AAMC Chemistry
Consider the response rate defined by means of the following equation.
Rate = k[A]m[B]n
The standard reaction is stated to be 2d order whilst:
A. K is constant, m=0, and n=0.
B. K is regular, m=1, and n=1.
C. Ok is constant, m=2, and n=2.
D. Okay=2, and m and n are variable. - ANSAnswer Key: B The reaction is second order if the
sum of the exponents m and n equals 2. In rate laws, k is the fee steady. The sum of the
exponents equals 2 when m equals 1 and n equals 1. Thus, B is the correct answer.
A student plans to add HCl to a solution containing Pb(NO3)2(aq). To decide how lots Pb2+ will
precipitate from solution while the HCl is brought, the scholar desires to know which of the
subsequent?
A. Ka for HCl
B. Ka for HNO3
C. Ksp for PbCl2
D. Keq for the response Pb2+ + 2 e- Pb - ANSAnswer Key: C To determine how plenty Pb2+ will
precipitate, the student must realize the solubility of PbCl2, that is related to the solubility
product steady Ksp of PbCl2. Thus, C is the fine solution.
Which of the following electron configurations of nitrogen is the maximum strong?
A. 1s22s12p4
B. 1s22s22p3
C. 1s12s22p4
D. 1s22s22p23s1 - ANSAnswer Key: B The atomic wide variety of nitrogen is 7. The electrons
will start to fill the lower electricity ranges first; therefore, the 1s stage will fill first with electrons,
accompanied through 2s with electrons, after which the 2p level with the ultimate 3 electrons.
The resulting electron configuration is 1s22s22p3. Thus, B is the exceptional solution.
When a susceptible acid (HA) is titrated with sodium hydroxide in the presence of an indicator
(HIn), the pH at which a color exchange is located relies upon at the:
A.Very last concentration of HA.
B. Final attention of HIn.
C. Preliminary awareness of HA.
D. PKa of HIn. - ANSAnswer Key: D The indicator will exchange shade over a particular pH
variety. The range at which the shade change takes region depends on the point at which HIn is
transformed to In-, and this relies upon at the pKaof the indicator which is answer D.
, Approximately how many moles of Al3+ are reduced whilst zero.1 faraday of fee passes through
a mobile for the duration of the manufacturing of Al? (Note: Assume there's extra Al3+ to be had
and that Al3+ is decreased to Al metal handiest.)
A. Zero.033 mol
B. 0.050 mol
C. 0.067 mol
D. 0.10 mol - ANSAnswer Key: A
A faraday is same to 1 mole of electric charge. Because each aluminum ion gains 3 electrons,
0.1 faraday of charge will lessen 0.1/three moles of aluminum, or 0.033 moles of aluminum.
The following response happens spontaneously.Cd(s) + 2 H+(aq) → Cd2+(aq) + H2(g) Which of
the following has the best electron affinity?
A. Cd(s)
B. H+(aq)
C. Cd2+(aq)
D. H2(g) - ANSAnswer Key: B The response equation shows the reduction of H+ with the aid of
Cd. Because the H+ accepts the electron effortlessly from Cd, it can be determined that H+ has
the highest electron affinity.
Phosphorus seems immediately beneath nitrogen inside the periodic table. The boiling factor of
ammonia, NH3, is better than the boiling factor of phosphine, PH3, under general situations.
Which of the following statements excellent explains the difference in the boiling points of those
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 decrease boiling factors.
D. Ammonia bureaucracy more potent intermolecular hydrogen bonds than phosphine. -
ANSAnswer Key: D A substance boils when enough heat has been supplied to conquer the
intermolecular forces. The truth that ammonia has a higher boiling point than phosphine shows
that ammonia requires greater heat to overcome the intermolecular forces than does phosphine
and consequently the intermolecular forces in ammonia are more potent than those in
phosphine.
A gas that occupies 10 L at 1 atm and 25oC will occupy what extent at 500 atm and 25oC?
A. Exactly zero.020 L
B. Somewhat greater than zero.02 L due to the space occupied by means of the man or woman
gasoline molecules
C. Somewhat extra than zero.02 L because of the repulsions among the man or woman gas
molecules
D. Somewhat extra than zero.02 L because of the improved number of collisions with the edges
of the box - ANSAnswer Key: B The perfect gasoline law makes the assumption that molecules
have no quantity. This assumption is good enough while the gas is at 1 atm, however when the
strain is extended to 500 atm the quantity of the gasoline molecules is no longer negligible.
Consider the response rate defined by means of the following equation.
Rate = k[A]m[B]n
The standard reaction is stated to be 2d order whilst:
A. K is constant, m=0, and n=0.
B. K is regular, m=1, and n=1.
C. Ok is constant, m=2, and n=2.
D. Okay=2, and m and n are variable. - ANSAnswer Key: B The reaction is second order if the
sum of the exponents m and n equals 2. In rate laws, k is the fee steady. The sum of the
exponents equals 2 when m equals 1 and n equals 1. Thus, B is the correct answer.
A student plans to add HCl to a solution containing Pb(NO3)2(aq). To decide how lots Pb2+ will
precipitate from solution while the HCl is brought, the scholar desires to know which of the
subsequent?
A. Ka for HCl
B. Ka for HNO3
C. Ksp for PbCl2
D. Keq for the response Pb2+ + 2 e- Pb - ANSAnswer Key: C To determine how plenty Pb2+ will
precipitate, the student must realize the solubility of PbCl2, that is related to the solubility
product steady Ksp of PbCl2. Thus, C is the fine solution.
Which of the following electron configurations of nitrogen is the maximum strong?
A. 1s22s12p4
B. 1s22s22p3
C. 1s12s22p4
D. 1s22s22p23s1 - ANSAnswer Key: B The atomic wide variety of nitrogen is 7. The electrons
will start to fill the lower electricity ranges first; therefore, the 1s stage will fill first with electrons,
accompanied through 2s with electrons, after which the 2p level with the ultimate 3 electrons.
The resulting electron configuration is 1s22s22p3. Thus, B is the exceptional solution.
When a susceptible acid (HA) is titrated with sodium hydroxide in the presence of an indicator
(HIn), the pH at which a color exchange is located relies upon at the:
A.Very last concentration of HA.
B. Final attention of HIn.
C. Preliminary awareness of HA.
D. PKa of HIn. - ANSAnswer Key: D The indicator will exchange shade over a particular pH
variety. The range at which the shade change takes region depends on the point at which HIn is
transformed to In-, and this relies upon at the pKaof the indicator which is answer D.
, Approximately how many moles of Al3+ are reduced whilst zero.1 faraday of fee passes through
a mobile for the duration of the manufacturing of Al? (Note: Assume there's extra Al3+ to be had
and that Al3+ is decreased to Al metal handiest.)
A. Zero.033 mol
B. 0.050 mol
C. 0.067 mol
D. 0.10 mol - ANSAnswer Key: A
A faraday is same to 1 mole of electric charge. Because each aluminum ion gains 3 electrons,
0.1 faraday of charge will lessen 0.1/three moles of aluminum, or 0.033 moles of aluminum.
The following response happens spontaneously.Cd(s) + 2 H+(aq) → Cd2+(aq) + H2(g) Which of
the following has the best electron affinity?
A. Cd(s)
B. H+(aq)
C. Cd2+(aq)
D. H2(g) - ANSAnswer Key: B The response equation shows the reduction of H+ with the aid of
Cd. Because the H+ accepts the electron effortlessly from Cd, it can be determined that H+ has
the highest electron affinity.
Phosphorus seems immediately beneath nitrogen inside the periodic table. The boiling factor of
ammonia, NH3, is better than the boiling factor of phosphine, PH3, under general situations.
Which of the following statements excellent explains the difference in the boiling points of those
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 decrease boiling factors.
D. Ammonia bureaucracy more potent intermolecular hydrogen bonds than phosphine. -
ANSAnswer Key: D A substance boils when enough heat has been supplied to conquer the
intermolecular forces. The truth that ammonia has a higher boiling point than phosphine shows
that ammonia requires greater heat to overcome the intermolecular forces than does phosphine
and consequently the intermolecular forces in ammonia are more potent than those in
phosphine.
A gas that occupies 10 L at 1 atm and 25oC will occupy what extent at 500 atm and 25oC?
A. Exactly zero.020 L
B. Somewhat greater than zero.02 L due to the space occupied by means of the man or woman
gasoline molecules
C. Somewhat extra than zero.02 L because of the repulsions among the man or woman gas
molecules
D. Somewhat extra than zero.02 L because of the improved number of collisions with the edges
of the box - ANSAnswer Key: B The perfect gasoline law makes the assumption that molecules
have no quantity. This assumption is good enough while the gas is at 1 atm, however when the
strain is extended to 500 atm the quantity of the gasoline molecules is no longer negligible.
