AAMC FL 4
Limestone does NOT decompose while 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 more than TΔS - ANSB.
The response does now not arise (isn't spontaneous). This indicates that the ΔG =ΔH-TΔS
>zero. From inspection of the response, it is able to be concluded that ΔS>zero. Consequently,
ΔH>TΔS explains why the reaction does now not arise
When limestone is heated at some stage in Step 1, an equilibrium is installed. Which of the
subsequent expressions is the equilibrium regular for the decomposition of limestone?
A. [CaO]
B. [CaCO3]
C. [CO2]
D. [CaO] x [CaCO3] - ANSC.
From the law of mass action, an equilibrium regular expression includes a ration of products to
reactants with exponents decided from the stoichiometry of the response. Furthermore, solids
are excluded from equilibrium steady expressions. CO2 (g), because the best non-solid fabric
within the response, is the handiest substance that appears inside the equilibrium consistent
expression.
During Reaction 2, did the oxidation country of N alternate?
A. Yes; it changed from -three to -four
B. Yes; it modified from 0 to +1
C. No; it remained at -three
D. No; it remained at +1 - ANSC.
The part of the Reaction 2 that involves nitrogen is the protonation of ammonia (NH3 + H-->
NH4+). Acid-base reactions doe now 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 suitable extent?
A. 22.4 L
B. 44.Eight L
C. 67.2 L
D. 89.6 L - ANSA.
The amount of Gas X became given as 1 mole. One mole of gasoline occupies 22.4 L at STP.
,Why become it crucial that the cuvettes containing the glucose oxidase and the blood sample
were equal in terms of optical houses?
A. To permit the assessment of the absorption spectra
B. To reduce the absorption in the glass partitions
C. To decrease the uncertainty in the wavelength
D. To boom the absorption inside the answers - ANSA.
The equal optical homes of the cuvettes ensure that the absorbed radiation is due best to the
presence of glucose in the blood and no longer because of the distinction within the absorption
features of the partitions.
What is the approximate power of a photon inside the absorbed radiation that yielded the data in
Table 1?
A. 1 eV
B. 2 eV
C. 3 eV
D. 4 eV - ANSB. ( I selected C)
The photon energy is E=hc/λ = 19.Eight x 10^-26 J.M/ (625 x 10^-9) = three.1 x10^-19 J , so
approximately 2 eV
According to Table 1, what's the concentration of the glucose inside the blood from which the
diluted sample changed into taken?
A. 60 mg/dL
B. Ninety mg/dL
C. 120 mg/dL
D. One hundred fifty mg/dL - ANSD.
From Table 1, the glucose awareness in the diluted sample is (o.20/0.24) x 6.Zero mg/dL = 5.0
mg/dL.
The blood then has a glucose concentration of 30 x 5.0 mg/dL= one hundred fifty mg/dL.
Suppose a blood pattern examined above the variety (6.Zero mg/dL) of the standards used in
the test. What modification will provide a greater unique analyzing via information interpolation
in preference to extrapolation the use of the identical standards?
A. Increase the enzyme concentration.
B. Increase the oxygen stress.
C. Decrease the content of the oxygen acceptor
D. Dilute the pattern with extra solvent. - ANSD.
,By including solvent, the attention of glucose could be lowered, and the resulting absorbance
will fall inside the variety of the requirements. This is effortlessly accomplished, and the resulting
calculations that account for the dilution aren't tough.
Which of the subsequent motives pleasant explains why it's far possible to separate a 1:1
mixture of one-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 factor of 1-chlorobutane is extensively better than that of one-butanol
D. The boiling point of one-chlorobutane is extensively decrease than that of 1-butanol - ANSD.
The reality that 1-chlorobutane can have a boiling factor this is significantly lower than that of
1-butanol can be rationalized from chemical concepts. The molecules have similar molecular
weights, but 1-butanol has a hydroxyl practical institution that could participated in hydrogen
bonding. Hydrogen bonding is a in particular robust pressure of intermolecular appeal.
Which of the following oxidative ameliorations is not going to arise?
A. A number one 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 - ANSB.
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-three-methylpent-2-ene
B. E-3-methylpent-2-ene
C. Z-three-ethylbut-2-ene
D. E-3-ethylbut-2-ene - ANSA.
By IUPAC policies, first perceive the longest unbroken chain of carbon atoms. Next, quantity the
carbon atoms on this chain starting from the cease that offers the C=C the lowest numbers. The
double bond is identified through the part of the carbon atoms from the lowest numbers ed (2),
after which the methyl group is assigned on the 3-role. The stereochemical designator for the
double bond is X because the best precedence agencies arise at the equal side of the double
bond. (On Z equal side = 8 devices =five devices make an angle that may vary from 0° to one
hundred eighty° . The magnitude of the resultant vector A + B CANNOT have the value of:
A. 2 devices
B. 5 devices
C. 8 units
D. 12 devices - ANSA.
, The importance of A+B is as small as three units and as huge as thirteen units. The magnitude
of 2 units is smaller than the smallest viable significance of vector A+B.
What is the impact produced with the aid of the PRK method designed to correct
nearsightedness?
A. The density of the cornea is elevated
B. The radius of curvature of the cornea is accelerated
C. The index of refraction of the cornea is elevated
D. The thickness of the cornea on the apex is elevated - ANSB.
According to the passage, to accurate nearsightedness, the laser beam is directed onto the
valuable part of the cornea, resulting in a flattening of the cornea. This method that the radius of
curvature of the cornea is extended.
What is the significance of the electrical area within the electric discharge produced within the
excimer laser tube?
A. 2.0 x 10^6 V/m
B. 4.0 x 10^five V/m
C. 6.0 x 10^four V/m
D. 8.0 x 10^3 V/m - ANSA.
The value of the electrical discipline produced with the aid of eight.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.Four Hz
B. Four.0 Hz
C. 25 Hz
D. 250 Hz - ANSB.
The frequency is 1/250 ms = four Hz
The use of pulsed laser radiation inside the PRK procedure, instead of non-stop laser radiation,
allows the cornea to:
A. Soak up extra radiation
B. Change its index of refraction
C. Boom the vicinity uncovered to radiation
D. Preserve a lower average temperature - ANSD.
Limestone does NOT decompose while 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 more than TΔS - ANSB.
The response does now not arise (isn't spontaneous). This indicates that the ΔG =ΔH-TΔS
>zero. From inspection of the response, it is able to be concluded that ΔS>zero. Consequently,
ΔH>TΔS explains why the reaction does now not arise
When limestone is heated at some stage in Step 1, an equilibrium is installed. Which of the
subsequent expressions is the equilibrium regular for the decomposition of limestone?
A. [CaO]
B. [CaCO3]
C. [CO2]
D. [CaO] x [CaCO3] - ANSC.
From the law of mass action, an equilibrium regular expression includes a ration of products to
reactants with exponents decided from the stoichiometry of the response. Furthermore, solids
are excluded from equilibrium steady expressions. CO2 (g), because the best non-solid fabric
within the response, is the handiest substance that appears inside the equilibrium consistent
expression.
During Reaction 2, did the oxidation country of N alternate?
A. Yes; it changed from -three to -four
B. Yes; it modified from 0 to +1
C. No; it remained at -three
D. No; it remained at +1 - ANSC.
The part of the Reaction 2 that involves nitrogen is the protonation of ammonia (NH3 + H-->
NH4+). Acid-base reactions doe now 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 suitable extent?
A. 22.4 L
B. 44.Eight L
C. 67.2 L
D. 89.6 L - ANSA.
The amount of Gas X became given as 1 mole. One mole of gasoline occupies 22.4 L at STP.
,Why become it crucial that the cuvettes containing the glucose oxidase and the blood sample
were equal in terms of optical houses?
A. To permit the assessment of the absorption spectra
B. To reduce the absorption in the glass partitions
C. To decrease the uncertainty in the wavelength
D. To boom the absorption inside the answers - ANSA.
The equal optical homes of the cuvettes ensure that the absorbed radiation is due best to the
presence of glucose in the blood and no longer because of the distinction within the absorption
features of the partitions.
What is the approximate power of a photon inside the absorbed radiation that yielded the data in
Table 1?
A. 1 eV
B. 2 eV
C. 3 eV
D. 4 eV - ANSB. ( I selected C)
The photon energy is E=hc/λ = 19.Eight x 10^-26 J.M/ (625 x 10^-9) = three.1 x10^-19 J , so
approximately 2 eV
According to Table 1, what's the concentration of the glucose inside the blood from which the
diluted sample changed into taken?
A. 60 mg/dL
B. Ninety mg/dL
C. 120 mg/dL
D. One hundred fifty mg/dL - ANSD.
From Table 1, the glucose awareness in the diluted sample is (o.20/0.24) x 6.Zero mg/dL = 5.0
mg/dL.
The blood then has a glucose concentration of 30 x 5.0 mg/dL= one hundred fifty mg/dL.
Suppose a blood pattern examined above the variety (6.Zero mg/dL) of the standards used in
the test. What modification will provide a greater unique analyzing via information interpolation
in preference to extrapolation the use of the identical standards?
A. Increase the enzyme concentration.
B. Increase the oxygen stress.
C. Decrease the content of the oxygen acceptor
D. Dilute the pattern with extra solvent. - ANSD.
,By including solvent, the attention of glucose could be lowered, and the resulting absorbance
will fall inside the variety of the requirements. This is effortlessly accomplished, and the resulting
calculations that account for the dilution aren't tough.
Which of the subsequent motives pleasant explains why it's far possible to separate a 1:1
mixture of one-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 factor of 1-chlorobutane is extensively better than that of one-butanol
D. The boiling point of one-chlorobutane is extensively decrease than that of 1-butanol - ANSD.
The reality that 1-chlorobutane can have a boiling factor this is significantly lower than that of
1-butanol can be rationalized from chemical concepts. The molecules have similar molecular
weights, but 1-butanol has a hydroxyl practical institution that could participated in hydrogen
bonding. Hydrogen bonding is a in particular robust pressure of intermolecular appeal.
Which of the following oxidative ameliorations is not going to arise?
A. A number one 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 - ANSB.
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-three-methylpent-2-ene
B. E-3-methylpent-2-ene
C. Z-three-ethylbut-2-ene
D. E-3-ethylbut-2-ene - ANSA.
By IUPAC policies, first perceive the longest unbroken chain of carbon atoms. Next, quantity the
carbon atoms on this chain starting from the cease that offers the C=C the lowest numbers. The
double bond is identified through the part of the carbon atoms from the lowest numbers ed (2),
after which the methyl group is assigned on the 3-role. The stereochemical designator for the
double bond is X because the best precedence agencies arise at the equal side of the double
bond. (On Z equal side = 8 devices =five devices make an angle that may vary from 0° to one
hundred eighty° . The magnitude of the resultant vector A + B CANNOT have the value of:
A. 2 devices
B. 5 devices
C. 8 units
D. 12 devices - ANSA.
, The importance of A+B is as small as three units and as huge as thirteen units. The magnitude
of 2 units is smaller than the smallest viable significance of vector A+B.
What is the impact produced with the aid of the PRK method designed to correct
nearsightedness?
A. The density of the cornea is elevated
B. The radius of curvature of the cornea is accelerated
C. The index of refraction of the cornea is elevated
D. The thickness of the cornea on the apex is elevated - ANSB.
According to the passage, to accurate nearsightedness, the laser beam is directed onto the
valuable part of the cornea, resulting in a flattening of the cornea. This method that the radius of
curvature of the cornea is extended.
What is the significance of the electrical area within the electric discharge produced within the
excimer laser tube?
A. 2.0 x 10^6 V/m
B. 4.0 x 10^five V/m
C. 6.0 x 10^four V/m
D. 8.0 x 10^3 V/m - ANSA.
The value of the electrical discipline produced with the aid of eight.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.Four Hz
B. Four.0 Hz
C. 25 Hz
D. 250 Hz - ANSB.
The frequency is 1/250 ms = four Hz
The use of pulsed laser radiation inside the PRK procedure, instead of non-stop laser radiation,
allows the cornea to:
A. Soak up extra radiation
B. Change its index of refraction
C. Boom the vicinity uncovered to radiation
D. Preserve a lower average temperature - ANSD.
