Exam Review key

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Spring 2015 Honors Chemistry Exam Review
How are diffusion and effusion similar? Different?
Similar – result = motion from high concentration to low concentration – caused by random motion of
particles.
Different – effusion is through small holes (lid not tight, goretex)
List the five principles of kinetic gas theory.
1) Gas particles are small and far apart
2) In constant random motion
3) All collisions are elastic
4) There are not attractions between particles
5) Kinetic Energy depends on the temperature
How is a real gas different from an ideal gas?
Real gases have attractions
A balloon is 2.5 L at 1.2 atm and will break at 5.7L. At what pressure will it break?
(2.5L)(1.2atm) = (5.7L)P
P=.53 atm
A container starts at .9 atm and 20C. What is the pressure at -30C?
(.9atm)/293K = P/243K
P=.75 atm
A balloon is 2.5L at 20C. What will its volume be at 75C?
(2.5L)/293K = V/348K
V=2.97 L
A balloon is 1.7L at 15C and 1.1 atm. What is the volume at STP?
(1.7L)(1.1atm)/288K = (1atm)V/273K
V=1.77 L
A balloon has 2.2 moles of helium at STP. What is its volume?
(1atm)(V)=(2.2mol)(.0821)(273K)
V=49.31 L
There is a 57.8g sample of oxygen. What is its volume at STP?
57.8g O2(1 mol O2/32.00g O2) = 1.81 mol O2
(1atm)(V) = (1.81mol)(.0821)(273K)
V=40.57 L
How many moles of hydrogen are in a sample 41L at 39C and 2.3 atm?
(2.3atm)(41L) = mol(.0821)(312K)
Mol = 3.68 mol
Partial Pressures & examples
Partial pressures are parts of the total pressure.
Since N2 is 78% of the atmosphere, it is 78% of total pressure. If we are at 1.5 atm, PN2 = 1.17 atm
Phase diagram – effects of temp and pressure on phase
See page 381. As we increase pressure, we can boil water at a higher temp – or water boils at a lower
temp if the pressure drops (elevation)
Energy changes associated with heating and phase changes
Heating is increasing temp, therefore increasing KE – no change to PE
Phase change is a change in PE – no change in KE
Define Molar, Molal, and Normal.
Molar = moles/liter of solution
Molal = moles/kg of solvent
Normal = moles H+/OH- /liter of solution
What is the molarity and normality of 78 g of H3PO4 dissolved in 2.3 L of solution?
78 g H3PO4 (1 mol H3PO4 /98 g H3PO4) = .80 g H3PO4
.80 g H3PO4 /2.3L = .35 M H3PO4
.35 M H3PO4 * 3H/ H3PO4 = 1.05 N H3PO4
What is the molality of 56 g of NaCl in 786 mL of water?
56 g NaCl(1 mol NaCl/58.44 g NaCl) = .96 mol NaCl
.96 mol NaCl/.786L = 1.22 m NaCl
If 30 mL of 5 M HCL is mixed with 70 mL of water, what is the new concentration?
(5M)(30mL)=C(100mL)
C=1.5 M
Define dilute, concentrate, saturated, and super saturated, electrolyte, non-electrolyte
Dilute – not much dissolved
Concentrated – a lot dissolved
Saturated – as much dissolved as possible at the current temperature
Super saturated – more dissolved than would normally be possible at current temperature
Define solute, solvent, solution
Solute – what is dissolved
Solvent – what does the dissolving
solution – mixture of solvent and solute
Compare solutions, colloids, and suspensions
Solutions – everything in same phase
Colloids – two different phases, but does not separate (settle)
Suspensions – two different phases, but does separate
Explain “like dissolves like”
Only things with similar charge can dissolve each other (polar dissolves polar & ionic)
What factors impact rate of solution
Heat
Agitation
Surface Area
Define heat of solution
Change in energy as a mole of solute is dissolved in a particular solvent
Calculate heat of solution – a 200 mL sample of water goes from 19.1 C to 22.3 C when 46 g KCl is dissolved.
(4.18J/1g1C)(200g)(3.2C) = 2675.2J
46 g KCl (1 mol KCl/74.55 g KCl) = .62 mol KCl
-2.6752 kJ/.62 mol
-4.31 kJ/mol KCl
Contrast hydrate and anhydrous
Hydrate has water attached to the compound
Anhydrous are dried out hydrates
Explain impacts of heat and pressure on gases and solids in solutions
Heat decreases the solubility of gases and increases the solubility of solids
Pressure increases the solubility of gases and has no impact on the solubility of solids
How many moles of NaCl are in 500 ml of 2 M NaCl?
(2 mol NaCl/L)(.5L) = 1 mol NaCl
How many grams of Fe(NO3)3 are needed to make 3 L of 4 M Fe(NO3)3?
(4 mol Fe(NO3)3/L)(3L)(241.88 g Fe(NO3)3/mol Fe(NO3)3) = 2902.56 g Fe(NO3)3
Explain how to make 3.7 L of 6 M HCl.
(6 mol HCl/1 L)(3.7L)(36.45g HCl/1 mol HCl) = 809.41 g HCl
1) Measure about 2 L water
2) Add 809.41 g HCl and dissolve
3) Add water until 3.7 L
What is the new concentration if 200 ml water is added to 100 ml of 6M HCl?
(6M)(100 mL) = C(300 mL)
C=2M
How many mL of 3 M HCl are needed to react with 75 mL of 3 M Mg(OH)2?
2HCl + Mg(OH)2-> MgCl2 + 2H2O
(3 mol Mg(OH)2/1 L Mg(OH)2)(.075 L Mg(OH)2)(2 mol HCl/1 mol Mg(OH)2)(1 L HCl/3 mol HCl) = .15 L
HCl
How many moles of particles (ions) will be formed when 3 moles of FeCl3 dissociates?
FeCl3 –-> Fe+3 + 3Cl-, so 4 particles (ions)
3 mol FeCl3 * 4 ions = 12 mol ions
Explain how something can be very soluble, but a weak electrolyte.
Acetic acid is a weak acid (does not dissociate well), but does dissolve well.
Electrolytes are made by ionic compounds dissociating, so acetic acid would be a weak electrolyte even
though it can dissolve well.
If 75 g C6H12O6 is added to 300 mL of water, what will the new freezing point be? (Kf = -1.86C/m)
75 g C6H12O6 (1 mol C6H12O6 /180.18 g C6H12O6) = .42 mol C6H12O6
.42 mol C6H12O6 /.3Kg water = 1.4 m C6H12O6
(1.4 m C6H12O6)(-1.86 C/m) = -2.6 C
If 225 g Na2SO4 is added to 400 mL of water, what will the new freezing point be? (Kf = -1.86C/m)
225 g Na2SO4 (1 mol Na2SO4 /142.05 g Na2SO4) =1.58 mol Na2SO4
1.58 mol Na2SO4 /.4 kg water = 3.95 m Na2SO4
(3.95 m Na2SO4)(3 particles)(-1.86C/m) = -22.04C
What is the concentration of Fe(OH)2 if 40 ml of 0.1 M HCl was needed to react with 20 ml of Fe(OH)2?
2HCl + Fe(OH)2 -> FeCl2 + 2H2O
(.1 mol HCl/LHCl)(.04 L HCl)(1 mol Fe(OH)2/2 mol HCl)/.02L Fe(OH)2
.1 M Fe(OH)2
What is specific heat? What is the specific value for water?
Specific heat is the energy needed to raise 1 gram of material 1C
4.18J/1g1C
What is the specific heat of a 70 g metal block if the block starts at 90 C and is put into a cup of 100 ml water
and the temperature goes from 17.2 C to 21.9 C.
(4.18J/1g1C)(100g)(4.7C) = 1964.6J (water)
-1964.6J/70g/-68.1) = .41J/1g1C
Calculate the Heat of Formation – problem #3 page 524.
S + 3/2 O2 -> SO3
-395.2 kJ
SO3
-> SO2 + ½ O2 +99.1 kJ
S + O2 -> SO2
-296.1 kJ
Define enthalpy, entropy, and free energy.
Enthalpy – total heat energy in a sample
Entropy – degree of randomness or chaos in a sample
Free energy – the result of both enthalpy and entropy
What is the relationship between enthalpy, entropy, and free energy?
ΔG = ΔH – TΔS
How are activation energy and activated complex related?
The activation energy is used to make the activated complex
What two things can happen to an activated complex?
It can break back into the reactants or it can break in the direction of the products
Explain the Collision Theory.
Reactants must collide with enough energy and the right orientation in order to react
Name the four factors of reaction rate we can control and explain how/why they impact reaction rate.
Heat – increases collisions and provides some of the activation energy needed
Concentration – more concentration = more particles = more collisions
Surface Area – reactions happen on the surface, so more surface area = more places to react
Catalyst – reduces activation energy needed
Draw energy diagrams for both an exothermic and endothermic reactions. Label E, E’, Ea, Ea’, and where the
activated complex would exist.
Write a possible reaction mechanism for 2H2 + O2 -> 2H2O
H-H + H-H + O=O
1) H-H + H-H-O=O
2) H-H + H + H-O=O
3) H-H + H-O=O-H
4) H-H + H-O + O-H
5) H-O-H-H + O-H
6) H-O-H + H + O-H
7) H-O-H + H-O-H
What is the Arrhenius definition of an acid and a base?
Acid – creates H+ ion, Base – creates OH- ion
What is the Bronsted/Lowery definition of an acid and a base?
Acid – Proton donor, Base – Proton receptor
What is the pH of .5 M H2SO4?
-log[1], pH = 0
What is the pH of .007 M Fe(OH)3?
-log[.021], pOH = 1.67, pH = 12.33
Label the acid/base pairs in the following reaction
H3O+ + NH3 -> NH4+ + H2O
Acid
Base Acid
Base
Name the following and give a pH range
NaOH
Sodium
Hydroxide
>7
HBr
Hydrobromic
Acid
<7
Mg(OH)2
Magnesium
Hydroxide
>7
H3PO4
Phosphoric
Acid
<7
Hydrosulfuric Acid
H2 S
<7
Aluminum Hydroxide
Al(OH)3
>7
Write formulas for the following and give a pH range
Sulfurous Acid
H2SO3
<7
Cesium Hydroxide
CsOH
>7
How does limestone help prevent the acidification of lakes?
Limestone (CaCO3) reacts with acid creating a salt, water and carbon dioxide, neutralizing the acid’
Name, explain, and give an example of each of the four reaction types
Synthesis – two or more reactants make one product
2Ca + O2 -> 2CaO
Decomposition – a reactant breaks down into two or more smaller parts
CuCO3 -> CuO + CO2
Single replacement – one reactant takes the place of a part of the other reactant
FeCl3 + Al -> AlCl3 + Fe
Double replacement – part of each reactant trade places with the other
AgNO3 + NaCl -> AgCl + NaNO3
Label the following reactions as one of the following. Then balance the equation.
Mg + 2H2O -> Mg(OH) 2 + H2
Single replacement
2FeCl3 + 3Mg(OH) 2 -> 2Fe(OH) 3 + 3MgCl2
Double replacement
H2SO4 + 8HI -> H2S + 4I2 + 4H2O
Other
H2O2 -> H2 + O2
Decomposition
Write a word equation for the following equations
2H2 + O2 -> 2H2O
Hydrogen and oxygen produce water
NaHCO3 + HCl -> NaCl + CO2 + H2O
Sodium bicarbonate and hydrochloric acid produce sodium chloride, carbon dioxide, and water
Use the activity series (pg. 266) to determine if these reactions happen, if there is something needed, and what
the products would be
Ca + H2O ->Ca(OH)2 + H2
Cr + H2O -> Cr(OH)2 + H2
Steam
Cu + H2O -> no rxn
Na + FeCl3 -> NaCl + Fe
Cu + FeCl3 -> no rxn
Using heat to decompose CuCO3 into CuO and CO2
heat
CuCO3 ---------------------- CuO + CO2
heat
For the following reactions – 1) identify the type of reaction, 2) balance the reaction, 3) determine phases of
reactants and product, then write the 4) ionic and 5) net ionic equations.
2(NH4)3PO4 + 3FeCl2 -> 6 NH4Cl + Fe3(PO4)2
2(NH4)3PO4(aq) + 3FeCl2(aq) -> 6NH4Cl(aq) + Fe3(PO4)2(s)
6(NH4)+ + 2PO4-3 + 3Fe+2 + 6Cl- -> 6(NH4)+ + 6Cl- + Fe3(PO4)2
3Fe+2 + 2PO4-3 -> Fe3(PO4)2
3Zn + 2H3PO4(aq) -> Zn3(PO4)2(s) + 3H2(g)
3Zn + 6H+ + 2PO4-3 -> Zn3(PO4)2(s) + 3H2(g)
3Zn + 6H+ + 2PO4-3 -> Zn3(PO4)2(s) + 3H2(g)
If 78 g NaHCO3 is used in the reaction of NaHCO3 + HCl -> NaCl + CO2 + H2O, how much of each product
will be made?
78 g NaHCO3 (1 mol NaHCO3 /84.01 g NaHCO3)(1 mol NaCl/1 mol NaHCO3)(58.44 g NaCl/1 mol NaCl)
= g 54.26NaCl
78 g NaHCO3 (1 mol NaHCO3 /84.01 g NaHCO3)(1 mol CO2 /1 mol NaHCO3)(48.01 g CO2 /1 mol CO2) =
40.86 g CO2
78 g NaHCO3 (1 mol NaHCO3 /84.01 g NaHCO3)(1 mol H2O /1 mol NaHCO3)(18.02 g H2O /1 mol H2O) =
16.73 g H2O
If 18 g of Li reacts with 500 g water, how much LiOH and H2 will be produced?
_____ g LiOH
_____ L H2
2Li + 2 H2O -> 2LiOH + H2
18 g Li (1 mol Li/6.94 g Li)(2 mol LiOH/2 mol Li)(23.96 g LiOH/1 mol LiOH) = 62.14 g LiOH
18 g Li (1 mol Li/6.94 g Li)(1 mol H2/2 mol Li)(22.4 L H2/1 mol H2) = 29.05 L H2
If 102 g Fe and 57 L Cl2 are used, which is the limiting reagent for the following reaction?
2Fe + 3 Cl2 -> FeCl3
102 g Fe(1 mol Fe/55.85 g Fe)(2 mol FeCl3/2 mol Fe)(162.2 g FeCl3/1 mol FeCl3) = 296.23 g FeCl3
57 L Cl2 (1 mol Cl2 /22.4 L Cl2)(2 mol FeCl3/3 mol Cl2)(162.2 g FeCl3/1 mol FeCl3) = 275.16 g FeCl3
What is the percent yield if 30 g Mg makes 42 g MgO?
2Mg + O2 -> 2MgO
30 g Mg(1 mol Mg/24.31 g Mg)(2 mol MgO/2 mol Mg)(40.31 g MgO/1 mol MgO) = 49.74g MgO
% yield = 42 g Mg/49.74 g MgO
% yield = 84.44%
Complete the following organics table
Pentanoic acid
3 octanol
Propanone
Pentanal
Pentanal or
Some Penatanone
1,4 dibromo hexane
2,4 nonadiol
3 heptanone
5bromo
3,3dichloro octane
2 decyl alcohol
C4H7COOH
C8H17OH
C3H6O
C5H10O
C5H10O
C6H12Br2
C9H18(OH)2
C7H14O
C8H15BrCl2
CH3CH2CH2CH2COOH
CH3CH2CHOHCH2CH2CH2CH2CH3
CH3COCH3
CH3CH2CH2CH2CH2CH2CH3
CH3CH2CH2CH2 CHO
CH3CH2COHCH2CH3
CH3BrCH2CH2CHB2CH2CH3
CH3CHOHCH2CHOHCH2CH2CH2CH2CH2
CH3CH2COHCH2CH2CH2CH3
CH3CHCl2CH2CH2CBrHCH2CH2CH3
C10H21OH
CH3CHOHCH2CH2CH2CH2CH2CH2CH2CH3
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