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Saxony Abitur Chemistry study materials
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Saxony Abitur Chemistry practice questions

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Question 1Physical chemistry, amounts and reactions

Assign the Oxidationszahl of Mn in MnO4−.

  1. +7 Correct answer.
  2. +2
  3. +4
  4. −1

Explanation

Oxygen is −2 each (4×−2=−8); ion charge −1 ⇒ Mn + (−8) = −1 ⇒ Mn = +7. Permanganate redox titrations are classic quantitative redox tasks.

Question 2Physical chemistry, amounts and reactions

Standard potentials: E°(Zn2+/Zn)=−0.76 V, E°(Cu2+/Cu)=+0.34 V. What is E°cell for Zn|Zn2+||Cu2+|Cu?

  1. 1.10 V Correct answer.
  2. 0.42 V
  3. −1.10 V
  4. 0.76 V

Explanation

E°cell = E°(cathode) − E°(anode) = 0.34 − (−0.76) = 1.10 V. Copper is reduced; zinc is oxidised.

Question 3Physical chemistry, amounts and reactions

During electrolysis of molten NaCl, what is produced at the cathode?

  1. Na metal (Na+ is reduced) Correct answer.
  2. Cl2 gas only
  3. O2 from oxide impurities exclusively
  4. Solid NaCl crystals

Explanation

Cathode = reduction: Na+ + e− → Na. Anode: 2 Cl− → Cl2 + 2 e−. Elektrolyse forces non-spontaneous redox - contrast with galvanic cells.

Question 4Physical chemistry, amounts and reactions

Faraday's law: how many moles of electrons are needed to deposit 0.50 mol Cu from Cu2+?

  1. 1.0 mol e− Correct answer.
  2. 0.50 mol e−
  3. 2.0 mol e−
  4. 0.25 mol e−

Explanation

Cu2+ + 2 e− → Cu ⇒ 0.50 mol Cu needs 1.0 mol electrons. Amount calculations are standard LK electrochemistry.

Question 5Physical chemistry, amounts and reactions

Which metal is more noble according to the Spannungsreihe if E°(Ag+/Ag)=+0.80 V and E°(Fe2+/Fe)=−0.44 V?

  1. Silver - higher (more positive) standard potential Correct answer.
  2. Iron - because it rusts more easily and is therefore nobler
  3. Neither; nobility is unrelated to E°
  4. Iron - lower potential always means nobler

Explanation

More positive E° means greater tendency for the ion to be reduced / the metal is nobler. Fe dissolves more readily; Ag is nobler.

Question 6Physical chemistry, amounts and reactions

Balance in acidic solution: MnO4− + Fe2+ → Mn2+ + Fe3+. How many Fe2+ are oxidised per MnO4−?

  1. 5 Correct answer.
  2. 1
  3. 3
  4. 7

Explanation

Mn goes +7→+2 (gains 5 e−); each Fe2+→Fe3+ loses 1 e− ⇒ 5 Fe2+ per MnO4−. Classic redox stoichiometry for titration.

Question 7Physical chemistry, amounts and reactions

What is the function of a Salzbrücke in a galvanic cell?

  1. Maintain charge balance by ion flow between half-cells Correct answer.
  2. Supply electrons to the external circuit
  3. Increase E° by adding a third electrode reaction
  4. Prevent any ion movement so concentrations stay fixed forever

Explanation

Without a salt bridge (or porous junction), charge separation quickly stops the cell current. Electrons still flow in the wire; ions flow in the bridge.

Question 8Physical chemistry, amounts and reactions

Nernst equation (25 °C) for a metal ion/metal electrode involves which concentration term?

  1. E = E° − (0.059/n) log(1/[Mn+]) for M^{n+} + n e− ⇌ M(s) Correct answer.
  2. E = E° + 0.059 log(n) only, independent of concentration
  3. E is always exactly E° at any concentration
  4. E = Kw / [Mn+]

Explanation

For M^{n+} + n e− ⇌ M, Q = 1/[M^{n+}], so E = E° − (0.059/n) log(1/[M^{n+}]) = E° + (0.059/n) log[M^{n+}]. LK Potenzialberechnungen use this.

Question 9Physical chemistry, amounts and reactions

Korrosion of iron in aerated water is best described as:

  1. A short-circuited galvanic process: Fe oxidised anodically while O2 is reduced cathodically Correct answer.
  2. A purely physical melting of iron
  3. Reduction of Fe to Fe2−
  4. Electrolysis requiring an external power supply by definition

Explanation

Rusting is electrochemical: anodic Fe → Fe2+ + 2 e−; cathodic O2 + 2 H2O + 4 e− → 4 OH− (in neutral/alkaline aerated water). Saxon Alltag/Technik links often use Korrosion.

Question 10Physical chemistry, amounts and reactions

In the Daniell cell, which way do electrons flow in the external wire?

  1. From Zn electrode to Cu electrode Correct answer.
  2. From Cu electrode to Zn electrode
  3. Through the salt bridge only
  4. They do not flow; only Zn2+ ions travel in the wire

Explanation

Zn is oxidised (anode) so electrons leave the Zn electrode toward the Cu cathode where Cu2+ is reduced.

Question 11Physical chemistry, amounts and reactions

Which statement about Elektrolyse is correct?

  1. An external voltage forces a non-spontaneous redox reaction Correct answer.
  2. Electrolysis always has E°cell > 0 without external power
  3. The anode is where reduction occurs during electrolysis
  4. No redox occurs; only heating happens

Explanation

Elektrolyse uses electrical work to drive reactions with negative spontaneous cell potential (e.g. water splitting, metal winning).

Question 12Physical chemistry, amounts and reactions

Oxidising agent (Oxidationsmittel) in a redox reaction is the species that:

  1. Is reduced (accepts electrons) Correct answer.
  2. Is oxidised (donates electrons)
  3. Must be an elemental metal
  4. Always contains oxygen atoms

Explanation

The Oxidationsmittel causes oxidation of another species by accepting electrons - so the agent itself is reduced. Donator-Akzeptor language matches the Saxon Reaktionskonzept.

Question 13Physical chemistry, amounts and reactions

Standard potentials: E°(Cl2/Cl−)=+1.36 V, E°(Br2/Br−)=+1.07 V. Can Cl2 oxidise Br− to Br2 under standard conditions?

  1. Yes, because E°cell = 1.36 − 1.07 = +0.29 V > 0 Correct answer.
  2. No, because bromine is above chlorine in the PSE
  3. Yes, but only if E°cell is negative
  4. No, because both are halogens

Explanation

Spontaneous: higher E° couple is reduced. Cl2 oxidises Br− with E°cell=+0.29 V. Classic Spannungsreihe application.

Question 14Physical chemistry, amounts and reactions

During electrolysis of CuSO4(aq) with inert electrodes, what is typically observed at the anode in acidic solution emphasis?

  1. O2 evolution from water oxidation (when sulfate is not discharged) Correct answer.
  2. Cu metal plating
  3. Solid sulfur deposition as the main product
  4. H2 evolution as the only possible anodic product

Explanation

At the anode, water/OH− oxidation to O2 often wins over sulfate oxidation; at the cathode Cu2+ is reduced to Cu. Students must compare possible electrode reactions.

Question 15Physical chemistry, amounts and reactions

A galvanic cell has E°=1.10 V. If Q increases strongly toward products, Nernst predicts E:

  1. Decreases (log Q term grows) Correct answer.
  2. Always increases without bound
  3. Becomes independent of Q
  4. Equals Kw

Explanation

E = E° − (RT/nF) ln Q; larger Q lowers E until E=0 at equilibrium (Q=K). Concentration cells and non-standard conditions appear at LK depth.

Question 16Physical chemistry, amounts and reactions

Why can aluminium not be won by electrolysis of aqueous Al3+ solutions under ordinary conditions?

  1. Water/H+ is reduced preferentially to H2; Al3+ reduction requires anhydrous melts (e.g. cryolite process) Correct answer.
  2. Al3+ cannot exist in any electrolyte
  3. Aluminium ions are too noble and plate out instantly from water
  4. Electrolysis never produces metals

Explanation

Competing cathode reaction: water reduction. Industrial Al production electrolyses Al2O3 in molten cryolite - Technik/Alltag link in Saxon materials (Wahlbereich Aluminiumgewinnung).

Question 17Physical chemistry, amounts and reactions

What does Le Chatelier's Prinzip vom kleinsten Zwang predict when product concentration is increased at constant T?

  1. The equilibrium shifts toward reactants to partially counteract the added product Correct answer.
  2. Kc increases automatically
  3. The equilibrium shifts toward more products
  4. The reaction stops permanently

Explanation

Adding product stresses the system; the reverse reaction is favoured until a new equilibrium is reached. Kc itself stays constant if T is unchanged.

Question 18Physical chemistry, amounts and reactions

For the equilibrium H2(g) + I2(g) ⇌ 2 HI(g), a mixture at equilibrium has [H2]=0.20, [I2]=0.20, [HI]=1.20 mol/L. Calculate Kc.

  1. 36 Correct answer.
  2. 6.0
  3. 0.028
  4. 1.2

Explanation

Kc = [HI]^2 / ([H2][I2]) = (1.20)^2 / (0.20×0.20) = 1.44 / 0.040 = 36.

Question 19Physical chemistry, amounts and reactions

For N2 + 3 H2 ⇌ 2 NH3 (all g), how does increasing total pressure by decreasing volume shift equilibrium?

  1. Toward NH3 (fewer gas moles: 4 → 2) Correct answer.
  2. Toward N2/H2 (more gas moles)
  3. No shift because Kc is pressure-independent in all cases
  4. Always toward the endothermic side only

Explanation

Le Chatelier: higher pressure favours the side with fewer gas moles. For Haber-Bosch that is the product side. (Kc in concentration units may stay numerically related via T; the position still shifts.)

Question 20Physical chemistry, amounts and reactions

A catalyst is added to a system at equilibrium. What happens?

  1. Forward and reverse rates increase equally; equilibrium position and Kc remain unchanged Correct answer.
  2. Kc increases because the reaction goes further
  3. Only the forward rate increases, so yield rises permanently
  4. Equilibrium is destroyed and the reaction goes to completion

Explanation

Catalysts lower Ea for both directions. They shorten the time to equilibrium but do not change Kc or the Gleichgewichtslage - essential Vernetzung for Saxon Bewertungskompetenz on industrial catalysis.

Question 21Physical chemistry, amounts and reactions

For PCl5(g) ⇌ PCl3(g) + Cl2(g), Kc = 0.042 at a given T. At equilibrium [PCl5]=0.20 and [PCl3]=0.10 mol/L. Find [Cl2].

  1. 0.084 mol/L Correct answer.
  2. 0.42 mol/L
  3. 0.021 mol/L
  4. 2.1 mol/L

Explanation

Kc = [PCl3][Cl2]/[PCl5] ⇒ 0.042 = (0.10·[Cl2])/0.20 ⇒ 0.042 = 0.50·[Cl2] ⇒ [Cl2] = 0.084 mol/L.

Question 22Physical chemistry, amounts and reactions

CaCO3(s) ⇌ CaO(s) + CO2(g). What is the Kc expression (solids omitted as usual)?

  1. Kc = [CO2] Correct answer.
  2. Kc = [CaO][CO2]/[CaCO3]
  3. Kc = [CaCO3]/[CO2]
  4. Kc = [CaO]/[CaCO3]

Explanation

Pure solids are omitted from Kc; only CO2(g) appears, so Kc = [CO2] (or Kp = p(CO2)). This pattern appears in thermal decomposition / Kalkbrennen contexts.

Question 23Physical chemistry, amounts and reactions

The solubility product of AgCl is Ksp = [Ag+][Cl−] = 1.8×10^−10. In pure water, approximate [Ag+].

  1. 1.3×10^−5 mol/L Correct answer.
  2. 1.8×10^−10 mol/L
  3. 3.6×10^−10 mol/L
  4. 1.8×10^−5 mol/L

Explanation

For AgCl(s) ⇌ Ag+ + Cl−, [Ag+]=[Cl−]=s and s^2 = 1.8×10^−10 ⇒ s = √(1.8×10^−10) ≈ 1.34×10^−5 mol/L. Löslichkeitsgleichgewichte are a Saxon Wahlbereich.

Question 24Physical chemistry, amounts and reactions

How many grams of CaCO3 (M=100.1 g/mol) are needed to produce 11.2 L CO2 at STP (Vm=22.4 L/mol) via CaCO3 → CaO + CO2?

  1. 50.0 g Correct answer.
  2. 100 g
  3. 25.0 g
  4. 11.2 g

Explanation

n(CO2)=11.2/22.4=0.500 mol; mole ratio 1:1 ⇒ n(CaCO3)=0.500 mol; m=0.500×100.1≈50.0 g. Combined gas-volume and mass stoichiometry.

Question 25Physical chemistry, amounts and reactions

For an exothermic equilibrium, raising temperature at constant pressure typically:

  1. Decreases K and shifts toward reactants Correct answer.
  2. Increases K and shifts toward products
  3. Leaves K unchanged but always increases yield
  4. Stops the reverse reaction

Explanation

Le Chatelier: heat is like a 'product' of an exothermic reaction; adding heat favours the endothermic reverse direction, so K falls.

Question 26Physical chemistry, amounts and reactions

Limiting reagent: 4.0 mol H2 reacts with 1.0 mol N2 for N2 + 3 H2 → 2 NH3. What is the maximum n(NH3)?

  1. 2.0 mol Correct answer.
  2. 4.0 mol
  3. 2.67 mol
  4. 1.0 mol

Explanation

For N2 + 3 H2 → 2 NH3, 1.0 mol N2 requires 3.0 mol H2. With 4.0 mol H2 available, N2 is limiting. Maximum n(NH3) = 2 × 1.0 = 2.0 mol; 1.0 mol H2 remains unused.

Question 27Physical chemistry, amounts and reactions

Percentage yield: theoretical mass of product is 25.0 g; actual isolated mass is 18.5 g. What is the percentage yield?

  1. 74% Correct answer.
  2. 18.5%
  3. 25%
  4. 135%

Explanation

% yield = (actual/theoretical)×100% = (18.5/25.0)×100% = 74%. Ausbeute discussions appear with Gleichgewichte and industrial processes in Saxon Bewertungskompetenz tasks.

Question 28Physical chemistry, amounts and reactions

At equilibrium for 2 SO2 + O2 ⇌ 2 SO3, removing SO3 continuously will:

  1. Drive the forward reaction to produce more SO3 (Le Chatelier) Correct answer.
  2. Increase Kc at constant temperature
  3. Stop O2 from reacting
  4. Favour the reverse reaction only

Explanation

Removing product shifts equilibrium toward products. The Kontaktverfahren uses conditions and removal strategies to improve effective conversion - a standard Saxon industrial example.

Question 29Physical chemistry, amounts and reactions

A solution is prepared by dissolving 0.100 mol glucose in water to a final volume of 0.500 L. What is the molar concentration?

  1. 0.200 mol/L Correct answer.
  2. 0.100 mol/L
  3. 0.500 mol/L
  4. 0.050 mol/L

Explanation

c = n/V = 0.100 / 0.500 = 0.200 mol/L. Concentration arithmetic underpins titration and equilibrium ICE setups.

Question 30Physical chemistry, amounts and reactions

For the equilibrium CO(g)+H2O(g)⇌CO2(g)+H2(g), Kc=10.0 at T. If initially [CO]=[H2O]=0.100 M and no products, what is [CO] at equilibrium? (Let x = extent)

  1. 0.024 M Correct answer.
  2. 0.076 M
  3. 0.100 M
  4. 0.050 M

Explanation

Kc = x^2/(0.100−x)^2 = 10 ⇒ x/(0.100−x)=√10≈3.162 ⇒ x=3.162(0.100−x) ⇒ x=0.3162−3.162x ⇒ 4.162x=0.3162 ⇒ x≈0.076. [CO]_eq=0.100−0.076=0.024 M.

Question 31Physical chemistry, amounts and reactions

Which statement correctly distinguishes Kinetik from Gleichgewichtslage?

  1. Kinetics describes how fast equilibrium is approached; K describes the composition once rates are equal Correct answer.
  2. A large K always means a fast reaction
  3. A small Ea always means products dominate at equilibrium
  4. Catalysts change K but not rate

Explanation

Saxon Abitur tasks often mix rate and equilibrium ideas; candidates must keep them separate. Industrial compromises (T, p, catalyst) exploit both.

Question 32Physical chemistry, amounts and reactions

What is the Ionenprodukt of water at 25 °C?

  1. Kw = [H3O+][OH−] = 1.0×10^−14 Correct answer.
  2. Kw = [H3O+]/[OH−] = 1.0×10^−7
  3. Kw = [H3O+] + [OH−] = 1.0×10^−14
  4. Kw = 7

Explanation

Autoprotolyse: 2 H2O ⇌ H3O+ + OH− with Kw = 1.0×10^−14 at 25 °C. In pure water [H3O+]=[OH−]=1.0×10^−7. Explicit in Saxon curriculum notes on Autoprotolyse.

Question 33Physical chemistry, amounts and reactions

For a weak acid HA with Ka = 1.8×10^−5 and c=0.10 M, approximate [H3O+] (x^2/c ≈ Ka).

  1. 1.3×10^−3 M Correct answer.
  2. 1.8×10^−5 M
  3. 0.10 M
  4. 4.2×10^−3 M

Explanation

x = √(Ka·c) = √(1.8×10^−6) ≈ √1.8 × 10^−3 ≈ 1.34×10^−3 M. Standard weak-acid approximation used in Saxon pH tasks.

Question 34Physical chemistry, amounts and reactions

In a strong acid–strong base titration, the Äquivalenzpunkt pH is approximately:

  1. 7 (at 25 °C), because salt of strong acid/base does not hydrolyse Correct answer.
  2. Always 3
  3. Always 11
  4. Equal to pKa of a weak acid regardless of the titration type

Explanation

NaCl-type salts from strong acid + strong base give neutral solutions at equivalence. Titrationskurven are standard Säure-Base content.

Question 35Physical chemistry, amounts and reactions

Henderson-Hasselbalch: for a buffer with [A−]/[HA]=1 and pKa=4.76, what is pH?

  1. 4.76 Correct answer.
  2. 7.00
  3. 9.52
  4. 0

Explanation

pH = pKa + log([A−]/[HA]) = 4.76 + log(1) = 4.76. Buffer / Pufferlösungen is a Saxon Wahlbereich and LK depth topic.

Question 36Physical chemistry, amounts and reactions

Which conjugate pair forms a buffer?

  1. CH3COOH / CH3COONa Correct answer.
  2. HCl / NaCl
  3. NaOH / NaCl
  4. HCl / NaOH

Explanation

A buffer needs a weak acid and its conjugate base (or weak base/conjugate acid) in appreciable amounts. Strong acid + salt does not buffer effectively.

Question 37Physical chemistry, amounts and reactions

At the half-equivalence point of a weak acid titration, which relation holds?

  1. pH = pKa (because [HA]=[A−]) Correct answer.
  2. pH = 7 always
  3. pH = 0
  4. pOH = pKa

Explanation

Half-neutralisation: half of HA converted to A−, so [HA]≈[A−] and HH gives pH=pKa. Useful for reading pKa from a Titrationskurve.

Question 38Physical chemistry, amounts and reactions

Identify the conjugate base of HSO4−.

  1. SO4^2− Correct answer.
  2. H2SO4
  3. OH−
  4. H3O+

Explanation

Removing one proton from HSO4− yields SO4^2−. Ampholyte behaviour of hydrogen sulfate is a common exam nuance.

Question 39Physical chemistry, amounts and reactions

Which indicator transition range is most suitable for a strong acid–strong base titration ending near pH 7?

  1. An indicator whose Umschlagbereich includes ~pH 6–8 (e.g. Bromthymolblau) Correct answer.
  2. Methyl orange only (Umschlag ~3–4.4) as the unique correct choice always
  3. Any indicator with Umschlag below pH 2
  4. No indicator can work because equivalence is exactly pH 0

Explanation

Choose an indicator whose colour change overlaps the steep jump of the Titrationskurve. For strong–strong, that jump crosses neutrality.

Question 40Physical chemistry, amounts and reactions

Why is the equivalence-point pH > 7 for titration of a weak acid with strong base?

  1. The salt's anion is a weak base (A− + H2O ⇌ HA + OH−) Correct answer.
  2. Strong base is always left over at equivalence by definition
  3. Weak acids cannot be fully neutralised
  4. Indicators force the pH upward

Explanation

At equivalence, essentially only A− remains; its hydrolysis produces OH−, so the solution is basic. This shapes the Titrationskurve and indicator choice.

Question 41Organic chemistry and macromolecules

Mild oxidation of a primary Alkanol typically yields:

  1. An Alkanal (aldehyde), then further a Carbonsäure on stronger oxidation Correct answer.
  2. Only an alkene via elimination exclusively
  3. A tertiary alcohol
  4. An alkane by hydrogenation

Explanation

Primary alcohols → aldehydes → carboxylic acids depending on conditions/oxidant strength. Secondary → ketones; tertiary resist mild oxidation - standard Saxon organic redox.

Question 42Organic chemistry and macromolecules

What is Veresterung?

  1. Reaction of a Carbonsäure with an Alkanol to form an Ester and water (equilibrium) Correct answer.
  2. Polymerisation of ethene only
  3. Complete combustion of an ester
  4. Reduction of a ketone to a secondary alcohol

Explanation

RCOOH + R′OH ⇌ RCOOR′ + H2O. Equilibrium and catalysis (H+) are classic; Duft-/Aromastoffe Wahlbereich uses ester examples.

Question 43Organic chemistry and macromolecules

Electrophilic addition of HBr to ethene produces:

  1. Bromoethane (CH3CH2Br) Correct answer.
  2. Ethanol
  3. Ethyne
  4. Polyethylene only

Explanation

AE (elektrophile Addition): π-bond attacks H−Br; product is the Haloalkan. Reaktionstypen Substitution/Addition/Eliminierung are explicit Saxon organic foci.

Question 44Organic chemistry and macromolecules

SN1 vs SN2: which substrate favours SN2 most?

  1. Primary Haloalkan with a good nucleophile in polar aprotic solvent Correct answer.
  2. Tertiary Haloalkan in protic solvent with weak nucleophile
  3. Benzene undergoing nitration
  4. Ethene adding Br2

Explanation

SN2 is backside attack favoured at unhindered (primary) carbons. Tertiary substrates favour SN1. Mechanism classification is LK organic reasoning.

Question 45Organic chemistry and macromolecules

Complete combustion of C2H5OH: how many O2 molecules per ethanol molecule in the balanced equation?

  1. 3 Correct answer.
  2. 2
  3. 1
  4. 6

Explanation

C2H5OH + 3 O2 → 2 CO2 + 3 H2O. Stoichiometry of organic combustion is frequently embedded in materialgestützte Aufgaben.

Question 46Organic chemistry and macromolecules

Eliminierung of HBr from a bromoalkane (alcoholic KOH, heat) typically yields:

  1. An alkene Correct answer.
  2. A primary alcohol only
  3. A carboxylic acid
  4. An aromatic ester

Explanation

Eliminierung (E) removes HX to form a C=C bond - one of the three named Reaktionstypen alongside Substitution and Addition.

Question 47Organic chemistry and macromolecules

Isomers: C2H6O can be ethanol or dimethyl ether. What differs?

  1. Connectivity of atoms (Konstitutionsisomerie) leading to different functional groups and properties Correct answer.
  2. Only the radioactive isotopes present
  3. Nothing: they are identical compounds
  4. Only the crystal lattice type of a salt

Explanation

Konstitutionsisomere share molecular formula but differ in Strukturformel - ethanol (Alkanol) vs methoxymethane (ether) is the textbook pair.

Question 48Organic chemistry and macromolecules

In esterification equilibrium, removing water as it forms will:

  1. Shift equilibrium toward ester (Le Chatelier) Correct answer.
  2. Decrease Kc at constant temperature
  3. Prevent any ester from forming
  4. Convert the mechanism into radical polymerisation

Explanation

Product removal drives Veresterung forward. Links organic synthesis to Gleichgewicht - typical Vernetzung in Saxon tasks.

Question 49Organic chemistry and macromolecules

How many structural isomers exist for C4H10?

  1. 2 (n-butane and isobutane/methylpropane) Correct answer.
  2. 4
  3. 1
  4. 8

Explanation

Butane and 2-methylpropane are the two Konstitutionsisomere of C4H10 - basic isomer counting before functional-group isomers.

Question 50Structure, bonding and inorganic chemistry

In the Stoff-Teilchen-Konzept, what essential distinction must candidates keep clear?

  1. Observable macroscopic Stoff properties versus explanations on the Teilchenebene (atoms, ions, molecules) Correct answer.
  2. Whether a reaction is endothermic or exothermic
  3. Whether a compound is organic or inorganic only
  4. Whether a calculation uses moles or grams

Explanation

The Stoff-Teilchen-Konzept requires a consistent Perspektivwechsel: phenomena are described on the Stoffebene and explained by Teilchenbau, Anordnung and Wechselwirkungen. Confusing the two levels is a classic Klausur error.