circuit
    4. Closed circuit
Answer key
  1. 0.267 A I = V/R 12 / 45 = 0.267
  2. 150 V 6 × 25 = 150
  3. 156 C Q = It 1.5 × 104 = 156
  4. 10 W P = V²/(R₁ + R₂) 20² / (32 + 8) = 10
  5. 198 C 6 × 33 = 198
  6. 2.342 A I = V/R₁ + V/R₂ 26/32 + 26/17 = 2.342
  7. 2156/93 Ω 44 × 49 / (44 + 49) = 2156/93
  8. 183 Ω 86 + 11 + 86 = 183
  9. 26 kWh E(kWh) = P(W) × t(h)/1000 2000 × 13 / 1000 = 26
  10. 1.4 A 0.6 + 0.8 = 1.4
  11. 650 W 50 × 13 = 650
  12. Insulator Insulator: A material that strongly resists charge flow under ordinary conditions.

Related ready-made worksheets: Grade 11 Electricity & circuits worksheet · Grade 11 Electricity & circuits (Physics) worksheet

03

PS Physical Sciences

Analytical chemistry

Concentration, titration, chromatography and yield.

Short lesson: Concentration and dilution

Study guide for Grades 9 to 12: 4 key ideas, including concentration, dilution and volume; 2 worked examples and 3 practice questions. Count solute amount before and after adding solvent.

  • Concentration. Molar concentration c = n/V uses solution volume in liters.
  • Dilution. In a simple dilution without reaction or solute loss, c₁V₁ = c₂V₂.
  • Volume. Final solution volume is not the same as the volume of added solvent.
  • Units. Convert milliliters to liters for c = n/V; ratios may use matching volume units.

Worked example

Find concentration for 0.15 mol in 0.50 L of solution.

c = 0.15/0.50 = 0.30.

Answer: 0.3 mol/L

Read the full study guide

Worksheets

A practice worksheet and a review worksheet, 12 and 12 questions.

Worksheets (PDF)Answer keys (PDF)
Practice worksheet: 12 questions

  1. The theoretical dry pure product mass is 92 g; the actual dry pure product mass is 48 g. Find percentage yield. Round to 3 decimal places if needed. Answer in %.

  2. A model solution contains 0.7 mol in a final volume of 1.16 L. Find its molar concentration. Round to 3 decimal places if needed. Answer in mol/L.

  3. A 130 g sample is 99% desired substance by mass. Find the desired substance mass. Round to 3 decimal places if needed. Answer in g.

  4. A supplied valid-range calibration is S = 19c + 0.4. The signal is 8.4. Find c in the calibration's arbitrary concentration units. Round to 3 decimal places if needed.

  5. An aqueous solution has concentration 1.5 mol/L. A model aliquot has volume 123 mL. Find the solute amount. Round to 3 decimal places if needed. Answer in mol.

  6. Measured from the same origin, a spot travels 31 mm and the solvent front 56 mm. Find Rf. Round to 3 decimal places if needed.

  7. A 58 mL aliquot at 1.6 mol/L is diluted to 638 mL total without reaction or loss. Find the final concentration. Round to 3 decimal places if needed. Answer in mol/L.

  8. A model solution contains 36 g solute and 281 g solvent. Find the solute mass percentage. Round to 3 decimal places if needed. Answer in %.

  9. In the supplied complete neutralization equation H₂A + 2OH⁻ → A²⁻ + 2H₂O, how much OH⁻ reacts with 0.47 mol H₂A? Round to 3 decimal places if needed. Answer in mol.

  10. A sample is 80% desired substance by mass and has mass 25 g. Find the desired substance mass.

  11. Five repeated measurements of the mass of the same sample (g) are: Trial 1 = 5.96; Trial 2 = 7.17; Trial 3 = 7.15; Trial 4 = 7.17; Trial 5 = 7.15. Leave out the outlier (trial 1) and find the mean of the other four. Round to 2 decimal places.

  12. What is the solvent in a dilute aqueous salt solution?

    1. The dissolved salt.
    2. The container.
    3. Water.
Answer key
  1. 52.174 % yield % = actual/theoretical × 100 48/92 × 100 = 52.174
  2. 0.603 mol/L c = n/V 0.7/1.16 = 0.603
  3. 128.7 g component mass = sample mass × purity/100 130 × 99/100 = 128.7
  4. 0.421 c = (S − b)/m for S = mc + b (8.4 − 0.4)/19 = 0.421
  5. 0.185 mol n = cV; V(L) = V(mL)/1000 1.5 × (123/1000) = 0.185
  6. 0.554 Rf = spot distance / solvent-front distance 31/56 = 0.554
  7. 0.145 mol/L c₂ = c₁V₁/V₂ 1.6 × 58/638 = 0.145
  8. 11.356 % mass % = solute mass / solution mass × 100 36/(36 + 281) × 100 = 11.356
  9. 0.94 mol H₂A + 2OH⁻ → A²⁻ + 2H₂O; n(OH⁻) = 2n(H₂A) 2 × 0.47 = 0.94
  10. 20 g 0.80 × 25 = 20 g.
  11. 7.16 g (7.17 + 7.15 + 7.17 + 7.15) ÷ 4 ≈ 7.16
  12. Water. Aqueous means water is the solvent.
Review worksheet: 12 questions

  1. A supplied valid-range calibration is S = 12c + 0.3. The signal is 3.1. Find c in the calibration's arbitrary concentration units. Round to 3 decimal places if needed.

  2. An aqueous solution has concentration 0.7 mol/L. A model aliquot has volume 365 mL. Find the solute amount. Round to 3 decimal places if needed. Answer in mol.

  3. Measured from the same origin, a spot travels 113 mm and the solvent front 117 mm. Find Rf. Round to 3 decimal places if needed.

  4. A model solution contains 23 g solute and 194 g solvent. Find the solute mass percentage. Round to 3 decimal places if needed. Answer in %.

  5. A 232 g sample is 84% desired substance by mass. Find the desired substance mass. Round to 3 decimal places if needed. Answer in g.

  6. A model solution contains 2 mol in a final volume of 0.21 L. Find its molar concentration. Round to 3 decimal places if needed. Answer in mol/L.

  7. A 68 mL aliquot at 0.9 mol/L is diluted to 136 mL total without reaction or loss. Find the final concentration. Round to 3 decimal places if needed. Answer in mol/L.

  8. The theoretical dry pure product mass is 163 g; the actual dry pure product mass is 95 g. Find percentage yield. Round to 3 decimal places if needed. Answer in %.

  9. In the supplied complete neutralization equation H₂A + 2OH⁻ → A²⁻ + 2H₂O, how much OH⁻ reacts with 0.3 mol H₂A? Round to 3 decimal places if needed. Answer in mol.

  10. Five repeated measurements of the mass of the same sample (g) are: Trial 1 = 17.9; Trial 2 = 17.41; Trial 3 = 17.36; Trial 4 = 17.38; Trial 5 = 17.37. Which trial is most likely an outlier?

    1. Trial 2
    2. Trial 3
    3. Trial 4
    4. Trial 1
    5. Trial 5

  11. How many moles are in 0.20 L of a 0.40 mol/L solution?

  12. How many moles are in 25.0 mL of 0.100 mol/L titrant?

Answer key
  1. 0.233 c = (S − b)/m for S = mc + b (3.1 − 0.3)/12 = 0.233
  2. 0.256 mol n = cV; V(L) = V(mL)/1000 0.7 × (365/1000) = 0.256
  3. 0.966 Rf = spot distance / solvent-front distance 113/117 = 0.966
  4. 10.599 % mass % = solute mass / solution mass × 100 23/(23 + 194) × 100 = 10.599
  5. 194.88 g component mass = sample mass × purity/100 232 × 84/100 = 194.88
  6. 9.524 mol/L c = n/V 2/0.21 = 9.524
  7. 0.45 mol/L c₂ = c₁V₁/V₂ 0.9 × 68/136 = 0.45
  8. 58.282 % yield % = actual/theoretical × 100 95/163 × 100 = 58.282
  9. 0.6 mol H₂A + 2OH⁻ → A²⁻ + 2H₂O; n(OH⁻) = 2n(H₂A) 2 × 0.3 = 0.6
  10. Trial 1 Trial 1 (17.9 g) is far from the other four values.
  11. 0.08 mol n = cV = 0.40 × 0.20 = 0.080.
  12. 0.0025 mol 0.100 × 0.0250 = 0.00250 mol.

Related ready-made worksheets: Grade 11 Analytical chemistry worksheet

04

PS Physical Sciences

Physical chemistry

Energy changes, reaction rates, equilibrium and gases.

Short lesson: Dynamic chemical equilibrium

Study guide for Grades 10 to 12: 4 key ideas, including dynamic, concentrations and constant; 2 worked examples and 3 practice questions. Do not confuse equal rates with equal amounts.

  • Dynamic. At equilibrium the forward and reverse rates are equal, while both directions continue.
  • Concentrations. Macroscopic concentrations remain constant under unchanged conditions, but need not be equal to one another.
  • Constant. For a specified reaction, the equilibrium constant depends on temperature.
  • Expression. Write the appropriate activities; simple school concentration expressions often omit pure solids and liquids.

Worked example

At dynamic equilibrium, what is equal?

Equal opposing rates maintain constant macroscopic composition.

Answer: The forward and reverse rates.

Read the full study guide

Worksheets

A practice worksheet and a review worksheet, 12 and 12 questions.

Worksheets (PDF)Answer keys (PDF)
Practice worksheet: 12 questions

  1. For A ⇌ B, use Kc = [B]/[A]. Equilibrium concentrations are [A] = 1.9 and [B] = 2.6 mol/L. Find Kc. Round to 3 decimal places if needed.

  2. An ideal gas has initial pressure 100 kPa and volume 3.7 L. At fixed amount and temperature its volume becomes 8.5 L. Find the new pressure. Round to 3 decimal places if needed. Answer in kPa.

  3. For one reaction as written, a bond model requires 565 kJ/mol to break bonds and releases 625 kJ/mol on forming bonds. Estimate ΔH. Round to 3 decimal places if needed. Answer in kJ/mol.

  4. Use PV = nRT and R = 8.314 kPa·L/(mol·K). Find the ideal-gas volume for 2 mol at 367 K and 114 kPa. Round to 3 decimal places if needed. Answer in L.

  5. A reactant concentration decreases from 1.43 to 1.18 mol/L in 56 s. Find its positive average disappearance rate. Round to 3 decimal places if needed. Answer in mol/(L·s).

  6. The given standard reduction potentials are 0.57 V for the cathode and -0.27 V for the anode. Find E°cell. Round to 3 decimal places if needed. Answer in V.

  7. In the dilute school model, [H₃O⁺] = 2 × 10^−3 mol/L. Calculate pH. Round to 3 decimal places if needed.

  8. At 25 °C, use the dilute model pH + pOH = 14. A solution has pH 3.2. Find pOH. Round to 3 decimal places if needed.

  9. What changes the equilibrium constant of a specified reaction in the usual school model?

    1. Changing the ink color of the equation.
    2. Changing temperature.
    3. Adding a catalyst at the same temperature.

  10. Why should amounts be recalculated after a significant addition before using the ratio expression?

  11. In an explicitly first-order rate law r = k[A], doubling [A] while k stays fixed multiplies r by what factor?

  12. Given reduction potentials +0.80 V at the cathode and −0.20 V at the anode, find E°cell.

Answer key
  1. 1.368 Kc = [B]/[A] for A ⇌ B 2.6/1.9 = 1.368
  2. 43.529 kPa P₂ = P₁V₁/V₂ 100 × 3.7/8.5 = 43.529
  3. -60 kJ/mol ΔH ≈ bonds broken − bonds formed 565 − 625 = -60
  4. 53.53 L V = nRT/P; R = 8.314 kPa·L/(mol·K) 2 × 8.314 × 367/114 = 53.53
  5. 0.004 mol/(L·s) r = (c₁ − c₂)/Δt (1.43 − 1.18)/56 = 0.004
  6. 0.84 V E°cell = E°cathode − E°anode 0.57 − (-0.27) = 0.84
  7. 2.699 pH ≈ −log₁₀([H₃O⁺]/(1 mol/L)) 3 − log₁₀(2) = 2.699
  8. 10.8 pOH ≈ 14 − pH at 25 °C 14 − 3.2 = 10.8
  9. Changing temperature. Temperature affects the equilibrium constant.
  10. Added acid or base changes the quantities of the conjugate pair. Initial concentrations need not remain the relevant ratio.
  11. 2 The rate is directly proportional to [A] in the given law.
  12. 1 V 0.80 − (−0.20) = 1.00 V.
Review worksheet: 12 questions

  1. For one reaction as written, a bond model requires 328 kJ/mol to break bonds and releases 277 kJ/mol on forming bonds. Estimate ΔH. Round to 3 decimal places if needed. Answer in kJ/mol.

  2. Use PV = nRT and R = 8.314 kPa·L/(mol·K). Find the ideal-gas volume for 2 mol at 294 K and 106 kPa. Round to 3 decimal places if needed. Answer in L.

  3. A reactant concentration decreases from 0.98 to 0.87 mol/L in 23 s. Find its positive average disappearance rate. Round to 3 decimal places if needed. Answer in mol/(L·s).

  4. For A ⇌ B, use Kc = [B]/[A]. Equilibrium concentrations are [A] = 1.6 and [B] = 0.5 mol/L. Find Kc. Round to 3 decimal places if needed.

  5. An ideal gas has initial pressure 78 kPa and volume 8.6 L. At fixed amount and temperature its volume becomes 8 L. Find the new pressure. Round to 3 decimal places if needed. Answer in kPa.

  6. The given standard reduction potentials are 0.36 V for the cathode and -0.51 V for the anode. Find E°cell. Round to 3 decimal places if needed. Answer in V.

  7. In the dilute school model, [H₃O⁺] = 6 × 10^−4 mol/L. Calculate pH. Round to 3 decimal places if needed.

  8. At 25 °C, use the dilute model pH + pOH = 14. A solution has pH 9.9. Find pOH. Round to 3 decimal places if needed.

  9. For A ⇌ B with Kc = [B]/[A], [B] = 0.60 and [A] = 0.20 mol/L. Calculate Kc.

  10. Why must pressure and volume units match the chosen R?

  11. At 25 °C with pH + pOH = 14, find pOH for pH 5.

  12. What does a catalyst lower in the simple energy-pathway description?

    1. The atomic numbers.
    2. The activation barrier.
    3. The amount of energy conserved.
Answer key
  1. 51 kJ/mol ΔH ≈ bonds broken − bonds formed 328 − 277 = 51
  2. 46.119 L V = nRT/P; R = 8.314 kPa·L/(mol·K) 2 × 8.314 × 294/106 = 46.119
  3. 0.005 mol/(L·s) r = (c₁ − c₂)/Δt (0.98 − 0.87)/23 = 0.005
  4. 0.313 Kc = [B]/[A] for A ⇌ B 0.5/1.6 = 0.313
  5. 83.85 kPa P₂ = P₁V₁/V₂ 78 × 8.6/8 = 83.85
  6. 0.87 V E°cell = E°cathode − E°anode 0.36 − (-0.51) = 0.87
  7. 3.222 pH ≈ −log₁₀([H₃O⁺]/(1 mol/L)) 4 − log₁₀(6) = 3.222
  8. 4.1 pOH ≈ 14 − pH at 25 °C 14 − 9.9 = 4.1
  9. 3 0.60/0.20 = 3 for this idealized expression.
  10. R has units that connect the other quantities; mismatched units create a wrong numerical result. Dimensional consistency is essential.
  11. 9 14 − 5 = 9.
  12. The activation barrier. It provides an alternative pathway.

Related ready-made worksheets: Grade 11 Physical chemistry worksheet

05

LS Life Sciences

Cells & microscopy

Cells, organelles, microscopes and cell division.

Short lesson: Cell cycle and mitosis

Study guide for Grades 9 to 12: 4 key ideas, including cycle, replication and mitosis; 2 worked examples and 3 practice questions. Separate DNA replication from chromosome separation.

  • Cycle. The cell cycle includes growth, DNA replication and division-related stages.
  • Replication. DNA is copied before mitosis; mitosis itself organizes and separates the copied chromosomes.
  • Mitosis. In a typical stable diploid model, mitosis preserves chromosome number in daughter nuclei.
  • Control. Division is regulated; an unregulated increase in cells is not the normal purpose of every cell type.

Worked example

When is DNA copied for a typical mitotic division?

Replication precedes mitotic chromosome separation.

Answer: Before mitosis, during S phase.

Read the full study guide

Worksheets

A practice worksheet and a review worksheet, 12 and 12 questions.

Worksheets (PDF)Answer keys (PDF)
Practice worksheet: 12 questions

  1. A printed image is 69 mm long at 40× magnification. Find the actual length in micrometers. Round to 3 decimal places if needed. Answer in µm.

  2. Which structure is found in plant cells but not in animal cells: the mitochondrion or the chloroplast?

    1. the mitochondrion
    2. the chloroplast

  3. Supplied plant-tissue data show initial mass 2.4 g and final mass 3.8 g. Find signed percentage mass change. No experiment is required. Round to 3 decimal places if needed. Answer in %.

  4. At diffusion equilibrium, do molecules stop moving?

    1. Only solute particles disappear.
    2. Yes, every particle stops.
    3. No.

  5. A cell is modeled as a cube of side 7 µm. Find surface area divided by volume. This geometry is a simplified model. Round to 3 decimal places if needed. Answer in µm⁻¹.

  6. Does every mature cell divide at the same rate?

    1. Only if cells are the same color.
    2. Yes, all once per hour.
    3. No.

  7. A printed image is 9 mm long at 40× magnification. Find the actual length in micrometers. Round to 3 decimal places if needed. Answer in µm.

  8. Which structure is found in plant cells but not in animal cells: the cell wall or the cell membrane?

    1. the cell wall
    2. the cell membrane

  9. Supplied plant-tissue data show initial mass 11.3 g and final mass 11.8 g. Find signed percentage mass change. No experiment is required. Round to 3 decimal places if needed. Answer in %.

  10. One side of a simplified model has 30 solute units in 10 volume units. Find its concentration.

  11. A cell is modeled as a cube of side 23 µm. Find surface area divided by volume. This geometry is a simplified model. Round to 3 decimal places if needed. Answer in µm⁻¹.

  12. A diploid parent cell has 8 chromosomes. In the standard mitosis model, how many chromosomes does each daughter nucleus receive?

Answer key
  1. 1725 µm actual size (µm) = image size (mm) × 1000 / magnification 69 × 1000/40 = 1725
  2. the chloroplast Cell walls, chloroplasts and a large central vacuole are typical of plant cells; both plant and animal cells have a nucleus, mitochondria, ribosomes and a cell membrane.
  3. 58.333 % (final − initial)/initial × 100 (3.8 − 2.4)/2.4 × 100 = 58.333
  4. No. Random motion continues with no net concentration-driven transfer.
  5. 0.857 µm⁻¹ SA/V = 6s²/s³ = 6/s for a cube 6/7 = 0.857
  6. No. Division rates differ by cell type and conditions.
  7. 225 µm actual size (µm) = image size (mm) × 1000 / magnification 9 × 1000/40 = 225
  8. the cell wall Cell walls, chloroplasts and a large central vacuole are typical of plant cells; both plant and animal cells have a nucleus, mitochondria, ribosomes and a cell membrane.
  9. 4.425 % (final − initial)/initial × 100 (11.8 − 11.3)/11.3 × 100 = 4.425
  10. 3 30/10 = 3 solute units per volume unit.
  11. 0.261 µm⁻¹ SA/V = 6s²/s³ = 6/s for a cube 6/23 = 0.261
  12. 8 Mitosis preserves the chromosome number in this model.
Review worksheet: 12 questions

  1. Supplied plant-tissue data show initial mass 8.3 g and final mass 8.3 g. Find signed percentage mass change. No experiment is required. Round to 3 decimal places if needed. Answer in %.

  2. A printed image is 39 mm long at 50× magnification. Find the actual length in micrometers. Round to 3 decimal places if needed. Answer in µm.

  3. Which structure does a typical bacterial cell have: the mitochondrion or the cell membrane?

    1. the mitochondrion
    2. the cell membrane

  4. A cell is modeled as a cube of side 19 µm. Find surface area divided by volume. This geometry is a simplified model. Round to 3 decimal places if needed. Answer in µm⁻¹.

  5. What is a selectively permeable membrane?

    1. A barrier that always blocks everything.
    2. A wall that ignores molecule properties.
    3. A barrier that permits some substances more readily than others.

  6. Starting with one cell, two complete rounds of division produce how many cells if none die?

  7. Supplied plant-tissue data show initial mass 13.9 g and final mass 14 g. Find signed percentage mass change. No experiment is required. Round to 3 decimal places if needed. Answer in %.

  8. A printed image is 3 mm long at 500× magnification. Find the actual length in micrometers. Round to 3 decimal places if needed. Answer in µm.

  9. Which structure is found in plant cells but not in animal cells: the ribosome or the large central vacuole?

    1. the ribosome
    2. the large central vacuole

  10. A cell is modeled as a cube of side 2 µm. Find surface area divided by volume. This geometry is a simplified model. Round to 3 decimal places if needed. Answer in µm⁻¹.

  11. Osmosis specifically describes net movement of what in these examples?

    1. Water across a selectively permeable membrane.
    2. Any solid object.
    3. Only DNA through a nucleus.

  12. Which outcome fits the standard role of mitosis in growth and repair?

    1. Creating four gametes in all cases.
    2. Halving chromosome number for every daughter cell.
    3. Producing genetically similar daughter nuclei, apart from mutations.
Answer key
  1. 0 % (final − initial)/initial × 100 (8.3 − 8.3)/8.3 × 100 = 0
  2. 780 µm actual size (µm) = image size (mm) × 1000 / magnification 39 × 1000/50 = 780
  3. the cell membrane Bacteria are prokaryotes: they have ribosomes and a cell membrane but no nucleus, mitochondria or Golgi apparatus.
  4. 0.316 µm⁻¹ SA/V = 6s²/s³ = 6/s for a cube 6/19 = 0.316
  5. A barrier that permits some substances more readily than others. Permeability differs by substance and transport mechanism.
  6. 4 Each cell divides into two: one cell becomes two, then those two become four cells.
  7. 0.719 % (final − initial)/initial × 100 (14 − 13.9)/13.9 × 100 = 0.719
  8. 6 µm actual size (µm) = image size (mm) × 1000 / magnification 3 × 1000/500 = 6
  9. the large central vacuole Cell walls, chloroplasts and a large central vacuole are typical of plant cells; both plant and animal cells have a nucleus, mitochondria, ribosomes and a cell membrane.
  10. 3 µm⁻¹ SA/V = 6s²/s³ = 6/s for a cube 6/2 = 3
  11. Water across a selectively permeable membrane. The moving substance is water, and membrane selectivity matters.
  12. Producing genetically similar daughter nuclei, apart from mutations. Mitosis differs from the chromosome-reduction process of meiosis.

Related ready-made worksheets: Grade 11 Cells & microscopy worksheet

06

LS Life Sciences

Human biology & physiology

Body systems: breathing, circulation, digestion and control.

Short lesson: Circulation, vessels and blood components

Study guide for Grades 8 to 11: 4 key ideas, including arteries, veins and capillaries; 2 worked examples and 3 practice questions. Classify vessels by direction relative to the heart.

  • Arteries. Arteries carry blood away from the heart; oxygen content is not the defining rule.
  • Veins. Veins carry blood toward the heart; some contain valves.
  • Capillaries. Capillaries provide exchange pathways between blood and tissues.
  • Components. Red blood cells, white blood cells, platelets and plasma have different roles.

Worked example

What defines an artery?

Direction relative to the heart is the defining rule.

Answer: It carries blood away from the heart.

Read the full study guide

Worksheets

A practice worksheet and a review worksheet, 12 and 12 questions.

Worksheets (PDF)Answer keys (PDF)
Practice worksheet: 12 questions

  1. A fictional circulation model has 61 beats/min and 87 mL/beat. Find its flow in mL/min. These are supplied model values, not diagnostic ranges. Round to 3 decimal places if needed. Answer in mL/min.

  2. A model surface is divided into 8 equal parts of 3 area units each. Find total area.

  3. Why is “arteries always carry oxygenated blood” an unsafe general rule?

  4. How do many endocrine signals travel?

    1. Through body fluids such as blood.
    2. Only through wooden tubes.
    3. Only outside the body.

  5. Why is a classroom blood-glucose example not a personal medical instruction?

  6. What does an antibody bind in the simplified model?

    1. A matching antigenic feature.
    2. Only air molecules in all cases.
    3. Any object regardless of structure.

  7. A fictional circulation model has 52 beats/min and 64 mL/beat. Find its flow in mL/min. These are supplied model values, not diagnostic ranges. Round to 3 decimal places if needed. Answer in mL/min.

  8. Why are thin exchange barriers useful?

    1. They prevent every gas from moving.
    2. They shorten diffusion distance.
    3. They make gas molecules disappear.

  9. Why should a calculated cardiac output from a simplified worksheet not be used to judge an individual's health?

  10. What is a synapse in a basic nervous-system model?

    1. A junction where a neuron communicates with another cell.
    2. Only a bone connection.
    3. A whole organ with no cells.

  11. Which would fit a simple negative-feedback response to an increase?

    1. No relation to the measured quantity.
    2. A response always doubling the increase.
    3. A response tending to decrease the regulated quantity.

  12. Why should the immune system not be reduced to antibodies alone?

Answer key
  1. 5307 mL/min flow = beats per minute × volume per beat 61 × 87 = 5307
  2. 24 8 × 3 = 24 area units.
  3. Pulmonary arteries carry blood toward the lungs with lower oxygen content than pulmonary veins. Use vessel direction as the consistent definition.
  4. Through body fluids such as blood. Hormones can circulate to distant target cells.
  5. It cannot account for an individual's condition, medicines or clinical measurements. Keep the learning model distinct from care decisions.
  6. A matching antigenic feature. Binding depends on molecular compatibility.
  7. 3328 mL/min flow = beats per minute × volume per beat 52 × 64 = 3328
  8. They shorten diffusion distance. Shorter paths can increase diffusion rate.
  9. The model lacks clinical context, measurement accuracy and individual factors. This is educational arithmetic, not a diagnosis.
  10. A junction where a neuron communicates with another cell. Synapses mediate communication between cells.
  11. A response tending to decrease the regulated quantity. The response opposes the deviation in this simplified case.
  12. Cells, barriers, signaling and other components also contribute. Name another relevant defense component.
Review worksheet: 12 questions

  1. A fictional circulation model has 51 beats/min and 71 mL/beat. Find its flow in mL/min. These are supplied model values, not diagnostic ranges. Round to 3 decimal places if needed. Answer in mL/min.

  2. Why does a large exchange surface help gas transfer?

    1. It changes oxygen to a metal.
    2. It provides more area for diffusion.
    3. It removes all need for concentration differences.

  3. A supplied control model has target 10 units and measurement 13 units. What is measurement minus target?

  4. Which is part of many chemical synapses?

    1. Only sound waves in air.
    2. A released chemical messenger interacting with receptors.
    3. A new chromosome created at every message.

  5. Which vessels are associated with exchange with tissues?

    1. Only solid bone canals.
    2. Capillaries.
    3. Only the largest arteries.

  6. A supplied fictional chart shows response level 3 units initially and 12 later. What is the factor increase?

  7. A fictional circulation model has 72 beats/min and 57 mL/beat. Find its flow in mL/min. These are supplied model values, not diagnostic ranges. Round to 3 decimal places if needed. Answer in mL/min.

  8. What helps maintain a diffusion gradient at lungs?

    1. Only bone length.
    2. Only skin color.
    3. Ventilation and blood flow.

  9. Why should a thermostat analogy not be treated as a complete model of human physiology?

  10. Why is saying all hormonal responses are slow too absolute?

  11. Which blood component is closely associated with oxygen transport?

    1. Red blood cells.
    2. Only platelets.
    3. Only dissolved salts.

  12. Which is an example of a physical barrier in innate defense?

    1. Only a worksheet score.
    2. Intact skin.
    3. Only a memory antibody.
Answer key
  1. 3621 mL/min flow = beats per minute × volume per beat 51 × 71 = 3621
  2. It provides more area for diffusion. More area supports greater transfer under comparable conditions.
  3. 3 13 − 10 = +3 units.
  4. A released chemical messenger interacting with receptors. Neurotransmitters bind to appropriate receptors.
  5. Capillaries. Thin capillary walls support exchange.
  6. 4 12/3 = 4; these units are model data, not clinical interpretation.
  7. 4104 mL/min flow = beats per minute × volume per beat 72 × 57 = 4104
  8. Ventilation and blood flow. Air and blood movement replenish and remove gases.
  9. Real organisms have interacting signals, changing conditions and multiple control pathways. Analogies illuminate some features but omit others.
  10. Hormones differ in mechanisms and response timing. The textbook contrast is a generalization, not a universal law.
  11. Red blood cells. Hemoglobin in red cells carries much of the blood's oxygen.
  12. Intact skin. Skin is one component of barrier protection.
07

LS Life Sciences

Genetics & inheritance

DNA, inheritance, genotypes and variation.

Short lesson: DNA base pairing and transcription

Study guide for Grades 9 to 12: 2 worked examples and 2 practice questions. Bases pair by shape and hydrogen bonding: adenine with thymine (uracil in RNA) and cytosine with guanine. Work one base at a time.

DNA: A-T, C-G · RNA: A-U, C-G

Bases pair by shape and hydrogen bonding: adenine with thymine (uracil in RNA) and cytosine with guanine. Work one base at a time.

Worked example

A DNA template strand reads AAGCCT. Which mRNA sequence is transcribed from it, read base by base in the same order?

AAGCCT → UUCGGA (A-U, T-A, C-G, G-C)

Answer: UUCGGA

Read the full study guide

Worksheets

A practice worksheet and a review worksheet, 12 and 12 questions.

Worksheets (PDF)Answer keys (PDF)
Practice worksheet: 12 questions

  1. In the simple Aa × Aa inheritance model with independent outcomes, the probability of aa is 1/4. Among 648 offspring, find the expected aa count, not a guaranteed outcome. Round to 3 decimal places if needed.

  2. Using base-pairing rules, which sequence pairs with the DNA strand ATTTGC, read base by base in the same order?

    1. UAAACG
    2. ATTTGC
    3. TAAACG
    4. AAAACG

  3. In pea plants, the allele for a tall stem (T) is dominant over the allele for a short stem (t). Two plants, Tt × tt, are crossed. What percentage of the offspring are expected to have the genotype Tt?

  4. A fictional diploid organism has 10 chromosomes in a normal somatic nucleus. In the standard meiosis model, how many chromosomes are in a haploid gamete? Round to 3 decimal places if needed.

  5. For Aa × Aa, what fraction of offspring are expected to be aa in the simple model?

  6. Why should a codon table be used rather than assuming every base encodes one amino acid?

  7. In a canonical double-stranded DNA sample, adenine is 16% of all bases. Find the guanine percentage, assuming A = T and C = G. Round to 3 decimal places if needed. Answer in %.

  8. In meiosis II, what usually separates?

    1. Homologous species.
    2. Sister chromatids.
    3. Only all mitochondria.

  9. For RW × RW in that flower model, what percentage is expected to be pink?

  10. Why is a sequence-answer question ambiguous if it ignores strand direction?

  11. In the simple Aa × Aa inheritance model with independent outcomes, the probability of aa is 1/4. Among 688 offspring, find the expected aa count, not a guaranteed outcome. Round to 3 decimal places if needed.

  12. Using base-pairing rules, which sequence pairs with the DNA strand ATCTATGAT, read base by base in the same order?

    1. ATCTATGAT
    2. UAGAUACUA
    3. TAGATACTA
    4. TAGATAATA
Answer key
  1. 162 expected aa count = offspring count × 1/4 648 × 1/4 = 162
  2. TAAACG ATTTGC → TAAACG (A-T, C-G)
  3. 50 % Tt × tt → Tt, Tt, tt, tt → 2/4 = 50%
  4. 5 haploid chromosome count = diploid count/2 10/2 = 5
  5. 1/4 Each parent contributes a with probability one half, giving one quarter.
  6. Codons are triplets, and some specify stop signals rather than amino acids. The mapping is not one base to one amino acid.
  7. 34 % G% = (100 − 2A%)/2 (100 − 2 × 16)/2 = 34
  8. Sister chromatids. The second division separates sister chromatids.
  9. 50 % Two of four equally likely genotype combinations are RW.
  10. A complementary sequence and a reverse complement can be written in different orders. The writing convention must be stated.
  11. 172 expected aa count = offspring count × 1/4 688 × 1/4 = 172
  12. TAGATACTA ATCTATGAT → TAGATACTA (A-T, C-G)
Review worksheet: 12 questions

  1. In the simple Aa × Aa inheritance model with independent outcomes, the probability of aa is 1/4. Among 496 offspring, find the expected aa count, not a guaranteed outcome. Round to 3 decimal places if needed.

  2. A DNA template strand reads ATCTCTTA. Which mRNA sequence is transcribed from it, read base by base in the same order?

    1. GAGAGAAU
    2. TAGAGAAT
    3. ATCTCTTA
    4. UAGAGAAU

  3. In pea plants, the allele for purple flowers (P) is dominant over the allele for white flowers (p). Two plants, PP × pp, are crossed. What percentage of the offspring are expected to have the genotype Pp?

  4. A fictional diploid organism has 82 chromosomes in a normal somatic nucleus. In the standard meiosis model, how many chromosomes are in a haploid gamete? Round to 3 decimal places if needed.

  5. What does haploid mean?

    1. One chromosome set.
    2. No DNA.
    3. Exactly one chromosome in every organism.

  6. A sample has trait values 2, 3, 4, 5 and 6 units. Find its arithmetic mean.

  7. Why can the same gene be present in two cell types while their protein production differs?

  8. In a canonical double-stranded DNA sample, adenine is 28% of all bases. Find the guanine percentage, assuming A = T and C = G. Round to 3 decimal places if needed. Answer in %.

  9. Which describes phenotype?

    1. Only a pair of allele letters.
    2. A chromosome count in every example.
    3. An observed characteristic in context.

  10. The same sample has 20% A and 20% T. What percentage is G if the remainder is equally C and G?

  11. In the simple Aa × Aa inheritance model with independent outcomes, the probability of aa is 1/4. Among 424 offspring, find the expected aa count, not a guaranteed outcome. Round to 3 decimal places if needed.

  12. A DNA template strand reads TTTGTACT. Which mRNA sequence is transcribed from it, read base by base in the same order?

    1. AAACAUGA
    2. TTTGTACT
    3. AAACATGA
    4. AAACCUGA
Answer key
  1. 124 expected aa count = offspring count × 1/4 496 × 1/4 = 124
  2. UAGAGAAU ATCTCTTA → UAGAGAAU (A-U, T-A, C-G, G-C)
  3. 100 % PP × pp → Pp, Pp, Pp, Pp → 4/4 = 100%
  4. 41 haploid chromosome count = diploid count/2 82/2 = 41
  5. One chromosome set. The term refers to sets, not a universal chromosome count.
  6. 4 The sum is 20, divided by five.
  7. Gene expression can be regulated differently in each cell type. Presence of a gene is not identical to its expression level.
  8. 22 % G% = (100 − 2A%)/2 (100 − 2 × 28)/2 = 22
  9. An observed characteristic in context. Phenotype refers to trait expression, which may be influenced by genotype and environment.
  10. 30 % 100 − 40 = 60%, divided equally gives 30% G.
  11. 106 expected aa count = offspring count × 1/4 424 × 1/4 = 106
  12. AAACAUGA TTTGTACT → AAACAUGA (A-U, T-A, C-G, G-C)

Related ready-made worksheets: Grade 11 Genetics & inheritance worksheet

08

LS Life Sciences

Ecology & populations

Ecosystems, food webs, populations and biodiversity.

Short lesson: Reasoning with food chains

Study guide for Grades 9 to 12: 2 worked examples and 2 practice questions. Arrows show the direction energy flows: from the organism eaten to the organism that eats it.

producer → primary → secondary → tertiary consumer

Arrows show the direction energy flows: from the organism eaten to the organism that eats it. In a simple chain, fewer prey usually means fewer predators, and fewer predators usually means more prey in the short term.

Worked example

In this simplified food chain: grass → grasshopper → frog → snake → hawk. At which trophic level is “grass”? (Producers are level 1.)

Count from the producer: grass → grasshopper → frog → snake → hawk.

Answer: 1

Read the full study guide

Worksheets

A practice worksheet and a review worksheet, 12 and 12 questions.

Worksheets (PDF)Answer keys (PDF)
Practice worksheet: 12 questions

  1. A supplied survey counts 21 plants in 7 non-overlapping quadrats, each 1 m². Find the mean density in the sampled area. Round to 3 decimal places if needed. Answer in plants/m².

  2. A fictional population starts at 739. During one interval: 62 births, 22 arrivals, 38 deaths and 18 departures. Find the final count. Round to 3 decimal places if needed.

  3. In this simplified food chain: grass → grasshopper → frog → snake → hawk. At which trophic level is “hawk”? (Producers are level 1.)

  4. A supplied food-web model transfers 11% of 839 kJ to the next level. Find transferred energy. The stated efficiency is not a universal ecological constant. Round to 3 decimal places if needed. Answer in kJ.

  5. Supplied fictional records give M = 61 initially marked, C = 24 in a later sample, and R = 10 marked in that sample. Use N ≈ MC/R under the stated assumptions. No animals are to be handled. Round to 3 decimal places if needed.

  6. Births plus immigration total 30 and deaths plus emigration total 45. What is net change?

  7. Using that mean of 5 plants/m² for a comparable 20 m² area, estimate the plant count.

  8. A supplied survey counts 185 plants in 6 non-overlapping quadrats, each 0.5 m². Find the mean density in the sampled area. Round to 3 decimal places if needed. Answer in plants/m².

  9. A fictional population starts at 353. During one interval: 14 births, 29 arrivals, 9 deaths and 30 departures. Find the final count. Round to 3 decimal places if needed.

  10. In this simplified food chain: willow shrubs → moose → wolf. At which trophic level is “willow shrubs”? (Producers are level 1.)

  11. A supplied food-web model transfers 21% of 1266 kJ to the next level. Find transferred energy. The stated efficiency is not a universal ecological constant. Round to 3 decimal places if needed. Answer in kJ.

  12. Supplied fictional records give M = 62 initially marked, C = 81 in a later sample, and R = 40 marked in that sample. Use N ≈ MC/R under the stated assumptions. No animals are to be handled. Round to 3 decimal places if needed.

Answer key
  1. 3 plants/m² density = total count / total sampled area 21/(7 × 1) = 3
  2. 767 N₂ = N₁ + births + immigration − deaths − emigration 739 + 62 + 22 − 38 − 18 = 767
  3. 5 Count from the producer: grass → grasshopper → frog → snake → hawk.
  4. 92.29 kJ transferred energy = available energy × stated efficiency/100 839 × 11/100 = 92.29
  5. 146.4 N ≈ MC/R 61 × 24/10 = 146.4
  6. -15 30 − 45 = −15 individuals.
  7. 100 5 × 20 = 100 under the representativeness assumption.
  8. 61.667 plants/m² density = total count / total sampled area 185/(6 × 0.5) = 61.667
  9. 357 N₂ = N₁ + births + immigration − deaths − emigration 353 + 14 + 29 − 9 − 30 = 357
  10. 1 Count from the producer: willow shrubs → moose → wolf.
  11. 265.86 kJ transferred energy = available energy × stated efficiency/100 1266 × 21/100 = 265.86
  12. 125.55 N ≈ MC/R 62 × 81/40 = 125.55
Review worksheet: 12 questions

  1. A supplied survey counts 105 plants in 12 non-overlapping quadrats, each 0.5 m². Find the mean density in the sampled area. Round to 3 decimal places if needed. Answer in plants/m².

  2. A fictional population starts at 382. During one interval: 50 births, 28 arrivals, 16 deaths and 37 departures. Find the final count. Round to 3 decimal places if needed.

  3. Supplied fictional records give M = 34 initially marked, C = 73 in a later sample, and R = 19 marked in that sample. Use N ≈ MC/R under the stated assumptions. No animals are to be handled. Round to 3 decimal places if needed.

  4. In this simplified food chain: grass → grasshopper → frog → snake → hawk. At which trophic level is “grasshopper”? (Producers are level 1.)

  5. A supplied food-web model transfers 7% of 1345 kJ to the next level. Find transferred energy. The stated efficiency is not a universal ecological constant. Round to 3 decimal places if needed. Answer in kJ.

  6. What does emigration mean?

    1. Every birth.
    2. Entering it.
    3. Leaving the population area.

  7. A sample estimate is 120 while a later complete count is 100. What is the absolute error?

  8. A supplied survey counts 83 plants in 11 non-overlapping quadrats, each 0.5 m². Find the mean density in the sampled area. Round to 3 decimal places if needed. Answer in plants/m².

  9. A fictional population starts at 723. During one interval: 63 births, 20 arrivals, 59 deaths and 32 departures. Find the final count. Round to 3 decimal places if needed.

  10. Supplied fictional records give M = 54 initially marked, C = 26 in a later sample, and R = 25 marked in that sample. Use N ≈ MC/R under the stated assumptions. No animals are to be handled. Round to 3 decimal places if needed.

  11. In this simplified food chain: grass → rabbit → fox. What is the role of “fox” in this chain?

    1. producer
    2. secondary consumer
    3. primary consumer
    4. tertiary consumer

  12. A supplied food-web model transfers 5% of 650 kJ to the next level. Find transferred energy. The stated efficiency is not a universal ecological constant. Round to 3 decimal places if needed. Answer in kJ.

Answer key
  1. 17.5 plants/m² density = total count / total sampled area 105/(12 × 0.5) = 17.5
  2. 407 N₂ = N₁ + births + immigration − deaths − emigration 382 + 50 + 28 − 16 − 37 = 407
  3. 130.632 N ≈ MC/R 34 × 73/19 = 130.632
  4. 2 Count from the producer: grass → grasshopper → frog → snake → hawk.
  5. 94.15 kJ transferred energy = available energy × stated efficiency/100 1345 × 7/100 = 94.15
  6. Leaving the population area. Emigration is outward movement.
  7. 20 The absolute difference is 20.
  8. 15.091 plants/m² density = total count / total sampled area 83/(11 × 0.5) = 15.091
  9. 715 N₂ = N₁ + births + immigration − deaths − emigration 723 + 63 + 20 − 59 − 32 = 715
  10. 56.16 N ≈ MC/R 54 × 26/25 = 56.16
  11. secondary consumer Level 3 in this chain: secondary consumer.
  12. 32.5 kJ transferred energy = available energy × stated efficiency/100 650 × 5/100 = 32.5

Related ready-made worksheets: Grade 11 Ecology & populations worksheet