Chem and Phys

conservation of energy equation

1/2mvi^2 + mghi = 1/2mvf^2 + mghf

Tollens' test is intended to identify

"reducing sugars," or sugars with the capacity to serve as reducing agents.

reducing sugar

-any monosaccharide w/ a hemiacetal ring (since they can be oxidized, they are considered reducing agents)

the final stage of glycolysis (phosphoenolpyruvate [PEP] → pyruvate) must be bypassed by gluconeogenesis. Thus, why gluconeogenesis has a two-step pathway split up between

1. the mitochondria and cytosol, in which pyruvate carboxylase converts pyruvate to oxaloacetate in the mitochondria by adding a COO− group. 2. Oxaloacetate is briefly converted to malate for transport out of the mitochondria, where it is then converted immediately back to oxaloacetate. At this point, in the cytosol, PEP carboxykinase converts oxaloacetate to PEP.

The human range of hearing (frequency)

20 Hz to 20 kHz. Anything above 20 kHz is Ultrasound

In one trial of this experiment, significant impurities were detected in the extracted caffeine. Based on the results of the experiment, which of the following would be the expected melting point range for this extracted sample? (Pure caffeine has a melting point of 235°C and a density of 1.23 g/cm3.) A. 195°C - 220°C B. 245°C - 267°C C. 233°C - 236°C D. 190°C - 195°C

A is correct. A melting point range includes the temperature when the first crystal of a compound starts to melt and the temperature when the compound is entirely melted. For a pure compound, the melting point range is narrow. Therefore, melting point determination is sometimes used to identify an unknown pure compound. However, the presence of impurities in a compound lowers and broadens the melting point range. According to the passage, the melting point of pure caffeine is 235°C, making choice A the best answer. B: This melting point range is above the melting point of pure caffeine. C, D: These melting point ranges are quite narrow, which is not the case when significant impurities are present.

What is the pH of a 0.010 M perchloric acid solution? A. 2 B. 4 C. 7 D. 12

A is correct. Perchloric acid is a strong acid that completely dissociates in aqueous solution, so the hydrogen ion concentration is 1.0 x 10-2 M. pH = -log[H+] = -log[10-2] = 2. B: This choice might be tempting if perchloric acid were incorrectly assumed to be weak. Remember, perchloric acid (HClO4) is a strong acid. C: This is the pH of pure water (neutral pH) at standard temperature. A significant amount of strong acid in water would not be expected to produce a neutral solution. D: This pH is basic, not acidic.

In comparison to piperazine, the inductive effect will cause the pKb of methyl-piperazine to be: (piperazine has a methyl) A. lower, because of electron donation by the methyl substituent. B. lower, because of electron withdrawal by the methyl substituent. C. higher, because of electron donation by the methyl substituent. D. higher, because of electron withdrawal by the methyl substituent.

A is correct. The pKb is a measure of basicity, where a smaller pKb value corresponds to a stronger base. Methyl-piperazine is a stronger base than piperazine because of the inductive effect generated by the electron-donating N-methyl group of methyl-piperazine. This effect increases the availability of electron charge density available to donate, thereby increasing the strength of N-methyl piperazine to act as a Lewis base. B, D: Alkyl groups, such as the N-methyl group of methyl-piperazine, are generally electron-donating substituents. C: Methyl-piperazine is a stronger base than piperazine because of the inductive effect generated by the electron-donating N-methyl group of methyl-piperazine. For this reason, the pKb of methyl-piperazine should be somewhat smaller than that of piperazine.

Participant 1 rides a fourth roller coaster as shown below. What is the minimum ramp height H if the ride at the top of the loop maintains the minimum speed needed to stay on the track throughout the loop? (pic on other side) A. 10 m B. 20 m C. 30 m D. 35 m

At the top of the loop, the gravitational and the normal forces (if any) point downward toward the center of the loop. Therefore, the net force causing centripetal acceleration is the sum of the gravitational force and the normal force. When the centripetal force is the minimum amount needed for the ride to stay on track, the normal force is zero. Note: At the top, the normal force can be zero but the gravitational force cannot. Let's designate the initial height where the ride starts as H. At the very beginning, the energy of the ride is the gravitational potential energy. Therefore, Ei = mgH. At the top of the loop, the ride includes both gravitational and potential energy. Therefore, Ef = mgH + ½ mv2 = mg (2r) + ½ mv2. The net acceleration at top (centripetal acceleration) = v2/r. Since we concluded that the normal force at the top is zero, the net acceleration at the top is the gravitational acceleration, g. Therefore, v2 = gr. Substituting this equation into the 'Ef' equation, we obtain the following: Ef = 2mgr + ½ mgr = (5/2) mgr To obey conservation of energy, Ei = Ef. Therefore: mgh = (5/2) mgr. We conclude that H is 5r/2. Since r is 8 m, H is 5(8 m)/2 = 20 m. A, C, D: These answers result from miscalculation.

Following the procedure of the experiment, how much total caffeine could you expect to extract? A. 6.50 g B. 9.75 g C. 11.75 g D. 12.5 g passage: To extract caffeine, three No Doz® tablets (3.3 g each) were crushed and added to 100 ml of distilled water, and boiled for ten minutes. Suspended solids were removed and the supernatant was transferred to a separatory funnel. Next, 100 ml of dichloromethane (DCM, density = 1.33 g/cm³) was added to the separatory funnel and the mixture was shaken, which resulted in formation of two layers. Each layer was separately collected, and the aqueous layer was poured back into the separatory funnel for the extraction to be repeated. One hundred ml of dichloromethane was added to the separatory funnel, and a second extraction was performed. After the second extraction, the layers containing caffeine from both extractions were mixed together. The resulting solution was boiled gently until a white solid formed. To determine how much caffeine was expected to dissolve in each layer and how much caffeine was extracted in total, the following formulae were used to determine the partition coefficient (Ψ) of caffeine:

B is correct. Caffeine is largely nonpolar, so it will be more soluble in dichloromethane than in water. Therefore, it will dissolve more in the dichloromethane layer. Using the partition coefficient equations and the caffeine properties provided by the passage, we must add up the amount of caffeine dissolved in each layer in both the first extraction and the second extraction. An alternative to this calculation is recognizing that the entire sample of caffeine is dissolved into both of those two layers, so the two extractions should give a yield similar to the original sample, which was roughly 10 g.

The conversion between glucose and pyruvate strongly favors the formation of pyruvate, and yet the gluconeogenic pathway is able to utilize several shared enzymes with glycolysis to create glucose from pyruvate. It is able to do this primarily because: A. the formation of glucose from the gluconeogenic precursors is strongly favored by the increasing entropy of the reaction. B. the formation of glucose, fructose-6-phosphate, and PEP through gluconeogenesis-specific enzymes push the equilibrium of reactions catalyzed using shared enzymes to favor gluconeogenesis. C. the activity of the enzymes with two or more functions is reversed when those enzymes are altered by kinases. D. enzymes in gluconeogenesis bypass reversible, equilibrium-state steps common to both pathways.

B is correct. Production of PEP, glucose, and fructose 6-phosphate by gluconeogenesis-specific enzymes that bypass irreversible steps of glycolysis push the equilibrium of reversible enzymes that function both in glycolysis and gluconeogenesis in the direction of glucose production. A: While the universe as a whole demonstrates increasing entropy, the drivers of glycolysis and gluconeogenesis are reliant on changes in enthalpy through breaking and forming high-energy bonds, rather than entropic changes. C: While some enzymes that regulate the glycolytic and gluconeogenic pathways, such as phosphofructokinase-2 and fructose 2,6-bisphosphatase, are bi-functional (the nature of their action depends upon their phosphorylation state), this is not true of the enzymes of glycolysis or gluconeogenesis. D: These steps are irreversible, not reversible, steps of glycolysis.

Anhydrous methanol has a greater eluting strength than pentane when used as solvents in a TLC procedure. Why?

Because Anhydrous is polar and will compete with the polar adsorbent on the TLC plate for binding to the compound, and carry it away.

A polar compound will exhibit a smaller retention factor on a TLC plate than a less polar compound. Why?

Because a polar compound will bind to the polar adsorbent (dipole -dipole interaction) and won't move, causing a smaller Rf value (retention factor)

Aspartame (N-L-alpha-aspartyl-L-phenylalanine 1-methyl ester) is a very well-known artificial sweetener found in the large majority of non-sugar containing food products. This compound is classified as a(n): A. ketone. B. phenol. C. dipeptide. D. amino acid.

C is correct. Aspartame contains two amino acids as mentioned above in the formula; thus, it is a dipeptide. Aspartame (shown below) is the methyl ester of the dipeptide containing aspartic acid and phenylalanine. Of course, you do not need to have any outside knowledge about the structure of aspartame, but you should be able to recognize the structure below as a dipeptide. A: Aspartame does not contain a ketone group. B: Aspartame does not contain a phenol, or a benzene ring directly bound to an -OH group. D: Aspartame is a dipeptide, not a single amino acid.

When situated in the substrate affinity pocket of PI3Kα, the morpholine group oxygen of ZSTK47 is most likely to interact with the side chain of what amino acid residue? A. Aspartate B. Cysteine C. Lysine D. Tyrosine

C is correct. From the passage: "one of the morpholine groups of ZSTK474 extends into the substrate affinity pocket of the enzyme, where its oxygen acts as a hydrogen bond acceptor for a primary amine of the enzyme." Of the choices given, only the ε-amino group of lysine could be the amino group, acting as a hydrogen bond donor, described in the passage. The structure of lysine at physiological pH is shown below. A, B, D: None of these amino acids contain amino groups as part of their side chains. Aspartate contains a carboxylate group, cysteine contains a thiol, and tyrosine has an aromatic ring bound to an -OH group as its side chain.

What is the normality of a 0.015 M solution of phosphoric acid? (KNOW THIS!!! I WAS LUCKY) A. 0.015 N B. 0.030 N C. 0.045 N D. 0.060 N

C is correct. The chemical formula of phosphoric acid is H3PO4. Normality, in the context of acids, refers to the number of moles of protons per liter of solution (in other words, to the "molarity of protons"). Normality can be calculated by multiplying the molarity of the solution by the number of protons per molecule of acid (here, 3). (0.015 M solution) x (3 protons per molecule) = 0.045 N A: This answer may result from confusing molarity and normality. B, D: These answers could result from improperly remembering the formula for phosphoric acid.

A certain metabolic process in the liver produces NADH as a part of the process. If this process is up-regulated, which of the following effects associated with gluconeogenesis is most likely to follow? A. Intracellular levels of oxaloacetate will increase. B. Plasma glucose concentrations will increase significantly. C. The rate of gluconeogenesis in the liver will decrease. D. Plasma glucose concentrations will decrease significantly.

C is correct. The key to this question is that we do not know which process the question stem refers to; we only know that it produces NADH. We are given no reason to assume that this process is the same as that described in the question stem, in which "the conversion of lactate to pyruvate is coupled with the reduction of NAD+ to NADH." Since we cannot assume that this is the process being discussed, the only information available for our use is that this process is producing larger-than-usual amounts of NADH. On the MCAT, it is very important to think about the most direct effect of the situation at hand (here, increased NADH levels), instead of thinking about potential factors that could have caused this situation. In other words, assume that the situation in the question stem has already occurred, and we now need to think about what will happen next. If NADH levels have already been increased, then NAD+ levels will be unusually low, since the production of NADH naturally involves the depletion of NAD+. With low NAD+ levels, the future production of OAA and pyruvate (two important gluconeogenic substrates) will be decreased, since pyruvate production is coupled with the conversion of NAD+ to NADH, and since OAA is produced from pyruvate. Thus, the rate of glucose production via gluconeogenesis will decrease. A: Because pyruvate carboxylase is inhibited by increased intracellular NADH levels, the rate of conversion of pyruvate to OAA will decrease. B, D: A decrease in gluconeogenesis may reduce the plasma levels of glucose somewhat, but without more information about the availability of dietary glucose or glycogen stores, neither of these choices can be concluded.

Which of the following compounds is amphiprotic? A. Acetic acid, HC2H3O2 B. Sodium acetate, NaC2H3O2 C. Sodium bicarbonate, NaHCO3 D. Sodium carbonate, Na2CO3

C is correct. The prefix "amphi-" means "both." Therefore, an amphiprotic species is one that can act as both an acid or a base. Sodium bicarbonate (choice C) dissolves in aqueous solution to produce sodium ions and bicarbonate ions. The former ion is neither acidic nor basic, but the bicarbonate ion, HCO3-, can act as a Bronsted-Lowry acid by loss of a hydrogen ion and can act as a B-L base by accepting a hydrogen ion to form carbonic acid, H2CO3. A: Acetic acid acts as a weak acid by losing a hydrogen ion, but it is not capable of accepting a hydrogen ion to any significant extent. B: Sodium acetate dissolves in aqueous solution to form sodium ions and acetate ions, the latter of which can accept hydrogen ions to form acetic acid and therefore acts as a base. However, the hydrogen atoms in acetate, C2H3O2-, are covalently bonded to a carbon atom and do not ionize to any significant extent in aqueous solution. Therefore, acetate does not act as an acid. D: Sodium carbonate dissociates into sodium ions and carbonate ions in aqueous solution. The later can accept hydrogen ions to form bicarbonate, meaning that carbonate acts as a base and does not have any hydrogen atoms to act as an acid.

The frequency used in U/S imaging must be greater than: A. 1 kHz. B. 10 kHz. C. 20 kHz. D. 40 MHz.

C is correct. Ultrasound is defined as sound with a frequency above the human range of hearing ("ultra" = "above"). To answer this question, then, we need to know what the human range of hearing is. This range is 20 Hz to 20 kHz, so anything greater than 20 kHz qualifies as ultrasound.

A large steel water storage tank with a diameter of 20 m is filled with water and is open to the atmosphere (1 atm = 101 kPa) at the top of the tank. If a small hole rusts through the side of the tank, 5.0 m below the surface of the water and 20.0 m above the ground, assuming wind resistance and friction between the water and steel are not significant factors, how far from the base of the tank will the water hit the ground? A. 5.0 m B. 10.0 m C. 20.0 m D. 30.0 m

C is correct. We first need to determine the velocity of the water that comes out of the hole, using Bernoulli's equation. P1 + ρgy1 + 1/2 ρv12 = P2 + ρgy2 + 1/2 ρv22 The atmospheric pressure exerted on the surface of the water at the top of the tank and at the hole are essentially the same. Additionally, since the opening at the top of the tank is so large compared to the hole on the side, the velocity of water at the top of the tank will be essentially zero. We can also set y1 as zero, simplifying the equation to: 0 = ρgy2 + 1/2 ρv22 -gy2 = 1/2 v22 Inserting the value for gravitational acceleration (g = -10 m/s2), and the level of the hole below the surface (y2 = 5.0 m) into the equation, we get: (10 m/s2)(5.0 m) = 1/2 v2 50 = 0.5 v2 102 = v2 10 m/s = v This velocity will be in the horizontal direction, and we will assume that the water acts like a projectile. Now, we need to determine the time that it will take the water to fall to the ground. We can assume that the initial vertical velocity of the water is zero and the displacement of the water is -20 m, because the hole is 20.0 m above the ground. The kinematic equation is: d = vit + 1/2 gt2 -20 = 0 + 1/2 (-10) t2 -20 = -5 t2 4 = t2 2 s = t Finally, we can determine the range, or displacement in the horizontal direction. Since the acceleration in the horizontal direction is zero, the equation becomes d = vit d = (10 m/s)(2 s) d = 20 m A, B, D: These answers result from miscalculation.

triple bond formula

CnH2n-2

Cysteine

Cys, C

Conversion of pyruvate into glucose requires enzymes present in: A. the interstitial fluid only. B. the mitochondria only. C. the cytosol only. D. both the mitochondria and the cytosol.

D is correct. Conversion of pyruvate to glucose requires its initial conversion into oxaloacetate, in a reaction catalyzed by pyruvate carboxylase in the mitochondria. Oxaloacetate (OAA) is then decarboxylated and phosphorylated by cytosolic or mitochondrial forms of phosophoenolpyruvate carboxykinase (PEPCK). After transport of either OAA in the form of malate or PEP directly from the mitochondria, the remainder of gluconeogenesis takes place in the cytosol. A: The interstitial fluid is located outside of the cells themselves. It does not contain gluconeogenic enzymes. B, C: Both of these answers are partially correct, but neither is complete, since the conversion of pyruvate into glucose requires enzymes present in both locations.

Which of the following statements correctly describe the methods used in the experiment? (TLC was used) I. The retention factor in a TLC procedure depends on the solvent system, temperature, and the adsorbent. II. A polar compound will exhibit a smaller retention factor on a TLC plate than a less polar compound. III. Anhydrous methanol has a greater eluting strength than pentane when used as solvents in a TLC procedure. A. I only B. I and II only C. II and III only D. I, II, and III

D is correct. Roman numeral (RN) I is accurate, as retention factor is the distance migrated by the compound divided by the total distance traveled by the solvent. TLC depends on the differential affinity of a compound for the stationary vs. the mobile phase. Therefore, depending on how polar the solvent and the adsorbent are, a compound will move on a TLC plate at a certain rate. Temperature also affects the rate of movement. RN II is also true. A polar compound will be attracted to the adsorbent (via dipole-dipole interactions) and, therefore, will move slower on a TLC plate, whereas a nonpolar compound will have more affinity for the mobile phase and will move faster. Finally, RN III may include a less familiar concept ("eluting strength"), but it is true as well. Eluting strength depends on how strongly a compound adsorbs onto the adsorbent. Since typical adsorbents are highly polar, eluting strength increases with increasing solvent polarity. Methanol is more polar than pentane and therefore has a greater eluting strength. A schematic of a TLC plate is shown below. A: Both RN II and RN III are also correct. B: RN III is also correct. C: RN I is also correct.

How many grams of hydrogen gas are required to completely react with 32 g of oxygen to form hydrogen peroxide? A. 0.5 g B. 1.0 g C. 1.5 g D. 2.0 g

D is correct. The formation reaction for hydrogen peroxide is: H2 (g) + O2 (g) → H2O2 (l) 32 g O2 x (1 mol/32 g) x (1 H2/1 O2) x 2 g/mol = 2 g H2 A, B, C: These answers result from miscalculation.

A scientist wished to prepare a buffer for an experiment to be conducted at pH 9.7. Which of the following organic acids would be the best choice for this experiment? A. Acetic acid (pKa = 4.76) B. Carbonic acid (pKa = 6.35) C. Tricine (pKa = 8.05) D. Taurine (pKa = 9.06)

D is correct. To construct the best possible buffer, we should choose the organic acid with the pKa closest to the pH at which the experiment will take place (9.7). This gives us answer choice D. Note that an ideal buffer should have a pKa within 1 pH unit of the expected experimental conditions. A, B, C: These pKa values are too far from the ideal range of +/- 1 pH unit.

Bronsted-Lowry base

a molecule or ion that is a proton acceptor

Tollens' test typically involves exposure of a carbohydrate to a solution of CuO in ammonia (NH3). Glucose yields a positive Tollens' test, but sucrose does not. Which of the following best explains this fact? A. Sucrose contains a hemiacetal group, while glucose does not; this classifies sucrose as a reducing sugar. B. Sucrose contains a hemiacetal group, while glucose does not; this classifies glucose as a reducing sugar. C. Glucose contains a hemiacetal group, while sucrose does not; this classifies sucrose as a reducing sugar. D. Glucose contains a hemiacetal group, while sucrose does not; this classifies glucose as a reducing sugar.

D is correct. Tollens' test is intended to identify "reducing sugars," or sugars with the capacity to serve as reducing agents. Specifically, sugars with hemiacetal groups can undergo mutarotation, allowing them to be oxidized by CuO. The process of mutarotation requires ring opening, which occurs at a hemiacetal group. Thus, sugars with hemiacetal groups can be oxidized and can thus function as reducing sugars. We do not need to have the structures of glucose and sucrose memorized to answer this question. Instead, simply note that the question stem tells us that glucose yields a positive Tollens' test. Glucose must therefore contain a hemiacetal group, whereas sucrose (which gives a negative test result) must not. The hemiacetal group on glucose, shown below, marks it as a classic reducing sugar. Remember, a hemiacetal consists of a carbon atom directly attached to one -OR and one -OH group. The same carbon atom is also attached to a hydrogen atom and an R group. A, C: According to the question stem, sucrose gives a negative Tollens' test. As such, it is not a reducing sugar. B: Sucrose does not contain a hemiacetal group. Also, the lack of a hemiacetal group would not classify a molecule as a reducing sugar.

What is the pH of a 0.010 M sodium hydroxide solution at 25°C? A. 1 B. 2 C. 7 D. 12

D is correct. You should immediately notice that choice D is the only option that is basic. To answer this question using math, however, first note that sodium hydroxide is a strong base that completely dissociates in aqueous solution. Therefore, the hydroxide ion concentration of this solution is also 0.010 M, or (in scientific notation) 1 x 10-2 M. Taking the negative logarithm of the hydroxide concentration gives a pOH of 2. Since pH + pOH = 14 (at 25ºC), we can calculate the pH by subtracting the pOH from 14, which yields pH = 14 - 2 = 12. A, B: These choices are highly acidic and can be eliminated immediately, since we are dealing with a solution of strong base. Choice B is the pOH, not the pH. C: This is the pH of pure water, which is considered neutral.

first law of thermodynamics

Energy can be transferred and transformed, but it cannot be created or destroyed.

electron donating groups

Groups that push (donate) electron density towards another functional group through sigma or pi bonds. left = EDG right = EWG

phosphoric acid

H3PO4 (weak acid)

perchloric acid

HClO4 (Strong Acid)

Hückel's rule:

Having (4n + 2) π-electrons, where n is an integer. Aromatic compounds are conjugated cyclic molecules with a planar structure that also satisfy an additional criterion known as

Carnot efficiency

Ideal maximum percentage of input energy that can be converted to work in a heat engine.

It is found that in the absence of molecular oxygen, the resulting imidazolinone-containing reactant is not fluorescent. According to Figure 1, what best explains this inability to fluoresce? A. The 5-membered ring is not conjugated with the aromatic phenol ring of tyrosine. B. The tyrosine side chain in the final chromophore remains deprotonated. C. The reactant lacks conjugation among any double bonds. D. The more thermodynamically favorable serine keto tautomer is formed.

In step 3 of Figure 1, the Cα-Cβ bond of tyrosine is oxidized to a double bond in a reaction which consumes molecular oxygen. This double bond places the 5-membered ring into an aromatic system in conjugation with the aromatic phenol ring of the tyrosine side chain. From the second paragraph, we know that the "imidazolinone structure, together with the tyrosine phenolic group, form a conjugated system of π-electrons that cause excitation and emission in ... GFP-like proteins." Without the molecular oxygen needed for this oxidation, such conjugation will not arise. Its absence could explain the failure of the molecule to fluoresce. B: The protonation of the tyrosine side chain is not dependent on the presence of molecular oxygen. C: The molecule shown is not entirely without conjugation—the tyrosine side chain is aromatic, so conjugation does exist. However, the molecule lacks the specific conjugation between the imidazolinone structure and the tyrosine phenolic group required for excitation and emission in GFP fluorophores. D: Serine does not exist as a tautomer.

two important gluconeogenic substrates

OAA and Pyruvate (coupled with formation of NADH from NAD+)

Ultrasound definition

Sound waves of such high frequency that they are not detectable by the human ear

In 250 mL of the MH solution with the most favorable solubility profile, how many moles of nicotinamide (MW = 122 g/mol) are present? (flip for table) A. 3.1 x 10-2 B. 3.1 x 10-1 C. 1.6 x 10-2 D. 1.6 x 10-1

Table 1 shows that there are 40 mg/mL of total hydrotrope in each MH solution. The solution with the most favorable solubility is that with an N:B:C ratio of 15:20:5. In this solution, the concentration of nicotinamide (N) is 15 mg/ml. Since the question states there are 250 mL of solution, the amount of nicotinamide must be 250 x 15 mg, or 3750 mg. We can estimate this value as 4000 mg, which is equal to 4 g. 4 g nicotinamide x (1 mol nicotinamide / 122 g) = approximately 3.33 x 10-2 This value is closest to answer choice A.

All of the following can be concluded about the heat engines tested EXCEPT: (flip for diagram) A. if the efficiency of engine 1 increases, the interior area of its heat cycle in Figure 2 will decrease. B. without a change in temperature, no net work could be extracted from the heat cycle. C. isothermal expansion is followed by an adiabatic expansion in Figure 2. D. isothermal compression is followed by an adiabatic compression in Figure 2.

The formula for the efficiency of any system is output work/input energy. Therefore, the efficiency of engine 1 is W/QH. Efficiency is increased by increasing W. Since work in Figure 2 is the interior area of the cycle, if W increases, the interior area should increase as well. This makes choice A, which states the opposite, correct for this EXCEPT question. B: In Figure 2, process 1 is at a higher temperature than process 2. If these two processes occur at the same temperature, process 1 will be on top of process 2. Subsequently, the interior area of the cycle will reduce to zero. Since the interior area is W, no net work could be extracted if TH and TC are equal. C: Process 1 is an isothermal expansion, as that part of the graph is at a constant temperature of 900 K and the volume increases from 20 m3 to 35 m3. Process 2 is an adiabatic expansion, because there is no heat exchange (as mentioned in paragraph two of the passage) and the volume increases from 35 m3 to 43 m3. Therefore, an isothermal expansion is followed by an adiabatic expansion. D: Process 3 is an isothermal compression because that part of the graph is at a constant temperature of 300 K and the volume decreases from 43 m3 to 26 m3. Process 4 is an adiabatic compression because there is no heat exchange (as mentioned in paragraph two of the passage) and the volume decreases from 26 m3 to 20 m3. Therefore, an isothermal compression is followed by an adiabatic compression in Figure 2.

The molecule below is: I. aromatic. II. antiaromatic. III. pyrrole. A. I only B. II only C. I and III only D. II and III only

The molecule depicted is furan, one of the classic aromatic heterocyclic compounds. Furan is a useful molecule to recognize for the MCAT. II: The molecule has 6 pi electrons, making it aromatic; this makes II incorrect. III: A pyrrole ring is nitrogen-based, not oxygen-based. The structure of pyrrole is shown below. Even if you did not know this structure (or anything about pyrrole), if you could identify the given compound as furan, you would know that it could not also be pyrrole. B: RN II is incorrect, while RN I is correct. C: RN III is incorrect. D: RN II and RN III are incorrect, while RN I is correct.

Retention Factor (Rf)

a ratio used to characterize and compare components of samples in liquid chromatography

Questions asking about imidazol ring =

asking about conjugation or lack thereof

pyrrole

building block of porphyrin (aromatic)

Smaller IC50 value

the greater the toxicity

sucrose structure

glucose + fructose

low wavelength =

high energy radiation

pyruvate carboxylase is inhibited by

increased intracellular NADH levels, causing the rate of conversion of pyruvate to OAA to decrease.

the presence of impurities in a pure compound

lowers and broadens the melting point range.

No UV absorption change in different solution =

no significant structural changes occur

The relationship between pH and pOH

pH + pOH = 14

Bronsted-Lowry acid

proton donor

IC50 (uM) definition

the amount of the drug it takes to kill 50% of the cells

shared enzymes will favor a certain direction of reaction due to

the formation of reactants which push the equilibrium of reactions catalyzed by reversible enzymes forward

IC50 (uM) =

the lower the concentration/number, the more toxic the compound is

normality

the number of equivalents of a substance dissolved in a liter of solution

Normality, in the context of acids, refers to

the number of moles of protons per liter of solution (in other words, to the "molarity of protons").

attenuate

to reduce in force or degree; to weaken

If NADH levels have already been increased, then NAD+ levels will be _______ _______, since the production of NADH naturally involves _______ ______ ____ _______. With low NAD+ levels, the future production of OAA and pyruvate (two important gluconeogenic substrates) will be _________.

unusually low, the depletion of NAD+. decreased.

an ideal buffer should have a pKa

within 1 pH unit of the expected experimental conditions.

Which one of the engines in the study has the highest efficiency? (table on other side) A. Engine 1 B. Engine 2 C. Engine 3 D. Engine 4

η (efficiency) = W/QH Table 1 gives us the values for QH and QC. Therefore, it is easier to rewrite the above equation by taking into account the first law of thermodynamics (conservation of energy), which tells us that QH = QC + W, using Figure 1 of the passage. In the above equation, we replace W with QH - QC: η = (QH - QC) /QH = 1 - (QC/QH) Next, efficiency of every engine can be calculated: Engine 1: η = 1 - (QC/QH) η = 1 - ((3.81 x 106 J) / (5.68 x 106 J)) η = 0.33 Engine 2: η = 1 - (QC/QH) η = 1 - [(8.30 x 106 J) / (9.21 x 106 J)] η = 0.10 Engine 3: η = 1 - (QC/QH) η = 1 - [(2.32 x 105 J) / (5.32 x 105 J)] η = 0.56 Engine 4: η = 1 - (QC / QH) η = 1 - [(1.20 x 105 J) / (3.10 x 105 J)] η = 0.61 Thus, engine 4 has the highest efficiency (0.61).

Engine efficiency equation

η (efficiency) = W/QH η = Qh - Qc/Qh η = 1 - (Qc/Qh) the smaller the amount of heat that leaves the engine, the better (implies that heat energy was transferred into work)