FL 4 C/P-done

Question 1

Photon energy formula is ___.

Answer 1

E = hc/λ

Question 2

___nm in 1 m

Answer 2

10^-9

Question 3

Q7: Glucose concentration

Answer 3

From Table 1, the glucose concentration in the diluted sample is (0.20/0.24) × 6.0 mg/dL = 5.0 mg/dL. The blood then has a glucose concentration of 30 × 5.0 mg/dL = 150 mg/dL.

Question 4

Which of the following reasons best explains why it is possible to separate a 1:1 mixture of 1-chlorobutane and 1-butanol by fractional distillation? (which boiling point is higher/lower and why)

Answer 4

D. The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol.

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 participate in hydrogen bonding. Hydrogen bonding is a particularly strong force of intermolecular attraction.

Question 5

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 5

B) Oxidation of tertiary alcohols is difficult because it involves C–C bond breaking.

Question 6

Oxidative transformation of primary, secondary, and tertiary alcohol and aldehyde. What do they become and is it possible?

Answer 6

Primary…aldehyde
Secondary…ketone
Tertiary…not possible without breaking C-C bond (there is no reactive C-H bond adjacent the C-OH group)
Aldehyde to carboxylic acid

Question 7

Two vectors of 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 7

A) The magnitude of A + B is as small as 3 units (when A and B are anti-parallel and make an angle of 180°) and as large as 13 units (when A and B are parallel and make an angle of 0°). The magnitude of 2 units is smaller than the smallest possible magnitude of vector A + B.

The range of values that a vector addition can produce is anywhere between A+B and A-B. Doing those gives you a range between 3 and 13. Since the lowest value you could possibly get is 3, the vector addition could not produce a value of 2.

Question 8

The refraction index of the cornea is ______ due to the PRK technique because it is an intrinsic property of the tissue.

Answer 8

unchanged

Question 9

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 9

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.

Question 10

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 10

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.

Question 11

The intermolecular forces that exist among the molecules of NH3 gas are:

A.dipole–dipole forces only.
B.London dispersion forces only.
C.both dipole–dipole and London dispersion forces.
D.neither dipole–dipole nor London dispersion forces.

Answer 11

C) Since NH3 is a permanent dipole, it will exhibit dipole–dipole intermolecular forces in addition to the London dispersion forces exhibited by all molecules.

Question 12

What molecules have dipole-dipole forces?

Answer 12

Polar molecules

Question 13

NH3 acts as a weak base in water with Kb = 1.76 × 10–5 at 25°C. The corresponding equilibrium is shown below.

NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)

At 25°C, the equilibrium concentration of the NH4+ ion in a 10 M aqueous solution of NH3 would be closest to which of the following?

Answer 13

B) The concentration of NH4+ at equilibrium can be estimated using Kb and the equilibrium constant expression. Kb = 1.76 × 10–5 = [NH4+][OH–]/[NH3]. Plugging in 10 M for [NH3], which is a good approximation since very little NH3 ionizes, and noting that [NH4+] = [OH–], it is possible to solve for [NH4+]: 1.76 × 10–5 = [NH4+]2/10 → 1.76 × 10­–4 = [NH4+]2 → 1.32 × 10–2 = [NH4+].

Question 14

The longest wavelength corresponds to the ______ photon energy, which corresponds to the transition between the closest energy levels.

Answer 14

The longest wavelength corresponds to the SMALLEST photon energy, which corresponds to the transition between the closest energy levels.

Question 15

Amide structure

Answer 15

R1(C=O)NR2R3

Question 16

Amine structure

Answer 16

R3N

Question 17

Imine structure

Answer 17

R2C=NR

Question 18

Carbamate

Answer 18

R1O(C=O)NR2R3

Question 19

Ping-pong/double displacement mechanism

Answer 19

No ternary complex is formed (no 3 things together)

Question 20

Beer’s Law

Answer 20

A = εbc

Question 21

A buffer has a buffering capacity that is ±___ pH unit away from the pKa, which means that only two of the experiments would have used Tris.

Answer 21

1

Question 22

If a thin thread is placed between a screen and a bright source of light, a pattern of parallel dark and bright fringes appears on the screen. The phenomenon best explaining the formation of this pattern is:

A.refraction.
B.polarization.
C.diffraction.
D.reflection.

Answer 22

C) The thin thread disrupts the propagation of light by impeding light to pass through it. Diffraction causes the initially plane-parallel wave-fronts of light to change direction and partially enter the shadow region behind the thread due to its narrowness. The overlapping of different wave-fronts on the screen causes the pattern of dark and bright fringes on the screen according to the phase difference between the wave-fronts that interfere there.

Question 23

If a thin thread is placed between a screen and a bright source of light, a pattern of parallel dark and bright fringes appears on the screen. The phenomenon best explaining the formation of this pattern is:

A.refraction.
B.polarization.
C.diffraction.
D.reflection.

Answer 23

C) The thin thread disrupts the propagation of light by impeding light to pass through it. Diffraction causes the initially plane-parallel wave-fronts of light to change direction and partially enter the shadow region behind the thread due to its narrowness. The overlapping of different wave-fronts on the screen causes the pattern of dark and bright fringes on the screen according to the phase difference between the wave-fronts that interfere there.

Question 24

Which forms of guanine and thymine are favored under physiological conditions?

A.The keto form of guanine and the keto form of thymine
B.The keto form of guanine and the enol form of thymine
C.The enol form of guanine and the keto form of thymine
D.The enol form of guanine and the enol form of thymine

Answer 24

A) Based on the pKa of the protons on the nitrogen atoms in the rings in Figure 2, the keto state is preferred for both guanine and thymine at physiological pH (7.2). The nitrogen atoms are protonated because pH 7.2 is approximately 2 pH units less than the pKa.​

Question 25

What atom remains after the β– decay of P32?

A.Si
B.Al
C.S
D.Cl

Answer 25

C) 32P undergoes the β– decay according to the reaction 32 15P → 32 16S + e– + energy.

Question 26

The ionic form of which metal atom is NOT likely to be found in the pocket of a catalytically active DNA polymerase?

A.Zn
B.Fe
C.Mg
D.Na

Answer 26

D) Sodium does not readily form a divalent cation, essential for DNA polymerase catalytic activity as shown in Figure 1.

Question 27

Roughly to what height would a 5 kg stone need to be raised in order to have the same stored energy as the energy stored in the defibrillator’s capacitor (400J)?

A.4 m
B.8 m
C.16 m
D.32 m

Answer 27

The gravitational potential energy is mass × gravitational acceleration × height. Equating 400 J = 5 kg × 9.8 m/s2 × height and solving for height leads to 400 J/(49 kg × m/s2) ≈ 8 m.​

Question 28

One company sells a defibrillator for home use that uses a 9-volt DC battery. The battery is rated at 4.2 A•hr (amp•hour). Roughly how much charge can the battery deliver?

A.4.2 C
B.38 C
C.15,000 C
D.136,000 C

Answer 28

C) The definition of current is flow of charge per unit time. Thus, charge equals current multiplied by time, hence 4.2 A × 1 hr = 4.2 A × 3600 s = 15,120 C ≈ 15,000 C.

Question 29

Suppose a defibrillator successfully returns a baby’s heart to normal beating. Suppose further that 20 g of blood enters the heart at 25 cm/s and leaves 0.10 s later at 35 cm/s. What is the estimated average force on the 20 g of blood as it moves through the baby’s heart?

A.0.020 N
B.0.20 N
C.20 N
D.2000 N

Answer 29

A) According to Newton’s second law, the average force is equal to the mass of blood multiplied by the average acceleration of the blood. The average acceleration is (35 cm/s – 25 cm/s)/0.10 s = 100 cm/s2 = 1 m/s2. The average force is 20 g × 1 m/s2 = 0.020 kg × 1 m/s2 = 0.020 N.

Question 30

The electric field inside each of the conductors that forms the capacitor in the defibrillator is zero. Which of the following reasons best explains why this is true?

A.All of the electrons in the conductor are bound to atoms, and thus there is no way for an external electric field to penetrate atoms with no net charge.
B.Free electrons in the conductor arrange themselves on the surface so that the electric field they produce inside the conductor exactly cancels any external electric field.
C.Free electrons in the conductor arrange themselves on the surface and throughout the interior so that the electric field they produce inside the conductor exactly cancels any external electric field.
D.All electrons in the conductor, both free and bound, arrange themselves on the surface so that the electric field they produce inside the conductor exactly cancels any external electric field.

Answer 30

Conductors contain both atom-bound electrons and free electrons. Free electrons arrange themselves on the surface of conductors, and their collective electric field produced inside the conductor cancels any external electric field. The resulting electric field inside the conductor is zero.