MCAT Chemical and Physical Foundations of Biological Systems Quiz: Analyze Evaluate Scientific Explanations Predictions

What this quiz covers

This quiz focuses on Analyze Evaluate Scientific Explanations Predictions, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. A sustained net ionic current is expected because higher temperature always increases ion concentration.
2. No sustained net current is expected because equal concentrations imply no chemical potential gradient for KCl at steady state. (correct answer)
3. A sustained net current is expected because the hot side has higher $RT$ and therefore higher activity at the same concentration.
4. No sustained net current is expected because ions cannot diffuse through a polymer film under any conditions.

Explanation: This question tests understanding of electrochemical equilibrium and the role of chemical potential gradients. The chemical potential for an ideal solute is μ = μ° + RT ln a, where activity a ≈ concentration for dilute solutions. At the same concentration (0.10 M) on both sides, the activity is identical, so the only difference in chemical potential comes from the RT term. However, for ionic transport to create a sustained current, there must be a net driving force - either a concentration gradient or an applied voltage. While thermal gradients can create transient charge separation (Seebeck effect), they cannot sustain a net ionic current in a closed system at steady state because this would violate electroneutrality. Both compartments must remain electrically neutral, preventing sustained unidirectional ion flow. Choice C incorrectly suggests higher RT creates higher activity at the same concentration, misunderstanding that RT appears outside the logarithm. Choice A wrongly claims temperature increases ion concentration. When evaluating ionic transport, remember that sustained currents require maintained driving forces that don't violate fundamental constraints like electroneutrality.

Question 2

1. It decreases, because increasing $P_{\text{CO}_2}$ increases $[\text{CO}_2(\text{aq})]$ and can increase $[\text{H}^+]$. (correct answer)
2. It increases, because CO$_2$ is a base that consumes H$^+$ when dissolved.
3. It is unchanged, because Henry's law affects only gases, not aqueous equilibria.
4. It decreases only if the vessel is open to the atmosphere; in a sealed vessel pH cannot change.

Explanation: This question tests the ability to analyze and evaluate scientific predictions about pH changes when CO₂ dissolves in water. The key principle is that dissolved CO₂ forms carbonic acid through the equilibrium CO₂ + H₂O ⇌ H⁺ + HCO₃⁻, which releases H⁺ ions and lowers pH. In this sealed vessel system, Henry's law determines that [CO₂(aq)] = kH × PCO₂ = 3.3×10⁻² M/atm × 0.20 atm = 6.6×10⁻³ M. Since each dissolved CO₂ can contribute at most one H⁺, this increases [H⁺] and decreases pH below the initial 7.0. Answer A correctly predicts pH decreases because increasing PCO₂ increases dissolved CO₂ concentration, which increases H⁺ concentration. Answer B incorrectly claims CO₂ is a base - this is a common misconception confusing CO₂ with carbonate ions. To evaluate such predictions, check whether the dissolved species acts as an acid (releases H⁺) or base (consumes H⁺), then apply equilibrium principles to predict pH changes.

Question 3

1. Water moves from Side 2 to Side 1, and $\Delta\pi<0$.
2. Water moves from Side 1 to Side 2, and $\Delta\pi>0$. (correct answer)
3. Water moves from Side 1 to Side 2, and $\Delta\pi<0$.
4. No net water movement occurs because sucrose does not dissociate ($i=1$).

Explanation: This question tests the ability to evaluate predictions about osmotic flow using the osmotic pressure equation Δπ = iRTΔC. The fundamental principle is that water moves from regions of lower solute concentration (lower osmotic pressure) to higher concentration (higher osmotic pressure) across semipermeable membranes. With Side 1 at 0.10 M and Side 2 at 0.20 M sucrose, the concentration difference ΔC = 0.10 M creates an osmotic pressure difference Δπ = π₂ - π₁ = iRT(C₂ - C₁) > 0. Water moves from the dilute Side 1 to the concentrated Side 2 to equalize concentrations. Answer B correctly predicts water movement from Side 1 to Side 2 with Δπ > 0. Answer A reverses both the flow direction and sign - this is a common error confusing which side has higher pressure. To evaluate osmotic predictions, identify the concentration difference, calculate Δπ = iRTΔC, and remember water flows toward higher solute concentration.

Question 4

1. It is half, because a longer pathlength spreads light over a larger volume.
2. It is unchanged, because absorbance depends on concentration only.
3. It doubles, because absorbance is proportional to pathlength at fixed $c$. (correct answer)
4. It quadruples, because absorbance is proportional to $\ell^2$.

Explanation: This question tests the ability to evaluate predictions using Beer-Lambert law A = εℓc. The key principle is that absorbance depends linearly on three factors: molar absorptivity ε (constant for a given compound and wavelength), pathlength ℓ, and concentration c. When only pathlength changes from 1.0 cm to 2.0 cm while keeping ε and c constant, absorbance doubles: A₂ = ε(2ℓ)c = 2(εℓc) = 2A₁. This makes physical sense - light travels through twice as much absorbing solution, encountering twice as many molecules. Answer C correctly predicts doubling because absorbance is proportional to pathlength at fixed concentration. Answer D incorrectly suggests a quadratic relationship - this confusion may arise from intensity relationships but doesn't apply to absorbance. To evaluate spectrophotometry predictions, identify which Beer-Lambert variables change and apply the direct proportionality A ∝ ℓ when ε and c are constant.

Question 5

1. $\text{rate}_X/\text{rate}_Y = 10$, because $[\text{H}^+]$ is 10-fold higher at pH 3 than at pH 4. (correct answer)
2. $\text{rate}_X/\text{rate}_Y = 1/10$, because lower pH means fewer hydroxide ions to drive hydrolysis.
3. $\text{rate}_X/\text{rate}_Y = 2$, because a 1-unit pH change doubles $[\text{H}^+]$.
4. $\text{rate}_X/\text{rate}_Y = 1$, because pH affects equilibrium but not reaction rate.

Explanation: This question tests the ability to evaluate predictions about pH effects on acid-catalyzed reaction rates. The key principle is that for reactions with rate = k[H+][S], the rate is directly proportional to hydrogen ion concentration. Since pH = -log[H+], a one-unit pH decrease represents a 10-fold increase in [H+]: at pH 3.0, [H+] = 10⁻³ M, while at pH 4.0, [H+] = 10⁻⁴ M. With identical substrate concentrations, the rate ratio equals the [H+] ratio: rateX/rateY = (10⁻³)/(10⁻⁴) = 10. Answer A correctly predicts a 10-fold ratio because [H+] is 10-fold higher at pH 3 than pH 4. Answer C incorrectly suggests only doubling - this misunderstands the logarithmic pH scale where each unit represents a 10-fold change. To evaluate pH-dependent rate predictions, convert pH differences to [H+] ratios using the 10-fold rule per pH unit, then apply the rate law proportionality.

Question 6

1. It decreases, because shorter wavelength photons have less energy and eject slower electrons.
2. It increases, because photon energy increases as $\lambda$ decreases, increasing $K_{max}$. (correct answer)
3. It is unchanged, because stopping potential depends only on the metal's work function.
4. It becomes negative, because higher photon energy reverses electron flow direction.

Explanation: This question tests the ability to evaluate predictions about the photoelectric effect using Einstein's model. The key principle is that photon energy E = hc/λ increases as wavelength decreases, and the maximum kinetic energy of ejected electrons equals photon energy minus work function: Kmax = E - φ = eVs. When wavelength decreases from 500 nm to 400 nm, photon energy increases from 2.48 eV to 3.10 eV (using E = 1240 eV·nm/λ). Since the work function φ remains constant for the same metal, the increased photon energy produces higher Kmax and thus higher stopping potential Vs. Answer B correctly predicts increased stopping potential because photon energy increases as λ decreases. Answer A incorrectly claims shorter wavelengths have less energy - this reverses the E ∝ 1/λ relationship. To evaluate photoelectric predictions, remember that decreasing wavelength means increasing photon energy, which increases the maximum kinetic energy and stopping potential for the same metal.

Question 7

1. The system follows Fickian diffusion with constant $D$ under these conditions (correct answer)
2. Membrane thickness increases with $\Delta C$, canceling changes in flux
3. Diffusion is driven by pressure gradients, not concentration gradients
4. The solute is actively transported, producing proportional flux at steady state

Explanation: This question tests the ability to analyze and evaluate explanations for diffusion behavior using Fick's first law. The observation is that flux (J) varies linearly with concentration difference (ΔC), passing through the origin. According to Fick's first law, J = -D(ΔC/Δx), which predicts exactly this linear relationship when D and Δx are constant. Choice A correctly identifies this as Fickian diffusion with constant diffusion coefficient. Choice D incorrectly invokes active transport, which would not necessarily produce a linear relationship through the origin and contradicts the passive diffusion context. To evaluate diffusion explanations, check whether the proposed mechanism matches the mathematical relationship observed and whether assumptions (like constant D) are reasonable.

Question 8

1. Trial 2 has a larger rate constant $k$ and a faster initial rate (correct answer)
2. Trial 2 has a smaller rate constant $k$ because molecules move too quickly to react
3. Both trials have identical $k$ because $A$ is constant
4. Trial 2 has a larger $k$ only if the reaction is endothermic

Explanation: This question tests the ability to analyze and evaluate predictions based on the Arrhenius equation for temperature effects on reaction rates. The Arrhenius equation k = A·exp(-Ea/RT) shows that rate constant k increases with temperature because the exponential term becomes less negative. At higher temperature, a larger fraction of molecules have sufficient energy to overcome the activation barrier, increasing both k and the initial reaction rate. Choice A correctly predicts both effects, while choice B incorrectly suggests k decreases with temperature. To evaluate kinetic predictions, verify that higher temperature always increases k for elementary reactions with positive activation energy, regardless of whether the reaction is endothermic or exothermic.

Question 9

1. The force doubles because $F\propto r$
2. The force halves because $F\propto 1/r$
3. The force becomes one-fourth because $F\propto 1/r^2$ (correct answer)
4. The force is unchanged because the charges are unchanged

Explanation: This question tests the ability to analyze and evaluate claims about electrostatic forces using Coulomb's law. The student's claim that doubling distance doubles force contradicts Coulomb's law F = k|q₁q₂|/r², which shows force varies inversely with the square of distance. When distance doubles from r to 2r, force becomes F' = k|q₁q₂|/(2r)² = F/4, decreasing to one-fourth the original value. Choice C correctly identifies this inverse square relationship, while the student's claim would require a direct proportionality (F ∝ r). When evaluating force laws, verify whether the proposed relationship matches the established physical law, particularly noting whether relationships are direct, inverse, or involve powers.

Question 10

1. Resistance decreases because higher temperature increases electron speed
2. Resistance increases because $\rho$ increases with temperature (correct answer)
3. Resistance is unchanged because $L$ and $A$ are constant
4. Resistance increases only if the wire's mass increases

Explanation: This question tests the ability to analyze and evaluate predictions about temperature effects on electrical resistance. For metals, resistivity ρ typically increases with temperature due to increased electron-phonon scattering. Since resistance R = ρL/A and geometry (L, A) is held constant, resistance increases proportionally with resistivity. Choice B correctly predicts this increase, while choice A incorrectly suggests resistance decreases with temperature in metals. Choice C incorrectly assumes resistance depends only on geometry, ignoring the temperature dependence of resistivity. When evaluating resistance predictions, consider both geometric factors and material properties like temperature-dependent resistivity.

Question 11

1. $A=1.0$ because dilution increases light transmission and thus increases absorbance.
2. $A=0.25$ because concentration halves while path length and $\varepsilon$ stay constant. (correct answer)
3. $A=0.50$ because absorbance depends only on the number of chromophores, not concentration.
4. $A=0$ because dilution eliminates absorption at 280 nm.

Explanation: This question tests the skill of analyzing and evaluating scientific explanations and predictions in spectroscopy using the Beer-Lambert law. The law A = ε ℓ c predicts absorbance scales linearly with concentration c, so twofold dilution halves c, halving A if ℓ and ε are constant. Initial A=0.50 at 0.200 mM becomes A=0.25 post-dilution to 0.100 mM. The correct answer B logically follows because halving c directly halves A, aligning with the linear relationship. Distractor A is incorrect as it confuses dilution with increased transmission leading to higher A, misconstruing A = -log(T) where lower c increases T and decreases A. For similar questions, apply proportionality A ∝ c. Verify by calculating new A from dilution factor for consistency.

Question 12

1. pH increases at 50°C because water becomes less ionized
2. pH decreases at 50°C because $K_w$ increases, increasing $[\text{H}^+]$ (correct answer)
3. pH stays 7.0 at all temperatures because neutrality requires pH 7.0
4. pH decreases at 50°C because $[\text{OH}^-]$ becomes greater than $[\text{H}^+]$

Explanation: This question tests the ability to analyze and evaluate predictions about temperature effects on water autoionization. For the endothermic autoionization reaction, increasing temperature shifts equilibrium toward products (H⁺ and OH⁻), increasing Kw. Since [H⁺] = √Kw for pure water, [H⁺] increases with temperature, lowering pH below 7.0. Choice B correctly predicts this pH decrease, while choice C incorrectly assumes neutral water always has pH 7.0. Choice A incorrectly suggests water becomes less ionized at higher temperature. When evaluating temperature effects on equilibria, apply Le Chatelier's principle considering whether the reaction is endothermic or exothermic.

Question 13

1. Flask 2 has a larger $[B]/[A]$ because catalysts favor product formation
2. Flask 2 has a smaller $[B]/[A]$ because catalysts stabilize reactants
3. Both flasks have the same $[B]/[A]$, but Flask 2 reaches it faster (correct answer)
4. Both flasks have the same $[B]/[A]$ only if the reaction is exothermic

Explanation: This question tests the ability to analyze and evaluate predictions about catalyst effects on chemical equilibrium. Catalysts lower activation energy for both forward and reverse reactions equally, increasing the rate of equilibrium attainment but not changing the equilibrium position. Since catalysts don't change ΔG°, they don't affect the equilibrium constant or the final [B]/[A] ratio. Choice C correctly predicts identical equilibrium compositions with faster equilibration in the catalyzed flask. Choice A incorrectly suggests catalysts favor products, confusing kinetics with thermodynamics. To evaluate catalyst effects, remember they affect only reaction rates, not equilibrium positions.

Question 14

1. Solution 2 has the larger boiling point elevation because it has the larger van 't Hoff factor (correct answer)
2. Solution 1 has the larger boiling point elevation because glucose has more covalent bonds
3. Both solutions have the same $\Delta T_b$ because their molality is the same
4. Solution 2 has smaller $\Delta T_b$ because ions lower vapor pressure less than molecules

Explanation: This question tests the ability to analyze and evaluate predictions about colligative properties using boiling point elevation. The equation ΔTb = iKbm shows that boiling point elevation depends on the total particle concentration through the van 't Hoff factor. Glucose has i = 1 (no dissociation), while MgCl₂ has i = 3 (dissociates into Mg²⁺ and 2Cl⁻). With equal molalities, the MgCl₂ solution has three times the particle concentration and thus three times the boiling point elevation. Choice A correctly identifies this relationship based on van 't Hoff factors. When evaluating colligative property predictions, count total particles after complete dissociation.

Question 15

1. Heat is used to overcome intermolecular forces, increasing potential energy rather than temperature (correct answer)
2. No heat is absorbed during melting because temperature does not change
3. The substance's specific heat becomes zero at the melting point
4. The thermometer reads constant because pressure is constant

Explanation: This question tests the ability to analyze and evaluate explanations for phase transition behavior. During melting, added heat energy breaks intermolecular forces rather than increasing molecular kinetic energy, so temperature (related to average kinetic energy) remains constant. This heat goes into increasing potential energy as molecules separate, called latent heat of fusion. Choice A correctly explains this energy partitioning, while choice B incorrectly claims no heat is absorbed during the phase change. Choice C incorrectly invokes specific heat, which applies to single-phase heating. To evaluate phase transition explanations, distinguish between kinetic energy changes (temperature) and potential energy changes (phase transitions).

Question 16

1. The reaction is zero-order in A because doubling [A] doubles the rate
2. The reaction is first-order in A because doubling [A] doubles the rate (correct answer)
3. The reaction is second-order in A because doubling [A] doubles the rate
4. The rate law cannot depend on [A] because [B] was held constant

Explanation: This question tests the ability to analyze experimental data and evaluate scientific explanations about reaction kinetics. The relationship between reactant concentration and reaction rate reveals the reaction order: if doubling the concentration doubles the rate, this indicates a first-order dependence (rate ∝ [A]¹). In this experiment, when [A] increased from 0.10 M to 0.20 M (a factor of 2) while [B] remained constant, the rate also increased by a factor of 2, demonstrating that the rate is directly proportional to [A]. Therefore, the reaction is first-order in A, making choice B correct. Choice C incorrectly suggests second-order kinetics, which would require the rate to quadruple when [A] doubles (since rate ∝ [A]²). When evaluating kinetic data, always check if the rate change matches the concentration change raised to the proposed order: for nth order, doubling concentration increases rate by 2ⁿ.

Question 17

1. Increasing ionic strength increases the magnitude of electrostatic repulsion, decreasing aggregation.
2. Increasing ionic strength screens electrostatic repulsion, increasing aggregation. (correct answer)
3. Aggregation depends only on temperature, not on ionic strength, under these conditions.
4. NaCl decreases turbidity by reducing light scattering, independent of aggregation.

Explanation: This question tests the skill of analyzing and evaluating scientific explanations and predictions related to colloidal stability and ionic strength effects. The principle involves electrostatic repulsion between charged particles in colloids, where increasing ionic strength screens these repulsions via the Debye-Hückel theory, promoting aggregation. In this scenario, turbidity measurements serve as a proxy for aggregation in suspensions of negatively charged particles with varying NaCl concentrations. The correct answer B logically follows because higher NaCl concentrations increase turbidity, indicating greater aggregation due to screened repulsions, supporting the hypothesis. Distractor A is incorrect as it misrepresents the effect of ionic strength, confusing screening with enhanced repulsion, a common misconception in electrolyte solutions. To check similar questions, verify if experimental trends align with predicted outcomes from physical principles like charge screening. Always connect measurable quantities, such as turbidity, to underlying molecular interactions for robust evaluation.

Question 18

1. Temperature decreases noticeably because bond breaking dominates in neutralization.
2. Temperature increases by several degrees because heat released is absorbed by a few grams of solution. (correct answer)
3. Temperature remains constant because the reaction goes to completion with no heat exchange.
4. Temperature increases only if a catalyst is present to release heat.

Explanation: This question tests the skill of analyzing and evaluating scientific explanations and predictions in thermochemistry for exothermic reactions. The principle of energy conservation states that exothermic ΔH releases heat, increasing temperature in insulated systems, with ΔT ≈ |q| / (m c), where q = n ΔH. Neutralization of 0.001 mol each produces q ≈ 57 J in ~4 g solution, yielding ΔT ~ several °C. The correct answer B logically follows as heat release raises T noticeably in the small mass, consistent with calorimetry. Distractor A is incorrect assuming endothermic, misconstruing neutralization as bond-breaking dominant, ignoring net exothermicity. In similar problems, estimate ΔT order-of-magnitude. Verify by calculating q and comparing to system heat capacity for predictions.

Question 19

1. The rate-determining step likely involves two $\text{NO}_2$ species (e.g., a bimolecular collision of $\text{NO}_2$). (correct answer)
2. The rate-determining step must involve CO because CO is a reactant in the overall equation.
3. The reaction is zero-order overall because the exponent sum is 2.
4. The rate law proves the balanced equation is incorrect and should include $2\text{CO}$.

Explanation: This question tests the skill of analyzing and evaluating scientific explanations and predictions linking rate laws to reaction mechanisms. The principle states that the rate law reflects the slowest (rate-determining) step, so rate = k [NO2]^2 suggests a bimolecular RDS involving two NO2 molecules. The overall equation includes CO, but the rate law omits [CO], implying CO acts after the RDS. The correct answer A logically follows as it proposes a plausible RDS consistent with the observed order. Distractor B is incorrect insisting CO must be in the RDS, a misconception that rate laws always match overall stoichiometry, ignoring multi-step mechanisms. For related questions, match exponents to molecularity of proposed steps. Check consistency by ensuring post-RDS steps don't affect rate.

Question 20

1. Repulsive forces dominate, increasing measured pressure relative to ideal.
2. Attractive forces dominate, decreasing measured pressure relative to ideal. (correct answer)
3. The gas must have dissociated into ions, increasing the number of particles.
4. The temperature must have increased, lowering $Z$ below 1.

Explanation: This question tests the skill of analyzing and evaluating scientific explanations and predictions about real gas behavior using the compressibility factor. The principle of van der Waals theory explains Z<1 at high pressure due to intermolecular attractions reducing wall collisions, thus lowering measured P relative to ideal. Here, Z drops from 1.00 to 0.92 as pressure increases, indicating non-ideal behavior. The correct answer B logically follows because attractions dominate, decreasing P and Z, supporting the hypothesis. Distractor A is incorrect as it attributes Z<1 to repulsions, confusing with Z>1 conditions at very high pressures, a common van der Waals misconception. For analogous scenarios, recall attractions lower Z, volume effects raise Z. Evaluate by comparing Z to 1 and linking to dominant intermolecular forces.