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Limestone does NOT decompose when heated to 900 K because, at 900 K, ΔH is: A. Positive and less than TΔS B. Positive and greater than TΔS C. Negative and less than TΔS D. Negative and greater than TΔS - ANSWER B. The reaction does not occur (is not spontaneous). This indicates that th...

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  • 12 februari 2025
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  • aamc fl 4 correct 100
  • aamc fl 4
  • limestone does not decompose when heated to 900 k
  • when limestone is heated during step 1 an equilib

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Voorbeeld van de inhoud

AAMC FL 4 Correct 100%
Limestone does NOT decompose when heated to 900 K because, at 900 K, ΔH is:

A. Positive and less than TΔS
B. Positive and greater than TΔS
C. Negative and less than TΔS
D. Negative and greater than TΔS - ANSWER B.
The reaction does not occur (is not spontaneous). This indicates that the ΔG =ΔH-TΔS
>0. From inspection of the reaction, it can be concluded that ΔS>0. Consequently,
ΔH>TΔS explains why the reaction does not occur

When limestone is heated during Step 1, an equilibrium is established. Which of the
following expressions is the equilibrium constant for the decomposition of limestone?

A. [CaO]
B. [CaCO3]
C. [CO2]
D. [CaO] x [CaCO3] - ANSWER C.
From the law of mass action, an equilibrium constant expression involves a ration of
products to reactants with exponents determined from the stoichiometry of the reaction.
Furthermore, solids are excluded from equilibrium constant expressions. CO2 (g), as
the only non-solid material in the reaction, is the only substance that appears in the
equilibrium constant expression.

During Reaction 2, did the oxidation state of N change?

A. Yes; it changed from -3 to -4
B. Yes; it changed from 0 to +1
C. No; it remained at -3
D. No; it remained at +1 - ANSWER C.
The part of the Reaction 2 that involves nitrogen is the protonation of ammonia (NH3 +
H--> NH4+). Acid-base reactions doe not involve oxidation state changes. Furthermore
the oxidation states of N in NH3 is -3 not 0.

If all of Gas X (from Step 6) is held in a sealed chamber at STP, what will be its
appropriate volume?

A. 22.4 L
B. 44.8 L
C. 67.2 L
D. 89.6 L - ANSWER A.
The quantity of Gas X was given as 1 mole. One mole of gas occupies 22.4 L at STP.

,Why was it important that the cuvettes containing the glucose oxidase and the blood
sample were identical in terms of optical properties?

A. To enable the comparison of the absorption spectra
B. To reduce the absorption in the glass walls
C. To decrease the uncertainty in the wavelength
D. To increase the absorption in the solutions - ANSWER A.
The identical optical properties of the cuvettes ensure that the absorbed radiation is due
only to the presence of glucose in the blood and not due to the difference in the
absorption features of the walls.

What is the approximate energy of a photon in the absorbed radiation that yielded the
data in Table 1?

A. 1 eV
B. 2 eV
C. 3 eV
D. 4 eV - ANSWER B. ( I chose C)

The photon energy is E=hc/λ = 19.8 x 10^-26 J.m/ (625 x 10^-9) = 3.1 x10^-19 J , so
about 2 eV

According to Table 1, what is the concentration of the glucose in the blood from which
the diluted sample was taken?

A. 60 mg/dL
B. 90 mg/dL
C. 120 mg/dL
D. 150 mg/dL - ANSWER D.

From Table 1, the glucose concentration in the diluted sample is (o.20/0.24) x 6.0 mg/dL
= 5.0 mg/dL.
The blood then has a glucose concentration of 30 x 5.0 mg/dL= 150 mg/dL.

Suppose a blood sample tested above the range (6.0 mg/dL) of the standards used in
the experiment. What modification will provide a more precise reading by data
interpolation as opposed to extrapolation using the same standards?

A. Increase the enzyme concentration.
B. Increase the oxygen pressure.
C. Decrease the content of the oxygen acceptor
D. Dilute the sample with additional solvent. - ANSWER D.

By adding solvent, the concentration of glucose will be lowered, and the resulting
absorbance will fall within the range of the standards. This is easily accomplished, and
the resulting calculations that account for the dilution are not difficult.

,Which of the following reasons best explains why it is possible to separate a 1:1 mixture
of 1-chlorobutane and 1-butanol with fractional distillation?

A. Both 1-chlorobutane and 1-butanol are polar
B. Both 1-chlorobutane and 1-butanol are nonpolar
C. The boiling point of 1-chlorobutane is substantially higher than that of 1-butanol
D. The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol -
ANSWER D.
The fact that 1-chlorobutane will have a boiling point that is substantially lower than that
of 1-butanol can be rationalized from chemical principles. The molecules have similar
molecular weights, but 1-butanol has a hydroxyl functional group that can participated in
hydrogen bonding. Hydrogen bonding is a particularly strong force of intermolecular
attraction.

Which of the following oxidative transformations is unlikely to occur?

A. A primary alcohol to an aldehyde
B. A tertiary alcohol to a ketone
C. An aldehyde to a carboxylic acid
D. A secondary alcohol to a ketone - ANSWER B.

Oxidation of tertiary alcohols is difficult because it involved C-C bond breaking

According to the IUPAC, which is the systematic name for the hydrocarbon shown?

A. Z-3-methylpent-2-ene
B. E-3-methylpent-2-ene
C. Z-3-ethylbut-2-ene
D. E-3-ethylbut-2-ene - ANSWER A.

By IUPAC rules, first identify the longest unbroken chain of carbon atoms. Next, number
the carbon atoms in this chain starting from the end that gives the C=C the lowest
numbers. The double bond is identified by the portion of the carbon atoms from the
lowest numbers ed (2), and then the methyl group is assigned at the 3-position. The
stereochemical designator for the double bond is X because the highest priority groups
occur on the same side of the double bond. (On Z same side)

Two vector magnitudes |A| = 8 units and |B| =5 units make an angle that can vary from
0° to 180° . The magnitude of the resultant vector A + B CANNOT have the value of:

A. 2 units
B. 5 units
C. 8 units
D. 12 units - ANSWER A.

, The magnitude of A+B is as small as 3 units and as large as 13 units. The magnitude of
2 units is smaller than the smallest possible magnitude of vector A+B.

What is the effect produced by the PRK technique designed to correct
nearsightedness?

A. The density of the cornea is increased
B. The radius of curvature of the cornea is increased
C. The index of refraction of the cornea is increased
D. The thickness of the cornea at the apex is increased - ANSWER B.

According to the passage, to correct nearsightedness, the laser beam is directed onto
the central part of the cornea, resulting in a flattening of the cornea. This means that the
radius of curvature of the cornea is increased.

What is the magnitude of the electric field in the electrical discharge produced in the
excimer laser tube?

A. 2.0 x 10^6 V/m
B. 4.0 x 10^5 V/m
C. 6.0 x 10^4 V/m
D. 8.0 x 10^3 V/m - ANSWER A.

The magnitude of the electric field produced by 8.0 kV across 4.0 mm is 8000 V/ 0.004
m = 2.0 x 10^6 V/m.

What is the frequency of the pulses that deliver laser radiation to the cornea?

A. 0.4 Hz
B. 4.0 Hz
C. 25 Hz
D. 250 Hz - ANSWER B.

The frequency is 1/250 ms = 4 Hz

The use of pulsed laser radiation in the PRK procedure, as opposed to continuous laser
radiation, allows the cornea to:

A. absorb more radiation
B. change its index of refraction
C. increase the area exposed to radiation
D. maintain a lower average temperature - ANSWER D.

The pulsed laser radiation interacts with the cornea for a shorter time than a continuous
laser radiation, thus less heat is transferred to the cornea. This allows the cornea to
maintain a lower average temperature by cooling off between pulses

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