Dawn - CP - Flashcards

Question

``` The H-C-H bond angle in dichloromethane (DCM) is expected to be: A. Greater than 109.5, but less than 120 B. Equal to 109.5 C. Less than 109.5 D. Equal to 120 ```

Answer 1

Less than 109.5 because the steric clashing between the Cl will cause the H's to be pushed together

Answer 2

A system that is connected by p orbitals with delocalized electrons in molecules with alternating single and pi bonds

Answer 3

1. Resonace - rearranging of valence electrons from one atom to another 2. Constitutional isomers - structural isomers, that have completely differnt functional groups 3. Conformational isomers - same molecular formula, same connections but are comprised of bonds that are freely rotatable 4. Epimerization - Mutarotation - change in specific rotation of a chiral compound. The sugars ring opens up and rearranges the orientation at the anomeric carbon

Answer 4

if something is stable and does not change upon mixing it means it is in kinetic control - if something changes to a new state it as a preference and is likely in thermodynamic

Answer 5

- More conjugated double bonds

Answer 6

Ratio of Kcat/Km

Answer 7

wavelength = V/f

Answer 8

- The ideal gas model tends to fail at lower temperatures or higher pressures, when intermolecular forces and molecular size becomes important - It also fails for most heavy gases, such as many refrigerants, and for gases with strong intermolecular forces, notably water vapor

Answer 9

towards you, higher f, higher pitch, smaller wavelength

Answer 10

- Want to use complementary colors - Y(560-600) --> V (400-500) - O (600-640)--> B (450-480) - R (640-700) --> G (450-560)

Answer 11

it would be 1/Km which is the X intercept

Answer 12

- DOES NOT AFFECT HOMODIMERS!! - If the dimers are of different molecular weight, THEN they will make different bands, other wise all one band - Applies negative charge to molecule

Answer 13

- Ordered mechanism - One thing binds then another binds - Random - either substrate could bind first - Ternary because both substrates occupy the catalytic cleft

Answer 14

1. Oxidation of aldehydes = carboxylic acids 2. Reduction of ketones = alcohols 2. Oxidation of ketones = cannot be oxidized further! bc C=O bond already possesses two C-C bonds and cannot be broken easily

Answer 15

Must have two alcohol groups

Answer 16

B - read closely - From the passage we learn that CoQ takes electrons from Complex I and II and passed them to Complex III. Since it takes electrons from Complex I, it is reduced and it oxidizes Complex I. Later, when it passes electrons to Complex III, CoQ is oxidized and reduces Complex III.

Answer 17

1. O-H (3200-3600) 2. C=O (~1700) 3. C=C (~1650) 4. CtriplebondC, CtriplebondN (2100-2300) NOTE: C-H is 2950 and will NEVER be the answer

Answer 18

Means negative delta G

Answer 19

Voltage must be 0!

Answer 20

A spectator species is unchanged in the net ionic equation | i.e. appears the same on both sides of the reaction and is not consumed

Answer 21

C! bc During the ionization of transition metals, electrons from 4s subshell orbitals are generally removed before those from 3d subshell orbitals. This is because it is also generally true that, when occupied by electrons, electrons of 4s subshell orbitals are higher energy than those in 3d subshell orbitals.

Answer 22

The combination of CO2 and H2O to make carbonic acid (H2CO3) - Carbonic acid and its conjugate base (bicarbonate) play a large role in maintaining the pH of the human bod - Makes pH more acidic!

Answer 23

B bc The highest oxidation state for sulfur is +6, corresponding to the loss of all of its valence electrons

Answer 24

Amide hydrogens!

Answer 25

Reducing activation energy of the reaction

Answer 26

KMnO4 - Oxidize alcohols to carboxylic acids PCC - mild oxidizing reagent when added to a primary alcohol in non-aqueous solution, it converts the alcohol to an aldehyde (KMnO4 gets it all the way to carboxylic acid)

Answer 27

I. Blood flow to the tissue | II. Availability of oxygen to the tissue

Answer 28

D - Looking for a high proton to neutron ratio! | - Positive beta decay, also known as positron emission,

Answer 29

1. High - steepest slope = stiffest | 2. Low - least steep = not stiff

Answer 30

PE (spring) = 1/2kx^2

Answer 31

B - bc nonionizing radiation doesnt break bonds - Irradiating the proteins with X-rays or gamma rays certainly can denature them, but these are forms of ionizing radiation.

Answer 32

A is correct. Volume flow rate is proportional to the radius raised to the fourth power. An increase of 10% means the rate of flow will increase to 1.14 of the original rate. We need not do any calculations here, since only choice A says that the flow rate will increase.

Answer 33

Fnet = ma = Fapplied – Fk, where Fk is the force of kinetic friction. (50 kg)(1 m/s2) = 100 N - Fk 50 N = 100 N - Fk Fk = 50 N To calculate work, we use W = Fd cos θ. W = (50 N)(4 m)(cos 180º) = -200 Nm = -200 J

Answer 34

Amines and alcohols are not reactive on their own so to add them you need to activate it (heat - dehydrate)

Answer 35

"Calcium sulfate = CaSO4 | Calcium sulfite = CaSO3"

Answer 36

C - its the only method of separation, the rest are methods of analysis

Answer 37

"1. Amide 2. Acid 3. Alcohol 4. Ketone 5. Aldehyde 6. Amine 7. Ester 8. Ether 9. Alkane "

Answer 38

(+1, +1, -1)

Answer 39

"C - Elements will often adopt a configuration with a complete octet – the configuration found in the noble gases. This complete octet is preferred for the outermost valence electrons."

Answer 40

``` "C - 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: √(3002 + 7002) = √(90000 + 490000) = √(580000) = √(58 x 104) = (√58) x 102 ``` The square root of 58 falls between 7 and 8, so the overall value of our answer falls between 700 and 800, meaning that choice C must be correct (the actual value is 761). "

Answer 41

"B is correct. Newton’s third law can be expressed as: FA on B = −FB on A Here, the force of the earth pulling on the book is equal to and opposite of the book pulling up on the earth. Thus, choice B is correct. A: Remember that Newton’s third law applies very narrowly to the interactions between two objects. While the table does exert a normal force upwards on the book, that normal force is not what Newton’s third law refers to. Notice that in the equation above, the reaction force of A on B is strictly limited to B on A. C: The question stem defines mg as the gravitational force on the book. As such, since the book is resting on top of the table, mg would not be the gravitational force exerted by the table on the earth unless the table were entirely massless (which is not stated). D: This statement, like choice A, is true. However, the question stem is asking for the Newton's third law corollary to ""The book experiences a gravitational force of mg due to the earth's gravity."" The two objects we should be focusing on are thus the book and the earth — not the table."

Answer 42

" C is correct. The MCAT will expect you to be familiar with N2 as a very inert gas. It makes up approximately 80% of the air you breathe, yet has no significant chemical reactions with your lungs – or with anything other than nitrogen-fixing plants. This information implies that nitrogen is very inert (unreactive). As such, it would serve as a good artificial atmosphere when working with reagents that might react with oxygen or other gases. A, B: Although N2 does have relatively high electron density, the strength and stability of the triple bond means it doesn’t react strongly as many other compounds. D: Atmospheric CO2 levels are primarily determined by the seasons (when the northern hemisphere enters spring and summer, many plants grow leaves and pull CO2 out of the atmosphere) and by humans burning fossil fuels, which puts CO2 into the atmosphere. Nitrogen and carbon dioxide do not react readily with each other."

Answer 43

Strong reudcing agent that conversta aldehydes to primary alcohols

Answer 44

A - bc the bond is a S-S bond but S is serine in the amino acids

Answer 45

From species with less positive reduction potentials to those with more positive reduction potentials (thus, O2 must have the hgihest standard reduction potential)

Answer 46

"- Total internal reflection can only result when a ray of light begins in a higher-index material and reaches a boundary with a lower-index one (e.g. starting in water and moving towards air). - ie: start in water n=1.3 move to air n =1 "

Answer 47

"A is correct. One quick way to determine the orbital hybridization around the central atom is to simply count up the number of bonds and lone pairs. For example, ammonia has three bonds and one lone pair around its central nitrogen atom, for a total of four regions of electron density. This orbital hybrid therefore needs four orbitals to hybridize: s, p, p, and p. Thus, ammonia is an sp3 hybrid. We can narrow it down to choices A and B. Next, XeF4 has two lone pairs and four bonds around the central xenon atom. We know this because the central Xe atom has eight valence electrons. It gets four of these eight electrons from the two lone pairs and four from the bonds — one electron from each bond. Alternatively, we could try drawing the Lewis structure of XeF4, knowing that 8 electrons must come from the Xe atom and 7 from each of the F atoms, for a total of 8 + 7(4) = 36 electrons. Drawing XeF4 with Xe as the central atom and single bonds between the Xe atom and each F atom produces a structure with only 32 electrons, so we must add two lone pairs to the central Xe atom. Two lone pairs and four bonds gives us six total regions of electron density. So we need six hybrid orbitals to hold those electrons: s, p, p, p, d, and d. That’s an sp3d2 hybrid."

Answer 48

"Directly related to the energy required to remove a hydrogen atom from the compound - More energy = slower reaction because harder to remove "

Answer 49

"- Relative stabilization energy | - The more negative the number the more stabilizition "

Answer 50

A is correct. The first part of each answer describes whether the fluorine group is stabilizing or destabilizing. From Table 1, we see that the fluorine group has an RSE of +2.9 and is therefore destabilizing (eliminate B and D). Next, we must explain this destabilizing influence. Carbocations are typically unstable due to their high concentration of positive charge. If a substituent can contribute some electron density to the carbocation, this positive charge will be "balanced" and the species stabilized. In contrast, electron-withdrawing substituents pull even more electron density away from the carbocation, decreasing its stability further. Since fluorine is destabilizing, and since it is highly electronegative, we can deduce that it must be electron-withdrawing.

Answer 51

Ff=uFn which is equal to the graviational force pulling something down

Answer 52

"D -- First we need to find the time it will take the dart to travel the 5 m to the target, then we can find out the height of the dart at that time. - To do this we need the horizontal velocity. vx of the dart = v0 cos 45 = 10√2/2 m/s Time to travel 5 m = (5 m)/(10√2/2) = 1/√2 s vy of the dart = v0 sin 45 = 10√2/2 m/s Now we can use one of our kinematics equations to find out the height of the dart at that exact time. Δy = vyt + ½ at2 Δy = 10√2/2 (1/√2) + ½ (-10)(1/√2)2 = 5 + (-5)(1/2) = 5 – 5/2 = +2.5 m Thus, the dart will impact the target 2.5 m above the bullseye."

Answer 53

"1. Scalar - no direction only magnitude - length, area, volume - Speed, distance - Mass, density - Pressure, temperature - Energy, work, power 2. Vector - direction plus magnitude - Displacement, velocity, acceleration - Momentum - Force - Lift, drag, thrust - Weight "

Answer 54

moves to a different location, due to rotation

Answer 55

"D is correct. - The only force acting parallel to the plane is the relevant component of the gravitational force (the x-component). - This force is equal to Fgravsin(θ), or mgsinθ. - Since this is the only force that will drag the gurney along the plane, it is also our net force. - From Newton’s second law, remember that Fnet = ma. Substituting in our equation yields mgsinθ = ma, or gsinθ = a."

Answer 56

"D is correct. - We have two downward torques: that of the 10-N applied force and that of the weight of the board (20 N), which is exerted at the board’s center of mass, 15 cm from the point of rotation. τtotal downward = (10 N)(0.3 m)(sin 90°) + (20 N)(0.15 m) (sin 90°) τtotal downward = 3 N∙m + 3 N∙m = 6 N∙m - The upward torque, then, must also be equal to 6 N∙m. - We are told that the upward force is exerted 10 cm from the point of rotation, meaning that we can calculate it as follows: τ upward = F ∙ d ∙ sin (θ) 6 N∙m = (F)(0.1 m)(sin 90°) 6 N∙m = 0.1F F = 60 N"

Answer 57

"- A molecule serves as a substrate for its target enzyme, as well as a regulatory molecule of the enzyme's activity. O2 is a homotropic allosteric modulator of hemoglobin. - AKA what gives the sigmoidal shape of oxygen binding curve "

Answer 58

"B is correct. Hydrostatic pressure for liquids is linear because as depth changes, the density of the liquid remains constant. Gases, however, have densities that change according to the forces applied to them. Gases are compressible, while liquids and solids are not. A: For gases, higher temperatures result in behavior that is more ideal."

Answer 59

"D is correct. With decreasing temperature, air is able to hold less H2O. Since colder air can hold less total H2O, this means the same relative humidity would result in less total water in the air. For example, if warm air could hold 100 g of water in a given volume, then 50% relative humidity would be 50 g of water in the air. And if cold air could only hold 40 g of water in the same given volume, then 50% relative humidity would be 20 g of water in the air. So with the same relative humidity, the cold high-altitude air has less mass of water."

Answer 60

"C is correct. The decay of a substance can be calculated by knowing its half-life, the time it takes for half of the original amount to decay. The final paragraph states that about 93% of the decay occurs in the first 100 minutes. This leaves about 7% of the sample left, which would indicate that close to 4 half-lives (0.54 = 0.0625 = 6.25% remaining) have passed at the 100-minute mark. Thus, 1 half-life = 100/4 = 25 minutes. The hospital sample is 2000 g and must be replaced once it reaches 250/2000, or 1/8 of its original radioactivity. Using the shortcut that the amount remaining = (1/2)n, where n represents the number of half-lives that have passed, we can calculate 1/8 = (1/2)3. Therefore, 3 half-lives will pass before the sample drops below 250 g. At 25 minutes apiece, 25 min (3) = 75 minutes."

Answer 61

"Smallest pH because spectrophotometric absorbance is used to measure the conc of metal ion remaining in aquesous solution following extraction - This value is maximized when the extend fo the extraction is minimized "

Answer 62

"B is correct. If the frequency of the merry-go-round is 0.1 Hz, then the period is 10 s. We can calculate the speed of the merry-go-round from V = 2πr/T = 2(3.14)(5.0 m)/(10 s) = 3.14 m/s. Using this in the centripetal force equation: Fc = mv2/r = (50 kg)(3.14 m/s)2/(5.0 m) Approximate the square of the speed at 10 m2/s2. Fc = (50)(10)/(5) = 100 N"

Answer 63

"- Suicide inhibition occurs when an enzyme binds the inhibitor (structurally a substrate analogue) and forms an irreversible complex with it, usually through a covalent bond. - This can involve the inhibitor being chemically modified by the enzyme during the normal course of catalysis to produce a reactive group that is specifically responsible for the formation of the irreversible inhibitor-enzyme complex."

Answer 64

"1. Amines are derivatives of NH3 2. Amides are acid derivatives (or more rarely, anionic amines) - (i) the acid derivative, R C ( = O ) N R 2 , or (ii) the conjugate BASES of ammonia, for instance potassium amide, K N H 2 , which is unknown in aqueous solution, but known in liquid ammonia 3. Imines are nitrogen derivatives of carbonyl compounds. - R C ( = N R ' ) R , formed by condensation of a carbonyl and an amine:"

Answer 65

The half equivalence point on a titration curve

Answer 66

"MA = hypotenuse / height | a 30º-60º-90º right triangle, with side ratio of x : x√3 : 2x. Thus:"

Answer 67

C is correct. In order for the gurney to slide down the ramp, the force pulling it downward (mgsinƟ) must be greater than the static frictional force (μsFN = μsmgcosƟ). The net force on each gurney is thus Fnet = (mgsinƟ) - (μsmgcosƟ). Since net force is equal to the product of mass and acceleration, we can rewrite this equation as ma = (mgsinƟ) - (μsmgcosƟ), where an acceleration greater than 0 means the gurney will slide down the ramp. The mass of the gurney is present in all terms and can be canceled, meaning that it is not a factor in whether the gurney slides. Thus, both gurneys have an equal likelihood of slipping down the ramp, regardless of the fact that they have different total masses.

Answer 68

A is correct. Hydrochloric acid is a strong acid and completely dissociates in aqueous solution. In this solution, the hydronium ion concentration is 10.6 M, which can be approximated as 10 M to make the math easier. The pH is the -log of the hydronium ion concentration: -log[10] = -log[101] = -1. While the typical pH range is normally thought of as ranging from 0 to 14, if the concentration of hydronium ion is greater than 1 M, negative pH values are possible. It is also possible to have pH values greater than 14, i.e. if the hydroxide concentration is greater than 1 M.

Answer 69

"- Silica is highly polar | - Something that is nonpolar will move faster (larger Rf) "

Answer 70

"1. Ester - RCOOR’. Esters are made by the reaction between a carboxylic acid with an alcohol. 2. Ester - two alkyl groups, aryl groups, or an alkyl and an aryl group are connected to both side of an oxygen atom(ROR’)"

Answer 71

"1. C terminal to N terminal - Solid phase peptide synthesis | 2. N terminal to C terminal - Ribosome protein synthesis (more common) "

Answer 72

"1. Inert to acids and bases 2. aromatic hydrocarbon 3. very stable "

Answer 73

"D is correct. Watch your units! We’re given one number in m3/s and another in m2. To get an answer in the units of m/s, just divide the two numbers in the question: Speed = (4 x 10-4 m3/s) / (5 x 10-6 m2) = (4/5) x (10-4/10-6) = 0.8 x 102 m/s = 8 x 101 m/s This can also be solved using the equation flow rate = cross-sectional area x velocity."

Answer 74

"C is correct. Pyruvate dehydrogenase is the enzyme that mediates pyruvate decarboxylation. This step is the critical link between anaerobic and aerobic respiration. With this link broken, cells would be limited to anaerobic respiration. Thus, a decrease in aerobic respiration makes sense in this context. A: Lactic acid is a byproduct of anaerobic respiration. With pyruvate decarboxylation blocked, the cell would likely see an increase in such products. B, D: Acetyl-CoA and NADH are products of pyruvate decarboxylation. Blocking this enzyme would lower these concentrations."

Answer 75

"D - Group 16, directly under oxygen, so the structure of hydrogen sulfide would be similar to that of water: bent. The Lewis dot structure of H2S also predicts a bent structure due to the presence of the two lone electron pairs on the sulfur atom. A: Neither hydrogen sulfide nor water is tetrahedral. A common example of a tetrahedral molecule is methane (CH4). B: Trigonal planar molecules have three substituents bound to the central atom, which is not true of hydrogen sulfide. Boron trifluoride (BF3) is a good example of a trigonal planar species. C: Hydrogen sulfide is not pyramidal. One of the most common examples of a trigonal pyramidal structure on the MCAT is ammonia (NH3)."

Answer 76

D is correct. This question is asking us for the definition of a competitive inhibitor, which binds to the active site and blocks the substrate from attaching. The mechanism of a generic competitive inhibitor is depicted below.

Answer 77

"- Recrystalization works by having a solvent that makes the product of interest highly soluble in warm temperatures and insoluble in cold. - The MP tells you the purity (ie: if its close to what you want its likely more pure) - If a compound is always insoluble then it will never dissolve and cannot recrystalize - if comppound is always soluble then we cannot reform the desired solid even when we have cooled the solvent "

Answer 78

"1. Vacuum distillation - used when you have compounds with high BP to help lower the ambient pressure to make them more prone to vaporization - AKA temps that are hard to get in lab 2. Fractional distillation - use when you have compounds with a BP difference less than 50 - CANNOT be used to separate solvent from solute - DO have to repeat to purify 3. Simple distillation - used when compounds with BP difference GREATER than 50 - CAN be used to separate solvent from solute - DONT have to repeat to purify "

Answer 79

- Absorbance is measured as a function of the incident energy that is absorbed directly into the chemical bonds

Answer 80

"I. SDS confers a uniform charge density to all proteins in the sample. - Long hydrophobic tail II. SDS is amphipathic, allowing it to denature most globular proteins."

Answer 81

"- Highly charged environments because they are hydrophobic globular cytosolic proteins - Would case transmembrane proteins to denature "

Answer 82

"The presence of highly conjugated systems | - ie: double bonds or triple bonds "

Answer 83

"- Negative charge | - Migrate towards the anode WHICH is (+) in electrophoresis! (cathode is -) "

Answer 84

The SI units for length, mass, and time are often described as “MKS units,” or meters, kilograms, and seconds

Answer 85

- Shows that the velocity is increasing

Answer 86

D is correct. Time in flight does not depend on mass; in other words, if two objects of different masses are simultaneously dropped from the same height, they will hit the ground at the same time. Additionally, though the refrigerator is thrown horizontally while the other two objects are dropped, all projectiles begin with a vertical velocity of zero.

Answer 87

"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 gravitational 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."

Answer 88

vf2 = vi2 + 2aΔx

Answer 89

B is correct. To answer this question, we must remember the equation Fgrav = GMm/r2. Since Earth has a mass that is 100 times that of the moon and a radius that is four times the moon’s radius, it will have an Fgrav that is (100)/(42) = (100/16) times greater. This fraction is closest to 6.

Answer 90

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.

Answer 91

"1st - objects at rest will stay at rest and objects in motion will stay in motion as long as no net force is acting on it 2nd - F=ma 3rd - for every reaction there is an equal and opposite reaction "

Answer 92

The horizontal components of the tension force

Answer 93

u = Fgrav/Fnormal

Answer 94

"- Always pointed towards the center of the circle | - Velocity vector points at tangent from the circle so angle is 90 which means 0 work is done "

Answer 95

Mg + Mv2/R (Combo of centripetal force and gravity)

Answer 96

"- Total energy and mechanical energy are both conserved. - If friction and air resistance do not affect this process, we can assume that all forces involved are conservative. (The fact that gravitational PE is converted to KE is irrelevant, and happens whenever an object falls or slides downwards.) In cases like this, (KE + PE)initial = (KE + PE)final, so mechanical energy is conserved. - TOTAL energy is ALWAYS conservered"

Answer 97

(kg∙m2)/s3.

Answer 98

J/s and you multiply by the seconds

Answer 99

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%.

Answer 100

"Work = PdeltaV | - Conversion = 101.325"

Answer 101

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.

Answer 102

- Hydroxyl groups can hydrogen bond, thus if there are more present then the hydrogen bonding INCREASES the BP of a liquid

Answer 103

- One of the fatty acid salts was unsaturated, and a small percentage isomerized under the reaction conditions. - If one of the R groups in the triacylglycerol contained a carbon–carbon double bond and if isomerization of the double bond occurred during the saponification reaction, four fatty acids would be obtained instead of three.

Answer 104

- Its triple bond! | - Not polar (bc N's are identical)

Answer 105

1. Molecular - actual composition of the molecule (C6H12O6) | 2. Empirical - Relative # of atoms of each element it contains (CH2O)

Answer 106

1. STP = 1atm, 22.4L, 0C | 2. Standard conditions = 1 atm, 25C, solutions are 1M

Answer 107

- Depends only on the # of moles of gas present, not the identity of the gas

Answer 108

Octahedral | - An ion consisting of a central metal ion and six ligands is expected to have an octahedral geometry.

Answer 109

ACIDS 1. Arrhenius - H+ in H2O 2. Bronsted - H+ donor 3. Lewis - e- acceptor BASE 1. Arrhenius - OH in H2O 2. Bronsted - H+ acceptor 3. Lewis - e- donor

Answer 110

1. Galvanic cell converts chemical energy into electrical energy 2. Electrolytic cell converts electrical energy into chemical energy - BOTH - flow from anode to cathode

Answer 111

A. Acetylation - Adds more C-C B. Hydrogenation - Adds Hydrogen to a C-C multiple bond C. Elimination- create additional C-C bonds D. Substitution - Add something in (doesnt change saturation)

Answer 112

High levels of estrogen and progesterone inhibit the production of milk - Levels drop after birth to help stimulate milk production

Answer 113

A. Pepsin - Hydrolysis (breaking) of peptide bonds to make smaller peptides and AA B. Amylase- Hydrolysis of starches to sugars (produced by salivary glands and panceas C. Hexokinase - Phosphorlyation of sugars D. Lactase - hydrolysis of lactose to glucose and galactose

Answer 114

Secondary, tertiary (and by default quaternary)

Answer 115

Cyrstalization (precipitate) out of solution

Answer 116

1. Cyclic 2. Planar 3. Conjugated 4. Obeys Huckel's rule (# of pi electrons satisfies 4n+2 rule --> n = 0 or any positive integer, n = number of pi electrons)

Answer 117

The magnetic field can only change the direction not the speed (how fast its moving) - the velocity can change but the speed cannot

Answer 118

- heterocyclic - 5 membered aromatic ring - 4 carbon atoms and one oxygen

Answer 119

- When they become charged - Adding base to mixture = deprotonate the strongest acid = it will separate out - Adding acid = strongest base

Answer 120

- nonpolar solvents are more effective when attempting monosubstitution - polar solvents are typically used for polysubstitution.

Answer 121

A. Lactose - Glucose + galactose B. Maltose- 2 glucose C. Sucrose - fructose + glucose D. Lactulose - Galactose + fructose

Answer 122

fructose is 5 membered ring with 2 CH2OH and glucose is 6

Answer 123

``` A. Lactose - beta anomer (beta acetal linkage to galactose) All others have alpha linkages B. Maltose C. Sucrose D. Lactulose ```

Answer 124

- test to indicate reducing sugars - Reduces all monosaccharides (such as fructose or lactose) and some disaccharides - Tests for presence of hemiacetals - (AKA + benedicts test = brownish red precipitate)

Answer 125

- Extremely resistant to Nu attack under basic or neutral conditions - This is bc the alkoxide ions constitute relatively poor leaving groups

Answer 126

Cholesterol - All steroid hormones are derivatives of cholesterol - In general, steroid hormones can be distinguished from peptides by their suffixes, which tend to be either “-ol,” “-one,” or “-en.”

Answer 127

1. Start numbering carbons at carbonyl 2. Prefix is from total # of carbons with -oic acid always used as suffix 3. For unsaturated fatty acids you must say cis or trans 4. location of double bonds identified using lower # 5. TRIEN is inserted between prefix and suffix to show the alkenes present

Answer 128

B is correct. Phospholipids are amphipathic, meaning that they contain both hydrophobic and hydrophilic regions. For this reason, a phospholipid will interact more favorably with water than will entirely hydrophobic molecules such as cholesterol or triacylglycerols. A positive (or less negative) change in entropy is a hallmark of a more favorable dissolution. In other words, since the solvation of a phospholipid is more favorable than the solvation of a cholesterol molecule, it will have a more positive entropy.

Answer 129

enantiomeric excess = (observed optical rotation * 100) / (specific rotation)

Answer 130

- Wont rotate at all because 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. - If it had been SN2 it would have been opposite

Answer 131

A. conformational isomers - based on rotation around a single bond B. structural isomers - share molecular formula but have different bonding configurations C. geometric isomers - Presence of double bond or ring D. enantiomers - stereoisomers with one center changed

Answer 132

1. Epimer - molecules differ at a single stereogenic center, (which is not the anomeric carbon --> typically the alpha carbon) 2. Anomer - specific type of epimer that differs only at the anomeric carbon (ie: carbon attached to two O's) - Derived from the carbonyl carbon in the sugars straight chain aldehyde or ketone form

Answer 133

1. Diastereomers - use fractional distillation bc of distinct physical and chemical properties (BP, MP, solubility) 2. Enantiomers - use reaction with chiral reagent to make into diastereomer and then react

Answer 134

1. Increase molecular weight 2. Increase intermolecular forces (halogenate = makes more polar) 3. Decreased branching - increased branching decreases SA available for intermolecular attraction = lower BP

Answer 135

- more alkyl substitutents bond to the position (ie: tertiary > secondary > primary >methyl) - Thus TERTIARY compounds are favored for the reaction (carbocation stability effects more) - Unimolecular substitution - Polar protic solvent - Racemic mixture - Want weak Nu: (weak base)

Answer 136

E1 - Tertiary reagent - Mild or weak base - Polar protic solvents - Unimolecular E2 - Biomolecular - Needs strong base - LG must be anti to hydrogen

Answer 137

1. SN1 - Alkyl halide (electrophile) -Tertiary > secondary > primary >methyl - Want weak Nu: (weak base) in acidic conditons (sulfuric acid) - Unimolecular substitution - Polar protic solvent (alcohols) - Racemic mixture - Biggest barrier --> carbocation stability 2. SN2 - Alkyl halide (electrophile) - primary >> secondary > tertiary - Want strong Nu: (with negative charge) - bimolecular substitution - Polar aprotic solvent (DMSO, acetone) - Inversion - Biggest barrier --> steric hinderance

Answer 138

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.

Answer 139

- PCC oxidizes a primary alcohol to an aldehyde - Aldehyde is given suffix of -al - Secondary alcohols react with PCC to make ketones (end in -one) - NOTE: can then use K2Cr2O7 to oxidize aldehyde into carboxylic acid

Answer 140

- AKA make the proton less acidic - Looking for electron donating group - add electron density to make it more negatively charged and less likely to lose its proton

Answer 141

- Strong oxidizing agent 1. Primary - carboxylic acid 2. Secondary - ketones 3. Tertiary - cant?

Answer 142

an unstable intermediate with vacant orbitals.

Answer 143

- Sodium hydride is very strong but not a Nu: base, but it can deprotonate a primary alcohol - Then the deprotonated species can act as a Nu and attack the x-I to add it

Answer 144

Chicken foot with OH on leg | - Notoriously a weak acid = conj would be a strong base

Answer 145

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.

Answer 146

An alpha hydrogen is one that is bound to the carbon immediately adjacent to a carbonyl carbon

Answer 147

- Basically LAH | - Strong reducing agent that can reduce both carboxylic acids and aldehydes into primary alcohols

Answer 148

- Strong base! - Will attack most acidic site (typically tertiary) - Makes carboanion

Answer 149

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.

Answer 150

- "-oate" - Very bad LG bc it creates unstable RO- -

Answer 151

- Based on stability --> least stable = most reactive | - Resonance increases stability

Answer 152

- Conversion of one ester to another - Need alcohol bc -OH acts as Nu: and attacks ester carbonyl carbon - Replaces previous R group with its own

Answer 153

- Must be synthesized from carboxylic acid via Nu: acyl substitution - Contains a carbonyl group immediately adjacent to a heteroatom (generally EN)

Answer 154

``` Best LG (also most reactive) - I > Br > Cl > F - Note: this is opposite of stability (F is most stable) ```

Answer 155

- H2N-C=O-NH2 | - it is an amide

Answer 156

1. Cyclic amide = "lactam" | 2. Cyclic ester = "lactone"

Answer 157

- Cyanohydrin - The carbon (not nitrogen) atom attacks as a Nu: to attack carbonyl carbon - final cyanohydrin product consists of the former carbonyl carbon bound to -OH, -H, the original -R group from the aldehyde, and -C≡N.

Answer 158

- Get an amide product

Answer 159

1. Exergonic - a reaction is spontaneous ΔG < 0 (shown when -change in energy) 2. Exothermic - energy is released to surroundings

Answer 160

- Magnitude of the enthalpy of vaporization (ΔHv) is related to the strength of intermolecular forces holding molecules together in the liquid phase - AKA stronger IMF = larger ΔHv

Answer 161

- All EXCEPT HF | - HF is a weak acid that will only partially dissociate

Answer 162

- IF the graph is flat | - Where they cross doesnt mean anyhting?

Answer 163

- Shows that it is an endothermic reaction

Answer 164

1. Boyles PV=PV P inversely related to V 2. Charles V/T=V/T V directly proportional to T 3. Gay- lussac's P/T P directly proportional to T 4. Combined PV/T = PV/T 5. Ideal gas law - PV=nRT

Answer 165

B is correct. Dalton’s law states that the total pressure in a vessel is the sum of the partial pressures of the components. Moreover, the partial pressure of a gas is proportional to its mole fraction in the container. We can quickly find the total number of moles in the flask, which is 2.25 + 1.25 + 0.5 + 2 = 6 moles. Since exactly one-third of this is nitrogen, the total pressure is 3*300 mmHg, or 900 mmHg.

Answer 166

- Find energy diff needed on plateau and divide it by the grams of substance - units are kJ/kg typically

Answer 167

- When its vapor pressure equals the atmospheric pressure (1 atm = 760 mmHg) - Adding nonvolatile solute should increase BP

Answer 168

ΔTf = Kfmi, where Kf is a constant, m is molality, and i is the Van’t Hoff factor, or number of dissolved particles per molecule - NOTE: addition of solute decreases freezing point (opposite for BP - would increase it)

Answer 169

boiling point will increase, melting point (freezing point depression) will decrease, and vapor pressure will decrease.

Answer 170

C is correct. To find the fraction of the raft that is underwater, we can use its specific gravity, or the ratio of its density to that of water. Since pure water has a density of 1000 kg/m3, this raft’s specific gravity is 0.491. From this, we already know that 49.1% of the object will fall beneath the surface, but this question asks for the amount that sticks out above the water. 100% - 49.1% = 50.9%.

Answer 171

1 g/cm3, 1 g/mL, or 1 kg/L | - 1000g/L also works or 1000kg/m3

Answer 172

Gauge pressure (P) = densitygy - Absolute pressure = gauge + atmospheric pressure - Convert 101,500pa = 1atm

Answer 173

1. Adhesive = attractive forces between different particles 2. Cohesive = attractive forces between like particles - If meniscus is concave --> then fluid particles are more attracted to the glass sides of the tube than each other (adhesive stronger than cohesive)

Answer 174

1. Bernoulli's - Conservation of energy for fluids - Viscosity effects it more (friction) 2. Continuity - A1v1=A2v2 - Compressibility effects it more (volume flow rate)

Answer 175

v=squareroot(2gh)

Answer 176

Pressure in A must be smaller than Pressure in B

Answer 177

cm3 = 10^-6 m^3

Answer 178

Sum (multiple) all reactions together that the whole reaction contains

Answer 179

- change in H = gas- liquid if you get it wrong, know vaporization is an endothermic process so it must be +

Answer 180

- Power will increase as R decreases | - P=V^2/R

Answer 181

Q= piPr^4/8nl - Q = flow rate - P = Pressure - l = length - r = radius

Answer 182

Vrms = squareroot (3RT/M) | - M = molar mass

Answer 183

The spring constant of the rope can be determined by using Hooke’s law (F = -kx). - PEspring = ½ kx2

Answer 184

X (0.33) = remaining | - Then mutliple remaining X by 0.21

Answer 185

The hydrogen ion protonates the carbonyl oxygen, making the carbonyl carbon more susceptible to nucleophilic attack by the alcohol.

Answer 186

at the half equivalence point!

Answer 187

pH = pKa + log([A-]/[HA])

Answer 188

- Mole ratio of products tells you the relative stability of these products

Answer 189

Capacitor charge x Voltage of battery = Charge (C) | (1/2)capacitor charge x (Voltage of bat)^2) = J

Answer 190

The relative thermodynamic stability of isomeric organic compound

Answer 191

Kw = KaxKb | - given Kw is 1.0 x 10^-14 at 25C

Answer 192

the activated complex minus the energy of the reactants.

Answer 193

The shortest distance between repetition (ie: peaks) | - can also just say, how long does it take for the pattern to repeat

Answer 194

E = hc/λ = 6.6 × 10-34 J × (3 × 108 m/s)/(wavelength)

Answer 195

enzyme that catalyzes the transfer of electrons from one molecule, the reductant - Usually needs NADP or NAD+ or FAD

Answer 196

2 mol e– per mole of glucose consumed. The device measured 0.67 (2/3) μmol of electrons, indicating that 0.33 (1/3) μmol of glucose was consumed. This weighs 60 μg, based on its molar mass of 180 g/mol.

Answer 197

Cofactor oxidant! (need it for oxidizing agents)

Answer 198

- 4 fused rings!

Answer 199

Need to change the proportion of acidic buffer and basic! - To decrease pH, increase the acidic buffer component - To increase pH, increase the basic buffer component

Answer 200

- Water is released! | - 18 amu (Sum of atomic masses)

Answer 201

D because once the reaction reaches equilibrium, measurement of Vo will be impossible and the kinetic data will look the same regardless of substrate concentration

Answer 202

The addition of RC=O

Answer 203

- It means the group on the ring not th ring itself

Answer 204

- if + ions(cation) then you need a lewis base (- charged) | - if - ions (anion) then you need a lewis acid (+ charge)

Answer 205

- Basic pyrrole ring (5 sided heterocycle containing one nitrogen)

Answer 206

Frequency!

Answer 207

wavelength = speed of sound/frequency

Answer 208

``` Group 1 - Alkali metals Group 2 - Alkaline earth metals Group 3/d block - Transition metals Group 18- Noble gases Metaolids - B,Si, Ge, As, Sb, Te Other non metals ```

Answer 209

Molarity x 1/density

Answer 210

- Solubility constant - If one Ksp is smaller than the other, then the substance (with smaller Ksp) will not precipitate - Only rseponds to temp changes! (decrease temp = decrease Ksp)

Answer 211

Ksp = (x)^y(x)^z depending on the compounds

Answer 212

Greater! - If Acid A has a larger Ka than Acid B, Acid A must be more strongly acidic. A comparatively stronger acid will always have a comparatively weaker conjugate base, or one with a larger pKb.

Answer 213

The higher the pKa value the *weaker* the acid and the *smaller* the Ka and the *larger* the Kb

Answer 214

Where ever the pH is lowest you have the fewest Hydrogen ions = most protonated

Answer 215

HClO4 < H3PO4 < H2PO4- < PO43- - Don’t miss the inclusion of perchloric acid in the midst of three phosphorus-containing examples! - HClO4 is a strong acid and thus has the lowest pH in solution of all four choices (assuming, of course, that starting concentrations are equal). - Next, H3PO4 is a triprotic acid with a fairly stable conjugate base, although phosphoric acid certainly is not strong. - Subsequent deprotonation yields stronger and stronger conjugate bases, making PO43- the most basic of the listed species. - This tendency is very normal; intact polyprotic acids are always more acidic than their deprotonated conjugates.

Answer 216

B is correct. Think of auto-ionization in terms of equilibrium: heat + 2 H2O → H3O+ + OH-. The reaction will thus shift toward the product side in response to increased temperature. Interestingly, both [H+] and [OH-] will rise as a result, decreasing the pH and pOH to the same extent.

Answer 217

- Equimolar amounts of a weak acid and its conjugate

Answer 218

Frequency! - frequency always remains constant. Remember, the energy of a photon is given by E = hf. Since no energy is lost as a direct result of the transition between media, the frequency of this wave must remain at its previous value.

Answer 219

- One object is moving relative to another, if something is being held (right angel) still then you block the ability to detect

Answer 220

(period)(velocity of wave) = distance represents a 360 phase gap

Answer 221

λ = 2L/n, where n= harmonic #

Answer 222

B is correct. - Using the equation dB = 10 log(I/I0) , we can plug in the variables that are given - This yields 50 dB = 10 log(1/10^-12W/m2) . - Dividing both sides by 10 quickly simplifies this equation to 5 = log(1/10^-12W/m2) . or 10^5 = . Multiplying both sides by I0 gives us I = 10^-7.

Answer 223

f’ = f (vsound ∓ vreceiver / vsound ∓ vsource). - objects are moving CLOSER together overall, the perceived frequency of sound should increase - Can also do relative velocity + total/ total

Answer 224

1. Longitudinal - oscillation is foward and backwards in amplitude while its overall direction of propagation is forward - the amplitude of oscillation is parallel to the direction of the wave’s motion - Looks liek a coil or spring or little black bars 2. Transverse wave - amplitude perpendicular to its overall direction - Looks like normal sin wave

Answer 225

- Typically unpolarized light intensity will be cut in half - if it goes through a second that is oriented the same way it does not change - if the second filter is turned 90degrees then it blocks 100% of the light and the new intensity is 0

Answer 226

``` LOWEST E, LOWEST f, LONGEST wavelength 1. Radio wave 2. Microwaves 3. Infrared waves 4. Visible light (ROYGBIV) 5. UV light 6. X rays 7. Gamma rays HIGHEST E, HIGHEST f, SHORTEST wavelength ```

Answer 227

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.

Answer 228

- Ray will bend towards the normal - Angle of refraction will be smaller - lower n = moves faster through that medium

Answer 229

1. Light traveling from HIGH n to LOW n | 2. Angle of incidence must be relatively large

Answer 230

Image on opposite side of object

Answer 231

Concave with a negative focal length | - This is nearsightedness

Answer 232

More negative oxidation potential = worst reducing agent = more likely to reduce

Answer 233

1. Nonspontaneous = (-) overall Ecell | 2. Spontaneous = (+) overall Ecell

Answer 234

1. Electrolyic - Cell converts electrical energy into chemical energy - reduction takes place at the right electrode = cathode (-) - Oxidation takes place at the left electrode = anode (+) 2. Galvanic - cell converts chemical energy into electrical energy. - reduction take place at the left electrode = cathode (+) - Oxidation takes place at the right electrode = anode (-)

Answer 235

1. Ammeter- measures currently directly | 2. Voltmeter - measures potential (indirectly measures current)

Answer 236

- Point away!

Answer 237

Blue! - blue light is refracted the most - n = c/v,

Answer 238

F = k (Qq/r^2) | k = 1/4piE

Answer 239

- Electron-volt | - Amount of energy needed to move one electron (or any particle with an elementary charge) through a one-volt potential.

Answer 240

1. Parallel - voltage across is same, current is different if resistors are diff 2. Series - current is same, voltage is different

Answer 241

1.6x10^-19C/ion

Answer 242

1. Velocity | 2. Charge

Answer 243

- Typically you just have fingers as force, but must switch final direction for negative charges (ie; if ts pointing into the page it would have to point out)

Answer 244

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.

Answer 245

- Work done TO the system will allow heat transfer (typically only hot to cold, but work allows you to go backwards) - Like rolling ball up a hill

Answer 246

According to the third law of thermodynamics, the entropy of a pure, crystalline substance at 0 K is zero. This is known as absolute zero.

Answer 247

Kcat/Km so highest for the smallest Km and largest Kcat

Answer 248

5 carbon - 6 member ring (switched orientation of glucose)

Answer 249

Its adsorption to the water-alveolar interface decreases surface tension, decreasing the pressure difference required to inflate the airway.

Answer 250

v"I. ΔH > ΔG | II. 0 < ΔS"

Answer 251

"I. generate free radicals. II. excite electrons to higher energy levels. III. eject electrons from molecular orbitals."

Answer 252

Nucleic acids

Answer 253

The closer an electron is to the nucleus, the harder it is to eject. Because sp-hybridized orbitals have the most s character of all of the answer choices, they contain the electrons that are hardest to eject.

Answer 254

"D is correct. Since the crate is initially at rest, the Vi = 0 m/s, the acceleration a = 2.0 m/s2, and the displacement = 6 m. The appropriate kinematic equation is vf2 = vi2 + 2ad. Substituting the variables gives: ``` vf2 = 02 + 2(2)(6) vf2 = 24 vf = ~5 m/s" ```

Answer 255

"A: An alpha particle consists of two protons and two neutrons, having a mass of 4 amu. B: A beta particle is the nuclear equivalent of an electron, which has a mass of approximately 1/1800 of a proton. C: A positron is the antiparticle of an electron and has its same mass D: A gamma particle is a photon of electromagnetic energy, which does not have mass."

Answer 256

"1. Reaction is sterically hindered (3) 2. Has two steps and wants good leaving group bc it leaves behind an unstable crbocation 3. carbocation is planar and lacks stereochemistry so you get a racemic mix "

Answer 257

Higher to lower

Answer 258

"- H3PO3 | - H3PO4"

Answer 259

- When attractive forces of an ion molecule causes a thin shell of water molecules to surround it

Answer 260

Then the larger Ksp indicates that it is more soluble

Answer 261

"B is correct. This is a multi-step question, though each step is relatively straightforward. The gravitational potential energy at the runner’s height is: PE = (60 kg)(10 m/s2)(0.5 m) = 300 J Most of this energy is conserved as the runner hits the ground and his muscles capture the energy as spring potential energy, so the question only asks about the lost energy, amounting to 10%, or 30 J. Don’t worry about the mechanics of which foot lands first and how much of the kinetic energy each one absorbs. The question stem doesn’t give that kind of information, so there’s nothing for it anyway. So, an additional 30 J is needed per stride, and a stride occurs every 0.8 s. Thus: P = (30 J)/(0.8 s) = 40 W This is between answer choices B and C. So is it really 40 W? Well, no, 8 goes into 30 less than 4 times, so the answer should be lower. Also, g is closer to 9.8 than 10, so that also caused the value of energy, used in the power calculation, to be overestimated. 37 W, then, must be the match."

Answer 262

PE or KE= (1/2)kx^2

Answer 263

The electron-donating groups around the basic nitrogen on arginine make its conjugate acid more stable.

Answer 264

D is correct. Cell 2 can only accelerate while it is being pushed/in contact with cell 1, so statement I is false. Since the cells experience drag from the medium, cell 2 will decelerate after the collision, making statement II correct and statement III incorrect. The drag will cause cell 2 to continuously decelerate until it comes to rest.

Answer 265

The radius of the curvature

Answer 266

Each light ray will refract at each surface of the lens and will increasingly diverge from the others because of the shape of each surface, so only diverging rays can result

Answer 267

is absorbed and re-emitted by the atomic structure of the optically dense medium.

Answer 268

"1. shifts its location 2. Distance = smaller (closer) or larger (further) dot (more fuzzy if further) 3. Pin hole size = clarity = smaller = sharper = changes focal length"

Answer 269

"- must determine how many wavelengths will fit within the width of a bead, because there is one wavelength, λ of spatial length per wave. - The item gives the speed of light to be v = 3.0 × 108 m/s and the frequency to be f = 1015 Hz. - The wavelength λ of the light may be found from the relationship for wave speed v = λf, thus, λ= v/f = (3.0 × 108 m/s)/( 1015 Hz) = 3.0 × 10–7 m. - The number of waves that will fit along a diameter is found by dividing this wavelength into the diameter D = 5.0 × 10–5 m, so the number of waves is D/λ =(5.0 × 10–5 m)/( 3.0 × 10-7 m) = 1.7 × 102. Thus, B is the best answer"

Answer 270

Interaction with phosphate groups

Answer 271

"1. ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions. Ionic bonds are formed between a cation, which is usually a metal, and an anion, which is usually a nonmetal. 2. A covalent bond involves a pair of electrons being shared between atoms."

Answer 272

"- Initially at baseline - Drops as they reach minimum at ideal bond length - Continue to move closer = repel each other and bring PE back above baseline "

Answer 273

- Carbon that is bound to four substituents with no double bonds

Answer 274

"Tetrahedral! | - sulfur has four bonding groups and no lone pairs "

Answer 275

"1. Trigonal planar - Tertiary carbon has three bonding groups and no lone pairs 2.Trigonal pyramidal has three substituents and a lone pair "

Answer 276

"1. Protium 2. Deuterium 3 Tritium "

Answer 277

- Atom contains a positively charged nucleus surrounded by electrons that orbit the nucleus in defined paths, similar to planetary bodies that circle the sun

Answer 278

"- First find n = principle energy shell - Find l = (s=1,p=2,d=3, f=4) - Find m (can vary) "

Answer 279

"1. Diamagnetic - Contains no unpaired electrons (fully filled orbitals) 2. Paramagnetic - Contain at least one unpaired electorn "

Answer 280

effective nuclear charge = # of protons (aka nuclear charge) - # of electrons in all shells preceeding the one in question

Answer 281

"1. ultraviolet --> n = 1 2. Visible --> n = 2 3. Infrared --> n = 3"

Answer 282

"6! - Selenium has an atomic number of 42, the atom is oxidized so it would have 6 electrons but instead has 4 - The diagram includes 32 electrons because (2,8,18,4) for each of the shells. Since they are asking for n=4 then 34-28= 6"

Answer 283

"- MUST BALANCE: Ba(NO3)2 + H2SO4 --> BaSO4 + 2HNO3 | - so for every mole of Ba(NO3)2 you get 3 total moles of product and so you have 15 moles of product "

Answer 284

"1. Find the equation for combustion 2. Find moles - 900/44 mg/mmol = 18 mmol 3. Need to get hydrogen from hydrocarbon 360/18 = 20 mmol - dont need limiting because were looking at reactants not products 4. take mmol # and multiply by the number of atoms there for each substance C:18:36, H=40 and O = 20 5. put ratio in least terms (CHO = 18, 40, 36) - always small error bc burning isnt perfect 20:40:60 ~100 but you dont have to do oxygen always --> see if it eliminates the other answers! "

Answer 285

"- take mole fraction and multiply it by 100 moles (assume total bc its easiest unless given) - Find the mass of each by multiplying the moles by the respective molar masses - Use masses to find total mass and divide mass of individual piece by total"

Answer 286

mass of nitrogen/ mass of glutamine (28/146)x100 ~19%

Answer 287

Tells us that the object is moving backwards and the flat slope means that it is not accerlerating

Answer 288

"83 Minutes! - Orbits radius = 160 so diameter = 320pi ~1000 - 200m/s can be coverted to 0.2km/s - 1000/0.2 = 5000 seconds, which is about 83 minutes"

Answer 289

``` "40m! F=ma to find acceleration (150N =(7.5)a) = 20 - Then plug in to v^2 = v^2 + 2ad 40 = 0 + 2(20)(d) d = 40m" ```

Answer 290

"20m! - Vi = 10m/s but that is horizontal, Vi vertical = 0 - So acceleration = 10 - Use x = vt + 1/2at^2 --> 20 = 0 + (1/2)(10)(t^2) - t = 2 - Then 10m/s*2 = 20 m"

Answer 291

30 meters because the horixontal component of velocity does not matter so jake and phils baseball will fall at teh same rate. When jakes hits it has fallen 30 so so much phils

Answer 292

total energy! (not KE, or PE, or momentum)

Answer 293

Fgrav = Gm1m2/r^2

Answer 294

Static! must be at least the the minimum of Fs = usFn

Answer 295

"Kinetic friction is dependent only on the coefficient of friction and the normal force and does not take belocity into account - The normal force is constant throughout the descent, foas the path of the coaster is linear - Force of friction remains constant throughout "

Answer 296

"kg*m/s | J = force x Time"

Answer 297

"Fc = mg and Fn = 0 - When the cart reaches the top fo the loop, it must have a centripetal force pushing radially outward with a magnitude equal to the force of gravity to keep it from falling. at this point eh cart will not be pushing against the track. therefore, the centripetal force will equal the gravitational force (mg) while the normal force will be 0 "

Answer 298

"E=torquexangle - 400pN-nm equal to the torque (40) multiplied by the angle (which we are looking for) - get 3radians but it takes 2radians to make it around the circle, - Thus 1.5 revolutions is our answer! "

Answer 299

"torque = Ia - Moment of intertial for a ring is MR^2 - 100/2.5 = M = 40 kg"

Answer 300

sin theta = v^2/rg found by setting mv^2/r = mgsin theta

Answer 301

"To find the pilots weight at the bottom of the loop add gravitation force and centripetal force - mg + mv^2/r = 6800"

Answer 302

"1. Gravity 2. Elastic force 3. Electric "

Answer 303

Sum of KE and PE

Answer 304

W = PdeltaV

Answer 305

"2.33L! - Energy used for expansion must come from the energy of combustion minus the amount lost to friction. SInce the burning of hydrocarbons produces 1 kJ but lost 300J, 700J of energy contributes tot he expansion mentioned in the question. - Using conversion factor we see that this value ~7 L*atm. This can equal PdeltaV and then divide by P (which is 3 atm not 1) and you get 2.33 as the change in volume "

Answer 306

"- Must consider energy from rotation of pulley - so total energy = mgh + rotational KE - Overall decreases the height or distance something can be pulled "

Answer 307

"C is correct. Whew, this is a fairly heavy question! It might be one to skip over, mark, and save for later. To find the magnitude of this acceleration, we need to know the velocity at impact. For this, we can use vf2 = vi2 + 2ad. Initial velocity is 0 m/s, distance is 2 m, and acceleration is g, or 9.81 m/s2. ``` vf2 = (0 m/s)2 + 2(10 m/s2)(2 m) vf2 = 40 m2/s2 ``` We know that 62 = 36 and 72 = 49, so √40 must fall between 6 and 7. Let's call it 6.4. Now, acceleration = (final velocity - initial velocity) / time. a = (0 m/s - 6.4 m/s) / 0.09 s We get a negative value in the numerator, but all of our answer choices are positive, and the question asks for the person's deceleration. Since the negative sign simply denotes that he is decelerating, we can ignore it going forward. We can also round 0.09 s to 0.1 s. ``` a = 6.4 m/s / 0.1 s a = 64 m/s2 ``` Finally, note that our answer must be written in terms of g. Estimating g as 10 m/s2, 64 m/s2 is equal to 6.4g. However, the value of g is actually lower, so our real answer should be slightly higher than 6.4g. Choice C, 7g, is closest."

Answer 308

"A is correct. Let’s begin with the information that we have, most notably the kinetic energy value. Remember, KE = ½ mv2, so we can use this value to calculate the child’s total velocity immediately before impact. 720 J = ½ (40 kg)(v2) 720 J = 20 kg (v2) 36 = v2 v = 6 m/s Note that this is the child’s total final velocity, so it includes both a horizontal and a vertical component. Since air resistance is negligible and gravity acts only in the vertical direction, the child’s final horizontal velocity is the same as his horizontal velocity at the moment of the push. This value can be found using the cosine of the given angle (if in doubt, draw out the situation)! ``` vhorizontal = vtotal cos(60°) vhorizontal = 0.5vtotal = 0.5 (6 m/s) = 3 m/s" ```

Answer 309

"D is correct. The polarity of a sample affects its retention time, but it has no effect on streaking. A, B: In TLC, streaking can be caused by overloading the spot with sample or by using a sample that is too concentrated. C: If a TLC sample is impure, its multiple components will travel different distances along the plate, which could lead to a streaked appearance."

Answer 310

"Rate of effusion depends on root-mean-square speed, which is calculated using the formula vrms = √(3RT)/M. Here, T is temperature and M is molar mass. Since the identity of the gas is constant, the only factor changing is T. Therefore, for two samples of the same gas with identical pressures but different temperatures, the following relationship holds: rate1/rate2 = √(T1/T2) rate2/rate1 = √(800/200) = 2 This means that the new rate is twice as large, or 100% larger than the original rate."

Answer 311

"1. Serine (S) 2. Threonine (T) 3. Tyrosine (Y) 4. Histidine (H)"

Answer 312

"C is correct. Here, the central molybdenum atom is bound to six substituents. In such a configuration, the molecule takes the shape of an octahedron. The number of substituents, or bonding regions, around the atom is equal to the number of orbitals that hybridize: here, there are two 4d, one 5s, and three 5p atomic orbitals, for an overall hybridization of d2sp3. A: sp3 hybridization would be expected in a tetrahedral molecule, like CH4. B: This is very close to the correct answer, with one distinction! In sp3d2 hybridization, all of the orbitals have the same principal quantum number. In d2sp3 hybridization, the principal quantum number of the d orbitals is one less than the principal quantum numbers of the s and p orbitals, as is the case here. We see d2sp3 hybridization in the transition metals and sp3d2 hybridization in the nonmetals. D: dsp3 hybridization would be expected if the central atom had only 5 attached substituents. This would yield a trigonal bipyramidal rather than an octahedral geometry."

Answer 313

"- Need diverging lens! - Eyes near point is smaller than that of a healthy eye - Focal point is closer in than the retina "

Answer 314

``` "Converging lens/ concave mirror 1. do > 2f - RIS 2. 2f > do > f - RIL 3. do < f - VUL Diverging lens/ convex mirror VUS for all " ```

Answer 315

"1. Real image = +i - Mirror = image on same side - Lens = Image on opposite 2. Virtual image = -i - Mirror = image opposite side - Lens = image same side "

Answer 316

"- Stereoisomers - same atomic connectivity, but diff rotation 1. Enantiomers - nonsuperimposable image - Each has opp config at ALL chiral centers - Optical rotation of equal mag BUT opp sign - Racemic mixture = 50:50 mix = achiral = no optical activity - Everything else identical - Separate using resolution! 2. Diastereomers - nonsuperimposable NON mirror images - Opp config for AT LEAST ONE but NOT all - unrelated to optical activity - Diff chem/phys props - Has cis/trans - Physically separate 3. Epimers - diasteromers that differ in config at ONLY one chiral center - all epimers are diasteromers but not all diastereomers are epimers"

Answer 317

C is correct. Gases behave most ideally under high temperatures and low pressures. The tricky part of this question is noticing that not all answer choices are given in the same units. Of the temperatures listed, the highest is 273°C (approximately 546 K). For the pressure values, remember that 1 atm is roughly equal to 101,500 Pa. 50 Pa (1 atm / 1.01 x 105 Pa) yields a value on the order of 10-4 atmospheres, so 50 Pa is the lowest pressure given as part of an answer choice as well.

Answer 318

"Want to find weakest bond - EN donor atom pulls electron density away from a hydrogen atom, causing it to take on a partial positive character. Another electronegative atom with a lone pair acts as a hydrogen bond acceptor. - The greater the electronegativity of the donor atom, the stronger the positive charge that accumulates on the hydrogen. - Thus, A is the most EN atom will have the strongest bond - Since the acceptor atom is oxygen in the remaining three choices (B, C, and D), we need only consider the effect of the electronegativity of the donor atom on the strength of the bond. Fluorine is the most electronegative atom on the list, followed by oxygen, then nitrogen. Therefore, choice D, which contains nitrogen, has the weakest bond. "

Answer 319

"- The gain of bonds to oxygen (or other EN substituents) or the loss of bonds to hydrogen (1) loss of an electron (2) increased oxidation state (3) loss of a C–H bond (e.g. alkane → alkene) (4) gain of a C–O or C–N bond (or any bond between carbon and a highly electronegative atom). "

Answer 320

"n = number of chiral centers 2^n - excluding meso compounds "

Answer 321

"(1) gain of an electron (2) decreased oxidation state (3) formation of a C–H bond (e.g. alkene → alkane) (4) loss of a C–O or C–N bond (or any bond between carbon and an electronegative atom)."

Answer 322

PO2 = mole fraction of O2 x Ptotal

Answer 323

"Case 1: Work equals zero when the displacement of an obect equals zero. (W=0 if d=0) Case 2: Work is zero if the applied force is zero, like a frictionless surface (W=0 if F=0) Case 3: Work equals zero if force is perpendicular to the motion. W= 0 if F x d x cos90º"

Answer 324

"A is correct. Because of the law of conservation of energy, the energy of the initial photon must equal the sum of the energies of the emitted photon and the electron. However, we are given only frequencies. Since frequency and energy are directly proportional, we can use E = hf to find the energies required to solve. First, though, we must find the frequency of the electron, which is equal to the difference between the two photon frequencies. 1.6 x 1019 Hz - 1.599 x 1019 Hz = 1016 Hz E = hf = (6.6 x 10-34 J∙s)(10 16 Hz) = 6.6 x 10-18 J This quantity is NOT equal to the kinetic energy of the electron. The work function represents this amount of energy. We must first convert lead’s work function to joules, then subtract the work function from our calculated energy to get the kinetic energy. To make this calculation easier, use 4 instead of 4.14: ``` WPb = (4 eV)(1.6 x 10-19 J) = 6.4 x 10-19 J KE = Etotal - WPb = 6.6 x 10-18 J - 6.4 x 10-19 J = 6 x 10-18 J" ```

Answer 325

"Use HH equation to find ration of conj base/ acid - According to the Henderson-Hasselbalch equation, then, log ([conjugate base]/[acid]) is about 4. This corresponds to a conjugate base-to-acid ratio of approximately 104, or 10,000, to 1."

Answer 326

"B is correct. For questions like these, begin with the information given. We know that the current is 5.0 A, or 5.0 coulombs/s. To find coulombs, we simply need to multiply by seconds. Four hours and ten minutes is equal to 250 minutes, or 15,000 seconds. (5 C/s)(15,000 s) = 75,000 C of charge transmitted (75,000 C) / (96,485 C/mol) ~ 0.8 moles of electrons transferred (as 75/96 is slightly larger than 75/100, or 0.75) According to the reduction half-reaction, only one molecule of H2 is created for every two electrons. Thus, the amount of hydrogen gas produced must be approximately 0.4 moles."

Answer 327

"An Ox RED cat | - Flow from anode (+) to cathode (-) since electrons are necessary for reduction "

Answer 328

Like other longitudinal waves, sound waves propagate due to vibratory changes in pressure or collision of adjacent particles within an elastic medium.

Answer 329

"KE = (3/2)KBT, where KB is the Boltzmann constant. | - Thus only temp effects the KE"

Answer 330

C is correct. Bile salts contain both polar elements and a hydrophobic cholesterol core. Because of this amphipathic structure, they contain both polar and nonpolar regions, acting as a detergent and solubilizing lipids in aqueous digestive contents by interacting with both hydrophobic (lipid) and aqueous phases

Answer 331

"I. the wavelength of excitatory radiation. | II. the electronic structure of the molecule being analyzed."

Answer 332

- Nonpolar hydrophobic

Answer 333

"To solve for height, we can use: Δy = v0yt + ½ at2, where v0y = 0 and a = -10 m/s2. Δy = ½ (-10 m/s2)(t2) Unfortunately, we do not know the time, but we can solve fairly easily using the formula Δx = vxt. (This simple formula works because the ball does not experience acceleration in the horizontal direction.) We must rearrange to get t = Δx/v x, or t = (18 m) / (30 m/s) = 3/5 s = 0.6 seconds. Plugging into the first equation yields Δy = ½ (-10)(0.6)2 = -1.8 m. Thus, the ball will end up exactly 2 - 1.8 = 0.2 m above the ground."

Answer 334

"According to PV = nRT, volume would approach zero. However, at a certain point, volume will not be able to decrease, because the particle volume cannot be compressed. As a result, the higher the pressure, the more Vreal will deviate from Videal, or Vpredicted.) If Vreal is significantly higher than expected, the PV term in PV/nRT will also be especially high, and PV/nRT will exceed 1."

Answer 335

"- A is correct. Fluorescence is the emission of a lower-energy photon from a fluorophore excited by the prior absorption of a higher-energy photon. With this in mind, the Aequorea fluorophore must be excited by the absorption of a photon of higher energy (and frequency) than that associated with green visible light. Of the choices listed, only ultraviolet light, choice A, is of higher energy than green light. - Even if we were not considering the green fluorescence of this organism, we could simply choose the highest-energy radiation as that which is most likely to promote fluorescence in general."

Answer 336

"n = c/v where c = speed of light and v = velocity of light through that medium - NOTE: frequency does not change "

Answer 337

Negatively charged!

Answer 338

Decreasing the concentration of the Mg2+ solution and increasing the concentration of the Cu2+ solution

Answer 339

"C is correct. Dipole-dipole forces are present only between polar molecules, and methane is nonpolar. A: Carbon atoms cannot cause hydrogen bonding. B: Whether dipole-dipole forces are present depends only on whether the entire molecule is polar or nonpolar, not on whether the molecule contains particular atoms. (Hydrogen bonding, a subcategory of dipole-dipole interactions, is the one possible exception. However, hydrogen bonding has nothing to do with Br atoms.) D: Van der Waals’ forces include London dispersion forces, which are always present."

Answer 340

"1. Induced dipole - occurs only when a polar molecule induces a nonpolar molecule to become temporarily polar. 2. Dipole-dipole forces - are attractive forces between the positive end of one polar molecule and the negative end of another polar molecule. 3. Hydrogen bonding - is a special type of dipole-dipole attraction between molecules, not a covalent bond to a hydrogen atom. It results from the attractive force between a hydrogen atom covalently bonded to a very electronegative atom such as a N, O, or F atom and another very electronegative atom."

Answer 341

``` "pH = logsquareroot (ka*HA) pKa = -log(ka) when concentrations are equal" ```

Answer 342

oxioreducatase

Answer 343

"1. One step = based on how many items are there 2. Bimolecular = reaction order of 2 3. Multistep - find the rate limiting step (aka slow step) "

Answer 344

``` "10= A6/[A3]^2 10 = 0.1/ X^2 = 0.5" ```

Answer 345

- Lowest molecular weight and weakest intermolecular forces

Answer 346

Its numerical value is given by (1 m)-1+(0.02 m)-1=1 D+50 D=51 D

Answer 347

Hydrolyze fatty acids

Answer 348

"1. Each drop down in pH = 10 fold increase 2. Each level up = 1/10 increase - so 3 different = 1000"

Answer 349

"1, 4 glycosidic bond = straight chain 1,6 glycosidic bond = branched - all alpha --> cant break down beta bonds in carbohydrates "

Answer 350

Must be +!

Answer 351

B. The absorption of ultraviolet light by organic molecules always results in electronic excitation. Bond breaking can subsequently result, as can ionization or bond vibration, but none of these processes are guaranteed to result from the absorption of ultraviolet light.

Answer 352

Aromatic side chain because it absorbs near the UV region = longer wavelengths = longer energy

Answer 353

Makes it more negative!

Answer 354

"hydroxyl! | - note: Tyrosine possesses a hydroxyl group in the aromatic ring, making it a phenol derivative"

Answer 355

"1. Pauli principle - carrying capacity in an orbital - No two electrons can be identical (spin up/down) 2. Aufbau principle - how electrons are added or removed from orbitals with diff energy - Electrons are added from lowest to highest energy - Valence e are in the highest shell and removed highest from lowest 4s removed before 3d 3. Hunds rule - How electrons are added or removed from orbitals with same energy - Electrons fill degenerate orbitals before pairing - Paramagnetic = 1 impaired e - Diamagnetic = all electrons are paired"

Answer 356

It is negative and the reaction is spontaneous

Answer 357

"1. Nonpolar covalent - same EN (equal sharing) 2. Polar covalent - Diff EN (unequal sharing) 3. Metallic covalent - Metal + metal 4. Coordinate covalent - Ligand + metal"

Answer 358

The coordination number of Cu2+, Because ammonia is neutral, the number 4 reflects only the number of ammonia molecules that bind to the central Cu2+ cation and does not indicate anything about its oxidation number.

Answer 359

"C. The solubility product constant expression for CaCO3 is Ksp = [Ca2+][CO32–]. Since equal quantities of Ca2+(aq) and CO32–(aq) are produced when CaCO3 dissolves, this expression reduces to 4.9 × 10–9 = x2, or 49 × 10–10 = x2. - Take the square root of both sides "

Answer 360

D because the number of charges in excess can be computed as +3.2 × 10 –9 C/1.6 × 10 –19 C = +2.0 × 10 10. This means that the rod has an excess of positive charge, created by removing a number of +2.0 × 10 10 electrons from the material, as it is not possible to add protons in a manner described in this question.

Answer 361

PE = 1/2kx^2

Answer 362

dB = 10log(I/I0)

Answer 363

always VUS

Answer 364

D because the chemical energy of the battery elements is used as electrical energy to set the charge carriers in motion through the resistor, where they experience drag from the crystal lattice of the resistive conductor and dissipate their energy as heat from the resistor

Answer 365

"m = -i/o | - virual images always upright "

Answer 366

Reducing agent donates electrons and is thus oxidized

Answer 367

The amount of a substance that will dissolve in water is described by the Ksp. The Ksp for a substance, AaBb, equals [A]a[B]b. If the amount of the compound present is in excess of the Ksp, then a precipitate would form to maintain the Ksp.

Answer 368

- Lower the value of Ksp = the lower the concentration of the cation and anions in aqueous solution

Answer 369

Equivalent mass (or equivalent weight) is the mass of an acid that yields one mole of hydrogen ions or the mass of a base that reacts with one mole of hydrogen ions.

Answer 370

When they exceed their Ksp

Answer 371

"Potential, resistance per unit length, and length | - Ohm’s Law, I = V/R. To determine R, the resistance per unit length, the length, and the potential V are needed"

Answer 372

"- Wont dissolve in water | - WIll dissolve in acid (if it dissolved in water too it would be an oxyacid)"

Answer 373

D because the formula for calcium chloride is CaCl2 and it produces twice as many Cl– ions as Ca2+ ions in solution when it dissolves.

Answer 374

B because the Doppler effect is used with ultrasound waves to provide fetal images. The Doppler effect relates the frequency of the ultrasound wave as detected by a moving detector to the frequency of the wave when the source is stationary, the speed of the source, and the speed of the detector.

Answer 375

Approximate radial size of an electorn cloud

Answer 376

if the enzyme (catalyst) is chiral then you can get a chiral product

Answer 377

The answer to this question is A because only sublimation of the water under reduced pressure will keep the mixture cold and will maintain the protein in its native state.