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From Flashcardlet > AP Chem Ch 4-5 > Flashcards

AP Chem Ch 4-5 Flashcards

0
Q

Hydration

A

Add water to something –> molecule bonds with water.

Heating, dehydration, reverses this.

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1
Q

Polar

A

Unequal distribution of charge. A net electronegative charge pointing in a direction

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2
Q

Solubility

A

The maximum amount of solute that dissolves into a solvent at equilibrium.
The measure of the amount of solute that can dissolve in a volume of solvent at a given set of conditions (temperature, pressure)

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3
Q

Solute

A

Thing being dissolved

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4
Q

Solvent

A

Thing being dissolved in – solute dissolved in the solvent.
Common solvent is water

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5
Q

Solubitility changes with temperature and stirring

A

Temperature up, solubility up
Stirring up, solubility up
Amount of solute up, solubility down- precipite

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6
Q

Solubility rules

A

See other flashcard deck

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7
Q

Super saturated

A

Helps make crystillation structure

Put solute into solvent, heat up and put lots in, then cool and makes crystal

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8
Q

Strong electrolytes

A

Soluble salts, strong acids and strong bases because they all three completely dissociate into ions when dissolved in water. Conduct electricity because of the ions

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9
Q

What do strong electrolytes do

A

100% dissolve in water. Carry an electrical current

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10
Q

Molarity

A

Number of mol / L

Moles of solute over the liters of the solution

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11
Q

If we have 11.5 g of NaOH and they molar mass is 40g/mol, what is the molarity when dissolved in 1.50 L of water?

A

11.5 g * 1 mol / 40 g = .288 mol
M = mol / L = .288/1.5 = .192 M
Completely dissociates in water

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12
Q

Example:
Say we have a solution of .50 M of Co(NO3)2
What is the concentration of each ion present

A

Co2+ –> M= .50

NO3 - –> M=.5*2=1.00

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13
Q

Example of dilution:

If we have 1 liter of 2.7 M of KMnO4, how could we make a 1.45 M solution and a .65 M solution?

A

Say we want .5 L solution of both 1.45 M and .65 M
M1V1=M2V2
2.7V = (1.45)(.5)
V= .27 L
Add .27 L of the 2.7 M solution and .23 L of water to make .5 L of the 1.45 M solution

2.7V= (.65)(.5)
V=.12 L
Add .12 L of the 2.7 M solution and .38 L of water to make .5 L of the .65M solution

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14
Q

What type of glassware is used to make series dilutions

A

Eurleymuer flask

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15
Q

Types of chemical reactions

A

Precipitation
Acid-base
Oxidation-reduction

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16
Q

What does silica gel dissolve/not dissolve in

A

Silica gel is not soluble in ethyl acetate

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17
Q

Aspirin solubility results

A

Soluble in ethyl acetate
Soluble in sodium bicarbonate
INSOLUBLE in HCl

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18
Q

Acetametaphine solubility results

A

Soluble in ethyl acetate, INSOLUBLE in sodium bicarbonate

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19
Q

Precipitation reaction

A

Formation of an insoluble substance
Filtrate –> solvent
Precipaite is formed

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20
Q

Example of precipaite reaction:

K2CrO4 (aq) + Ba(NO3)2 (aq)–> ?

A

K2CrO4 (aq) + Ba(NO3)2 (aq)–> 2KNO3 (aq) + BaCrO4 (s)

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21
Q

Types of formulas

A

Formula equation
Complete ionic equation
Net ionic equation

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22
Q

Formula equation

A

Overall reaction that includes everything going on

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23
Q

Comete ionic equation

A

Breaks down the aqueous materials into the ionic components

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24
Q

Net ionic equation

A

Cancels out the ions on both sides

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25
Q

Example of writing net ionic equation:

AgNO3 (aq) + KCl (aq)

A

AgNO3 (aq) + KCl (aq) –> AgCl (s) + KNO3 (aq)
Ag+ + NO3 - + K+ + Cl- –> AgCl(s) + K+ + NO3 –> complete ionic here
Cancel
Ag+ (aq) + Cl- (aq) –> AgCl(s)

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26
Q

Stoichiometry of precipitan reactions:
Example:
If we have 1.50 L of .1 M of AgNO3 and we add 8 grams of NaCl (.137 mol), what product will precipite and how much will form?

A

AgNO3 (aq) + NaCl (AQ) –> AgCl (s) + NaNO3 (AQ)
.1 M = x mol / 1.50 L.
X= .15 mol silver nitrate
NaCl is limiting reactant
So .137 mol of AgCl is formed, multiply by molar mass to get amount formed in grams

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27
Q

Arrhenius concept

A

Acids produce H+, bases produce OH-

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28
Q

Bronsted - Lowry definition

A

Acid is a proton donor

Base is a proton acceptor

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29
Q

Hydroxide with weak acid

A

Hydroxide is so strong as a bad that it completely reacts even with weak acids

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30
Q

Neutralization reactions:

What volume of .1M of HCl is required to neutralize 25 mL of a .350 M NaOH solutoon

A

HCl + NaOH –> H2O + NaCl (Aq)
1:1 ratio of HCl to NaOH –>
.350 M = x mol / .025 L
Get moles then need same moles of HCl and use M = mol / liter to do so

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31
Q

Titration

A

When you add a titrant –> base to acidic solution or an acid to a basic solution, observe how the pH changes as you add more of the titrant

The equivalence point is the point at which neutral pH. Endpoint when it is base if started acid or reverse if started base.

32
Q

Reduction

A

ADDING ELECTRONS

CHARGE IS REDUCED IN REDUCTION

33
Q

Oxidation

A

Giving away elections

The charge is increased –> less negative charge

34
Q

Oxidizing agent

A

Facilitates oxidation, but it is reduced

35
Q

Reducing agent

A

Facilitates reduction but it is oxidized

36
Q

Rules for assigning oxidation numbers

A

Florine always -1
Oxygen -2
Hydrogen +1

37
Q

What’s being reduced and what’s being oxidized:

Cr+ + Sn4+ –> Cr3+ + Sn2+

A

Cr is losing electrons, so it is being oxidized

Sn is gaining electrons, so it is being reduced

38
Q

How to balance redox reactions:
Example:
Bi(OH)3 + SnO2 2- –> SnO3 2- + Bi in an acidic solution

A

First identify what’s being reduced and what’s being oxidized
Bi is going from 3+ to 0 –> it is being reduced
Sn is going from 2+ to 4+ –> it is being oxidized

Now write the half reactions:
3e- + Bi(OH)3 –> Bi
Now make sure the hydrogens and oxygens are balanced as well as the charge on each side
3e- + Bi(OH)3 + 3H+ –> Bi + 3H2O

Now do the same for the other half reaction:
H2O + SnO2 2- –> SnO3 2- + 2e- + 2H+

Now multiply top by 2 and bottom by 3 to get it so the electrons cancel out and write the final equation
H+ may or may not cancel
If in basic, add OH- and form water to one side and keep the OH- on the other side

39
Q

Equilibrium

A

State where the rate of the forward reaction is equal to the rate of the reverse reaction

40
Q

Lattice

A

The pattern for which a solid will crystallize.
Energy is required to break this lattice apart. Need the break the intra molecular forces that keep the molecule together.

41
Q

Entropy

A

Measure of randomness – disorder and chaos occurring

42
Q

Miscible substances

A

Substances that can mix, like alcohol and water, are miscible. Measure of solubility cannot be applied for miscible

43
Q

Immiscible

A

Materials that do not mix together, like oil and water. Again solubility can’t be applied here

44
Q

Why do some things dissolve and others don’t?

A

When dissolved, solvent molecules surround solute molecules. Breaks the intermolecular bonds of the individual molecules (Na and Cl ions being pulled apart by water)

Need energy to break these bonds.
Happens because entropy wants the world to move toward disorder and mixing increases the entropy, causing a drive to increase entropy and therefore free energy

45
Q

Other factors to consider in solubility

A

Charge of the material being dissolved

Size of the molecule/compound – larger compound required more solvent per solute.

46
Q

Coulomb’s Law

A

Opposite charges attract while alike charges repel.
F= kq1q2 / r^2
As distance between two charges increases, electrostatic forces decrease.
Force proportional to magn of both charges

47
Q

Inverse square law

A

Force inversely prop to the square of the distance

48
Q

How does Coulomb’s law apply to Chem?

A

Ionization energy
The protons in the nucleus attract the electrons – removing one will require some amount of energy– it also results in an increase in positive charge and will exert a stronger hold on the remaining electrons.

49
Q

Ionization energy

A

The amount of energy required to remove electron.

50
Q

First vs second vs third ionization energies

A

First is removal of one highest energy electron.
Second is removal of he next highest energy electron
Third is removal of the next highest energy electron

51
Q

Trends with ionization energy

A

Group 1 wants to give up their electrons– lowest energy required to remove one
Group 17 doesn’t want to give up electrons – highest energy required to move one.
More energy required to remove inner electron than outer– easier to remove outer electrons.

So for magnesium, the first ionization energy will be bigger than the second because second is last electron in shell but the third ionization energy will be large because that goes inside the inner shell.

52
Q

Photoemission spectroscopy

A

Shine a frequency on an atom and measure when an electron comes off.
Relative number of electrons vs energy
Less energy for outer electron shells.
2p less energy than 2s which is way less than 1s for nitrogen

53
Q

Pressure

A

Force per area

Measured in mmHg, ATM, torr, pascals

54
Q

Barometer

A

Glass tube with vacuum on end–> height mercury rises at ATM is 760 mmHg.

55
Q

Why use mercury?

A

Because density – only rises 760 mm at ATM. Also a liquid at room temperature

56
Q

Manometer

A

Also used to measure pressure but with a gas container on one end and open on the other end.
Pressure of the gas = Patm - h for gas that is less and +h for when it is more.

57
Q

Pressure and volume relationship

A

P is proportional to 1/V
P = k/V

Pressure is inversely proportional to the volume.
P on Y axis and V on x axis produces 1/X curve.
V vs 1/P is linear.
Slope is k = PV. Constant PV
So, P1V1=P2V2

58
Q

Boyles Law

A

If moles and temperature are constant, then P1V1=P2V2

Only true for pressures around 1 ATM

59
Q

STP

A

0 Celsius, 273 kelvin, 1 ATM

60
Q

Volume and temperature

A

Directly proportional

61
Q

Charles Law

A

V= bT

V1/ T1 = V2/T2

62
Q

Absolute zero

A

0 kelvin. No kinetic energy, no volume.

63
Q

Avagadros Law

A

Gas at a constant t and constant p, volume directly proportional to number of moles of gas.
Molar volume of a gas at STP is always 22.42 L. Doesn’t matter the gas.
V=an

64
Q

Ideal gas Law

A

Combine avagadros, charles, Boyle to make

PV=nRT

65
Q

What will happen if water is boiled in a metal can and then the heat is turned off and the can is sealed?

A

Temperature and pressure down, so the can will be crushed as the water vapor molecules condense. Pressure from the atmosphere forces the pressure on the inside to collapse.

66
Q

Kinetic molecular theory

A
  1. Particles are so small compared with distances between them that the volume of the individual particles of a gas can be assumed to be zero.
  2. The particles are in constant motion. Collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas.
  3. Particles are assumed to exert no forces on each other. They are assumed neither to attract nor to repel each other.
  4. Average kinetic energy of a collection of gas particles is assumed to be directly proportional to the kelvin temperature of the gas.
67
Q

How does energy relate to temperature?

A

KE average = 3/2 RT

68
Q

Root mean square velocity

A

Urms = SQRT (average velocity squared) = SQRT (3RT/M)

M is the mass of a mole of gas in Kg
Use R = 8.314 J/molK

69
Q

Example:
What’s the root mean square velocity of helium at STP
molar mass is 4 grams / mol

A

SQRT (3RT/M) = SQRT (38.314273/.004) = 1305 meters per second

70
Q

Boltzmann distribution

A

Velocity is an average velocity
At a given temperature, the velocity varies
Higher temperature, higher average velocity and more spread out as well

71
Q

Diffusion

A

Describes mixing of gasses

72
Q

Effusion

A

The rate of a gas entering an evacuated chamber

73
Q

Grahams law of effusion

A

Relates the root mean square of gases when comparing rate of effusion between two gases

Rate of effusion for gas 1 / rate of effusion for gas 2 = SQRT molar mass of 2 / SQRT molar mass 1

E1/ E2 = SQRT M2 / SQRT M1

74
Q

Density relating to molar mass

A

Molar mass = dRT/P

Because n = m/molar mass
P= (m/molar mass)RT/V
P= dRT/ molar mass

75
Q

Dalton law of partial pressure

A

Total pressure of a mixture of gasses is a result of the sum of the individual pressures.
P total = P1 + P2 + …

76
Q

Example:
Say we have 46 L of He and 12 L of O2 both at 25 degrees Celsius and 1 ATM and they’re pumped into a 5 L tank also at 25 degrees Celsius
Find partial pressure of each gas and total pressure

A

Option 1. Calculate moles. Then use PV=nRT in new container to get total pressure and multiply P by mole ratio to get partial pressures.
Option 2. Use P1V1=P2V2. V2 is 5 for each gas.

77
Q

Mole fraction

A

X= moles of 1 / total moles

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