Is this an official English VWO examination?

No. It is an independent English-language MCQ study adaptation. The government's Landsexamen page identifies Dutch as the examination language, while language subjects and school project response choices require separate attention.

Does every VWO candidate study every topic here?

No. Subject packages, profiles and mathematics options differ. This bank includes selected foundations and extensions; use your current PTA and official subject programme to identify the work required for your route.

What assessment does the bank leave for separate practice?

Prepare full written responses, oral tests, practical assignments and the Profielwerkstuk under the current school or Landsexamen instructions. No additional universal standalone case-study component was verified.

Is age 19 a minimum for all VWO candidates?

No. The decree specifies age 19 by April 1 as a partial-exam admission alternative, alongside prior examination experience. Route-specific admission must be confirmed with the examination committee.

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Sample Sint Maarten Eindexamen VWO Practice Questions

Try these sample questions to review concepts for the Sint Maarten Eindexamen VWO exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 40+ question experience with AI tutoring.

1A satellite has a circular orbit 600 km above Earth. Use Earth's mass 5.97 * 10^24 kg, radius 6.37 * 10^6 m and G = 6.674 * 10^-11 N*m^2/kg^2. What is its orbital speed to three significant figures?

A.7.56 * 10^3 m/s

B.7.91 * 10^3 m/s

C.5.72 * 10^7 m/s

D.2.58 * 10^3 m/s

Explanation: Equating gravitational and centripetal forces gives v = sqrt(G*M/r). The orbital radius is 6.97 * 10^6 m, giving v = sqrt(5.72 * 10^7) = 7.56 * 10^3 m/s.

2A particle with charge +3.2 * 10^-19 C travels at 4.0 * 10^6 m/s perpendicular to a uniform 0.50 T magnetic field. What magnetic force magnitude acts on it?

A.6.4 * 10^-13 N

B.1.28 * 10^-12 N

C.3.2 * 10^-13 N

D.1.28 * 10^-18 N

Explanation: For perpendicular motion the magnetic force is |q|*v*B. Substituting gives 3.2 * 10^-19 * 4.0 * 10^6 * 0.50 = 6.4 * 10^-13 N.

3In a double-slit experiment, wavelength is 632.8 nm, slit separation is 0.250 mm and screen distance is 2.00 m. Using the small-angle approximation, what is the bright-fringe spacing?

A.5.06 mm

B.2.53 mm

C.10.1 mm

D.1.27 mm

Explanation: Fringe spacing is lambda*L/d. Converting both lengths to meters gives (632.8 * 10^-9 * 2.00)/(0.250 * 10^-3) = 0.0050624 m, or 5.06 mm.

4A photon has frequency 1.50 * 10^15 Hz. Use h = 6.626 * 10^-34 J*s and 1 eV = 1.602 * 10^-19 J. What is its energy?

A.6.20 eV

B.9.94 eV

C.3.10 eV

D.4.14 eV

Explanation: The photon energy is h*f = 9.939 * 10^-19 J. Dividing by 1.602 * 10^-19 J/eV gives 6.20 eV.

5For a radioactive sample with half-life 8.02 days, initial activity is 64.0 MBq. What activity remains after 24.06 days?

A.8.0 MBq

B.16.0 MBq

C.4.0 MBq

D.2.0 MBq

Explanation: The elapsed time is exactly three half-lives under the given data. Activity becomes 64.0*(1/2)^3 = 8.0 MBq.

6For 2 SO2(g) + O2(g) <=> 2 SO3(g), equilibrium concentrations are 0.20, 0.10 and 0.40 mol/L respectively. Using Kc = [SO3]^2/([SO2]^2*[O2]), what numerical value results?

A.40

B.8

C.100

D.4

Explanation: The numerator is 0.40^2 = 0.16 and the denominator is 0.20^2*0.10 = 0.004. Dividing gives the concentration-based value 40.

7A buffer's final concentrations are [CH3COOH] = 0.10 mol/L and [CH3COO-] = 0.20 mol/L. Use pKa = 4.74 and pH = pKa + log10([A-]/[HA]). What is the approximate pH?

A.5.04

B.4.44

C.4.74

D.5.74

Explanation: The base-to-acid ratio is 0.20/0.10 = 2. Therefore pH = 4.74 + log10(2) = 4.74 + 0.301 = 5.04.

8A reaction has rate law rate = k*[A]^2*[B]. At fixed temperature, [A] doubles while [B] halves. How does the rate change?

A.It doubles

B.It quadruples

C.It is unchanged

D.It halves

Explanation: Doubling [A] multiplies the rate by four because its exponent is two. Halving [B] multiplies it by one-half, giving an overall factor of two.

9Which substance is the final electron acceptor in the mitochondrial electron transport chain during aerobic respiration?

A.Oxygen, reduced to water

B.NAD+, reduced to NADH

C.Pyruvate, reduced to lactate

D.Carbon dioxide, reduced to glucose

Explanation: Oxygen accepts electrons at the end of the mitochondrial chain. Together with protons it forms water, allowing electron transport to continue.

10During DNA replication, what is the main role of topoisomerase ahead of the replication fork?

A.Relieving torsional strain caused by DNA unwinding

B.Synthesizing short RNA primers

C.Joining adjacent Okazaki fragments

D.Adding complementary nucleotides to the new DNA strand

Explanation: Unwinding DNA creates torsional strain ahead of the fork. Topoisomerases transiently break and reseal DNA to relieve that strain.

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