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CHEM1212 > LAB PRACTICAL > Flashcards

LAB PRACTICAL Flashcards

1
Q

Measurement estimates

A

one value beyond the last marking on measurement device

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2
Q

Pharmacokinetics

A

process by which drug is absorbed, metabolized, eliminated

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3
Q

Catalysis

A

increase in rate of reaction due to participation of catalyst which lowers activation energy

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4
Q

Order of reaction

A

determined by coefficient in balanced equation but must be determine EXPERIMENTALLY

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5
Q

Arrhenius Equation

A

k = Ae^_Ea/RT with R = 0.008314, Ea = activation energy

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6
Q

Greater activation energy means

A

slower rate

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7
Q

Chemical Kinetics Reaction Rate

A

Rate = k [Fe3+]m[I-]n (we determined m and n experimentally)

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8
Q

Chemical Kinetics Rate in terms of Triiodide ion

A

Rate =Δ[I3-] / Δt

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9
Q

What was Fe3+ conmibed with

A

I- and S2O3 2-

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10
Q

Chemical Kinetics Plot

A

y = ln(Rate) x = ln([I-]_ with slope = n

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11
Q

Methyl green

A

green color with seven methyl groups attached, parent dye has 0 groups

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12
Q

Rate of Decolorization [OH-]

A

[OH-] = 4.5 x 10^-2 x [PO4 3-] / [HPO4 2-]

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13
Q

Rate of Decolorization rate law

A

rate = k1[dye][H2O] + k2[dye][OH-]

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14
Q

Dye and Absorbance

A

[dye] = A / e L
-lnA = kobst -lnA0

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15
Q

Graph -lnA versus time

A

slope is observed rate constant

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16
Q

What is the slope of Kobs vs. [OH-] equal to

A

= K2, kobs = k1[H2O] + K2[OH-]
K1[H2O] remains constant and does not effect slope (acts as an intercept)

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17
Q

ΔG°= (2 equations)

A

ΔG° = ΔH° - TΔS°
ΔG° = -RTlnK

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18
Q

If K > 1

A

ΔG° < 0 and the process will proceed in the forward direction

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19
Q

lnK =

A

lnK = -ΔH° / RT + ΔS° / R

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20
Q

Ksp is independent or dependent on temperature

A

dependent

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21
Q

Plot lnK versus 1/T

A

slope is -ΔH°/R

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22
Q

Ice cream freezes

A

at a lower temperature than the freezing point of water so rock salt is added

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23
Q

m =

A

m = moles solute / kg solvent

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24
Q

is molality temperature dependent or independent

A

independent

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25
Q

Vant Hoff Factor

A

number of dissolved particles the solute contributes

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26
Q

m = (with VHF)

A

m = Δtf / I kf

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27
Q

Natural dye in solar cell experiment

A

TiO2 stained and dye absorbs light from sun to produce flow of electrons

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28
Q

Semi conducting material

A

TiO2, titanium dioxide

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29
Q

How to electrons flow in solar cell

A

from TiO2 coated electrode to the counter electrode producing electricity

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30
Q

What does the dye molecule release when it absorbs light

A

electrons

31
Q

Mediator

A

Iodide ion in the electrolyte

32
Q

Voltage

A

difference in energy levels of TiO2 and mediator

33
Q

Current is directly proportional to

A

amount of light or number of photons absorbed by the dye

34
Q

Power of solar cell

A

voltage x current in Watts

35
Q

Compound in raspberries that plays a role in solar cell

A

Anthocyanin (carbohydrate)
(REVIEW STRUCTURE)

36
Q

Chelation

A

bonding in ions and molecules to metal ions, involves presence of two separate bonds between polydentate ligand and single central metal atom
(REVIEW STRUCTURE)

37
Q

Mediator oxidation

A

I- + I2 <–> I3 -

38
Q

Mediator reduction

A

6e- + 3I2 <–> 2I3-

39
Q

Common uses of TiO2

A

toothpaste and food coloring

40
Q

Area of solar cell to produce 1 watt?

A

1 watt = Power Watt x Area

41
Q

Commercial vinegar contains

A

5-6% acetic acid

42
Q

Percent Error =

A

Absolute Value of Actual - Experimental / Actual x 100

43
Q

Molarity Calculations

A

M1V1 = M2V2

44
Q

Percentage (w/v) acetic acid in vinegar

A

mol / L x 1L / 1000 mL x 5.00 mL = mol
x mol x g / mol = x g / 5.00 mL = X x 100

45
Q

Examples of coordination compounds

A

hemoglobin, vitamin b12, chlorophyll

46
Q

Ammonia serves as

A

ligand

47
Q

Sulfate ion serves as

A

counter ion

48
Q

Copper coordination compound formula

A

Cux(NH3)y(SO4)z x H2O

49
Q

Color indicator in volumetric analysis of ammonia

A

methyl orange

50
Q

Beer Lambert Law

A

absorption of light as it passes through solution is proportional to the concentration of the absorbing species

51
Q

Empirical formula

A

Cu(NH3)4(SO4)

52
Q

Bonsted-Lowry acid

A

proton donor

53
Q

Strong acid

A

ionizes essentially to completion

54
Q

Larger Ka

A

stronger acid

55
Q

Equivalence point

A

final solution neutral (equal moles), where amount of base has been added, + 1 drop is light pink, vertical on curve

56
Q

End point

A

1 final drop turned light pink

57
Q

After reaching equivalence point

A

pH is high due to excess base

58
Q

Ka =

A

Ka = [A-][H3O+] / [HA]

59
Q

Ionization equation

A

[A-] = [H3O+] because 1:1 mole ratio

60
Q

Titration data M

A

MaVa=MbVb
Ma=MbVb / Va

61
Q

Percent ionization =

A

% = [H3O+] / Ma x 100

62
Q

Buffer solution

A

contains weak acid and conjugate base

63
Q

Henderson-Hasselbalch

A

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

64
Q

Halfway point

A

pH 1/2 = pKa

65
Q

1st derivative

A

max, equivalence point, intersection with x-axis

66
Q

2nd derivative

A

inflection point, change = 0

67
Q

Saturated solution

A

in state of dynamic equilibrium between dissolved, dissociated, ionic compound and undissolved solid

68
Q

The solubility of an ionic compound is

A

lower than a solution containing a common ion than in pure water

69
Q

Ksp Barium Nitrate

A

Ksp = [Ba2+][NO3]2

70
Q

Dynamic Equilibrium

A

rate of forward = rate of reverse

71
Q

Equilibrium constant

A

Keq = [products] / [reactants]

72
Q

Equilibrium constant graph

A

y axis = A/([Fe3+]I [SCN-]I)
x axis = A x ([Fe3+]I + [SCN-]I) / ([Fe3+]I [SCN-]I)
slope = -Keq

73
Q

Freezing point

A

temperature when solvent in solution and pure solid have same vapor pressure

74
Q

Larger Ka

A

more ionization and stronger

CHEM1212 (5 decks)

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