Chemistry Flashcards
(104 cards)
1
Q
peptide bond
A
peptide bond = amide bond
peptide bond is formed by the amine group attacking the carbonyl carbon
water is lost during peptide bond formation (hydrolysis)
2
Q
Metric System Units Mnemonic
A
King Kilo 10^3 Henry Hecto 10^2 Died Deca 10^1 Unexpectedly -UNIT- Drinking Deci 10^-1 Chocolate Centi 10^-2 Milk Milli 10^-3 Monday Micro 10^-6 Night Nano 10^-9
3
Q
sp2-hybridized carbon atoms
A
AX3 systems that possess exactly one doubly bonded atom and are therefore sp2-hybridized
4
Q
Michaelism Mentin
A
- describes relationship b/t reaction rate (Vo) and substrate concentration [S]. also allows us to estimate the affinity, Km, and Vmax
- assumes that the catalytic step is the rate-limiting step AND the [ES] rapidly reaches the steady state=stays constant
V = (Vmax [S)] / (KM + [S])
5
Q
Catalytic constant (kcat)
A
- How fast ES complex proceeds to E + P
- Number of catalytic cycles that each active site undergoes per unit time
- Rate constant of the reaction when enzyme is saturated with substrate
kcat = Vmax/[E]
6
Q
Absorption of ultraviolet light by organic molecules always results in what process?
A
- UV light absorption always results in excitation of bound electrons
- Bond breaking, bond vibration, ionization can occur, but not always
7
Q
Column chromatography
A
Silica gel is polar and stationary in the column
The mobile phase (e.g. benzene) can be adjusted for polarity to maximize separation.
- The least polar/most nonpolar compound will always elute first
- hydrocarbons are very nonpolar & elute first then ketones, alcohols, and carboxylic acids last
8
Q
The half-life of a radioactive material is
A
the time it takes for half of all the radioactive nuclei to decay into their daughter nuclei (which may or may not be radioactive)
9
Q
The half-life of a radioactive material is
A
the time it takes for half of all the radioactive nuclei to decay into their daughter nuclei (which may or may not be radioactive)
- The more unstable something is, the shorter the half-life
10
Q
Alpha decay
A
Alpha decay: (238/92)U → (4/2)He + (234/90)Th
- Ejection of a helium nucleus at relatively low speed
- Weakest form of radiaton
- Alpha particle: (4/2)a
- Alter the state of the original element in a way that effectively categorizes it as another element
11
Q
Beta decay
A
Beta decay: (234/90)Th → (0/-1)e) + (234/91)Pa
- Ejection of a high energy electron from the nucleus
- More energy than alpha particles
- Beta particle: (0/-1)B
- Alter the state of the original element in a way that effectively categorizes it as another element
12
Q
Gamma emission
A
Gamma emission: (222/86)Rn → (222/86)Rn + (0/0)Y
- Ejection of a high energy electron from the nucleus
- Most energy
- Gamma ray: (0/0)Y
13
Q
Half-life vs. [Stuff]
A
- After 1 half-life, the amount of the original stuff decreases by half.
- After 2 half-lives, the amount of the original stuff decreases by a factor of 4.
- After 3 half-lives, the amount of the original stuff decreases by a factor of 8.
14
Q
Why must a person sitting straight up either lean forward or slide their feet under the chair in order to stand up?
A
as the person is attempting to stand, the only support comes from the feet on the ground. The person is in equilibrium only when the center of mass is directly above their feet. Otherwise, if the person did not lean forward or slide the feet under the chair, the person would fall backward due to the large torque created by the combination of the weight of the body (applied at the person’s center of mass) and the distance along the horizontal between the center of mass and the support point
15
Q
How where do kinases act on ATP?
A
three phosphate groups, in order of closest to furthest from the ribose sugar, are labeled alpha, beta, and gamma
Kinases transfer the Gamma phosphate to other molecules
16
Q
Protein stability and melting point
A
higher melting point = more stable protein bc more energy is needed to unfold the protein
17
Q
Reduction of C=O
A
C-OH
18
Q
Venturi Effect
:
A
Oxygen pressure= oxygen static pressure (P) + oxygen flow pressure (ρv^2)/2.
- In the area of the mask openings, P air = P + (ρv^2)/2, thus Pair > P
- Air enters the mask because the static pressure of the air is larger than the static pressure of the oxygen in flow. This is the Venturi effect, and the mask is called the Venturi mask.
- Venturi tube can be used to determine the velocity of a fluid that is flowing within it
- Pitot tube, which is used to determine the velocity of a fluid flowing past it.
The continuity equation (Q = Av) states that for a fluid with constant flow rate Q, a decrease in cross-sectional area A is associated with an increase in velocity v.
-Bernoulli’s equation states that for a fluid at a constant height, an increase in velocity is associated with a decrease in pressure.
The decrease in pressure occurs when a fluid flows into a constricted region of a pipe is known as the Venturi effect. A venturi meter can be sued to determine the velocity of a fluid in a pipe.
19
Q
Bernoulli’s equation
A
The equation resulting from applying conservation of energy to an incompressible frictionless fluid:
P + ½ρv2 + ρgh = constant
- fast fluid flows have low static pressure
- High speed flow = high dynamic pressure
- For a fluid at a constant height, an increase in velocity is associated with a decrease in pressure
20
Q
DNA bonding and melting point
A
G-C base pairs have triple bonds- 1 sigma and 2 pi bonds
A-T base pairs have double bonds- 1 sigma and 1 pi bond
more pi stacking interactions = most stable thermal stability = high melting point of DNA rich in GC content
21
Q
relative vapor pressure
A
when the atmospheric pressure=vapour pressure of the substance that is when it will start to escape into the gas phase. Since the atmospheric pressure is lower, that also means that the boiling point is lower. Vapour pressure and boiling point are directly related.
22
Q
SI units
A
meter, kg, second, ampere, mole, kelvin, candela
23
Q
Volumetric flow rate (continuity equation)
A
(Q = Av) states that for a fluid with constant flow rate Q, a decrease in cross-sectional area A is associated with an increase in velocity v
OR
Q=V/t, where Q is flow rate, V is volume, and t is elapsed time
24
Q
continuity equation
A
- works under the assumption that the flow in will equal the flow out
- applies to any incompressible fluid
A1∗v1=A2∗v2 where v is the velocity of the fluid and A is the area of the cross section of the space the fluid is moving through
25
triacylglycerols are
neutral storage lipids that consist of three fatty acids ester-linked to a single glycerol
26
Cathode and Anode Mnemonic
PANiC
Positive is Anode
Negative is Cathode
An Ox= Anode Oxidation
Red Cat= REDuction CAThode
-anode shoots out e-
cathode takes in e-
27
Calculating the Gibbs Free Energy from EMF
ΔG° = −nFE°
| where n is the moles of electrons transferred, E° is the standard state EMF, F is Faraday's constant (10^5 C/mol x e-)
28
Calculate log easy
-log(1x10^-5) = 5
10^-5 = 1x10^-5
- log 2x10^-X = [X-1].7
- log 3x10^-X = [X-1].5
- log 5x10^-X = [X-1].3
- log 8x10^-X = [X-1].1
29
Current (Faraday's Law)
Current = coulombs of charge per second. I = q/t
Faraday's constant = coulombs of charge per mol of e- = total charge over total mols of e-. F = q/n.
q = It and q = nF, thus we get:
It = nF
Current x time = mols of e- x Faraday's constant.
30
Boyle's Law
Boyle's law: P1/V1=P2/V2
- constant temperature
**PV=nRT can be divided by T by both sides, leaving PV/T=nR ( a constant )
31
Charles's Law
Charles's Law states that V1/T1=V2/T2
| - constant pressure
32
Heisenber's Principle
states as we know more about momentum of a particle the less you know about its position and vice versa
it is impossible to determine both the position and the velocity of an object simultaneously.
33
Gay-Lussac's Law
P1/T1=P2/T2
| - volume is constant
34
Combined gas law
combined gas law:
(P1V1)/T1=(P2V2)/T2
35
Lewis base
a substance that donates an electron pair in forming a covalent interaction
36
Lewis acid
accepts a pair of electrons from a Lewis base (Lewis bases donate electrons)
37
Broasted-Lowry acid
donates a proton to a Bronsted base (Bronsted bases accept protons)
38
Brownsted-Lowry base
Bronsted bases accept protons
39
Arrhenius acid
increases H+ concentration in water (releases an H+)
40
acid-base reaction
H+ is being produced in the forward reaction, and consumed in the reserve reaction
41
Strong acids
HCl, HBr, HI, HNO3, H2SO4, HClO4
42
Strong bases
LiOH, NaOH, KOH,Ca(OH)2, Sr(OH)2, Ba(OH)2,
43
catalysts
increase the rate of chemical reaction, which is the amount of product formed per unit time
44
K > 1
ΔG° is negative and the reaction is spontaneous
45
K > 1
ΔG° is positive and the reaction is nonspontaneous
46
Coordinate covalent bond
Lewis acid–base interaction between a metal cation and an electron pair donor
47
Adding HCl to an acid-base rxn
NH3 is a stronger base than H2O, so HCl will most likely protonate NH3 creating NH4, thus decreasing [NH3]
48
Le Châtelier’s principle
Add reactant=shift right
Add product= shift left
Remove reactant= shift left
Remove product= shift right
With pressure, the reaction is shifted to the side with fewer moles of gas
49
- add reactants, we want to shift to the R to use up that extra product= equilibrium
- remove products, we shift to the right to increase production of products
if we add products, we want to shift to the R to decrease [reactant] = equilibrium
remove reactants, we shift to the right to increase production of reactants
ADD a product = shift reaction AWAY
| Remove a reactant = increase reactant production = shift towards lost side
50
Keq vs. Q
Keq utilizes the concentrations of species at equilibrium, Q tells us the reaction’s current position. Thus, if Q is equal to Keq, then the reaction is at equilibrium. However, if Q is less than Keq, it has not yet achieved equilibrium, and the reaction will proceed forward (towards the products). If Q is greater than Keq, then the reaction has gone past equilibrium and will proceed in the reverse direction (towards the reactants).
51
Which of the following atoms will be expected to have the smallest second ionization energy?
A.Na
B.C
C.O
D.Ca
D. Ca
Metals have lower ionization energies than non-metals as long as the ionization event involves a valence electron.
- Na is an alkali metal=has only one valence e- and has a large second ionization energy. Ca is an alkaline earth metal and has two valence electrons. It will therefore have the smallest second ionization energy of the four atoms listed, which include Na and two non-metals.
It's asking about the smallest SECOND ionization energy. Remember that atoms like having a complete octet. Calcium has two valence electrons, so when it is ionized for the first time, it only has one more valence electron between it and a complete octet. If you think of it like that, it wouldn't take a lot of energy for Ca to give up that last electron to reach noble gas configuration, so it has the smallest second ionization energy.
Edit:: after Na ionizes once, it's essentially a noble gas and it would take A LOT for it to lose an electron
Ionization energy is the amount of energy required to remove an electron from an atom or an ion. Ionization energy increases as you go up and to the right of the periodic table. When you remove the first electron from an atom, the amount of energy required is called the first ionization energy.
If you want to remove a second electron, the amount of energy required to remove the second electron is called the second ionization energy.
52
mass percent
(coefficientt of atom in the formula of the substance x molar mass) / molar mass of substance
ex) oxygen in water- (1 x 16) / 18 = 89%
53
Henderson-Hasselbach equation
calculate the pH of the solution
| pH = pKa + log([base]/[acid])
54
structure of the DNA double helix
Sugar-phosphate backbones form the exterior of the double helix
Interior helix:
- purines bind w/ pyrimidines
- Dan stands are antiparallel (opposite directions)
- base pairs (G-C) appear in equal concentrations
55
calculate the number of charges in excess
formula for charge is q = ne
q is the amount of charge (in Coulombs)
n is the number of electrons
e is the elementary charge (1.6 × 10^–19 C)
56
power consumed
P=PE /time = mgh/t
| PE= mgh = mass x gravitational constant x distance from earth's surface
57
Gativational potential energy and Work done against gravity
Wgrav = F x d = m x g x Δh = ΔPEgrav
58
Ways to calculate work
Work done by a constant force= F x d x cosθ = -pΔV = ΔE
Work (kinetic energy) Wnet= ΔKE
- net work on a system = change in KE of moving things
Work done against gravity Wgrav = F x d = m x g x Δh
KE=1/2mv^2
P=W/t = ΔE/t
If pressure is constant, W = P/ΔV (P= length of rectangle under the Pressure volume curve and width is ΔV)
(Springs and rubber bands) Elastic PE= U = 1/2kx^2 (k is the spring constant and x is the magnitude of displacement)
59
Ways to calculate work
Work done by a constant force= F x d x cosθ = -pΔV = ΔE
Work (kinetic energy) Wnet= ΔKE
- net work on a system = change in KE of moving things
Work done against gravity Wgrav = F x d = m x g x Δh
KE=1/2mv^2
P=W/t = ΔE/t
If pressure is constant, W = P/ΔV (P= length of rectangle under the Pressure volume curve and width is ΔV)
(Springs and rubber bands) Elastic PE= U = 1/2kx^2 (k is the spring constant and x is the magnitude of displacement)
60
Intensity of sound
β = 10 logI/I0
- β is sound level in decibels, I is intensity. I0 is 10-12 W/m2
- Intensity is power per area, or the rate of energy expenditure per area. The unit is W/m2
61
Intensity of sound ratio
β = 10 logI/I0
- β is sound level in decibels, I is intensity. I0 is 10-12 W/m2
- Intensity is power per area, or the rate of energy expenditure per area. The unit is W/m2
```
Ratio of sound intensities:
B2 = B1 + 10log (I2/I1)
(B2-B1) 10 = # = log I2/I1
take 10 to the power on both sides to cancel log
10^# = I2/I1
```
62
Lenses and glasses
Reading glasses ("farsighted") = converging lenses
normal glasses ("nearsighted") = diverging lenses
63
Half life equation
amount of original material = amount of original material left x 2 ^#half-lives = amount of original material left x 2 ^1/half life
64
protein secondary structure
represented by repeated patterns of hydrogen bonds between the backbone amide protons and carbonyl oxygen atoms.
65
resistors in parallel
1/Rtot=1/R1+1/R2+…
Req for 2 resistors in parallel is = R1*R2/(R1+R2)
66
the primary reason for chemical separation during a chromatographic separation
the relative amount of hydrogen bonding to the stationary phase will determine the relative rate of migration of the various components in the sample
67
Rf
the ratio of the distance travelled by the analyte relative to the solvent front during a chromatographic separation
68
stereospecific
provides essentially one stereoisomer product
| - SN2
69
stereoselective
provides some sort of mixture of chiral products
| - SN1
70
SN1
- bimolecular (2 steps)
- polar protic solvents (can H bond)
- favors 3>2>1>methyl
- rate = k[RL]
- racemic (2 inverse products
- strong nucleophile not required
- forms a carbocation (favors cation stability)
- barrier = carbocation stability
71
SN2
- unimolecular (1 step)
- polar aprotic solvents (can't H bond)
- favors methyl > 1 > 2> 3 C
- rate = k[Nu][RL]
- inverted products; optically active
- strong nucleophile favored
- barrier = steric hindrance
72
amide group
CO=NH2
73
amine
R-NH2
74
aldehyde
- CHO
| - C=O and C-H
75
ketone
-COC-
| C=O with r groups on both side
76
ester
- COOR'
- no H holding
- insoluble in water
77
imine
C=N
78
enamine
N-C=C
79
product formed for aldol condensation
β-hydroxy aldehyde or β-hydroxy ketone.
80
aldol addition product
hydroxy ketone or aldehyde
81
anhydride
-O=C-O-C=O-
82
first peak in the gas chromatograph (gc) trace
lowest molecular weight = weakest intermolecular forces of attraction = migrates the fastest = first peak
83
carboxylic acid derivatives as leaving groups
- the carbonyl oxygen is the one that is double bonded to the carbon and is not likely to be a good leaving group in this reaction
- oxygen on the -OH group is more likely to leave
84
focal length & strength
1/f = 1/o + 1/i = 2/r
Strength (Power) = 1/f
= S = 1/o + 1/i
- meters
O (the object) = 1m (this is the distance between the object and the lens of your eye) We are also given I (the image) is 2cm away on the other side of the lens (your retina - which is where the image forms). 2 cm =
2cm*1m/100 cm = 1/50 m = 0.02m
85
Ohm’s law
V=IR = I= V/R
current is equal to voltage divided by resistance. If current increases and resistance is constant, then voltage increases as well.
86
index of refraction of a medium
index of refraction of a medium is equal to the ratio of the speed of light in vacuum to the speed of light in the medium
87
intensity of electromagnetic radiation
the energy of electromagnetic radiation is directly proportional to the number of photons
the intensity of electromagnetic radiation is defined as energy emitted per unit time. Thus, intensity is directly proportional to the number of photons emitted.
88
THz to Hz
1 THz = 10^12 Hz
89
flow rate
mL/min
90
binding regions/amino acids
- hydrophobic region is that CH2 region
- Cross linking residues are cysteines
- Spacers would be the glycines
- You can tell the phosphorylated residue by the way that it is.
- Arg and Asp which facilitate cell adhesion.
91
Why is the velocity of blood flow slower in capillaries than in arteries?
the high number of capillaries in the body means that the total cross-sectional area of these vessels is larger than any other vessel type in the circulatory system. This causes the velocity of the blood to decrease
92
hybridization of an octahedral compound
- octahedral compounds have six σ bonds and no lone pairs. - the central atom requires the hybridization of six atomic orbitals, d2sp3
93
assumptions of the ideal gas law
a property of an “ideal” gas is that it is composed of particles that have negligible volume and do not exert intermolecular forces.
94
molecules made mostly of C and H
nonpolar- hydrophobic
95
the requirement of ATP for kinase activity implies . . .
the phosphate groups come from ATP
96
AX3 systems
- have 3 bonded atoms and no lone pairs
- preferred geometry for such a system is trigonal planar and the hybridization scheme that facilitates this geometry is sp2.
97
lipase
hydrolyze fatty acids
98
glycogen synthase
the main chain linkage of glycogen, which is an α-1,4-glycosidic bond
99
removal of phosphate prior to gel electrophoresis
result in a smaller band corresponding
100
atomic number and mass number
- number of electrons in a neutral atom is equal to the number of protons
- mass number of the atom (M) is equal to the sum of the number of protons and neutrons in the nucleus
- the number of neutrons is equal to the difference between the mass number of the atom (M) and the atomic number (Z) (Number of electrons)
101
Which is less likely to dissociate in water?
| HPO42−, H2SO4, H3PO4, HCL
HPO42− has a high negative charge and so dissociation of it will occur to the least extent
102
focal length
According to the thin lens equation, the distance between the lens and the image is 3F x 1F/(3F - 1F) = (3/2)F, where F is the focal length. The ratio sought is then equal to (3/2)F/(3F) = 1/2.
103
E° and spontaneity
spontaneous reaction = E° > 0
| non spontaneous reaction = E° < 0
104
transmission of mutations
only occurs in germ cell lines, NOT somatic
Ex) a mutation only expressed in tumor cells won't be passed down to kids
