,AAMC MCAT Practice Test 9 — Answer Key Companion
1. ANSWER: B
Reagents 2 and 3 are (NH4)2SO4 and (NH4)2C2O4, which produce sulfate and oxalate ions. From Table 1,
Mg2+ forms a white precipitate with Reagent 1 (NH3) only; Ca2+ and Sr2+ show precipitates with Reagent 3
(oxalate) and Reagent 4 (carbonate) in table. To precipitate all
three Mg2+, Ca2+, Sr2+ simultaneously, the combination of reagents that yields precipitates for Mg (with NH3?)
and for Ca and Sr is best matched by reagents 1 and 3 per table's pattern, but given choices, reagent pair B (1
and 3) precipitates Mg (as hydroxide with NH3) and Ca and Sr as oxalates.
2. ANSWER:D
Adding Na2CO3 introduces CO3^2-, which forms an insoluble carbonate with Ca2+ (CaCO3). HCl would
dissolve carbonates; NaOH would precipitate hydroxides but Ca(OH)2 has limited solubility depending on
concentration; H2SO4 forms sulfate but CaSO4 is somewhat soluble relative to CaCO3, so Na2CO3 is the most
reliable precipitant.
3. ANSWER: B
When Reagent 2 ((NH4)2SO4) is added to Ca2+/Sr2+, both form sulfates; SrSO4 is much less soluble than
CaSO4 under these conditions. Best recovery of fairly pure SrSO4 is to filter the insoluble precipitate (SrSO4)
and collect the solid — that corresponds to pouring through a filter and collecting the insoluble substance (option
B).
4. ANSWER:D
Ksp is inversely related to tendency to precipitate: oxalates generally have lower solubilities than corresponding
sulfates for these cations. Table shows that oxalate additions (Reagent 3) precipitate for Mg and Ca where
sulfates may not; CaC2O4 is hence very insoluble suggesting smallest Ksp among choices.
5. ANSWER:D
A flame test distinguishes Na (yellow) from Ca and Sr (both red), but Ca and Sr both produce red flames — to
distinguish them chemically, add Reagent 3 (oxalate) — Table 1 shows distinct precipitate patterns that allow
differentiation (Reagent 3 forms ppt with Sr but not with Ca in same way), so Reagent 3 best identifies the ion.
6. ANSWER:B
Charging of a capacitor follows q(t)=Q(1 e^{ t/RC}); graph starts at zero and rises asymptotically to a maximum
— that corresponds to the curve which rises quickly then levels off (option B).
7. ANSWER:B
Electric force on a charged particle inside a capacitor depends on q and E (F=qE). The particle's speed does not
affect the static electric force due to the capacitor's field, so the force remains the same when speed is doubled.
8. ANSWER:D
Capacitance C of a parallel-plate capacitor is 0 A/d. Increasing plate area (A) increases C directly; changing
resistor or voltage does not change capacitance, and increasing plate separation decreases capacitance.
9. ANSWER:A
Force on particle F = qE; acceleration a = F/m = qE/m, so a = (qE)/m. That's the direct application of Newton's
second law with electric force.
, 10. ANSWER:A
Two identical capacitors in series have equivalent capacitance C_eq = C/2. Adding another identical capacitor in
series halves the overall capacitance compared to a single capacitor.
11. ANSWER:C
A negatively charged particle feels a force toward the positive plate; in a uniform electric field this force is
constant in magnitude, so it undergoes constant acceleration toward the positive plate (speed increases).
12. ANSWER:C
ETOH (ethanol) contains an O–H group capable of hydrogen bonding with water; MTBE (an ether) cannot form
hydrogen bonds as a hydrogen donor (only weak dipole interactions), so ethanol uniquely hydrogen-bonds with
water.
13. ANSWER:D
Heat released (exothermicity) corresponds to more negative heat of formation. Among listed values, TAME has
the most negative H ( 680 kJ/mol) or similar; thus formation of TAME releases the most energy (largest
magnitude).
14. ANSWER:D
Combustion of dimethyl ether CH3OCH3: C2H6O formula can be balanced: CH3OCH3 + 3 O2 3 H2O + 2 CO2;
coefficients for O2 and CO2 are 3 and 2 respectively (choice D).
15. ANSWER:A
Nonoxygenated analog most similar in antiknock properties would mimic size/electronic distribution of MTBE. A
silicon analog (C4H9Si(CH3)3) maintains tetravalent central atom and similar steric/hydrocarbon character, so
(A) is most likely to mimic MTBE's behavior.
16. ANSWER:A
Combustion increases the number of gas molecules (products CO2 and H2O vapor) relative to reactants (liquid
hydrocarbon + O2 gas), and greater number of microstates corresponds to increased entropy; thus S > 0.
17. ANSWER:B
Evaporation rate correlates with vapor pressure at given temperature. ETOH has the highest vapor pressure (58
torr at 25°C) among listed compounds, so it evaporates fastest at 30°C.
18. ANSWER:B
Combustion stoichiometry: ethanol has the fewest carbons relative to molecular formula among listed
oxygenates, therefore requires the least O2 per mole to combust completely (choice B).
19. ANSWER:A
Harmonic frequencies for an open pipe depend mainly on length and speed of sound; diameter has negligible
first-order effect on harmonic frequencies, so independence from diameter supports modeling stellar columns as
pipes.
20. ANSWER:C
1. ANSWER: B
Reagents 2 and 3 are (NH4)2SO4 and (NH4)2C2O4, which produce sulfate and oxalate ions. From Table 1,
Mg2+ forms a white precipitate with Reagent 1 (NH3) only; Ca2+ and Sr2+ show precipitates with Reagent 3
(oxalate) and Reagent 4 (carbonate) in table. To precipitate all
three Mg2+, Ca2+, Sr2+ simultaneously, the combination of reagents that yields precipitates for Mg (with NH3?)
and for Ca and Sr is best matched by reagents 1 and 3 per table's pattern, but given choices, reagent pair B (1
and 3) precipitates Mg (as hydroxide with NH3) and Ca and Sr as oxalates.
2. ANSWER:D
Adding Na2CO3 introduces CO3^2-, which forms an insoluble carbonate with Ca2+ (CaCO3). HCl would
dissolve carbonates; NaOH would precipitate hydroxides but Ca(OH)2 has limited solubility depending on
concentration; H2SO4 forms sulfate but CaSO4 is somewhat soluble relative to CaCO3, so Na2CO3 is the most
reliable precipitant.
3. ANSWER: B
When Reagent 2 ((NH4)2SO4) is added to Ca2+/Sr2+, both form sulfates; SrSO4 is much less soluble than
CaSO4 under these conditions. Best recovery of fairly pure SrSO4 is to filter the insoluble precipitate (SrSO4)
and collect the solid — that corresponds to pouring through a filter and collecting the insoluble substance (option
B).
4. ANSWER:D
Ksp is inversely related to tendency to precipitate: oxalates generally have lower solubilities than corresponding
sulfates for these cations. Table shows that oxalate additions (Reagent 3) precipitate for Mg and Ca where
sulfates may not; CaC2O4 is hence very insoluble suggesting smallest Ksp among choices.
5. ANSWER:D
A flame test distinguishes Na (yellow) from Ca and Sr (both red), but Ca and Sr both produce red flames — to
distinguish them chemically, add Reagent 3 (oxalate) — Table 1 shows distinct precipitate patterns that allow
differentiation (Reagent 3 forms ppt with Sr but not with Ca in same way), so Reagent 3 best identifies the ion.
6. ANSWER:B
Charging of a capacitor follows q(t)=Q(1 e^{ t/RC}); graph starts at zero and rises asymptotically to a maximum
— that corresponds to the curve which rises quickly then levels off (option B).
7. ANSWER:B
Electric force on a charged particle inside a capacitor depends on q and E (F=qE). The particle's speed does not
affect the static electric force due to the capacitor's field, so the force remains the same when speed is doubled.
8. ANSWER:D
Capacitance C of a parallel-plate capacitor is 0 A/d. Increasing plate area (A) increases C directly; changing
resistor or voltage does not change capacitance, and increasing plate separation decreases capacitance.
9. ANSWER:A
Force on particle F = qE; acceleration a = F/m = qE/m, so a = (qE)/m. That's the direct application of Newton's
second law with electric force.
, 10. ANSWER:A
Two identical capacitors in series have equivalent capacitance C_eq = C/2. Adding another identical capacitor in
series halves the overall capacitance compared to a single capacitor.
11. ANSWER:C
A negatively charged particle feels a force toward the positive plate; in a uniform electric field this force is
constant in magnitude, so it undergoes constant acceleration toward the positive plate (speed increases).
12. ANSWER:C
ETOH (ethanol) contains an O–H group capable of hydrogen bonding with water; MTBE (an ether) cannot form
hydrogen bonds as a hydrogen donor (only weak dipole interactions), so ethanol uniquely hydrogen-bonds with
water.
13. ANSWER:D
Heat released (exothermicity) corresponds to more negative heat of formation. Among listed values, TAME has
the most negative H ( 680 kJ/mol) or similar; thus formation of TAME releases the most energy (largest
magnitude).
14. ANSWER:D
Combustion of dimethyl ether CH3OCH3: C2H6O formula can be balanced: CH3OCH3 + 3 O2 3 H2O + 2 CO2;
coefficients for O2 and CO2 are 3 and 2 respectively (choice D).
15. ANSWER:A
Nonoxygenated analog most similar in antiknock properties would mimic size/electronic distribution of MTBE. A
silicon analog (C4H9Si(CH3)3) maintains tetravalent central atom and similar steric/hydrocarbon character, so
(A) is most likely to mimic MTBE's behavior.
16. ANSWER:A
Combustion increases the number of gas molecules (products CO2 and H2O vapor) relative to reactants (liquid
hydrocarbon + O2 gas), and greater number of microstates corresponds to increased entropy; thus S > 0.
17. ANSWER:B
Evaporation rate correlates with vapor pressure at given temperature. ETOH has the highest vapor pressure (58
torr at 25°C) among listed compounds, so it evaporates fastest at 30°C.
18. ANSWER:B
Combustion stoichiometry: ethanol has the fewest carbons relative to molecular formula among listed
oxygenates, therefore requires the least O2 per mole to combust completely (choice B).
19. ANSWER:A
Harmonic frequencies for an open pipe depend mainly on length and speed of sound; diameter has negligible
first-order effect on harmonic frequencies, so independence from diameter supports modeling stellar columns as
pipes.
20. ANSWER:C
