MCAT TEST BANK 2026: 150 HIGH-YIELD MCQS WITH 100% VERIFIED ANSWERS
| GRADED A+ | GUARANTEED PASS!!
1.
Answer Key: B The reaction is second order if the sum of the exponents m and n equals 2.
In rate laws, k is the rate constant. The sum of the exponents equals
2 when m equals 1 and n equals 1. Thus, B is the correct answer.: Consider the reaction rate
described by the following equation.
Rate = k[A]m[B]n
The overall reaction is said to be second order when:
_
A
k is constant, m=0, and n=0.
B
k is constant, m=1, and n=1.
C
k is constant, m=2, and n=2.
D
k=2, and m and n are variable.
2.
Answer Key: C To determine how much Pb2+ will precipitate, the student must know the
solubility of PbCl2, which is related to the solubility product constant
Ksp of PbCl2. Thus, C is the best answer.: A student plans to add HCl to a solution
containing
Pb(NO3)2(aq). To determine how much Pb2+ will precipitate from solution when the HCl
is added, the student needs to know which of the following?
_
A
Ka for HCl
,B
Ka for HNO3
C
Ksp for PbCl2
D
Keq for the reaction Pb2+ + 2 e- Pb
3.
Answer Key: B The atomic number of nitrogen is 7. The electrons will begin to fill the
lower energy levels first; consequently, the 1s level will fill first with two electrons,
followed by 2s with two electrons, and then the 2p level with the last three electrons. The
resulting electron configuration is 1s22s22p3. Thus, B is the best answer.: Which of the
following electron configurations of nitrogen is the most stable? _ A
1s22s12p4
B
1s22s22p3
C
1s12s22p4
D
1s22s22p23s1
4.
Answer Key: D The indicator will change color over a specific pH range. The range at
which the color change takes place depends on the point at which HIn is converted to In-,
and this depends on the pKa of the indicator which is
answer D.: When a weak acid (HA) is titrated with sodium hydroxide in the presence of an
indicator (HIn), the pH at which a color change is observed depends on the:
,_
A
final concentration of HA.
B
final concentration of HIn.
C initial
concentration of
HA.
D
pKa of HIn.
5. A faraday is equal to one mole of electric charge. Because each aluminum ion gains 3
electrons, 0.1 faraday of charge will reduce 0.1/3 moles of aluminum,
or 0.033 moles of aluminum.: Approximately how many moles of Al3+ are reduced when
0.1 faraday of charge passes through a cell during the production of Al? (Note: Assume
there is excess Al3+ available and that Al3+ is reduced to Al metal only.)
_
A
0.033 mol
B
0.050 mol
C
0.067 mol
D
0.10 mol
6.
Answer Key: B The reaction equation shows the reduction of H+ by Cd.
Because the H+ accepts the electron readily from Cd, it can be determined that
, H+ has the highest electron affinity.: The following reaction occurs spontaneously.Cd(s) +
2 H+(aq) ’ Cd2+(aq) + H2(g) Which of the following has the highest electron affinity? _ A
Cd(s)
B
H+(aq)
C
Cd2+(aq)
D
H2(g)
7.
Answer Key: D A substance boils when enough heat has been supplied to overcome the
intermolecular forces. The fact that ammonia has a higher boiling point than phosphine
indicates that ammonia requires more heat to overcome the intermolecular forces than
does phosphine and therefore the intermolecular forces in ammonia are stronger than
those in phosphine.: Phos-
phorus appears directly below nitrogen in the periodic table. The boiling point of
ammonia, NH3, is higher than the boiling point of phosphine, PH3, under standard
conditions. Which of the following statements best explains the difference in the boiling
points of these two compounds?
_
A
Ammonia is a weaker base than phosphine.
B
The N-H bond is weaker than the
P-H bond. C
High molecular weight compounds generally have lower boiling points.
D
Ammonia forms stronger intermolecular hydrogen bonds than phosphine.
| GRADED A+ | GUARANTEED PASS!!
1.
Answer Key: B The reaction is second order if the sum of the exponents m and n equals 2.
In rate laws, k is the rate constant. The sum of the exponents equals
2 when m equals 1 and n equals 1. Thus, B is the correct answer.: Consider the reaction rate
described by the following equation.
Rate = k[A]m[B]n
The overall reaction is said to be second order when:
_
A
k is constant, m=0, and n=0.
B
k is constant, m=1, and n=1.
C
k is constant, m=2, and n=2.
D
k=2, and m and n are variable.
2.
Answer Key: C To determine how much Pb2+ will precipitate, the student must know the
solubility of PbCl2, which is related to the solubility product constant
Ksp of PbCl2. Thus, C is the best answer.: A student plans to add HCl to a solution
containing
Pb(NO3)2(aq). To determine how much Pb2+ will precipitate from solution when the HCl
is added, the student needs to know which of the following?
_
A
Ka for HCl
,B
Ka for HNO3
C
Ksp for PbCl2
D
Keq for the reaction Pb2+ + 2 e- Pb
3.
Answer Key: B The atomic number of nitrogen is 7. The electrons will begin to fill the
lower energy levels first; consequently, the 1s level will fill first with two electrons,
followed by 2s with two electrons, and then the 2p level with the last three electrons. The
resulting electron configuration is 1s22s22p3. Thus, B is the best answer.: Which of the
following electron configurations of nitrogen is the most stable? _ A
1s22s12p4
B
1s22s22p3
C
1s12s22p4
D
1s22s22p23s1
4.
Answer Key: D The indicator will change color over a specific pH range. The range at
which the color change takes place depends on the point at which HIn is converted to In-,
and this depends on the pKa of the indicator which is
answer D.: When a weak acid (HA) is titrated with sodium hydroxide in the presence of an
indicator (HIn), the pH at which a color change is observed depends on the:
,_
A
final concentration of HA.
B
final concentration of HIn.
C initial
concentration of
HA.
D
pKa of HIn.
5. A faraday is equal to one mole of electric charge. Because each aluminum ion gains 3
electrons, 0.1 faraday of charge will reduce 0.1/3 moles of aluminum,
or 0.033 moles of aluminum.: Approximately how many moles of Al3+ are reduced when
0.1 faraday of charge passes through a cell during the production of Al? (Note: Assume
there is excess Al3+ available and that Al3+ is reduced to Al metal only.)
_
A
0.033 mol
B
0.050 mol
C
0.067 mol
D
0.10 mol
6.
Answer Key: B The reaction equation shows the reduction of H+ by Cd.
Because the H+ accepts the electron readily from Cd, it can be determined that
, H+ has the highest electron affinity.: The following reaction occurs spontaneously.Cd(s) +
2 H+(aq) ’ Cd2+(aq) + H2(g) Which of the following has the highest electron affinity? _ A
Cd(s)
B
H+(aq)
C
Cd2+(aq)
D
H2(g)
7.
Answer Key: D A substance boils when enough heat has been supplied to overcome the
intermolecular forces. The fact that ammonia has a higher boiling point than phosphine
indicates that ammonia requires more heat to overcome the intermolecular forces than
does phosphine and therefore the intermolecular forces in ammonia are stronger than
those in phosphine.: Phos-
phorus appears directly below nitrogen in the periodic table. The boiling point of
ammonia, NH3, is higher than the boiling point of phosphine, PH3, under standard
conditions. Which of the following statements best explains the difference in the boiling
points of these two compounds?
_
A
Ammonia is a weaker base than phosphine.
B
The N-H bond is weaker than the
P-H bond. C
High molecular weight compounds generally have lower boiling points.
D
Ammonia forms stronger intermolecular hydrogen bonds than phosphine.
