Download L1 – CHEMISTRY FINAL REVIEW

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L1 – CHEMISTRY FINAL REVIEW
Name
key
Chemical Reactions:
1. Predict the products of the following reactions and write the complete balanced equations:
a. ferric iodide + fluorine  Ferric fluoride + Iodine (Single Replacement)
2 FeI3 + 3F2  2 FeF3 + 3 I2
b. burning of C3H8O  Carbon dioxide + water
(combustion)
C3H8O + 4.5 O2  3 CO2 + 4 H2O
c. calcium + water  Hydrogen + Calcium hydroxide (special single replacement)
Ca + 2 H(OH)  H2 + Ca(OH)2
d. heating of calcium peroxide  calcium + oxygen (decomposition)
2 CaO  2 Ca + O2
e. sodium phosphate + lithium sulfite sodium sulfite (aq) + lithium phosphate ( )(dble rep)
*according to our rules, this would not work but in fact lithium phosphate is insoluble.
No worries, exam will not have exceptions
2 Na3PO4 + 3 Li2SO3  3 Na2SO3 + 2 Li3PO4
f. nitric acid + copper (II)  Hydrogen ( ) + Copper II nitrate (single replacement)
*according to our rules, this would not work but in nitric acid is very strong so it does
work. No worries, exam will not have exceptions
2 HNO3 + Cu  H2 + Cu(NO3)2
g. decomposition of magnesium chlorate Magnesium chloride + oxygen( )(special decomp)
2 Mg(ClO3)2  2 MgCl2 + 3 O2
2. Write the molecular and net ionic equations for the following reactions:
Identify the spectator ions.
a. strontium hydroxide + hydrofluoric acid  strontium fluoride + water
molecular Sr(OH)2 + 2 HF  SrF2 + 2 H(OH)
net ionic
(OH)-
+ H+  H(OH)
spectator ions are Sr+2 and F-
b. magnesium chloride + silver nitrate  magnesium nitrate + silver chloride ( )
molecular MgCl2 + 2 AgNO3  Mg(NO3)2 + 2AgCl
net Ionic
Ag+ + Cl-  AgCl
spectator ions are Mg+2 and NO3-
2
3. Define the following and indicate units where applicable:
a. density –
b. molarity (M) –
c. molality (m) d. gfm or molecular mass –
e. mole ratio –
f. entropy –
g. enthalpy –
h. Gibb’s free energyi.
spontaneous reaction –
j.
endothermic –
k. exothermic –
l.
STP –
m. oxidation –
n. reduction –
o. oxidizing agent –
p. reducing agent –
q. molecular equation –
r. total ionic equation –
s. net ionic equation –
t. spectator ions –
u. FP & BP depression -
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Reaction Dynamics
4. Write the equation for the formation of sodium oxide (s), carbon dioxide (g) and water
vapor from the decomposition of sodium bicarbonate (s) which requires 129 kJ of heat.
NaHCO3(s) + 129kJ  Na2O(s) + CO2(g) + H2O
a. Is this reaction exothermic or endothermic?
endo
b. Is the enthalpy change (
H) of this reaction positive or negative?
a. Is the energy change favorable or unfavorable?
unfavorable
d. Is the entropy change in this reaction favorable or unfavorable?
How can you tell?
(+)
favorable
goes from orderly to disorderly (more products)
e. Would you expect this reaction to be spontaneous? Can’t predict
Why or why not?
f.
What conditions would have to be present in order for this reaction to be spontaneous?
favorable entropy is greader than unfavorable enthalpy
5. What values (positive or negative) are considered favorable for enthalpy and entropy?
H;
+
S
6. Under what conditions would reactions with the following enthalpy and entropy
values be spontaneous?
+
H, +
S
-
+ entropy>+enthalpy
H, -
S
+
-enthalpy>-entropy
H, -
S
-
none
H, +
S
already spontaneous
7. Predict whether the following reaction would be spontaneous. Explain your reasoning.
4NH3 (g) + 7 O2(g) + heat   4 NO2 (g) + 6 H2O (g)
Would not be spontaneous because it is endothermic (+enthalpy) and reactants are
smaller and in greater quantity (more disordered than the products (more orderly)
8. Balance the following using the half-reaction method. Identify the oxidizing agent, the
reducing agent, the atom oxidized and the atom reduced.
LEO says GER
+2 -2
a.
MnO +
+4 -2
+1 +5 -2
+1 +7 -2
+2
PbO2 +
HNO3 
HMnO4 +
Pb(NO3)2 +
+5 -2
+1 -2
H2O
Mn is oxidized; it is the reducing agent. Pb is reduced; it is the oxidizing agent
Mn+2  Mn+7
Pb+4  Pb+2
+2
+7
+4
Mn - 5 e  Mn
Pb + 2e-  Pb+2
2(Mn+2 - 5 e-)  Mn+7
5(Pb+4 + 2e- )  Pb+2
+2
+7
2 Mn - 10 e  2 Mn
5 Pb+4 + 10e-  5Pb+2
2 MnO +
5 PbO2 + 10 HNO3 
2 HMnO4 +
5 Pb(NO3)2 + 4 H2O
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Stoichiometry
9..How many grams of ammonium sulfate will form from the reaction of 102 g of ammonium
phosphide with 156 g of calcium sulfate? (151 g) Identify the limiting reagent.
How much reactant in excess is left over?
2 (NH4)3P + 3 CaSO4  3 (NH4)2SO4 + Ca3P2
(85g)
102g((NH4)3P
1
(136g)
(132g)
x 1 mol (NH4)3P x 3 mol (NH4)2SO4 x 132g(NH4)2SO4 = 237.6g
85g((NH4)3P
2mol(NH4)3P
1mol(NH4)2SO4
156g CaSO4 X 1mol CaSO4 X 3 mol (NH4)2SO4 X 132g(NH4)2SO4 = 151g
1
136g CaSO4
3mol CaSO4
1mol(NH4)2SO4
Limiting Reagent = Calcium sulfate
ammonium sulfate produced = 151g
156g CaSO4 X 1mol CaSO4 X 2mol(NH4)3P X 85g((NH4)3P = 65gneeded 102-65=37g
1
136g CaSO4
3mol CaSO4
1mol(NH4)3P
excess
10. In the reaction, 2 C4H10 + 13 O2  5 H2O + 330 kJ , how many grams of oxygen are
neccessary to produce 500 kJ of heat?
(630 g)
500 kJ X 13mol O2 X 32g O2 = 630g
1
330kJ
1mol
Gas Laws
11. Graham’s Law of Effusion:
RateA =
RateB
12. Compare rate of diffusion of CO2 to NO3.
11. Graham’s Law of Effusion:
=
rate of effusion of a gas is inversely
proportional to the square root of its
formula mass
molar mass B
molar mass A
molar mass B
molar mass A
7.874 = (1.19 x fast)
6.633
13 Boyle’s Law P1 V1 = P2 V2 Charles’ Law V1 T2 = V2 T1 Gay-Lusaac’s P1 T2 = P2 T1
14. If the volume of a gas is doubled, how will the pressure vary at constant temp.? decrease2x
15 If the K temp. of a gas is halved, how will the volume vary at constant pressure? volume half
16 If the K temp. of a gas is tripled, how will the pressure vary at constant volume? pressure tripled
17 Combined Gas Law:P1 V1 T2 = P2 V2 T1 B. Ideal Gas Law: P V = n R T
(R = .0821 L atm/mol K)
18 Under what conditions do real gases NOT behave like ideal gases? extremely high
pressures and extremely low temperatures.
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19 1 atm = 760 mm Hg = 30
in. Hg = 14.7 psi = 101.3
kpa
20. K = °C + 273
21. To what temperature must Oxygen gas contained in a 4.7L vessel at 20 oC and 80 kpa
(210.45 oC)
be heated to exert a pressure of 990 mm Hg?
T1 = 20 + 273 = 293°C
P1 = 80kpa x 760mmHg = 600mmHg
101.3kpa
T2 = ?
P2 = 990 mmHg
P1 T2 = P2 T1
600(x) = 990 (293)
x = 290070/ 600 =483.45 K
= 483.45 – 273 = 210.45°C
22. How many grams of Nitrogen gas will occupy a 2.32 L container at 30oC and 66 inches Hg?
PV=nRT
(5.88g)
P= 66”Hg x 1 atm = 2.2 atm
30”Hg
2.2(2.32) = n (.0821)(303)
V = 2.32 L
n = 5.104/24.88
n = 0.205 mole
T = 30 + 273 = 303°C
.205mol N2 x 28g N2 = 5.74 g N2
1mol N2
R = .0821 atm L/mol K
23. If 175g of Aluminum reacts with sulfuric acid at 20oC, what volume of Hydrogen gas will be
produced at a pressure of 16.2 psi?
(21.21L)
2 Al + 3 H2SO4  3 H2 + Al2(SO4)3
P = 16.2psi x 1atm = 1.10atm
14.7psi
V = ? H2
n= 175g Al x 1mol Al x 3mol H2 = 525 = 9.72 mol H2
27g Al 2 mol Al
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V = nRT/P
R = .0821 atm L/mol K
V = 9.72(.0821)(293) / 1.10
T = 20 + 273 = 293 K
V = 212.56 L
24. A volume of 75 ml of Nitrogen is collected in a eudiometer by water displacement. The water
level inside the tube is 27.2 mm higher than that outside. The temperature is 25 oC and the
barometric pressure is 741 mm Hg. What is the volume of the oxygen at STP?
(64.66 ml)
a. correction for water / mercury difference:
27.2mm / 13.6 = 2 mm
b. correction for difference in levels:
741 mm – 2 mm = 739 mm
c. correction for water vapor: (use table)
739 mm – 23.8 mm = 715.2 mm
d. solve the equation for volume at STP
:
P1 V1 T2 = P2 V2 T1
V1 = 75 ml
V2 = ?
P1 = 712.2 mm
P2 = 760 mm
T1 = 298 K
T2 = 273 K
V2 = P1 V1 T2 = (715.2 ) 75) (273) = 64.66 ml
P 2 T1
(760) (298)
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25. What is the density of a gas with a mass of 28 g and a volume of 31 L? What is its MW?
d = m/v = 28/31 = 0.9g/L
(.9 g/L)
(20.23 g/mol)
MW = d ÷ 1mol = 0.9g x 22.4L = 20.16 g/mol
22.4L
L
1mol
States of Matter/Phase Changes
26. Define the following terms: surface tension, viscosity, volatility, miscibility, vapor pressure,
boiling. How does intermolecular bonding relate to these characteristics?
Characteristic
Surface tension
Viscosity
Volatility
Miscibility
Vapor Pressure
bond strength
effect
greater the bond strength
greater the surface tension
greater the bond strength
greater the viscosity
greater the bond strength
the lower the volatility
has to do with “likes dissolve likes” not bond strength
the bond strength
the lower the vapor pressure
27. Determine the following characteristics for each substance below: Type of Intermolecular
Bonding, Mpt/Bpt, Phase, and Bond Strength.
a. NH3 – Hydrogen Bond (H bound to something very electronegative; O, F or N)
Strongest of the intermolecular bonds so NH3 will have the highest
MP/BP out of the three compounds.
b. SiH4 – Non- polar compound; therefore has London Dispersion Forces
(attractions)or known as indip-indip or temporary dipoles; the weakest of
the intermolecular bonds so SiH4 will have the lowest MP/BP out of the
three cmpds.
c. HBr- Polar molecule; dipole-dipole interactions between the molecules;
medium strength interactions.
28. What are some major differences between the following types of solids? In terms of Mpt/Bpt,
solubility, and conductivity. Metallic, Ionic, Polar Molecular, and Nonpolar Molecular. Give an
example of each.
Metallic bonding; Ex; Copper(s); Iron (s)-Very conductive; insoluble high MP/BP. Outer
electrons are free to “roam around”
Ionic – Ex; NaCl; soluble in polar and ionic solvents. High MP/BP; conductive as solutions
or molten
Polar Molecules- Ex; water; HBr; soluble in polar and ionic solvents weak conductors;
MP/BP depends on what type of
compounds you are comparing them with. With other molecular comps.
High, but compared to other types of cmpds. low.
Non-Polar Mc’s- Cl2; CH4; nonconductors and low MP/BP. Will dissolve in non-polar
Solvents only.
Network Covalent- Ex; graphite, diamond and quartz; they are big structures with very
strong covalent bond therefore they have high MP/BP and are insoluble
in water do not conduct
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29. Draw a phase diagram.
a. Label the following parts: solid, liquid, gas, triple point, critical point, normal melting
point, any sublimation point, the liquid/gas equilibrium line
Critical Point
Liquid
Liquid/gas equilibrium line
Solid
Gas
30. Draw a cooling curve for H2O.
Start with a temperature of 125oC and end with a temperature of -13oC.
a. Label the axes of the graph.
b. Label the beginning and ending temperatures, as well as the temperatures of any phase
changes.
c. Label all phase changes that occur throughout the process.
d. In parentheses, label all of the reverse phase changes (if heat was added instead of
removed).
e. Label all segments where the average kinetic energy of the substance is changing.
KE
100◦C
KE
0◦C
-13◦C
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Water and Solutions:
31. Draw a water molecule and indicate what makes it polar.
32. Describe the Van Der Waals forces between water molecules called dipole interactions
or Hydrogen bonds.
H-bond is a strong intermolecular bond between the slightly positive Hydrogen end
of one water molecule and the slightly neg. oxygen end of an adjacent water molecule.
33. Name 4 unique properties of water due to Hydrogen bonding. high surface tension;
capillary action; high specific heat (it takes a lot of energy to raise the temperature of
water); only substance in which the solid state is less dense than the liquid state; unusually
high BP when compared to other molecular compounds with similar molecular weight.
34. How is a saturation point an equilibrium point?
the rate at which the solute is dissolving
is equal to the rate at which the solution is recrystallizing.
35. Heating of ammonium chloride in solution lowers its solubility.
Write this thermochemical equation.
NH4Cl (s) + H2O  NH4Cl(aq) + Heat
Is this net endo or exo? exothermic
36. Heating potassium chloride makes it dissolve more.
On a solubility graph its curve would be __upsweeping
It would make a solution colder when it dissolves because
it has a net endothermic dissolving process.
37. What is the molality of a solution containing 1.70g of sodium nitrate in 162.6 g of water?
1.70gNaNO3 X 1moleNaNO3 = 0.02mole NaNO3
(.123m)
85g
.02mol NaNO3 X 1000gH2O = 0.123m
162.6g H2O
1kgH2O
38. What is the boiling point of a solution of 4.45 g of calcium chloride dissolved in 50 ml of water?
(kb = .512oC/m)
(101.23 oC)
BP = Constant (molality)
molality = moles solute/kg solution
CaCl2 breaks up into 3 particles when dissolved
4.45gCaCl2 X 1mole CaCl2 = 0.04moles CaCl2 X 1000gH2O = 0.80m X 3 = 2.4m total
111g CaCl2
50gH2O
1 kg H2O
2.4m (.512) = 1.23 + 100°C = 101.23°C