NS Physic Q Banks 1

The table below gives the specific rotation of plane-polarized light for two organic compounds. S-2 Bromobutane = +21 degree L-alanine= +11.45 degree In a solution containing both enantiomers of 2-bromobutane, the enantiomeric excess of the R form is 45%. Assuming that all measurements were obtained with a path length of 1 dm, what must be the observed angle of rotation? A. -9.45° B. +6.525° C. +9.45° D. +65.25°

A is correct. To solve this problem, use the following equation: enantiomeric excess = (observed optical rotation * 100) / (specific rotation). The specific rotation of S-2-bromobutane is given in the table; from this value, we can also deduce that the specific rotation of R-2-bromobutane is -21 degrees. By plugging in this number along with the given value for enantiomeric excess, the observed angle of rotation can be determined. B, C, D: All of these values are positive. Since the R form has a negative specific rotation and is present in excess, these answers do not make mathematical sense.

A hoop and a sphere, each of mass M, are rolled down a frictionless ramp. At the bottom of the ramp, which object will have the greater translational velocity, and why? A. The hoop, because its moment of inertia is higher B. The sphere, because its moment of inertia is lower C. Both will have the same velocity, since they are released from the same height. D. To answer this question, we first need to know the height of the ramp.

B is correct. An object's moment of inertia contributes to its rotational kinetic energy. Since a higher moment of inertia correlates to a higher rotational KE, it also means that translational KE must be lower, as the sum of these two values must equal the potential energy at the top of the incline. For the sphere, more of its mass is concentrated towards the center than for the hoop. As a result, the sphere's moment of inertia will be lower.

A chemist fills two separate beakers, one with 70% ethanol and the other with 70% ethanethiol. If subject to standard conditions, both beakers will have a pH of: A. less than 7, with ethanol having the lower pH of the two. B. less than 7, with ethanethiol having the lower pH of the two. C. greater than 7, with ethanol having the lower pH of the two. D. greater than 7, with ethanethiol having the lower pH of the two.

B is correct. Both ethanol (a two-carbon alcohol) and ethanethiol (a two-carbon sulfur-containing molecule) are acidic. In fact, the only difference between the two is that ethanethiol has an S atom bound to its acidic proton. Since sulfur is much larger than oxygen, the conjugate base of ethanethiol is better able to delocalize negative charge, increasing its stability. Remember, the more stable the conjugate base, the stronger the acid. C, D: Don't fall into the trap of thinking that ethanol is basic simply because it contains an -OH group! Alcohols (like ethanol) cannot easily abstract protons; in fact, they tend to weakly dissociate in aqueous solution.

A 10-kg crate is set on a cargo ramp with a 30° incline. If the crate is to remain in place, which of the following statements must be true? A. The ramp's coefficient of kinetic friction must be at least 0.6. B. The ramp's coefficient of static friction must be at least 0.6. C. The coefficient of kinetic friction cannot be less than one, so the ramp must be adjusted to a smaller angle in order for the crate to remain at equilibrium. D. The coefficient of static friction cannot exceed one, so the ramp must be adjusted to a smaller angle in order for the crate to remain at equilibrium.

B is correct. Since static friction opposes movement when an object is at rest, we need to determine a sufficient value for this coefficient. The equation for static friction is Fs = µsFN. To solve, we must first calculate the normal force, which is equal and opposite to the component of gravitational force that is perpendicular to the slope of the incline. FN = mgcos30 = (10 kg)(10 m/s2)(cos 30) = 86.6 N. In contrast, the component of gravity that acts in parallel to the ramp is equal to mgsin30, or (10 kg)(10 m/s2)(sin 30) = 50 N. Finally, we can place these values into our equation for static friction to determine the required coefficient. 50 N = µs(86.6 N), so µs = (50) / (86.6) or about 0.58. A, C: Kinetic friction acts only when an object is sliding along a surface. Here, we want the crate to remain still.

Consider a positively-charged particle that is experiencing a force due to an external electric field. Which of the following are conserved for the particle? I. Potential energy II. Kinetic energy III. Total energy IV. Momentum A. I only B. III only C. I and II only D. I, II, III, and IV

B is correct. Since the system (the charged particle) is experiencing an outside force, the only quantity conserved will be the total energy. Note that this value is always conserved except in extremely rare cases, such as nuclear fusion reactions. I, II: As the particle accelerates, potential energy (held as a result of its position) will decrease, while kinetic energy (held as a result of its velocity) will increase. IV: By definition, momentum is not conserved when external forces are acting on a system.

What is the conversion factor of L.atm to J ?

1 L∙atm = 101.325 J

What is an SN2 reaction and its requirement?

A Bimolecular nucleophilic substitution reaction that happens in one step and involve a "backside attack" by a strong nucleophile. Since the attacking atom must bind at the same time as the leaving group is removed, these reactions require an unhindered substrate.

After spending hours accurately synthesizing propyl propanoate, you realize that you were actually supposed to create a different ester molecule. Which of the following reagents, if added to your current solution, has the best chance of fixing the issue via transesterification? A. Octanol B. Ethyl butanoate C. Propanol D. Water

A is correct. Transesterification involves the conversion of one ester to another via reaction with an alcohol. The -OH acts as a nucleophile and attacks the ester's carbonyl carbon, replacing the previous R group on the ester with its own. Here, only choices A and C are alcohols. However, the R group on choice C is an propyl group, which will simply react with propyl propanoate to produce the same product as before. Since we know that you originally synthesized the wrong ester, octanol is the best choice. B: Ethyl butanoate is an ester, which will not react noticeably with another ester molecule. D: If anything, water will hydrolyze the propyl propanoate, which is not desirable here.

The reaction between propanal and cyanide forms a: A. cyanohydrin. B. primary amine. C. secondary amine. D. enamine.

A is correct. When cyanide reacts with propanal (an aldehyde), it forms a cyanohydrin. Interestingly, it is the carbon - not the nitrogen - atom that acts as a nucleophile to attack the carbonyl carbon. The final cyanohydrin product consists of the former carbonyl carbon bound to -OH, -H, the original -R group from the aldehyde, and -C≡N. D: An enamine is composed of a nitrogen atom adjacent to a carbon-carbon double bond, which does not exist here.

Which of these compounds contains the most acidic alpha proton? A. Propanal B. 2-hexanone C. 2-propanone D. Benzaldehyde

A is correct. When dealing with aldehydes and ketones, remember that the alpha protons (those bound to the carbon atom(s) adjacent to the carbonyl carbon) are especially acidic. Of the choices, propanal includes the alpha proton with the conjugate that is least destabilized by nearby electron-donating groups. B, C: These molecules are ketones, which have two alkyl groups bound to the carbonyl carbon. Since alkyl groups are electron-donating, these add excess electron density to the carbonyl carbon. This added "partial negative" charge destabilizes the also-negative conjugate bases of these species, making their alpha protons less acidic. D: Benzaldehyde does not have an alpha hydrogen.

A physical chemist is researching potential uses for the ion below. This structure represents: A. the kinetic enolate. B. the thermodynamic enolate. C. neither the kinetic nor the thermodynamic enolate. D. it is impossible to decide from the information given.

A is correct. The kinetic product is the one that is less thermodynamically stable, but is easier to synthesize due to a lower activation energy. In general, kinetic enolates are less substituted than their thermodynamic counterparts. Here, the double bond has formed in the position that is less sterically hindered (between carbons 1 and 6). B: If this were the thermodynamic enolate, it would have the double bond between carbons 1 and 2 (the more substituted position adjacent to what formerly was the carbonyl carbon).

Which of these particles will experience a force when subjected to a 6.5 T magnetic field? I. A neutron traveling at 6 × 102 m/s II. A stationary proton III. An electron moving at 4 × 104 m/s perpendicular to the field A. III only B. I and III only C. II and III only D. I, II, and III

A is correct. The magnetic force experienced by a particle in a magnetic field is given by F = qvB, where q denotes charge, v stands for velocity, and B gives the strength of the field. From this relationship, it is evident that a particle must possess both velocity and a charge (whether positive or negative) to be affected by a magnetic field. Only choice III includes both of these criteria.

A circular segment of gold wire is positioned in an external magnetic field of strength B that points out of the page. If the strength of the field is increased to 2B, what will happen to the wire? A. It will experience a clockwise current. B. It will experience a counterclockwise current. C. It will generate a magnetic field that also points out of the page. D. It will remain unchanged.

A is correct. This scenario relates to electromagnetic induction and to Lenz's law. This law, which brings to mind conservation of energy and even Le Châtelier's principle, states that a change to a magnetic field will always generate a current that counteracts that change. Here, the external field is becoming stronger, or more heavily positioned out of the page. To resist this change, it will induce a current that promotes a field pointing the opposite direction, into the page. The right-hand rule tells us that this current must travel clockwise. B, C: These would be true if the magnetic field were decreasing. D: This could only occur if the field were unaltered as well.

While standing in a packed bus, a child in the back thinks it is funny to nudge the person in front of him. That person stumbles forward a bit, then returns to his previous position. However, the damage is done; when the passenger fell forward, he nudges the person in front of him, and in this way, a wave of stumbling proceeds to the front of the bus. This wave could best be characterized as a: A. longitudinal wave. B. transverse wave. C. standing wave. D. None of the above are correct descriptions of this wave.

A is correct. This wave's oscillation is forward-and-backwards in amplitude, while its overall direction of propagation is forward. Because the amplitude of oscillation is parallel to the direction of the wave's motion, it is a longitudinal wave. B: A transverse wave has an amplitude perpendicular to its overall direction of propagation. If the boy forced the person in front of him to topple side-to-side, and if this toppling proceeded to other passengers in the forward direction, the resulting wave would be transverse.

WHICH way does a particle move in order for the magnetic field to exert a force on it?

A moving particle that is perpendicular to both the particles velocity and the magnetic field

Cefoperazone is a β-lactam antibiotic used to treat infections by resistant strains of Pseudomonas. In what way is the structure of this compound similar to that of α-acetolactone? A. Both contain small, strained rings that are highly reactive. B. Both contain at least one nitrogen atom. C. Both contain at least one cyclic ring. D. Both contain an ester functionality.

C is correct. Since these compounds are likely unfamiliar, we must use their names to discern their chemical structures. The antibiotic mentioned first is said to be a lactam, or cyclic amide. In contrast, the second molecule is a lactone; this term refers to cyclic esters. From this alone, we know that both compounds must include at least a single ring. A: This statement describes α-acetolactone, but we have no way of knowing whether it applies to cefoperazone. In reality, it does not. B: This is true of lactams, but not necessarily lactones. D: This choice is wrong for the opposite reason as choice B. Esters are present in lactones, but may be absent in lactams.

An engine is supplied with 1 kg of a hydrocarbon fuel with a heat of combustion of 50 MJ/kg. The engine can consume exactly 1 kg of fuel per minute. If this engine can produce 150 kW of power for one minute, the efficiency of the engine is: A. 0.3%. B. 2%. C. 20%. D. 100%

C is correct. The basic formula to use here is efficiency = (energy out) / (energy in). First, then, we must find the energy output, or the amount of work that was actually done by the engine. To do so, multiply the power (1.5 × 105 J/s) by the time interval given (60 s), yielding 9 × 106 J or 9 MJ. Since the engine used 1 kg of combustion material, the energy input is exactly 50 MJ. Dividing by this value yields 9 / 50, or 18%. A: This small value comes from directly dividing 150 × 103 W by 50 × 106 J. This actually gives us units of s-1, a clear sign that we did something wrong. B: The choice shown here likely results from a conversion error or one related to scientific notation.

Which of the following statements is true? I. Acetals cannot form under basic conditions. II. Hemiacetals and hemiketals exist in equilibrium with aldehydes and ketones. III. In multistep reactions, acetals and ketals often serve as protecting groups for vulnerable esters. IV. Ketals contain two -OR groups, while acetals contain only one. A. I only B. I and II only C. I, II, and IV only D. I, III, and IV only

B is correct. Statement I is true. Under basic conditions, only hemiacetals can be formed. Acetals, which result from the reaction of a hemiacetal with an alcohol, require acidic conditions. Statement II is also correct, as these species rapidly interconvert. III: Acetals and ketals serve as protecting groups for aldehydes and ketones, which are more reactive than esters. As comparatively inert groups, esters do not need protection as much as other functionalities. IV: Both ketals and acetals contain two -OR groups. In contrast, hemiacetals and hemiketals contain only one such group.

What is the effect of branching on Boiling point?

Branching, which decreases the surface area available for intermolecular attractions, tends to lower boiling point.

A chiral molecule in solution is found to rotate light in the clockwise direction. This species successfully undergoes a unimolecular nucleophilic substitution reaction. The product solution will rotate plane-polarized light: A. clockwise. B. counterclockwise. C. not at all. D. in a direction that can only be determined experimentally.

C is correct. In an SN1 reaction, the original stereochemistry of the molecule is lost when the carbocation intermediate (a planar structure) is formed. Thus, the product is present as a racemic mixture. Such a mixture includes each enantiomer in a 1:1 ratio, meaning that it promotes no net rotation of polarized light. B: If the described reaction were SN2, this would be the most likely answer.

Which of these statements accurately describes chirality or chiral compounds? A. D/L and R/S designation systems can be used interchangeably, where D = R and L = S. B. Enantiomers differ in their stereochemistry and in the magnitude of their specific rotation values. C. All meso compounds have multiple chiral centers but are not optically active overall. D. A chiral center always consists of a carbon atom surrounded by four different substituents

C is correct. This is the definition of a meso compound. Note that compounds with a single stereocenter cannot be meso, as these molecules require symmetry to yield a structure that is achiral as a whole. A: This is a common misconception. D/L is an entirely different naming convention than R/S. D/L is based on the location of functional groups in comparison to the small sugar glyceraldehyde, while R/S assigns priority based on atomic number. B: Enantiomers have specific rotation values that are identical in magnitude; they differ only in sign. D: Carbon is not the only atom that can serve as a chiral center.

Treatment of 4,5 dimethyl hexan-3-ol with sulfuric acid and methanol should result in: A. a single unique product. B. a mixture of stereoisomers. C. a mixture of products, not all of which are isomers. D. a mixture of structural isomers.

C is correct. Treatment of a tertiary alcohol with a weak nucleophile under acidic conditions is a perfect method by which to form a carbocation. However, the carbocation derived from this particular molecule can potentially undergo either elimination or substitution. Elimination will produce multiple products that are structural isomers, while substitution will form products that are stereoisomers. However, the elimination and substitution products will not be isomeric with respect to one another.

A 12-kg bag of clothes is lifted until it has 360 J of potential energy. Approximately how long will it take to hit the ground if dropped? A. 0.3 s B. 0.6 s C. 0.8 s D. 6 s

C is correct. Using PE = mgh and the potential energy given, we can first find the height of the bag. 360 J = (12 kg)(10 m/s2)(h) shows that the elevation must be 3 m. From there, we simply need to use d = v0t + ½at2 to find time, remembering that v0 is zero. 3 m = ½(10 m/s2)t2, and t2 = 0.6 s. You don't need to actually calculate the square root of 0.6 — you simply need to know that when you take the square root of a decimal, it becomes a slightly larger decimal. The only answer choice that fits is C.

Doppler effect

Change in frequency ( and also wavelength) of a wave due to a relative velocity difference between the observer and the source of the wave

What is Sodium borohydride?

Sodium borohydride would be perfectly adequate for selective reduction of the aldehyde, but it lacks the reducing power necessary to reduce the carboxylic acid.

What is Sodium hydride?

Sodium hydride is a strong base, but is not a hydride source and thus is not a reducing agent.

Sodium Hydride is known as

Sodium hydride is most notable as a strong base.

Does the trans or Cis form of 2-pentene usually have the higher boiling point?

The trans isomer of this alkene should melt at a higher temperature than the cis isomer. In general, this trend is true of alkenes, as the trans form (pictured on the left in this case) is better able to stack on top of like molecules. Better stacking equates to more surface area over which intermolecular forces may be exerted, making the substance more difficult to melt.

A sound wave with a frequency of 350 Hz is emitted from a speaker placed at x = 0 m. Another speaker, emitting a noise with the same frequency and decibel level, is positioned at x = d m as shown in the figure. What must the value of d be, if an observer standing at x > d hears a sound that has twice the amplitude of either individual sound? Assume that sound travels at 300 m/s.

This problem requires several steps. First, the question stem indicates that these waves display maximum constructive interference, meaning that they must be either 0 or 360 degrees out of phase. Additionally, we know from given information that the period of each wave is 1 / 350 s. Assume that the speakers are spaced a full wavelength apart. To determine how far this is in space, multiply the period by the velocity of the wave to get (1 / 350 s )(300 m/s), or 6/7 m. This is the distance that represents a 360° phase gap. In actuality, the speakers could be separated by any multiple of this distance. The only logical answer is 12/7 m, which denotes two full wavelengths.

Mg(OH)2 is slowly dissolved in 500 mL of 25oC water until the solution becomes fully saturated. Which of the following occurs when 10.0 mL of 0.1 M HCl is added? A. MgCl2 precipitates B. Mg(OH)2 precipitates C. Ksp for Mg(OH)2 increases D. [H2O] increases

This question is testing your understanding of solution chemistry, equilibrium, and solubility rules. In a saturated solution of Mg(OH)2, the maximum concentrations of Mg2+(aq) and OH-(aq) ions have been reached and are in equilibrium with the solubility product, as follows: Mg(OH)2(s) ó Mg2+(aq) + 2OH-(aq) When HCl is added to the solution, the pH of the solution decreases. The additional H+ ions react with the dissolved OH- ions forming water. Thus, D is the correct answer. Magnesium chloride is soluble in water, eliminating choice A. When HCl is added, the hydroxide concentration decreases shifting the equilibrium to the right making Choice B false. The solubility product, Ksp, is a constant that varies only with temperature making choice C incorrect.

How can you achieve trans-esterification?

Transesterification involves the conversion of one ester to another via reaction with an alcohol. The -OH acts as a nucleophile and attacks the ester's carbonyl carbon, replacing the previous R group on the ester with its own

Why cis and trans isomers do not exist in alkanes?

When a compound contains only single bonds. Substituents are able to freely rotate around such bonds, an action that is prevented in alkenes. In other words, bound atoms or groups are never "locked" in place when positioned around a single bond

Sodium nitrate and ammonium sulfate react at 230ºC to yield sodium sulfate, nitrous oxide, and water vapor. If the total pressure of the products is 1.8 atm, what is the partial pressure of the nitrous oxide? The reaction is as follows: 2 NaNO3 + (NH4)2SO4 → Na2SO4 (aq) + 2 N2O (g)+ 4H2O (g) (230—300° С) a) 0.5 atm b) 0.6 atm c) 0.9 atm d) 1.8 atm

a) 0.5 atm, incorrect, this would indicate that 2/7 of the pressure comes from 2 moles of nitrous oxide in the presence of 4 additional moles of water vapor and 1 mole of aqueous sodium sulfate (liquid). b) 0.6 atm, correct, 2 moles nitrous oxide and 4 moles of water vapor in the products indicate that 1/3 of the pressure is from the nitrous oxide c) 0.9 atm, incorrect, this answer incorrectly shows that 1/2 the pressure is from nitrous oxide, but the correct ratio is determined by the molar ratio of gas products d) 1.8 atm, incorrect, this would indicate that all pressure is from the nitrous oxide

What does charge dispersal means?

charge dispersal typically refers to the "spreading out" of charge across multiple atoms

Potassium cyanide

is a nucleophile and a weak base

WHAT IS the buoyant force?

the weight of the fluid displaced.

The structure of sucrose, a physiologically relevant disaccharide, is shown here. A sample of sucrose is treated with hydrochloric acid. The solution is then neutralized with a weak base, and Tollens' reagent is added to the reaction vessel. Which of the following is likely to occur next? A. A silver mirror will form on the surface of the vessel, since glucose is a reducing sugar. B. A silver mirror will form on the surface of the vessel, since sucrose is a reducing sugar. C. No silver mirror will form on the surface of the vessel, since fructose is not a reducing sugar. D. No silver mirror will form on the surface of the vessel, since sucrose is not a reducing sugar.

A is correct. Acid hydrolysis of sucrose will yield glucose and fructose, the two monomeric components of this disaccharide. One of these monosaccharides, glucose, is a reducing sugar. It would thus enable formation of a silver mirror, a positive result. B: Sucrose is not a reducing sugar, as it does not have a free aldehyde. The anomeric hydroxyl of the glucose molecule (on the left in the figure) is protected by the presence of the fructose molecule. Similarly, the anomeric hydroxyl on the fructose molecule is bound to a methyl group. Hence, no hemiacetals are present on sucrose, preventing the ring-opening process that is necessary to reveal the free aldehyde. C: Fructose actually is a reducing sugar. However, even if it were not, glucose would yield a positive result. D: It is true that sucrose is not a reducing sugar, but the question stem mentions that it is hydrolyzed before the addition of Tollens' reagent. Thus, it is the component monomers of sucrose that we must consider here.

A spring system is shown below, with m representing the mass attached to the spring, k representing the spring constant, and x denoting the distance stretched or compressed from the spring's equilibrium position. What expression gives the minimum velocity of the mass after the system is stretched and released? A. 0 B. 0.5kx2 C. x(k/m)1/2 D. (x2k)/m

A is correct. Despite the variable-heavy answers, no calculations are needed! In either a spring or a pendulum, a mass will reach a zero velocity at the extremes of its oscillation. In other words, velocity must be zero when the mass is changing directions. At these points, the mass has maximum potential and no kinetic energy.

If you wish to promote an E1 reaction, you should use: A. a tertiary substrate, a mild base, heat, and a polar protic solvent. B. a primary substrate, a strong base, heat, and a polar aprotic solvent. C. a tertiary substrate, a strong base, heat, and a polar protic solvent. D. a secondary substrate, a mild base, heat, and a polar aprotic solvent.

A is correct. E1 reactions are unimolecular eliminations. These reactions are analogous to SN1 mechanisms, as both are first-order and involve the formation of a carbocation. As such, both are heavily favored by tertiary reagents. Additionally, use of a mild or weak base tends to push a reaction toward E1, as E2 reactions (which are bimolecular) require moderately strong basic species. Finally, like SN1 processes, E1 reactions prefer protic solvents for carbocation stabilization. B: Primary substrates virtually never undergo E1 reactions. These molecules lack the electron-donating substituents that are essential for the stabilization of the high-energy carbocation intermediate.

Which fundamental law of physics prohibits an engine from having an efficiency that is greater than 100%? A. The first law of thermodynamics B. The second law of thermodynamics C. The theory of special relativity D. No law; this is just a current obstacle related to modern technology.

A is correct. It is the first law that states that energy cannot be created or destroyed, only converted between forms. In other words, we cannot possibly obtain a larger energy output than was originally supplied. Such a scenario would be required to yield an efficiency of greater than 100%. C: This theory deals with the relationship between time and space and is out of the scope of the MCAT.

A mixture of propylamine and butanoyl chloride should create: A. N-propylbutanamide. B. N-butylpropanamide. C. aryl butanamide. D. a primary amine.

A is correct. Mixing an acyl halide with an amine will promote a nucleophilic attack, yielding an amide as the product. Here, the two reagents will combine to form N-propylbutanamide, a seven-carbon amide. B: This represents an incorrect naming of the proper answer. As the butyl chain is longer, it should form the suffix, while "propyl" should serve as part of the prefix. C: "Aryl" denotes a cyclic structure. Our product should not contain a ring. D: The amide that results from this reaction will be secondary, as its nitrogen atom will be attached to both the carbonyl carbon and its own original hydrocarbon chain.

The image below depicts pyrrole, a nitrogenous molecule. All of the below statements are false regarding pyrrole EXCEPT: A. its nitrogen is sp2-hybridized. B. its nitrogen is sp3-hybridized. C. it lacks a heterocyclic ring. D. it is not aromatic because it doesn't satisfy Huckel's rule.

A is correct. Pyrrole is an aromatic, heterocyclic compound. Because its structure contains only two double bonds, it may not initially appear to satisfy the requirements for aromaticity. However, note that its nitrogen atom is attached to three substituents and carries a lone pair. In order to achieve conjugation (and thus a lower energy state), this lone pair is located in a p orbital, making the nitrogen atom sp2-hybridized. As a result, 6 (or 4n + 2) pi electrons are present in the cyclic structure, making it aromatic. B: If nitrogen were sp3-hybridized, the extra two lone pair electrons would be unavailable to add to the four already present in the ring. Pyrrole would not be aromatic, and both choices B and D would be true - an impossibility. C: Pyrrole, which contains a nitrogen atom in a five-membered ring, is heterocyclic.

A small metal racecar (pictured as the square below) is positioned at the top of a curved ramp and released. The mass of the car (m) is 2 kg, the length of the track (L) is 15 m, and the track's height (h) is 3 m, while µk = 0.1. (Note: assume the average frictional force during the drop is µmg.) With what velocity will the racecar leave the track? Assume that the car's wheels slide instead of rotating. A. 5.5 m/s B. 6.5 m/s C. 15 m/s D. 30 m/s

A is correct. The kinetic energy at the end of the track is equal to the initial potential energy (mgh) minus the energy lost to friction. Since that dissipated energy is the same as the work performed by the frictional force, it can be written as µkmgL. In other words, ½ mv2 = mgh - µkmgL. Inserting given values shows us that ½ mv2 = (2 kg)(10 m/s2)(3 m) - (0.1)(2 kg)(10 m/s2)(15), so KE = 30 J. The final step is to recognize that velocity is thus equal to √30, which is closest to 5.5 m/s. B: This value is slightly too large to be √30. To figure this out, just try 62; we get 36, which is already too large. D: This is vf2, not vf.

What is the difference between accuracy and precision?

Accuracy is how many arrows hit the center of the target, but precision refers to how large the spread of the data is. If all arrows do not hit the target but are spread close to each other, the data is precise,

Acetals and ketals serve as protecting groups for ___________and _____________, which are more reactive than esters.

Acetals and ketals serve as protecting groups for aldehydes and ketones, which are more reactive than esters.

What is the name of a ketone that is used as polar aprotic solvent ?

Acetone ( propanone)

A Doppler ultrasound device uses the Doppler effect to determine the speed of the blood traveling through a vessel. The tech is holding the probe at a right angle to the direction of blood flow. Assuming the actual flow is 10 cm/s, what frequency would the ultrasound device measure for sound waves bouncing off the blood? (Note: frequency of ultrasound wave emitted: 20 MHz; assume human tissue is equivalent to water; speed of ultrasound in tissue: 1500 m/s) A. 19.99 MHz B. 20 MHz C. 20.01 MHz D. 1500 MHz

B is correct. The Doppler effect relies on the notion that one object is moving relative to another object. In the classic example of a car approach while honking its horn, the pitch of the horn is higher (high frequency) when the car is approaching the listener and lower while the car is moving away from the listener (lower frequency). For an ultrasound device to detect a Doppler shift, the blood being measured must be moving towards the sensor or away from it. By holding the device at a right angle to the direction of blood flow, the technician is preventing the machine from detecting any Doppler shift. Thus the best answer is B. A: This would assume the blood is moving away from the detector. C: This would assume the blood is moving towards the detector.

Deca-1,3,5-triene contains how many more sigma carbon-carbon bonds than pi bonds? A. 3 B. 6 C. 9 D. 10

B is correct. This ten-carbon structure includes nine sigma bonds between carbons, as well as three pi bonds. Remember, double and triple bonds must contain one sigma bond before they "add" bonds made from pi orbitals! Since the question asks for the difference between sigma and pi quantities, the answer is 9 - 3 or 6. A: This value simply represents the number of pi bonds in this alkene. C: Similarly, this is the number of sigma carbon-carbon bonds, not the difference that the question asks for. D: This is the number of carbon atoms in deca-1,3,5-triene.

Which of the following functional groups is NOT a carboxylic acid derivative? I. An amide II. An ether III. A ketone IV. An acyl halide A. II only B. I and IV only C. II and III only D. I, III, and IV only

C is correct. A carboxylic acid derivative is a molecule that can be synthesized from a carboxylic acid via nucleophilic acyl substitution. Alternately, such a compound is one that contains a carbonyl group immediately adjacent to a heteroatom, generally one that is electronegative. Neither ketones nor ethers fit these requirements. (Note: do not confuse ethers with esters, which are classic examples of such derivatives!) I: An amide is a carboxylic acid derivative with an amino group next to the carbonyl carbon. IV: Acyl halides are the most reactive of the carboxylic acid derivatives. Each acyl halide contains a halogen (usually chlorine or bromine) in its alpha position.

Which of these solvents would best facilitate a bimolecular nucleophilic substitution reaction? A. Ethanol, due to its ability to hydrogen bond B. Water, due to its polar protic nature C. Acetone, due to its lack of an -OH or -NH bond D. Ammonia, due to its basic tendencies

C is correct. Acetone is a polar aprotic solvent, meaning that it cannot donate hydrogen bonds. SN2 reactions proceed best in the presence of these solvents. A, B, D: These solvents are polar protic. Due to their large dipoles, they are easily able to stabilize charged species, but their protic nature facilitates attraction to and solvation of the nucleophile. Since SN2 reactions require a highly reactive nucleophile, this is unfavorable.

Isaac, a chemistry student, stumbles across a closed vessel containing n-heptane. He wishes to react this compound in a way that increases its boiling point. Of the following, what should Isaac do? A. Use a Grignard reaction to further alkylate the compound, thus increasing its molecular weight B. Isomerize the compound to make it more highly branched without changing its molecular formula C. Utilize a free-radical reaction to halogenate the compound at one of its carbons D. Hydrolyze the compound into two smaller alkanes

C is correct. Boiling point is mainly determined by intermolecular forces and molecular weight. Since unsubstituted alkanes are able to exert only London dispersion forces, they typically display low boiling points. However, if Isaac were able to halogenate n-heptane, he would be giving it a polar bond between the halogen atom and one carbon. This would allow the product to experience dipole-dipole forces, raising its boiling temperature. A: As a straight-chain hydrocarbon, n-heptane will not react with a Grignard reagent. Grignard processes are typically used to alkylate compounds that contain carbonyl functionalities. B: Branching, which decreases the surface area available for intermolecular attractions, tends to lower boiling point. D: Higher (not lower) molecular weights make compounds more difficult to boil.

Compared to its analogous acyl chloride, a carboxylic acid is considered: A. more electrophilic due to resonance. B. more electrophilic due to the inductive effect. C. less electrophilic due to resonance. D. less electrophilic due to the inductive effect.

C is correct. Both chlorine and oxygen are technically capable of electronic delocalization. However, the resonance structure in which the oxygen atom forms a double bond with the carbonyl carbon is more significant than the corresponding structure in which Cl forms the same double bond. Thus, although both atoms are very electronegative, the dominant resonance contribution is more significant for the oxygen species. The resulting increase in electron density around the carbonyl carbon confers decreased electrophilicity. A, B: Acyl chlorides are more, not less, electrophilic than carboxylic acids. This explains why acyl halides are more often subject to nucleophilic attack. In fact, these molecules are the most reactive carboxylic acid derivatives.

James, a physics student, wants to choose the largest possible velocity for a theoretical projectile cannon. He should pick: A. 160 cm/hr. B. 80 µm/ms. C. 19 m/s. D. 0.500 km/min

C is correct. First, it is necessary to convert all of these values to the same units. Let's convert them to m/s, simply for the sake of using SI base units. Starting with choice A: 160 cm/hr × 1 hr/60 min × 1 m/100 cm × 1 min/60 s = 0.00044 m/s 80 µm/ms × 1000 ms/s × 1 m/106 µm = 0.08 m/s 0.500 km/min × 1 min/60 s × 1000 m/km = 8.3 m/s 19 m/s is by far the largest value of the four.

At a particular instant in time, an electron that is moving toward the right is held in equilibrium while experiencing two forces: one due to gravity and pointing straight downward, and one due to an external magnetic field. For this scenario to be logical, this field must point: A. directly upward. B. directly downward. C. out of the page. D. into the page.

C is correct. For this question, use the right-hand rule. Since the electron is traveling with a rightward velocity vector, first point your thumb toward the right. Next, because the magnetic force exists in equilibrium with the downward-pointing gravitational force, the magnetic force must be exerted in the upward direction. To account for this, ensure that your palm is facing upward. At this point, your fingers should point into the page; however, this rule is typically used for positive charges, so you must switch this final direction. The magnetic field must point out of the page instead. A: This is the direction of the magnetic force, not the field. B, D: These answers may have been obtained by assuming that the particle was positively, not negatively, charged.

A 10-kg sled rests on a 30° ramp with a coefficient of static friction of 0.5. An upward force is applied to the sled, parallel to the slope of the ramp, in incremental values until the sled begins to accelerate up the ramp. Approximately what minimum force is required to perform this action? A. 44 N B. 50 N C. 95 N D. 120 N

C is correct. The block will begin to accelerate once all opposing forces have been overcome. Here, these forces are static friction (which opposes motion) and gravity (which opposes upward movement). Since the applied force needs only to minimally surpass equilibrium, we can set our net force equation up as follows: Fnet = Fapplied - (Fg + Fs) = 0 N. This simplifies to Fapplied = Fg + Fs. Now, we need to find the component of the gravitational force that is actually acting on the block, along with the normal force. To do so, we can break our values into components. Gravitational force parallel to the plane is equal to mg sin 30º = (10 kg)(10 m/s2)(0.5) = 50 N. The maximum force exerted by static friction is given by µmg cos 30º, since normal force is equal and opposite to the y-component of the gravitational force. Thus, Fs = (µmg cos 30º) = (0.5)(10 kg)(10 m/s2)(0.866) = 43.3 N. Fapplied is then equivalent to 50 N + 43.3 N, or 93.3 N. Only C is close.

Tert-butanol is a common laboratory reagent that spontaneously degrades in solution to yield 2-methylpropene. To prevent this reaction and prolong the shelf life of tert-butanol, which of the following measures would be effective? I. Adding a small amount of pyridine to the storage vessel II. Storing the tert-butanol in a cold place III. Diluting the tert-butanol with a small amount of ethanol A. I only B. II only C. I and II only D. II and III only

C is correct. The degradation described in the question stem is an elimination reaction. Since the relevant carbon is tertiary, this reaction must proceed via a carbocation intermediate. However, for this to occur, the -OH substituent must act as a leaving group, a process best facilitated by protonation to form water. The addition of pyridine (a weak base) will decrease the chance of this protonation, inhibiting carbocation formation. Note also that elimination reactions are favored by heat, so chilling the solution will also slow the reaction. III: If anything, ethanol would protonate the tert-butanol -OH, enhancing its ability to act as a leaving group. In fact, ethanol is slightly more acidic that tert-butanol and may actually accelerate the process described in the question stem.

A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantity of a hydrocarbon material is combusted inside the cylinder to produce 1 kJ of energy, and if the volume of the chamber then increases to 1.5 L, what percent of the fuel's energy was lost to friction and heat? A. 15% B. 30% C. 85% D. 100%

C is correct. The energy used for expansion must come from the energy of combustion, minus the amount lost to friction. Since work = PΔV, we know that the expansion required (3 atm)(0.5 L), or 1.5 L∙atm. Using the conversion factor 1 L∙atm = 101.325 J, we see that this value is approximately 150 J. As the fuel was theoretically capable of producing a 1000-J energy output, the amount lost to the environment must be 1 - (150/1000), or about 85%. A: This is the percent devoted to useful work, not the percent lost to the outside environment. D: This value is impossible, since if all the energy were lost to friction, no volume change would be observed.

Unpolarized light with an intensity of 200 kW/m2 shines on a series of vertical polarizers, as shown below. The intensity of light incident on the eye after passing through both polarizers is recorded as the "reference intensity." If the second polarizer is then rotated by 90°, how much will the intensity of light reaching the eye change, compared to the reference intensity? A. The intensity will change by 33 kW/m2. B. The intensity will change by 50 kW/m2. C. The intensity will change by 100 kW/m2. D. The intensity will change by 200 kW/m2.

C is correct. The intensity of unpolarized light will be cut in half as it passes through a single polarizing filter, leaving an intensity of 100 kW/m2. Originally, because the second polarizer is oriented in the same direction as the first, it will not change the intensity of the light that reaches it, making our "reference intensity" equal to 100 kW/m2. Then, turning the second polarizer by 90° will cause the filters to be perpendicular, blocking 100% of the light. The new intensity will be recorded as 0 kW/m2, meaning that the intensity changed by 100 kW/m2 when compared to the reference intensity. D: This is the difference between the intensity of the unpolarized light and the final intensity. However, the question stem asked to compare the reference intensity with the final intensity.

Consider the figure below. Glucose and fructose can be described as: I. hexoses. II. ketoses. III. aldoses. IV. reducing sugars. A. I only B. I and II only C. I and IV only D. I, III, and IV only

C is correct. The sugar on the left is fructose, while that on the right is glucose. Both contain six carbons and are thus classified as hexoses. Additionally, both molecules are reducing sugars. Technically, this category includes sugars that can act as reducing agents, but it includes all monosaccharides with free aldehyde or ketone functional groups. (Note that ketones must convert to aldehydes via tautomerization before they can actually act as reductants.) II: Only fructose is a ketose, or a sugar that contains a ketone. III: Predictably, an aldose is a sugar that includes an aldehyde group. Only glucose, not fructose, is an aldose.

According to Newton's third law, one can conclude that: A. the acceleration of an object is given by the ratio of the force experienced to its mass. B. an object traveling at a constant velocity will continue at that velocity indefinitely, as long as no net force is acting upon it. C. an object resting on a surface will give rise to two equal and opposite forces. D. a box originally sliding at 3 m/s, with no net force acting upon it, will gradually slow to a stop.

C is correct. This is a rewording of Newton's third law that "forces come in pairs." In other words, since the resting object is experiencing no net force, an upward force must be present to cancel out its weight, which has a downward vector. A: This is Newton's second law. B: This is Newton's first law. D: This choice attempts to relate to Newton's first law, but is factually incorrect. If the sliding box is experiencing no net force, it would continue at a velocity of 3 m/s.

A small, flat plane cut entirely from diamond is submerged in a tank of fresh water. A ray of visible light travels through this water to contact the water-diamond interface, as shown. If z = 21°, what will be the angle of reflection with respect to the normal? Assume no total internal reflection takes place. A. 11.5° B. 21° C. 31° D. 69°

D is correct. For a ray that contacts a flat surface, the angle of incidence is equal to the angle of reflection. However, note that this question asks for the angle with respect to the normal! The diagram shows the angle "z" with respect to the surface itself. In contrast, the normal is the line that runs perpendicular, or at a 90° angle, to the diamond surface. For this reason, both the angle of incidence and that of reflection can be measured as (90 - 21) or 69° when compared to the normal. A: This value resembles that which would be found using Snell's law and an angle of incidence of 21°. This is wrong for multiple reasons: Snell's law is used only for refraction, and even if this question did not deal with reflection, 69° (not 21°) would need to be plugged in. B: Again, be sure to notice what the question is asking for. In general, in optics, angles are measured with respect to the normal; avoid assuming that the angles shown in diagrams should be used exactly as written. C: The angle given here was likely found by using Snell's law to calculate the ray's angle of refraction upon entering the diamond. While the angle used was at least correct (69°), this question asks about reflection.

Which of these molecules is a possible product of acid-catalyzed amide hydrolysis? A. Heptanal B. 2-methylhexanone C. Benzenamine D. Pentanoic acid

D is correct. In the acid-catalyzed hydrolysis of an amide, a proton first attaches to the carbonyl oxygen to add to the electrophilicity of that molecule. Next, water acts as a nucleophile to attack the carbonyl carbon. This allows the amine to function as a leaving group and forms a carboxylic acid. Note that such a product also forms during other hydrolytic reactions with acid derivatives, like esters. A, B: Neither aldehydes nor ketones are produced during the mentioned reaction. C: This molecule is an enamine, which is also not a typical product of acid-catalyzed amide hydrolysis. In fact, it does not even contain any oxygen molecules, which must be present to answer this question.

An unknown element decays twice, leaving behind three final products: a β+ particle, a γ particle, and a Na23 nucleus. The element was: A. F19. B. Ne22. C. Ne23. D. Mg23.

D is correct. In β+ decay, also known as positron emission, a proton is converted into a neutron while a positron (or β+ particle) is emitted. Gamma decay doesn't change the original element. If both of these forms of radioactive decay occurred, our answer must be an element with one more proton and one fewer neutron than Na23.

A slingshot propels a rubber ball horizontally from a 30 m platform at 15 m/s. Approximately how far from the base of the platform will the ball land? A. 14 m B. 21 m C. 30 m D. 38 m

D is correct. Let's start by listing our known values. The height of the platform, or vertical distance, is 30 m. The initial velocity is 15 m/s, but remember that this is entirely horizontal! The initial vertical velocity is 0 m/s and the acceleration can be estimated at 10 m/s2. First, use vertical components and the equation Δx = v0t + ½at2 to find time. 30 m = ½(10 m/s2)(t2) means that t2 = 6, so t ≈ 2.5 (you should know that the square root of 4 is 2 and the square root of 9 is 3; so the square root of 6, which is between 4 and 9 should be between 2 and 3; just estimate that as 2.5). Next, use horizontal components to find range, which is equal to the product of initial horizontal velocity and time. Range = (15 m/s)(2.5 s) = 37.5 m. That's closest to choice D.

With regard to leaving group alone, which of these species is the most unstable? A. Acetyl fluoride B. Acetyl chloride C. Acetyl bromide D. Acetyl iodide

D is correct. Of the four halogens included in these molecules, iodine serves as the best leaving group. To understand this concept, consider the periodic table. As a member of a much lower period than (for example) fluorine, iodine is a very large atom. As such, it can easily delocalize the negative charge gained when it exits as a leaving group. The better the leaving group, the more reactive the compound, and a more reactive molecule is by definition less stable. A: This represents the least reactive, and thus most stable, acyl halide listed.

What is a possible explanation for the high stability of acid anhydrides in comparison to esters? A. The leaving group on an ester is resonance-stabilized, while that of an anhydride is destabilized by a negative charge. B. The leaving group on an acid anhydride is resonance-stabilized, while that of an ester must carry a full negative charge on a single oxygen atom. C. The alkyl substituent of an alkoxide ion is unusually electron-withdrawing, giving esters especially good leaving groups. D. None of the above; acid anhydrides are actually less stable than esters.

D is correct. Remember, stability is the opposite of reactivity. Thus, this question is asking why anhydrides are less reactive than esters, which is simply not true. In fact, after acyl halides, anhydrides are the most reactive of the carboxylic acid derivatives. This trend relates to the anhydride's leaving group, which is a carboxylate ion. This species displays resonance between the two oxygen atoms. Since this makes the group stable in solution, it can easily leave and the anhydride can participate in a variety of reactions. A: Esters do not have resonance-stabilized leaving groups. In fact, their LGs are incredibly unstable on their own, making the compound as a whole unlikely to react. B: While this is partially true, it does not explain the question stem, which attempts to trick you into explaining a false statement. C: Alkyl groups are electron-donating, not electron-withdrawing

Monochromatic light with a wavelength of 640 nm is directed onto a screen with two slits. Light leaves these slits and shines on an optical screen, creating the effect seen below. A lab assistant measures the distance from one slit to a bright band, then measures the distance from the other slit to the same bright region. Which of the following could be the calculated difference between these two values? A. 320 nm B. 960 nm C. 1600 nm D. 1920 nm

D is correct. Since the area described is a bright region on the optical screen, it must result from constructive interference. This tells you that the two waves involved must be completely in phase, or differ by exact multiples of a wavelength. Given that the wavelength is 640 nm, the only logical differences between the distances traveled are 640 nm, 1280 nm, 1920 nm, etc. A, B, C: These values result from destructive, not constructive, interference.

Which solvent would be most suitable for the recrystallization of acetylsalicyclic acid (aspirin) immediately after its synthesis? Note that the known melting point for acetylsalicyclic acid is 136°C and the sample collected is displaying a melting point of 135.9°C. A. Solvent 1, in which acetylsalicyclic acid is insoluble at both 90°C and 15°C B. Solvent 2, in which acetylsalicyclic acid is highly (and equally) soluble at all temperatures C. Solvent 3, in which both acetylsalicyclic acid and its common impurities are fairly soluble at 15°C but insoluble at 90°C D. None of the above; recrystallization either would not work or is not necessary for the purification of this sample.

D is correct. Solvents 1, 2, and 3 will not work here. The ideal solvent would be one in which acetylsalicyclic acid is highly soluble in warm temperatures and insoluble in cold ones. However, note the melting point data, which shows that our sample has a narrow melting point within a tenth of a degree of the expected value. This implies that our obtained compound is very pure already and likely does not need to be recrystallized. A: If our compound is always insoluble in this solvent, it will never dissolve and recrystallization will be impossible. B: If acetylsalicyclic acid is always soluble in this solvent, we would have difficulty re-forming the desired solid even when we cooled the solvent mixture. C: We do not want the contaminants to share the properties of acetylsalicyclic acid. If we used this solvent, both our desired compound and its impurities would dissolve when the solution was heated, but would re-form a contaminated solid when the apparatus was later cooled.

A sample of deionized water is kept in a cylindrical beaker with a radius of 4 cm and a height of 10 cm. The density of the water is closest to: A. 5.024 × 10-4 kg/m3. B. 1 kg/m3. C. 502.4 g/cm3. D. 1000 g/L.

D is correct. The dimensions of the container are irrelevant; this question is simply asking for the density of water, which you should know in a variety of different units. For example, water has a density of about 1 g/cm3, 1 g/mL, or 1 kg/L. Since a single kilogram contains 1000 grams, 1000 g/L is also an accurate choice. A, C: Both of these answers mistakenly attempt to incorporate the volume of the container, which is 502.4 cm3. B: The density of water in these units is 1000 kg/m3, not 1 kg/m3

p-chlorophenol can most easily be synthesized by reacting phenol with: A. chlorine in water. B. chlorine and chromium trioxide in water. C. acyl chloride in water. D. chlorine in carbon disulfide.

D is correct. The process described is the halogenation of a phenol. When deciding which solvent to use for such a procedure, make sure to distinguish between monohalogenation and polyhalogenation. Here, since we are attempting monohalogenation, we should use a nonpolar solvent, such as carbon disulfide. A, C: Water, which is polar, is a better solvent for polysubstitution than for monosubstitution. The polar nature of this solvent favors the existence of charged species and transition states. B: Chromium trioxide is typically used for oxidation.

A billiard ball is submerged in a small tank of water at a depth of exactly 60 cm. However, the tank is located in Denver, Colorado, where the ambient pressure is roughly 0.83 atm. What is the absolute pressure experienced by the ball? A. 6000 Pa B. 6830 Pa C. 84,245 Pa D. 90,245 Pa

D is correct. The question asks for absolute pressure, which includes both gauge and atmospheric pressure. First, let's find the gauge pressure using the equation P = ρgy, where ρ denotes the density of the fluid and y refers to the depth of the object. Since all answer choices are in pascals, we need to use SI units to find our answer, meaning that we should use 0.60 m, not 60 cm, for y. The density of water is 1000 kg/m3 and gravity can be approximated as 10 m/s2, yielding a gauge pressure of (1000 kg/m3)(10 m/s2)(0.60 m) or 6000 Pa. Finally, we must add atmospheric pressure, which is 0.83 atm. Since 1 atm is equal to 101,500 Pa, 0.83 atm is equivalent to 84,245 Pa and our final answer is 90,245 Pa. Note that rounding is perfectly fine in a problem such as this one. A: This represents the gauge pressure. Absolute pressure must also include the contribution from ambient pressure. B: This choice results from an incorrect conversion between atmospheres and pascals.

An object is placed 40 cm to the left of a converging lens with a focal length of 20 cm. Where will the image appear? A. 20 cm to the left of the lens B. 40 cm to the left of the lens C. 20 cm to the right of the lens D. 40 cm to the right of the lens

D is correct. This question describes a converging lens and an object positioned more than one focal length away. In such a situation, a real and inverted image will form. Real images appear on the opposite side of the lens from the object, meaning that this image must fall on the right, not the left. To find an exact answer, use the formula . Plugging in 20 cm for f and 40 cm for do gives us a di value of 40 cm.

A variety of experiments are being conducted in a large tank containing an ideal fluid. For a spherical object with a volume of 0.84 m3 and a specific gravity of exactly 1.05, which is true? A. The object will rapidly sink beneath the surface of the fluid until it reaches the bottom of the tank. B. The object will sink beneath the surface, but will not hit the bottom of the tank. C. At least part of the object will project above the surface of the fluid. D. Not enough information is available to answer this question.

D is correct. This question is trying to trick us into assuming that the liquid in the tank is water. Since we do not know the actual identity of the fluid, we have no idea whether this object will sink or float. C: Even if this liquid were pure water, this choice would be incorrect. A specific gravity greater than one implies that an object will sink, not float.

Acid-base Extraction

Extraction, another separation technique, utilizes differences in solubility, particularly acid-base properties.

The pressure above a solid compound is slowly reduced. Which of the following is true? A) If the substance is H2O it will transition from solid directly into gas. B) The substance will transition into liquid phase as then gas phase as the pressure is reduced for all substances except H2O. C) The substance will transition from solid directly into gas for all substances except H2O. D) If the substance transitions from solid directly into gas it must be H2O.

For most substances, the solid phase is more dense than the liquid phase, so lowering the pressure will cause the solid phase to transition either to the liquid phase or the gas phase, depending on the temperature. Because this transition (melting vs. sublimation) is temperature-dependent, neither (B) nor (C) is always true. Similarly, both water and other substances are capable of subliming at certain temperature/pressure combinations, so (D) is not always true. You should also remember that although H2O is rare in having a less-dense solid phase, it is not unique. Thus (A) remains as the only option, and it is true. Because reductions in pressure favor the less-dense phase, ice will transition directly to gas as the pressure is reduced. water is interesting in that its solid phase (ice) is LESS dense than its liquid phase (liquid water). In other words, liquid water is more tightly packed together than solid ice. Now, imagine that you have liquid water. If you reduce the pressure, it will have less pressure pushing on it, so it is able to become less dense, turning it into SOLID ice. If you then reduce the pressure even more, it will become even LESS dense, turning it into gas. If this is still confusing, just take a look at the phase diagram for water. You'll see that, if pressure is on the y-axis, the section of the graph corresponding to gas is right above the section corresponding to solid. If you have a solid and you decrease the pressure, then, it'll turn to gas.

CHOLINE

IT IS AN ALCOHOL

What is lithium aluminum hydride (LAH)?

LAH is a strong reducing agent and is certainly capable of reducing both carboxylic acids and aldehydes.

What kind of force does work?

Remember, the equation for work is W = Fd cos θ . This relationship tells us that only the force exerted parallel to the object's motion contributes to the work

A 10 cubic cm cork floats in water with 1/2 of its volume submerged. How much of the cork will be submerged when in mercury (specific gravity = 13)? a) 1/20th b) 1/2 c) cork will completely submerge because of the mercury's greater cohesiveness than water d) 5/13

Specific gravity represents the ratio of the density of an object or fluid to the density of water. Here, mercury is said to have a specific gravity of 13 (water has a S. G. of 1). This means that mercury has a density 13 times that of water! In other words, it is very dense for a fluid. Now, our cork originally floated in water with half of its volume submerged, meaning that it has a specific gravity of 0.5 (half the density of water). Specific gravity is really awesome due to this exact trick: the fraction of an object that is submerged corresponds to its specific gravity divided by the specific gravity of the liquid it floats in. In water, 50% of the cork is submerged, so its specific gravity must be 0.5 because (0.5) / (1) = 50%. But what happens in mercury, where the S. G. is 13? Well, the specific gravity of the cork is still 0.5, as its density isn't going to change simply because we placed it in a different fluid. Now, the fraction submerged is (0.5) / (13) = 1/26, which is the exact answer to this question (the correct answer, 1/20, is rounded). This question is also nice because only one answer is logical. If half of the cork already appeared above the surface in water, a LOT more of its volume will float in a dense fluid like mercury. (This is why people float better in very salty water - because it is more dense than fresh water, meaning that we are comparatively "lighter per unit volume.") In other words, much less of the cork will be submerged. The only answer that is much less than 1/2 is 1/20.