Mcat chem/physics I got wrong (next step)

A researcher seeks to monitor the conversion of retinal to retinol using infrared spectroscopy. Which of the following will indicate the reaction is complete? A. The disappearance of peaks in the 3200-3500 cm-1region B. The disappearance of a peak in the 1700-1750 cm-1region C. The disappearance of a peak in the 1580-1640 cm-1region D. The appearance of a peak in the 1700-1750 cm-1region

B is correct. Retinol differs from retinal in that it contains -OH groups, but does not contain a carbonyl group (C=O). The carbonyl stretching frequency falls in the range of 1700-1750 cm-1, while the O-H stretching frequency is expected to fall in the range of 3200-3500 cm-1. The reaction is thus complete when the signal in the 1700-1750 cm-1 range has completely disappeared. A: This would indicate a lack of -OH groups, which is a characteristic of retinal, not retinol. C: This peak is indicative of C=C groups, which both molecules have and cannot be used to determine when retinal has been converted. D: This would indicate the presence of a carbonyl functionality, which is a characteristic of retinal, not retinol. Content Foundations: IR Spectroscopy Infrared (IR) spectroscopy is a method used to identify specific functional groups present on an unknown molecule. Group frequencies are vibrations that are associated with certain functional groups. A molecule can be identified when its vibrational frequencies on an IR spectrum are compared to known spectra values. Unlike UV-Vis spectroscopy, which use larger energy absorbance from electronic transitions, IR spectroscopy relies on the much smaller energy absorbance that occurs between various vibrational and rotational states. Only molecules that undergo a net change in the dipole moment during vibrational and rotational motion can absorb IR radiation. Diatomic molecules (e.g. O2, N2, and Br2) do not return IR signals because no net change in the dipole moment occurs. Molecules respond to the influx of energy by either stretching or bending. Stretching is a result of changing distances in a bond between two atoms on the molecule. Bending is any change in the angle between two bonds on the molecule. The various types of vibrations and rotations absorb at different frequencies within the infrared region, resulting in unique spectral properties for different molecular species. The most high-yield IR spectra for the MCAT are those of carbonyl groups (C=O) and hydroxyl groups (OH). The figure below shows the IR spectra for these important functional groups. On the MCAT, the theoretical details of IR spectroscopy are generally less important than why it is used, when it is best employed for analysis, and the key spectra results for biologically-relevant functional groups.

A mass of 10 kg is dropped from a height of 20 m. Ignoring air resistance, what is the maximum speed achieved by the mass? (assume g = 10 m/s2) A. 10 m/s B. 20 m/s C. 200 m/s D. 400 m/s

B is correct. The mass starts with gravitational potential energy: ΔPE = ΔKE or easier => vf2 = vo2 + 2a∆x PE = mgh = 10 kg x 10 m/s2 x 20 m = 2000 J The maximum speed is achieved just before impact when all of that potential energy is converted to kinetic: KE = ½ mv2 2000 J = ½ x 10 kg x v2 4000 = 10 v2 400 = v2 20 = v Thus, 20 m/s is the correct answer. You also could have arrived at the correct answer using the kinematics equation vf2 = vo2 + 2a∆x, where ∆x = 20 m, vo = 0 m/s, and a = 10 m/s2.

Which of the following is most likely to result in ankle injury? A. A 700-N normal force exerted by a flat horizontal surface on the feet B. An anterior force of 450 N at the pivot point of the ankle C. A 900-N normal force exerted by a flat horizontal surface on the feet D. An anterior force of 400 N at a distance of 10 cm from the pivot point of the ankle

D is correct. According to the passage, what matters is the anterior force, which pushes the ankle horizontally and causes either excessive translation or a twist (which requires torque). Only choice D involves a torque of any kind, as it includes an anterior force some distance away from the point of rotation. A, C: These describe standing still, with no motion at all. These forces are highly unlikely to cause injury, as they are essentially what an average- or large-sized person experiences while simply standing. B: Notice that this force acts at the pivot point, meaning that the length of the lever arm (r in the torque equation) is zero. As a result, this situation will not produce any torque. According to the passage, a force of this magnitude is unlikely to cause injury if no torque is involved. Content Foundations: Torque Torque (τ) is the rotational analog of force. Specifically, torque is caused by force applied to a lever arm at a certain distance from an object capable of rotating, known as a fulcrum. Torque can be defined as τ = F∙d∙sin(θ), where F is the force applied, d is the distance that the force is applied from the fulcrum, and θ is the angle between the lever arm and the force that is applied. Thus, there are three ways to increase the torque applied to an object: (1) increasing the force, (2) increasing the distance at which the force is applied from the fulcrum, and (3) adjusting the angle at which the force is applied to make it as close as possible to perpendicular to the lever arm. In equilibrium setups where objects are not moving, it may still be necessary to account for torque, if the potential for rotational motion exists. In such cases, the clockwise and counterclockwise torques will balance out (τCW = −τCCW).

During the force test with the mechanical stirrup, at what distance from the ankle joint was the force applied? A. 57 mm B. 60 mm C. 110 mm D. 133 mm "The standard talar test has a degree of subjectivity and is not as quantifiable as a new approach that uses mechanical isolation of the joint. By applying the force via a mechanical stirrup and combining the talar tilt test with the anterior drawer, consistent values have been measured. For an anterior force of 98 N and a torque of 13 Nm, the mean talar-tilt translation was 48°, and the mean anterior-drawer translation was 5.7 mm.Anterior forces of 500 N or higher lead to abnormal translation and result in injury. This can occur when a person is carrying a larger load, for example individuals who are overweight or obese. More common is when the ankle is subject to stress upon landing from a height, or turning at high speed. Injuries may also result from smaller anterior force values, if the torque is sufficiently high."

D is correct. Paragraph 3 discusses this test and tells us the associated applied force and torque values. For test day, you should know the equation for torque: τ = rFsin(Ɵ). The passage shows us that the force is applied perpendicular to the rotation of the ankle joint, so we can use sin (90) = 1. Now, we solve for distance (r) using r = τ/F = (13 Nm) / (98 N) = 0.133 m, or 133 mm. A, B, C: These answer choices are the result of miscalculation. Content Foundations: Torque Torque (τ) is the rotational analog of force. Specifically, torque is caused by force applied to a lever arm at a certain distance from an object capable of rotating, known as a fulcrum. Torque can be defined as τ = F∙d∙sin(θ), where F is the force applied, d is the distance that the force is applied from the fulcrum, and θ is the angle between the lever arm and the force that is applied. Thus, there are three ways to increase the torque applied to an object: (1) increasing the force, (2) increasing the distance at which the force is applied from the fulcrum, and (3) adjusting the angle at which the force is applied to make it as close as possible to perpendicular to the lever arm. In equilibrium setups where objects are not moving, it may still be necessary to account for torque, if the potential for rotational motion exists. In such cases, the clockwise and counterclockwise torques will balance out (τCW = −τCCW).

If a 65-kg man undergoes a turning acceleration of 5 m/s2 during a running turn, what is the magnitude of force experienced by the foot due to the ground? A. 325 N B. 650 N C. 750 N D. 1075 N

HARD Forces => 1) turning force (sideways on foot) => all forces have an equal and opposite force => turning force has a resistant force from ground 2)normal force C is correct. To turn while running, the foot must push off the ground, applying a shearing force while simultaneously supporting the weight of the body. When faced with problems like this, always consider the forces involved. Here, we must account for the normal force exerted by the ground on the foot; this is a vertical force which occurs as a result of the runner's weight. We also must consider the acceleration force, which (since the person is turning) is horizontal. These two force vectors are perpendicular and will form a right triangle. We are looking for the overall force experienced, so we must find the hypotenuse. Specifically, we need to find the hypotenuse of a triangle with legs of Fnormal = mg = (65 kg)(10 m/s2) = 650 N and Fturning = (65 kg)(5 m/s2) = 325 N. The combined vector will be bigger than either component alone, so eliminate choices A and B. To solve, let's round 650 N to 700 N and round 325 N to 300 N. With this in mind, this calculation can be approximated as:

A circuit is constructed with a 12-V battery and four identical resistors, each with a resistance of 16 Ω, hooked up in parallel. What is the total power dissipated by the circuit? A. 4 W B. 4 J C. 18 J D. 36 W

Issue => didn't memorize equation D is correct. The question asks us to solve for power, which is measured in watts; choices B and C use the wrong units and can be immediately eliminated. To choose between the remaining answers, we must use one of the three simple power equations: P = IV, P = I2R, or P = V2/R. (These equations are all derived from Ohm's law, so you can use whichever one is most convenient for the situation at hand.) We already have voltage, and we can solve for resistance: 1/Rtot = 1/R1 + 1/R2 + 1/R3 + 1/R4 1/Rtot = 1/16 + 1/16 + 1/16 + 1/16 1/Rtot = 4/16 = 1/4 Rtot = 4 Ω Next, we must solve for power: P = V2/R = (12)2/4 = 144/4 = 36 W A: This option represents the magnitude of the resistance (4 Ω), not the power. B, C: Power does not have units of joules (J). Joules are units for energy. Content Foundations: Circuits Circuits, as tested on the MCAT, are combinations of one or more resistors and/or capacitors connected by a conductive wire to which a battery is attached. The voltage differential of the battery (or "electromotive force") pushes current through the circuit. The three quantities of current, voltage, and resistance are linked together in Ohm's law, which is essential knowledge: V = IR. Power, which is conceptually equivalent to work over time, can be expressed as P = IV. Circuits can contain multiple resistors, which can be connected either in parallel or in series. For a circuit with resistors R1, R2, ... Rn wired in series, Itotal = I1 = I2 = ... = In, Vtotal = V1 + V2 + ... + Vn, and Rtotal = R1 + R2 + ... + Rn. If the resistors are wired in parallel, Itotal = I1 + I2 + ... + In, Vtotal = V1 = V2 = ... = Vn, and 1/Rtotal = 1/R1 + 1/R2 + ... + 1/Rn. Circuits can also contain capacitors, which store charge in two physically separated components. The charge stored by a capacitor is a function of its capacitance and voltage, as expressed by the equation Q = VC. When capacitors are connected in series, their capacitance adds reciprocally, like how resistance adds for resistors in parallel. When capacitors are connected in parallel, their capacitance adds directly, like resistors in series.

- you know that H2S is NOT a strong acid so the ionization constant must be much less than 1 - strong acid => Ka much greater than one A weak acid is an acid that ionizes only slightly in an aqueous solution. Acetic acid (found in vinegar) is a very common weak acid. Its ionization is shown below. A strong acid is an acid which is completely ionized in an aqueous solution. Hydrogen chloride (HCl) ionizes completely into hydrogen ions and chloride ions in water. Ionization Constant - AKA Dissociation Constant equilibrium constant for the ionization of a weak electrolyte The equilibrium constant for the ionization of a weak acid or weak base. the equilibrium constant for the formation of charged particles by an acid. Solution: The correct answer is B. The passage states that H2S is a weak acid. A weak acid is one that dissociates in water, but only to a very small extent. For example, the dissociation constant of the weak acid acetic acid is 1.8 x 10-5. This is much less than one, but is measurably non-zero. Answer choice B is the best choice.

To a first approximation, the ionization constant of H2S is: near zero. much less than 1. about 1. much more than 1. "In 1622, a Spanish ship carrying a cargo of silver crashed on a coral reef near Cuba and sank. The ship was laden with hardwood boxes of silver coins. The boxes came to rest on the ocean floor and began to decay. At first, aerobic microorganisms thrived but, as the concentration of oxygen decreased, these organisms died. Subsequently, sulfur-loving bacteria began to flourish. These sulfur bacteria consumed sulfate ions in seawater and excreted the weak acid H2S, as shown in Equation 1. SO42-(aq) + 2 H+(aq) + 4 H2(g) → H2S(aq) + 4 H2O(l) The excreted H2S then reacted with silver, which has a standard reduction potential of +0.80 V. One of the products was a black precipitate of Ag2S and the other was hydrogen gas, as shown in Equation 2. 2 Ag(s) + H2S(aq) → Ag2S(s) + H2(g) The hydrogen from this reaction provided additional food for the sulfur microorganisms and accelerated the corrosion of the silver coins. When the silver coins were completely coated with Ag2S, the corrosive reaction stopped. Because the seawater contained small amounts of CO2 (the solubility of CO2 is 0.145 g/100 g H2O at 25oC and 1.00 atm), bicarbonate ions were formed by the reaction shown in Equation 3. H2O(l) + CO2(g) => H+(aq) + HCO3-(aq) These bicarbonate ions combined with calcium to form the insoluble CaCO3, which crystallized, encapsulating the coins, sand, and decaying matter into rock-like clumps. The explorers who discovered the treasure found these rock-like structures."

Which of the following is closest to the bond angle between the carbons in a molecule of acetone? A. 90º B. 109.5º C. 120º D. 180º

break the word down Ace => Acetyl => C=O-CH3 tone => Ketone Both of these have 120 degree bonds

*need passage/diagram for this one* How many moles of captopril were present in the original analyte solution tested? A. 7.5 × 10-5 moles B. 1.5 × 10-4 moles C. 7.5 × 10-3 moles D. 1.5 × 10-2 moles "Potentiometric titration is a useful means of characterizing an acid. No indicator is used. Instead, the cell potential is measured across the analyte solution. When cell potential is plotted against titrant volume added, the equivalence point is the cell potential at the inflection point, the midpoint of the steep segment of the titration curve. For polyprotic acids, an acidic hydrogen will produce an inflection point only if it is not very weakly acidic and if its ionization constant differs from that of any other acidic hydrogen of the acid by at least a factor of 104. Captopril (molecular weight: 217.29 g/mol), shown in Figure 1, is a competitive inhibitor of angiotensin-converting enzyme (ACE). Students studying captopril were provided the following in vivo IC50 values (the minimum plasma concentration needed to inhibit 50% of target enzyme activity in vivo) for captopril inhibition of ACE under different pH conditions. Students then performed a potentiometric titration of captopril in order to determine the captopril content contained in a tablet formulation. Two tablets were ground and homogenized, producing 104.4 grams of fine powder. The powder was then dissolved in 100 mL of water and titrated with a solution of 2 x 10-2 M NaOH. The potentiometric titration curve obtained, along with a plot of the rate of change of potential during the titration, is shown in Figure 2.

didn't derive info in passage For polyprotic acids, an acidic hydrogen will produce an inflection point only if it is not very weakly acidic and if its ionization constant differs from that of any other acidic hydrogen of the acid by at least a factor of 104 B is correct. In a titration, the analyte — the substance whose quantity or concentration is to be determined — is reacted with a carefully-controlled volume of standard solution, of which the concentration is known. When analyzing a titration curve, look for the equivalence point(s), which are located halfway along the steep portion(s) of the curve. For a monoprotic acid at the equivalence point, Mbase × Vbase = Macid × Vacid = moles acid. In Figure 2, we see that the equivalence point of the reaction occurs when approximately 7.5 mL of 2 x 10-2 M NaOH solution was added. However, captopril is not a monoprotic acid; it contains both a carboxylic acid group and a thiol group and is thus diprotic. So which equivalence point is this? Look back to paragraph 2, which states that a hydrogen will produce a visible inflection (equivalence) point "only if it is not very weakly acidic, and if its ionization constant differs from that of any other acidic hydrogen of the acid by at least a factor of 104." Examining Figure 1, we see that captopril has a pka1 of 3.7 (the carboxyl group) and a pka2 of 9.8 (the thiol group). Since these values differ by approximately 6, ka1 and ka2 will differ by approximately 106. This means that the second proton is much weaker than the first, and only the first proton should produce an inflection point. Let's return to our original equation: MNaOH × VNaOH = (2 x 10-2 mol NaOH/L)(7.5 x 10-3 L) = 1.5 x 10-4 mol NaOH Remember, we have confirmed that the inflection point in question is the first equivalence point, and that our acid can be treated as a monoprotic species. Since 1 OH- ion is required to neutralize each proton, the molar amount of NaOH required to reach this first point must be equal to the molar amount of captopril present. Therefore, the total number of moles of captopril present in the original analyte solution was: (1.5 x 10-4 mol NaOH)(1 mol captopril/1 mol NaOH) = 1.5 x 10-4 mol captopril A, C, D: These answer choices are the result of miscalculation. Content Foundations: Titration Titration is the process of finding the concentration of an unknown solution (the analyte) by reacting it with a solution of known concentration (the titrant). The analyte is generally placed in an Erlenmeyer flask, while the titrant is placed in a burette so that the volume of solution added can be monitored. The titrant is added to the analyte until the endpoint is reached. Calculations are then performed to find the unknown concentration of the analyte. Titrations are typically performed for acid/base reactions but are not limited to them. At equivalence points during the titration, the number of acid or base groups added to the solution is equivalent to the number of base/acid groups in the original unknown solution. We can calculate our unknown concentration or volume using the formula NaVa = NbVb, where N and V are the normality (mol/L) and volume of the acidic and basic solutions, respectively. It is important to convert from molarity (M) to normality (N) for polyprotic acids and polyvalent bases. The flat regions of titration curves represent buffering solutions (a roughly equal mix of an acid/base and its conjugate), while the steep, near-vertical sections of the curve contain equivalence point(s), which indicate that enough of the titrant has been added to completely remove one equivalent (acid or base group) from each of the original molecules in the unknown solution. Species with multiple acid or base groups (e.g., H3PO4 or Ca(OH)2) will have multiple equivalence points during the titration. The final key point of any titration is the endpoint. To be successful, there must be some method for observing the endpoint of the reaction. The type of titration reaction that is being used will determine the method used for observing the endpoint. For example, in an acid-base titration, a specific pH value will be the endpoint (monitored by color-changing indicators), while for precipitation reactions, the endpoint is realized by the appearance of a precipitate. Regardless of the details of the reaction involved, the goal of titrations is always to use known volumes/concentrations to determine unknown volumes/concentrations.

Which of the following biological substances are likely derived from terpenes? I. Aldosterone II. Glucose III. Insulin IV. Estrogen A. I and II only B. I and III only C. II and III only D. I and IV only "Terpenes have been found to be essential building blocks of complex hormones and molecules, pigments, sterols, and even vitamins. Terpenes also play an incredibly important role by providing protection from bacteria and fungus. Terpenes have a basic structure of repeating isoprene, or (C5H8)n, units, and they are grouped according to the number of these repeating units. Monoterpenes contain 2 isoprene units; examples include menthol and pinene. Vitamin A1, a diterpene, contains 4 isoprene units. Research has revealed the Isoprene Rule, which states that adjacent isoprene units in terpenes are linked preferentially between carbon atoms located at opposite ends of the isoprene structural subunit. Head-to-head and tail-to-tail connections are rare exceptions to this rule."

passage stated => sterol (steroid hormone => anything ending in -ol -one or -en D is correct. Paragraph 1 states that terpenes are precursors to sterols. Terpenes, including those shown in Figure 1, are structurally similar to the fused 4-ring system in steroids and are the metabolic precursor to this important class of biomolecules. Two of the choices (aldosterone and estrogen) are steroid signaling molecules, and two are not. Note that the majority of steroid hormones have names that end in "-one," "-en," or "-ol." The structures of aldosterone and estrogen (specifically, estradiol, which is one form of estrogen) are shown below. Note the similarities in structure, especially the four fused rings.

A book rests horizontally on a table. The book experiences a gravitational force of mg due to the earth's gravity. According to Newton's third law: A. the book experiences a normal force of mg pushing up due to the table. B. the earth experiences a gravitational force of mg from the book. C. the table exerts a gravitational force of mg on the earth. D. the earth exerts a normal force up on the table equal to mg plus the weight of the table.

third law Fa=>b = -Fb=>a a= book b=earth b=earth a=book