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MCAT > 5. General Chemistry 1 > Flashcards

5. General Chemistry 1 Flashcards

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MCAT flashcards
1
Q

Z

A

atomic number = number of protons

bottom number

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2
Q

A

A

mass number = number of protons + neutrons

top number

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3
Q

isotopes

A

versions of an atom with varying neutrons, but same number of protons

all have the same Z (atomic number)

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4
Q

anion

A

negatively charged molecule

non-metals

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5
Q

cation

A

positively charged molecule

metals

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6
Q

group/family

A

a column on the periodic family is familiar

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7
Q

period

A

a row on the periodic table

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8
Q

alkali metals

A

group 1A

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9
Q

alkaline metals

A

group 2A

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10
Q

transition metals

A

middle 4 rows (D-block)

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11
Q

lanthanides

A

upper row in the F- block

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12
Q

actinides

A

lower row on the F-block

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13
Q

halogens

A

group 7A

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14
Q

noble gases

A

group 8A

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15
Q

S-block

A

first two columns

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16
Q

P-block

A

six columns on the right

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17
Q

D-block

A

middle of table

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18
Q

F-block

A

occur in s-block and d-block, rows 6 and 7

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19
Q

metals

A

larger atoms with loosely held elections (ionic bond with non-metals)

form cations

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20
Q

non-metals

A

smaller atoms with tightly held electrons, form anions, lower MP than metals, (form covalent bonds with non-metals)

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21
Q

pi bonds

A

extra bond formed in double bond. larger atoms from weaker pi bonds due to decrease size and overlap of p-orbitals

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22
Q

families similarities

A

CH4 and SiH4 will behave similarly as they are in the same family

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23
Q

Zeffective

A

as atoms increase in size, they are surrounded by more electrons (essentially how much the outer shell electrons feel the charge from the nucleus –> larger molecules lose electrons more easily due to less charge being felt)

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24
Q

periodic table trends

A

to top right corner: electron affinity, ionization energy, and non-metallic character

to bottom left corner: atomic radius, and metallic character

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25
Q

quantum numbers

A

Fe 1s22s22p63s23p63d64s2
Fe2+ 1s22s22p63s23p63d6
Fe3+ 1s22s22p63s23p63d5

Bi [Xe]4f145d106s26p3

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26
Q

Heisenberg Uncertainty and Pauli Exclusion Principles

A

the more we know about an election’s position, the less we know about its momentm

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27
Q

electron configuration

A

(charts with the up and down arrows)

note:anions move forward one box for each extra electron, and move back one box for each electron for cations

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28
Q

energy levels

A

think of them as steps. electrons can only be in one energy level or another, never in between (quantized)

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29
Q

photon-light emission

A

if an electron is struck with a photon that is not strong enough to raise its energy level, the photon would just pass through the molecule. if the photon is strong enough, the electron can go up and energy level, where it can then be lowered back to its original level later and release said photon.

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30
Q

work function

A

the ejection of a metal’s electron from the outer shell by bombarding it with energy (not the same as ionization energy)

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31
Q

kinetic energy of ejected electron

A

KE = E - phi

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32
Q

Energy of a photon

A

E = h f

v = f lambda
E = hv/ lambda = hc/lanbda
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33
Q

Two unique light sources are used to bombard a single metal sample with photons (φ = 349J). The first light source delivers photons with an energy of 700J at a rate of 1 x 105 photons per second. The second light source delivers 350 J photons at exactly twice that rate. If each light source is shone onto the surface of the metal for exactly one second, which of the following statements is true?

A

The second light source will eject twice as many electrons as the first. since both light sources are enough to exceed the work function. however the second light source delivers twice as many photons

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34
Q

A certain metal is known to have a work function of 500J. If a photon of 500J strikes the surface of the metal, what will be the result?

A

Because the energy of the photon exactly equals the work function nothing will happen. Theoretically, it is as if the electron is now “freed” from its attraction to the nucleus but cannot move because it lacks any excess energy to transfer into KE.

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35
Q

radioactive decay

A

process by which unstable atoms change their chemical composition after timee

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36
Q

alpha decay

A

loss of one Helium nucleus, which has a mass number of 4 and atomic number of two (loss of two protons and two neutrons)

(passes through paper)

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37
Q

beta decay

A

a neutron is changed into a proton with the election of an election

(passes through aluminum foil)

38
Q

electron capture

A

a proton is changed into a neutron via capture of an electron

39
Q

positron emission

A

a proton is changed into a neutron with expulsion of a positron

40
Q

gamma emission

A

either electron capture or positron emission. does not change number of nucleons

(passes through lead)

41
Q

neutron

A

proton and electron

42
Q

proton

A

neutron and positron

43
Q

half life problems

A

usually two of the three: t1/2, t or g, variable given

44
Q

bonding and anti-bonding

A

1) Anti-Bonding orbitals are higher in energy than Bonding orbitals
2) Bonding orbitals contain electrons that are “in phase” and are said to be “attractive”; anti-
bonding orbitals contain “out of phase” electrons that are said to be “repulsive.”
3) Know what drawings of bonding and anti-bonding orbitals look like. (up down energy diagram arrows onto a hexagon)

45
Q

covalent bonds

A

formed between two non-metals, involving electron sharing

46
Q

ionic bonds

A

formed between metal and nonmetal and are due to an electrostatic attraction.

47
Q

ionic character

A

difference in electronegativity between two atoms in a bond. (so a C-C bond has no ionic character, and theoretically francium fluoride would have the highest ionic character

48
Q

condosity

A

The “condosity” of a solution is the concentration (molarity) of an NaCl solution that will conduct electricity exactly as well as the solution in question. For example, for a 2M KCl solution, we would expect the condosity to be something more than 2.0. Why would we expect it to be above 2.0? Because potassium is more metallic than sodium. Thus, we know that it will be a better conductor. This means the NaCl solution will have to be slightly more concentrated to conduct as well as the KCl solution. Ex. what is the expected condosity of a 3M LiCl solution? Sodium is more metallic than lithium, so we would expect an NaCl solution to conduct electricity better than an equimolar LiCl solution. This means the NaCl solution would need to be less concentrated in order to conduct equally as well. This predicts a condosity of something less than 3.0 for a 3.0M LiCl.

49
Q

coulombs law

A

F = k q1 q2 / r^2

50
Q

bond length

A

distance between nuclei of the atoms forming the bond

51
Q

bond energy

A

energy stored in the bond. also, the same as energy required to break the bond. (stable compounds such as N2 have very high bond energy, while unstable compounds such as ATP have low bond energy) (does not mean high energy molecule)

52
Q

bond dissociation energy

A

amount of energy required to break a bond. same as the energy stored in a bond (same as bond energy)

53
Q

heat of combustion

A

amount of energy that is released when a molecule is combusted with oxygen, all covalent bonds are broken and reformed in a radical reaction. the higher the energy of the molecule the higher the heat of combustion.

54
Q

why do bonds form?

A

to create a lower energy state than was the unbounded form. forming bonds release energy

55
Q

coordinate covalent bonds

A

a covalent bond in which both electrons shared in the bond are donated by one atom

NH3 + HCl = NH4+ + Cl-

56
Q

empirical formulas

A

Empirical formulas represent the lowest possible number of moles of each element that can be present in a compound while still maintaining the same mole-to-mole ratio between the elements. CH6O

57
Q

molecular formula

A

A molecular compound is the actual number of moles of each element found in a specific compound. C6H12O6

58
Q

percent mass

A

(mass of one element/total mass of the compound)(100)

ex. percent mass of H in H2O is 2/18 * 100

note:deriving a formula from percent mass
1. change percent mass of each element into grams. 2. covert grams of each element into moles by dividing by molar mass.
3. lookout the element with the lowest number of moles. Calculate approximately how many times it
will divide into each of the other molar amounts for each of the other elements—this number will be the subscript for each element in the empirical formula. If the subscripts are not at their lowest common denominator, reduce to get the empirical formula.

59
Q

keq

A

keq >1, shift towards product. keq

60
Q

general ionic compunds

A

cation then anion, ex. calcium sulfate

61
Q

transition metals

A

iron(II)sulfate

62
Q

monatomic ions

A

ex. sulfide ions (ends with -ide)

63
Q

combination reaction

A

A+B=AB

64
Q

decomposition reaction

A

AB= A+B

65
Q

single displacement reatin

A

A+BC=AC+B

66
Q

double displacement reaction

A

AB+CD=AC+CD

67
Q

balancing reactions

A

balance carbons, then hydrogen, then oxygens, then the rest

68
Q

moles

A 
6.022E23
# atoms, PV=nRT, molecules, density (D=g/mL), molarity (M=moles/L), grams, ions
69
Q

limiting reagent

A

in a balanced equation in moles, the limiting reagent is the reaction you will run out of first, not the one with the least amount

70
Q

theoretical yield

A

calculated yield

71
Q

actual yield

A

actual yield produced

72
Q

percent yield

A

actual yield/theoretical yield * 100

to inc. yield: increase reactants, or shift to right with le chatter’s principle

73
Q

oxygen needed for combustion

A

+1.0 for each carbon

-0.5 for each oxygen

74
Q

Law of mass action

A

Keq = [products]^x/[reactants]^y
exponent is its coefficient in the balanced equation

note: pure liquids and solids are never included, only solutions and gases

75
Q

reaction quotient

A

The Equilibrium Constant can only be calculated at equilibrium.

If Q > K, the reaction will proceed to the left or reactants. If Q

76
Q

Le Chatlier’s Principle

A

Systems already at equilibrium that experience change will shift to reduce the effects of that change.

77
Q

Predict the effects of doing each of the following to a reaction at equilibrium:

A

to shift right: add reactants, remove products, increase initial pressure, inc them. Opposite to shift left

78
Q

reactions

A

reactants must collide with enough energy to overcome the energy of activation and must be in the correct spatial orientation

79
Q

rate

A

measured as the change in molarity of the reactants per second (M/s)

80
Q

How will increasing each of the following affect reaction rate: [reactants], [products], [catalyst], energy of activation (Ea), energy of the transition state, energy of the reactants, and temperature?

A

increasing reactants will inc rate, increasing products has no effect, increasing catalyst will increase rate, increasing Ea or energy of transition state increases rate, increasing temperature will increase rate,

81
Q

how to calculate rate laws

A

1) Find two trials where the [reactant] in question changed, but all other parameters did NOT (i.e., concentrations of the other reactants, temperature, pressure, etc., all remain constant)
2) Note the factor by which the reactant concentration changed
3) Note the factor by which the rate changed across those same two trials
4) Determine Y in the following equation: XY = Z ; where X = the factor by which the [reactant]
changed, Z = the factor by which the rate changed, and Y = the order of the reactant. Recall that any number raised to the zero power is equal to one.

82
Q

overall rate order

A

sum of exponents in the rate law

83
Q

rate order graphs

A

These graphs will only be linear when the reaction has only a single reactant OR when it is part of a multiple reactant reaction where the rate is independent of ALL other reactants (e.g., other reactant is zero order or is in excess).

84
Q

zero order

A

o Zero Order: [A] vs. time is linear (i.e., yields a straight line) with slope = -k

85
Q

first order

A

o First Order: ln[A] vs. time is linear with slope =-k

86
Q

second order

A

o Second Order: 1/[A] vs. time is linear with slope = k

87
Q

For a reaction with two reactants, A and B, a graph of 1/[A] vs. time is non-linear. Which of the following is known? 1) The reaction cannot be second order in A and independent of B, 2) Reactant B must be involved in the rate law, 3) Reactant B cannot be in excess.

A

The observation that a graph of 1/[A] is non-linear is somewhat inconclusive. The graph could have been non-linear because both reactants are impacting the rate (violating the rule that these graphs will only be linear if only one reactant species is impacting rate). However, it could also be non-linear if reactant A is the only reactant impacting rate, but reactant A is not second order. For example, suppose that reactant A is actually first order; the graph of 1/[A] would not be linear, but the graph of ln[A] would be. Statement 1 is the best choice because if reactant A was second order and reactant B did not impact rate, we would expect a linear plot. Statement 2 is NOT known. Reactant B could be in the rate law and that is why the graph is not linear; however, it is equally plausible that reactant B is NOT in the rate law and the graph of 1/[A] is not linear because A is first order, not second order. Statement 3 is NOT known either. As we have already explained, the graph could be linear either way. If reactant B is in excess then the graph could be non-linear because reactant A is actually first order, not second order. If reactant

B is not in excess (and is in the rate law) then the graph could be non-linear because both A and
B are impacting rate.

88
Q

For a reaction with two reactants, A and B, a graph of ln[A] vs. time is linear. How many of the following statements are true? 1) Reactant A must be first order, 2) Reactant B could be first order, 3) Reactant B cannot be impacting rate, 4) Reactant B must be in excess.

A

Getting a positive result (i.e., a line for one of the graphs) gives us much more information.
Because the graph of ln[A] is linear we know that statement 1 is true, reactant A is indeed first order. We also know, however, that no other reactants are impacting rate. Statement 2 is NOT known. Reactant B could be first order; however, if it is first order then we know it MUST be in excess because only reactant A is impacting rate. Reactant B could also not be first order. It could be zero order and not impacting rate at all—an alternative explanation for why only reactant A is impacting rate. Statement 3 is known for this exact moment and set of conditions. Reactant B cannot be impacting the rate; if it were, the graph of ln[A] would not have been linear. Statement 4 is NOT known but could be true. We know that reactant B is not impacting rate, but that could be either because it is in excess or because it is zero order.

89
Q

Describe how you could use the rate order graphs to determine the order of each reactant in a multi-reactant reaction experimentally in the lab: What would you need to measure? What would you do with the data?

A

To determine the order of each reactant experimentally you would only need to ensure that all other reactants were in excess and measure the concentration of the limiting reactant as a function of time. You would then use the measured concentration of the reactant in question in each of the types of graphs. If all other reactants are in excess, one of these graphs should be linear. You may be tempted to think that you wouldn’t get a positive result if the reactant you were testing was NOT impacting rate. However, that would produce a line for [A] vs. time (i.e., the reaction would zero order in that species). Remember that zero order reactants are not included in the rate law!

90
Q

multi step reaction rates

A

if there is a slow step, the slow step always detainees the rate, if slow step is first, the rate law can be written as if it were the only step, if the slow step is second, the rate law is the rate law of the slow step which will include an intermediate as one of the reactants

91
Q

catalysts

A

any substance that increases reaction rate without itself being consumed in the process. catalysts change the rate of reaction by providing a pathway with a lower activation energy.note: catalysts do not change equilibrium, Kew, enthalpy/entropy, gibbs free energy, or any other thermodynamic properties

MCAT (11 decks)

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