# Altius Question Sets Review

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### Specific heat is an intensive property that can be measured for any substance per unit mass. A student in the lab added 10 mL of ethanol to a calorimeter. An exothermic reaction provided 500 Joules of heat to the ethanol and calorimeter. The student used the heat capacity of the calorimeter and measured change in temperature to determine that 20% of the heat had been absorbed by the calorimeter. If the temperature of the ethanol increased by 20° C, what is the specific heat of ethanol? (density of ethanol = 0.789g/mL) A. 3.2 J/g°C B. 2.5 J/g°C C. 4.0 J/g°C D. 12.5 J/g°C

The UNITS in the answer choices themselves tell you how to calculate it without knowing any formula at all. REMEMBER that key point. The MCAT may give you a problem you know nothing about, but if you have units you can work backwards. We need to get an answer in terms of J/g°C, so first we look for Joules. The stems says 500J, but then tells us 20% went into the calorimeter and not into the ethanol itself. So, multiply 500 by 0.8 to get 400J. Now we look for grams; it is not given, but we are given mL and density in grams per mL. Many students get stuck at this point on a question of this type. If you look for grams and it is not there, check the passage. If it is not there either, then you KNOW that there must be a way to derive, calculate, estimate or even intuit the missing variable. Solve D=m/v for mass and you get 7.89g. The only variable left is degrees Celsius and it is given. Plug-n-chug from there and you should get answer B, 2.4 J/g°C.

### A man stands on the edge of a 45 meter cliff and throws a stone horizontally with an initial velocity of 40 m/s. How far from the bottom of the cliff will the stone land? A. 200 m B. 120 m C. 400 m D. 80 m

B; To know how far the stone will travel horizontally, you must know the time in the air. To find this, either estimate in your head, or solve X =1/2at2. In 3 seconds, the vertical velocity will go from 0 to 30 m/s, with an average velocity of 15 m/s. Multiplying this by 3 seconds gives exactly 45 meters, so the time in the air must be 3 seconds. The stone can only travel horizontally for as long as it is in the air, so the horizontal distance is 40m/s*3sec = 120m., or answer B

### Based on information in the passage (38 m/s = 85 miles/hr), a car traveling at 100 mph is traveling at approximately how many meters per second? A. 220 m/s B. 57 m/s C. 45 m/s D. 1760 m/s

C; Here, you should note that the speed of the water in the passage is given in both m/s and mph. This allows you to construct a rough conversion factor between the two. It appears that mph is approximately 2.2 times m/s. Using this, to go from to miles per hour to m/s we must divide by roughly 2.2. Answers A and D are false because they are larger than the original value in mph. Answer B is false because it is a little more than 100 divided by 2. Answer C looks like it could be approximately equal to 100/2.2

### The turbines are normally constructed of rolled steel, which as a modulus of elasticity (ME) of 2.9x10⁶ N/m². The shaft, however, experiences greater stress and must be made of titanium (ME= 5.0x10⁶ N/m²). Which of the following gives the ratio of the change in length to the original length for a titanium rod undergoing 5.0x10⁴ N/m² of tensile stress? A 1:100 B. 250:1 C. 1:10000 D. 2.9:5.1

44) A; The formula for Modulus of Elasticity is: ME = stress/strain and strain is equal to "change in dimension over original dimension." In other words, the problem is simply asking for strain. Rearranging the equation, we see that to calculate strain, we must divide stress by ME. This gives 5 x 104/5 x 106 = 0.01 or 1/100th. This means the distance the titanium will deform is one one-hundredth of its original length, or Answer A.

### A reaction known to have a very high energy of activation could also have which of the following properties? I. A fast reaction rate II. Delta-G less than zero III. A very large Keq A. I only B. II only C. I and II D. I, II and III

D; Having a very high energy of activation ONLY tells us about the relative rate of this unknown reaction and almost nothing else. A reaction with a very high Ea can have a fast rate as long as temperature is high, making Statement I true. The value of ΔG could indeed be negative (meaning the reaction is spontaneous), primarily because Ea is impacting rate and kinetics and thermodynamics do not relate to one another in any way EXCEPT for temperature, making Statement II true as well. Finally, a reaction with a high energy of activation could absolutely have a large equilibrium constant. This is fairly common, in fact. In many reactions the products are highly favored and there will be a lot of them compared to reactants once you reach equilibrium, but the high energy of activation means it will take a long time to get there. This makes D the correct answer. One important note is that although none of the factors in the three statements interact with rate, Gibb's free energy and equilibrium constant DO interact. If we have a large equilibrium constant, we know the reaction must be spontaneous, and if we have a Keq less than one we know it cannot be spontaneous. This is illustrated by the equation: ΔG = -RTlnK.

### All of the following statments accurately describe saturated and unsaturated fats, EXEPT: A. Carbon-carbon bond length varies in unsaturated fats, while in saturated fats it does not B. Saturated fats have higher freezing points than unsaturated fats C. Solid saturated fats have weaker Van der Waals forces than do solid unsaturated fats D. Saturated fats have more hydrogens than unsaturated fats of the same carbon chain length

C; Carbon-carbon bond length does vary between double and single bonds and thus because unsaturated fats have both, A is true. B is true because saturated fats do indeed have higher freezing points. Remember that freezing point is essentially the same as melting point, it is the phase change from liquid to solid. Saturated fats include such things as the fat in your steak and at room temperature these are solid or "frozen" while unsaturated fats such as vegetable oil are not. C is false and is thus the right answer. Saturated fats, because they stay solid longer as temperature rises, must have the stronger Van der Waals forces. Creating a double bond removes two hydrogens, so D must also be true.

### The DNA strand shown below is from the coding strand of a section of human DNA. Which of the following gives the matching pre-mRNA sequence? 5' ATTCG 3' A. 5' UAAGC 3' B. 3' UAAGC 5' C. 3' GCUUA 5' D. 3' AUUCG 5'

C ; To answer this correctly, you must differentiate "coding strand" from "template strand." The template strand is the one copied and the coding strand is the other strand, the complement to the template strand, which is NOT copied. As a result (visualize this in your mind) the new DNA strand OR the new pre-mRNA strand will be an exact copy of the coding strand (EXCEPT, in the case of pre-mRNA, T will be replaced with U). The next skill you need is to keep careful track of the 3' and 5' ends. Normally, we need the strands to run in opposite directions, so if the template strand was listed 5' to 3' the new strand would consist of the matching base pairs running 3' to 5'. However, note that here you are NOT given the template strand, but the coding strand. Thus, the coding strand and the new pre-mRNA strand will both run the same direction and will be identical except for replacing T with U. C is the only answer satisfying all of these requirements.

B

### Which of the following has the highest bond energy? A. C=O B. S=O C. P=O D. N=O

D; The bond with the highest bond energy is the one that is the most stable. Because the options are all double bonded to an oxygen, we can just look at the effect of the first species on the double bond. These species all vary in size, with the smallest atom being capable of getting the closest to the oxygen and thus forming the shortest, strongest double bond with the most pi orbital overlap. D is thus the best answer

### Are alkenes electrophiles or nucleophiles?

Nucleophiles!

The pi electrons in the double bond will attack electrophiles forming a new bond to one of the carbons and leaving a carbocation of the other. The carbocation is then quenched by a nucleophile

### When a student attempts to synthesize a carboxylic acid by adding Ca(OH)₂ to a reaction vessel containing an ester, the ester does react, but the student does not obtain the desired acid. Instead, he obtains an unexpected product. This product is most likely: A. The original ester B. a di-ester derivative with an alcohol substituted on the alpha carbon C. a carboxylic acid with an ester group substituted on the alpha carbon D. a hemiketal

D; Because you are adding a strong base, you can know that it must be attacking an acidic hydrogen if one is present. The alpha hydrogens here are very similar to the alpha hydrogens on an aldehyde which you should know has acidic alpha hydrogens. The MCAT will require you to make small leaps like this. If an aldehyde can self-condense in an aldol reaction could two esters do it? To what degree they really will or will not in the lab is unimportant. Draw out the structure formed if the alpha hydrogen is removed from one ester and it reacts with the carbonyl of another ester. When the carbanion from the first ester attacks the carbonyl of the second ester, it will turn the carbonyl into an alcohol (following protonation). The -OR group from the original ester will still be attached, creating a hemiketal and making answer D the best answer.

### Two amines, analine and ethylamine, are best separated by which of the following techniques? A. Thin layer chromatography B. Extraction C. Mass spectroscopy D. Distillation

D; Because you are adding a strong base, you can know that it must be attacking an acidic hydrogen if one is present. The alpha hydrogens here are very similar to the alpha hydrogens on an aldehyde which you should know has acidic alpha hydrogens. The MCAT will require you to make small leaps like this. If an aldehyde can self-condense in an aldol reaction could two esters do it? To what degree they really will or will not in the lab is unimportant. Draw out the structure formed if the alpha hydrogen is removed from one ester and it reacts with the carbonyl of another ester. When the carbanion from the first ester attacks the carbonyl of the second ester, it will turn the carbonyl into an alcohol (following protonation). The -OR group from the original ester will still be attached, creating a hemiketal and making answer D the best answer.

### Resistor A has a resistance of 2 ohms, while resistor B has a resistance of 2 ohm. A and B are attached in parallel within a circuit with an average current of 30 Amps. Which of the of the following statements is true? A. Because the resistors are in parallel, the sum of their inverses equals the total effective resistance. B. The current through B will be twice the current thru A. C. kirchoff's rule states that the total current into a node equals the total current out and thus the current through both A and B must be the same. D. The current through B will be 10 A and the current thru A will be 20 A

B ; This question tests your knowledge of Kirchoff's 1st rule along with the general principle that when charge is divided between two non-identical resistors in parallel, more current flows thru the resistor with less resistance and less thru the resistor with more resistance. The ratio of the current between the two is equal to the inverse of the ratio of their resistances in ohms. For example, if the ratio of the resistance of A to B is 2:1 ohms, then the ratio of current thru A to B is 1:2 Amps.

### A wave of light proceeds from object x ad strikes the front side of a convex mirror. An image is formed A. in front of the mirror and is upright B. behind the mirror and is inverted C. behind the mirror and is upright D. in front of the mirror and is inverted

C; Always remember the PRI/NVU rule. In this case, we can draw a convex mirror and predict that the light will be reflected outward or DIVERGED back toward the source. This will not create a focal point, so there must be a pseudo focal point that exists at the point behind the mirror where the two lines would cross if we traced them back. Because it is BEHIND the mirror it is NEGATIVE and therefore it must follow NVU and also be virtual and upright; making C the correct answer.

### Sound waves traveling through air from point A to point B transport which of the following? I. energy II. matter III. momentum IV. photons A. I only B. I and II C. I and III D. I, III and IV

A; All true waves transport energy, but there is no net transport of matter. The wave displaces the matter vertically or horizontally, but the matter rebounds after the wave passes. This makes I and III true and II false. Finally, only light, which is a transverse wave, transfers momentum (primarily because it transports photons).

1x10⁻⁸ W/m²

### An x-ray passes through air, enters a patient's arm, and exits the arm into the air on the opposite side. Which of the following attributes af the wave decreases as it passes from the patient's arm back into the air? A. velocity B. frequency C. wavelength D. none of the above

D; The arm is going to be denser than the air. This transverse wave will decrease in speed in the denser medium and speed back up in the air, making A fasle. Frequency NEVER changes, making B false. Given these two facts and the wave equation we see that wavelength must be shorter in the denser medium and longer in air, meaning it INCREASES, not decreases, as it passes back into the air. This makes D the only true answer.

### Cations formed from transition metals, such as lead, cobalt and iron, usually form insoluble compounds with other ions in solution. However, when they are mixed with nitrate or ammonium ions, no precipitates form. Which of the following provides the best explanation for this observation? A. Both nitrate and ammonium always form soluble compounds in aqueous solution. B. Nitrate is always soluble due to its ability to hydrogen bond with water anda ammonium doe snot react with the cations C. Ammonium is always soluble due to its ability to hydrogen bond with water and nitrate does not react with the cations D. Both nitrate and ammonium always form insoluble compounds in aqueous solution

B; Answer A may be tempting, especially since it is true. However, think carefully about what is actually happening here. Cations are added to a solution containing nitrate, a negatively charged anion, and ammonium, a positively charged cation. Two cations will never interact with each other. Thus, saying that ammonium compounds are always soluble does not logically apply because no compound would be formed anyway. Answer B is correct. The hydrogen bonding capability of nitrates IS a logical explanation for their solubility and ammonium, being a cation itself, would not react with other cations. You may question whether or not nitrate can form hydrogen bonds because there is no hydrogen directly attached to an F, O or N atom. This does mean that nitrate cannot be a hydrogen bond donor and therefore could not hydrogen bond with itself. However, it can act as a hydrogen bond acceptor. Any F, O or N with a lone pair can act as a hydrogen bond acceptor.

### There are several copper nitrates that differ only in the oxidation number of copper. Each of them creates a blue solution when dissolved in water proportional to the number of copper ions in solution. Which of the following would produce a solution with the lightest blue color? A. CuNO₃ : Ksp = 1x10⁻⁴ B. Cu(NO₃)₂ : Ksp = 1x10⁻¹ C. Cu(NO₃)₃ : Ksp = 1x10⁻⁶ D. All three would produce the same amount of blue color

A; If they form DIFFERENT numbers of ions, you must either calculate the actual solubility for each, OR you can usually use this shortcut: Raise each Ksp to the 1/x power, where x is the number of ions formed in solution. Do the math, and compare the resulting numbers, which will approximate the actual solubilities. In this case, Answer C appears to have the smallest Ksp, meaning it would be the least soluble and would thus produce the lightest blue color. However, when you use the shortcut above, you should get 1 x 10-1.5 for Answer C and 1 x 10-2 for Answer A, making A the best answer

### All of the following molecules produce a green solution proportional to the number of moles of copper ions dissolved in solution. Which of the following solutions will be the lightest shade of green? A. CuCO₃ B. CuBr C. CuO D. CuPO₄

D; We know B is probably wrong because halide ionic compounds are usually soluble. Even if you don't know this, you're still OK. We know carbonates and sulfates are very insoluble. We also know that the greater the charges on the ions, the harder they will be to pull apart in solution. So, the CuSO4 (2+/2- ions) will be harder to pull apart than the CuBr (1+/1- ions). The copper phosphate has two magnitude three charges, which will definitely be the hardest, making D the best answer.

### A 3.0 g sample of FeSO₄ is dissolved in water and 4.0g of K₂CO₃ is added. As the solution is mixed, a precipitate forms. Assuming that all the molecules react, what is the identity of the precipitate and the number of moles present? A. FeCO₃; 0.029 mol B. K₂SO₄; 0.0029 mol C. FeCO₃; 0.02 mol D. K₂SO₄; 0.002 mol

C; The two possibilities for a precipitate would be iron carbonate or potassium sulfate. Because carbonates are almost universally insoluble and sulfates are usually soluble, iron carbonate must be the precipitate and thus answers B and D are incorrect. To determine the number of moles, you must calculate the molecular weights of both species; then divide 3.0g and 4.0g by their respective molar masses. Again, using scientific notation will help you. You should get something around 0.02 and 0.03. Because one half of each species is recombining to form a new compound, you could never get more of the new compound than the least amount you have of the two compounds from which you are making it. Thus the moles must be 0.02, answer C.

### It is discovered that a small patch of skin cells on the face of certain male tree frogs secrete a hormone onto the surface of the skin. Called pheromones, these chemicals are thought to play a role in sexual excitation of female frogs. This response in the male frog is best described as a(an): A. nervous response, because it is highly specific B. endocrine response, because it results in secretion of a hormone. C. nervous and exocrine response, because both are involved. D. neuroexcitotary response, because it involves both nervous system and sexual arousal.

C ; C is the correct answer. There are no obviously wrong answers here, so you have to search for the "best" answer among the alternatives. Anytime you see stimulation of a very specific response or very specific set of cells being stimulated, you need to think "nervous" because it is the specific one and an endocrine response is much more broad. A could be correct, but leaves out the exocrine function. B is incorrect because secretion of the pheromone is exocrine, not endocrine.

### All of the following structures would be plausible locations for further study of the high-bood-pressure effect of this new disease, EXCEPT: I. adrenal cortex II. anterior pituitary III. posterior pituitary IV. parathyroid A. I only B. II only C. III only D. All of the above are plausible locations for further study

D ; You are looking for glands that secrete a hormone or hormones that could cause the high blood pressure effects described. The adrenal cortex secretes aldosterone, which raises blood pressure, so this is plausible. The posterior pituitary secretes ADH, which has a similar effect. The parathyroid secretes PTH, which increases calcium concentration in the blood. This would account for the erratic calcium levels in the blood and could increase the osmolarity and thus the pressure of the blood. The anterior pituitary does not secrete any hormones that directly change blood pressure, but ACTH stimulates the adrenal cortex to secrete both glucocorticoids and mineral corticoids such as Aldosterone. This makes all of them possible areas for study, answer D

### All of the following occurs as filtrate passes through the loop of Henle, EXCEPT: A. the filtrate is concentrated B. the medulla decreases in osmolarity C. active transport occurs D. salts diffuse across the ascending loop of Henly

B ; This shouldn't be a difficult question if you do them one-by-one as you've been taught. The filtrate is indeed concentrated, active transport IS used at various places (glucose recapture; ascending loop of Henle and others), and salts do diffuse across the loop of Henle for a short period of time on the way back up toward the cortex. B, however, is in reverse order. The Loop of Henle CONCENTRATES the medulla, which raises its osmolarity.

### A student wishes to study the distribution of actin in a human myocyte. Which of the following substances, radioactively labeled and injected into the cell, would make the best marker for microscopic identification? A. myosin B. tropomyosin C. calcium D. acetylcholine

B; This is a great MCAT question. It IS difficult, but if you slowly reason thru what actually happens in a muscle cell, if you truly understand muscles conceptually, it becomes clear. You should be able to narrow it down to either myosin or tropomyosin because they both bind to actin, but which one would be best? Tropomyosin is bound to actin until and unless calcium causes its displacement. Myosin is ONLY bound to actin for a microscopic period of time during the actual contraction of the muscle cell. In other words, in default state, in resting state, and probably (since it would be quite difficult to examine actin distribution DURING contraction) during the time period during which the student would be doing the examination, it is tropomyosin that is bound to actin, not myosin.

### A disease condition resulting in the loss of function of FSH would have all of the following effects, EXCEPT: A. incomplete follicle maturation B. reduced sperm production C. failed ovulation D. reduced proliferation of the uterine lining

D; FSH stimulates the follicle to mature in females, sperm production in males, and has nothing to do with proliferation of the uterine lining (estrogen and progesterone do this upon secretion from the corpeus luteum—always think of e & p as the preganancy maintaining hormones); thus answers A and B are false and D must be the correct answer. Because FSH does participate in ovulation, C is also false.

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