Here are some
of the answers to the homework.
The idea is to do the problem then check your answer. If you get it incorrect;
go back and attempt to find the mistake. It is better for your understanding
if you can find your own mistakes and learn from them.
Also, for school year HW (not summer work) see the HW Hints page if you
are stuck and need some help. But only do this AFTER you have seriously
attempted the problem for yourself without the hints.
----update 12/9/25----
-------------------------------------------------------------------------------------------------------------------------------------
SUMMER Practice Problems
Chapter 1
#1a) 4.0 ml with 2 sig. figs. and uncertainty +/- .1ml
#1b) 10.25 ml with 4 sig. figs. and uncertainty +/- .01ml
#1c) 15 ml with 2 sig. figs. and uncertainty +/- 1ml
#2) a) -2.22 % error b) 2.23 % error
#3) a)2 b)4 c)4 d)4 e)1 f)3 g)4 h)7 i)3
#4) a) 3.1x10(-4) b) 3.1x10(7) c) 3.6x10(4) d) 4.0x10(-6)
#5) a) 5x10(2) b) 4.8x10(2) c) 4.80x10(2)
#6) a) 102.623 b) 236.2 c) 1.9 d) 33.04
#8) Same . . .
#9) a) 351.1 K b) 248 K c) 0 K d) 1074 K
#10) a) 20.5 g water b) 41.8 g sugar
#11) 3.8 g/ml
Chapter 2
#1) look it up
#2) look it up
#3) a) 22p+, 20n b) 30p+, 34n c) 62p+, 14n d) 36p+, 50n e) 33p+, 42n f)
19p+, 22n
#4a) O with 8 on bottom left and 17 on top left
#4b) Co with 27 on bottom left and 60 on top left
#4c) Fe with 26 on bottom left and 57 on top left
#4d) I with 53 on bottom left and 131 on top left
#5) 207.2 amu and Lead
#6) look them up
#7) Sn and +2
#8) a) Te b) Rb) c) Ar d) At
#9) a) lose, Na+ b) lose, Sr+2 c) gain ,P-3 d) lose, Ba+2 e) gain, I-
f) gain, O-2 g) lose, Al+3 h) gain, S-2
#10) a) 26p+, 24e-, FeO b) 26p+, 23e-, Fe2O3 c) 56p+, 54e-, BaO d) 55p+,
54e-, Cs2O e) 16p+, 18e-, Al2S3 f) 15p+, 18e-, AlP g) 35p+, 36e-, AlBr3
h) 7p+, 10e-, AlN
#11) Ra and 142 neutrons
Writing Formulas and Naming Compounds (I put these in
order)
#1) sodium chloride, rubidium oxide, iron (III) bromide, chromium (III)
oxide, calcium bromide, cesium fluoride, calcium sulfide, aluminum iodide,
aluminum oxide, zinc (II) chloride, lithium nitride, silver (I) sulfide,
potassium chlorate, aluminum sulfate, barium sulfite, potassium permanganate,
strontium phosphide, calcium phosphate, lead (II) nitrate, barium chloride
dihyrate, potassium dichromate
#2) nitrogen triiodide, phosphorus trichloride, sulfur dioxide, iodine
trichloride, sulfur difluoride, dinitrogen tetrafluoride, phosphorus pentasulfide,
dinitrogen tetroxide
#3) hydrochloric acid, hydriodic acid, nitric acid, phosphoric acid, sulfuric
acid
#4) mercury (II) oxide, copper (I) iodide, copper (II) iodide, cobalt
(II) iodide, sodium carbonate, sodium hydrogen carbonate (or sodium bicarbonate),
acetic acid, ammonium nitrate, cobalt (III) sulfide, iodine monochloride,
lead (II) phosphate, potassium iodate, calcium hydroxide, cobalt (II)
sulfide, trisulfur tetranitride, sulfur hexafluoride, sodium thiocyanate,
barium chromate, ammonium hydrogen phosphate, ammonium dihydrogen phosphate,
strontium nitride, aluminum sulfite, tin (IV) oxide, sodium chromate,
nitrogen monoxide
#5) CsBr, BaSO4, ClF3, NH4Cl, BeO, ClO,
MgF2, SF2, SF6, NaH2PO4,
SiCl4, Li3N, Cr2(CO3)3,
SnF2, NH4CH3COO, NH4HSO4,
Co(NO3)3, Cu2S, KClO3, Li2C4H4O6
#6) Na2O, Na2O2, KCN, Cu(NO3)2,
SiF4, PbS, PbS2, Pb(SO4)2,
CuCl, ZnS, (NH4)2HPO4, HBr, HClO4,
SiO2, Na2SO4, Fe(NO3)3 •
5H2O
Chapter 3
#1) they should be balanced with lowest whole numbers, you can tell when
they are balanced because . . . they are balanced
#2a) (NH4)2CO3(s) --> 2NH3(g) + CO2(g) + H2O(g)
#2b) Zn(s) + 2HCl(aq) --> H2(g) + ZnCl2(aq)
#2c) N2(g) + 3H2(g) --> 2NH3(g)
#2d) 2H2O2(aq) --> 2H2O(l) + O2(g)
#2e) NaCl(aq) + AgNO3(aq) --> AgCl(s) + NaNO3(aq)
#2f) CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)
#2g) HCl(aq) + NaOH(aq) --> HOH(l) + NaCl(aq)
#2h) C6H12O6 + 6O2(g) --> 6CO2(g) + 6H2O(g)
#2b (the ones about where have you seen the reactions)
rxn b) 1st year lab #5
rxn e) possibly as a demo. and 1st year lab #2
rxn f) every time you use bunsen burner
rxn g) possibly as a demo. and 1st year lab #19
rxn h) generic reaction for respiration
#3a) 294.34 g/mol
#3b) 12.806 mol
#3c) 7.077x10(23) molecules
#3d) 2.41x10(23) atoms of O
#4) .0116 mol NH4+1
#5) 6.61 g
#6) 80.5 g
#7) 284.7 g/mol
#8a) CSCl2
#8b) C5H8O4NNa
#9) C2H6
#10) 49.95 % oxygen
#11a) 136 g FeCr2O7
#11b) 10.02 g O2
#11c) 8.828 g O2
#11d) 88.12 g Fe2O3
#11e) 89.76 % yield
#12a) 156 g Al
#12b) 197.9 g Na3AlO3
#12c) 3 mol NaOH
#12d) .06 mol H2
#13a) Cr2O3 will be left over
#13b) 15.1 g Cr2O3
#13c) 18.9 g Cr
#14) 51.8 g H2O
#15a) 18.02 g/mol and 44.01 g/mol
#15b) 2.93x10(23) molecules of CO2
#15c) 5.85x10(23) H atoms
#15d) Water is O with 2 unshaired pairs and 2 single bonds, each to H.
Carbon Dioxide is C with 2 double bonds, each to O, and each O with 2
unshared pairs
#15e) 5.18 g O
#15f) C3H6O2
#15g) 74.1 g/mol
#15h) C3H6O2
#15i) -235 kJ
#15j) -1450 kJ/mol
#15k) one is a carboxylic acid and the other is an ester
#15l) ester
*By the way, it's a good idea to redo the marathon problem a 2nd and even
a 3rd time. Good stuff.
Math Practice
#1) 82.1
#2) -107.2
#3) 1
#4) 2
#5) 27
#6) .044
#7) 1.8x10(-8)
#8) 4.70
#9) 1.70x10(-3)
#10) .034
#11) 1.2x10(-6)
(end of summer questions)
-------------------------------------------------------------------------------------------------------------------------------------
(Regular Homework from here on.)
Chapters 1 - 3
Page 23, 48, 71
Page 23
#49) 253 g acetic acid
#53a) 46.8 g NH4Br
#53b) 126 g H2O
#53c) 2.6 g NH4Br
Page 48
#31) a) Iodine Trichloride b) Dinitrogen Pentoxide c) Phosphorus Trihydride
(or Phosphine) d) Carbon Tetrabromide e) Sulfur Trioxide
#37) a) Potassium Dichromate b) Copper(II) Phosphate c) Barium Acetate
d) Aluminum Nitride e) Cobalt(II) Nitrate
#41) a) HNO2 b) Nickel(II) Iodate c) Au2S3 d) Sulfurous Acid e) NF3
Page 71
#37a) C5H8O4NNa
#41) C14H9Cl5
#45) 52.68 % H2O and 16
#63a) 2H2(g) + O2(g) --> 2H2O(g)
#63b) H2 is limiting
#63c) 4.15 mol H2O
#63d) 5.05 mol O2 excess
#75) 123 g glucose
#89) 3.657 g CaO and 2.973 g Ca3N2
Page 49, 69-70, 72
Page 49
#36a) Co(CH3COO)2 or Co(C2H3O2)2
#36b) BaO
#36c) Al2S3
#36d) KMnO4
#36e) NaHCO3
#39a) Hydrochloric Acid
#39b) Chloric Acid
#39c) Iron(III) Sulfite
#39d) Barium Nitrite
#39e) Sodium Hypochlorite
Page 69-70 Summary Problem (The answers are also in the book on page 70)
a) 47.9 amu
b) 1x10(13) Ti atoms
c) 1.04 g Ti
d) Ti2O3 and 143.74 g/mol
e) Ti + 2Br2 --> TiBr4
f) 148 g Br2
g) 69.0 g TiBr4 and 4.01 g Ti excess
h) 54.7 g TiBr4
i) CaTiO3
j) 14.2 kg mineral
Page 72
#44) CH4N and C2H8N2
#46) 51.172 % H2O and 3.825 g MgSO4
#64a) 2Al + 3S --> Al2S3
#64b) Al is limiting
#64c) 0.590 mol Al2S3
#64d) 0.48 mol S excess
#72a) about 0.75 g
#72b) Fe
#72c) O2
#72d) 3 g Fe
#72e) Fe2O3
#78) 2AB3 + 3C --> 2A + 3B2C
#81a) X+ 3Y --> XY3
#81b) 5 mol X and 10 mol Y
#81c) 3 mol product and 2 mol X and 1 mol Y
--------------------------------------------------------------------------
Chapter 6 & 7
Review of Electron Config. / Periodic Properties
WS
#1a) 1s2, 2s2, 2p4
#1b) 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p5
#1c) 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d2
#2a) [Ne] 3s2, 3p4
#3a) 2s2, 2p3 and N with 1 pair of e- and 3 single e-
#4a) 4
#4b) 7
#4c) 1
#5a) P3-
#5b) O2-
#5c) K1+
#16a) 15
#16b) -3
#16c) opposite spins
#17a) 1s2, 2s2, 2p6, 3s2, 3p6
#17b) 1s2, 2s2, 2p6
#19) 18
#20) 92
#21) Ga
#22) Br- or Se2- or As3- or Rb+ or Sr2+
Page 157
#1a) 7.09x10(14) / s or 7.09x10(14) Hertz
#1b) 4.70x10(-19) J
#1c) 283 kJ / mol
#27a) 3,0,0,-1/2 and OK
#27b) 2,2,1,-1/2 and No the 2nd level can't have "d" electrons
#31a) [Ne] 3s2, 3p3
#31b) [Ar] 4s2, 3d10, 4p3
#31c) [Kr] 5s2, 4d10, 5p2
#31d) [Kr] 5s2, 4d2
#31e) [Ne] 3s2, 3p1
#37a) ground
#37b) impossible (no 1p1)
#37c) excited
#37d) excited
#37e) excited
#41a) Ni
#49d) Sn2+ is 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10
and Sn4+ is 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 4d10
WS 32-1
#1b) attraction
#1c) greater . . . less . . . 1/4 as much
#2a) doubled
#2b) quadrupled
#2c) quadrupled
#2d) 16
PES worksheet
#1) 1s2, 2s2, 2p4 Oxygen and other one is 1s2, 2s2, 2p1 Boron
#2) a) Na b) 4th c) 1st d) closer to nucleus so stronger attraction and
less electron shielding
#3a) 1s, 2s, 2p, 3s, 3p
#3b) Argon has a greater nuclear charge (aka more protons)
#4) harder, greater nuclear charge
#5) easier, electron repulsion
Page 159
#53a) Sr, In, Te
#53b) Te, In, Sr
#53c) Sr, In, Te
#57a) K
#57b) O2-
#57c) Tl
#57d) Cu+
#67a)At
#67b)Ar
#67c)Be
#67d)O
#67e)Ne(and Li and Be)
#67f)V
#67g)Mn2+
#67h)Zn
#71a) True
#71b) True
#71b expansion) n4 to n2 energy per mol e- is -246 kJ/mol e- and n5 to
n2 energy is -275.5 kJ/mol e-
n4 to n2 energy per e- is -4.085x10(-19) J/e- and n5 to n2 energy is -4.575x10(-19)
J/e-
n4 to n2 frequency is 6.165x10(14) /s and n5 to n2 frequency is 6.905x10(14)
/s
n4 to n2 wavelength is 4.87x10(-7)m or 487nm and n5 to n2 wavelength is
4.34x10(-7)m or 434 nm
#71c) False, it has 14 electrons
#71d) False, group 13 has 1 unpaired electron
Page 190
#1a) C in middle, 4 single bonds to each Cl (with 3 pairs of e-)
#1b) N in middle (with 1 pair of e-), 3 single bonds to each Cl (with
3 pair of e-)
#1c) C in middle, 1 double bond to O (with 2 pairs of e-), 2 single bonds
to each Cl (with 3 pairs of e-)
#1d) S in middle (with 1 pair of e-), 3 single bonds to each O
(with 3 pairs of e-) AND entire thing in [ ] with -2 charge
#5a) O in middle (with 2 pairs of e-), 2 single bonds to each Cl (with
3 pairs of e-)
#5b) Pin middle (with 1 pair of e-), 3 single bonds to each F(with 3 pair
of e-)
#9a) same as #1c
#9b) single bond between two C, off of one C 3 single bonds to each H,
off of other C triple bond to N (with 1 pair of e-)
#9c) double bond between C's, each C has two single bonds to a H
#19a) B in middle, 3 single bonds to each F (with 3 pairs of e-) *Notice the B has a deficient octet w/ only 6 v e-
#19b) in lecture notes
#19c) C (with a single e-), tripple bond to O (with 1 pair of e-) AND entire thing in [ ] with +1 charge
#19d) Clin middle (with 1 single e-), 3 single bonds to each O (with 3
pair of e-)
Page 192
#27b) 0
#28a) -1
#29) structure I is better
#49a) sp
#49b) sp3d
#49c) sp2
#49d) sp3d2
#51a) sp3d
#51b) sp3
#51c) sp3
#51d) sp
#55a) 5 e- pairs, sp3d
#55b) 6 e- pairs, sp3d2
#55c) 6 e- paris, sp3d2
Page 191
#31a) linear
#31b) linear
#31c) trigonal planar
#31d) square planar
#35a) octahedron
#35b) T-shaped
#35c) distorted tetrahedron (or see-saw is what the book calls it)
#35d) tetrahedron
#41) angle 1 is 120°, angle 2 is 107°, angle 3 is 120°
#43a) has a dipole, 'cuz it is polar
#43b and c) no dipole because it is an ion
#43d) no dipole, it is nonpolar
#45a) non polar
#45b) polar
#45c) polar
#45d) polar (but it is also an ion, so a weird example)
#63) 9 sigma and 1 pi
#65a) 3 sigma and 1 pi
#65b) 3 sigma and 0 pi
#65c) 1 sigma and 2 pi
#65d) 2 sigma and 2 pi
#70) below is the entire chart:
AX2E2 2 2 bent sp3 Polar
AX3 3 0 triangular planar sp2 Nonpolar
AX4E2 4 2 square planar sp3d2 Nonpolar
AX5 5 0 triangular bipyramidal sp3d Nonpolar
#72a) C in middle
#72b) N in middle
#74) PH3 and H2S have a bond angle less than 109.5° because unshared
pair of e- exert greater repulsion than e- pairs in bonds
Ch 6-7 WS #1
#1) 2 e-
#2) 32 e-
#3) 7 orbitals
#4) probability of where electrons are located
#5a) 4s2
#5b) 6s2
#6) you should have drawn two complete "p" orbitals (two peanuts)
#7) 3,0,0,+1/2
#8) -275,500 J
#9) -4.580x10(-22) kJ
#10) different spins
#11) 1s2, 2s2, 2p6, 3s2, 3p6, 3d5
#12a) Xe in middle with 3 unshared pairs, and a single bond to each F
with 3 unshared pairs
#12b) N in middle with a single unshared electron and with a single bond
to an O (with 3 unshared pairs) and a double bond to other O (with 2 unshared
pairs). And a second resonance structure.
#12c) S in middle with a single bond to each O and 3 unshared pairs around
each O, and brackets around the entire thing with -2 superscript
#12d) Se in middle with 1 double bond to one S with has 2 unshared pairs,
and a single bond to the other two S which have 3 unshared pairs, and
show "resonance" with the other two versions.
#13) Linear, Bent, Tetrahedral, Triangular Planar
#14) sp3d, I'm not sure, sp3, sp2
#15) No dipole, Dipole, Ion, No Dipole
#16) Sc+3 and Sc+1 both have lost electrons, Sc+3 lost 3 e- (4s2 then
3d1) forming a stable (Noble, Octet, full outer shell) e- config. which
is low in energy. While Sc+1 lost 1e- forming an unstable e- config. which
is high in energy.
#17) The 4s2 electron "promotes" to the 3d5 (please draw orbital
diagrams to show it). This results in a lower energy system 'cuz of Hund's
Rule (e- prefer their own orbital). You can also say that the "paired"
electrons in a 4s2 repel each other and thus one of them jumps to the
empty orbital in the 3d.
#18) VSEPR states that "attached" atoms orientate themselves
as far away as possible and especially far from unshared e- on the central
atom. The two unshared pairs of e- on the O push the H atoms closer together.
(Draw Lewis structure to show it) So, the bond angle is less than 180,
about 105, and happens to be called "Bent."
#19) Both have same nucleus (7 protons), so same pull on the electrons.
N-3 has 3 extra electrons that repel making it larger.
#20) Both end in 3s. Mg has 1 more p+ in nucleus, pulling on e-, making
it smaller. Also you can discuss Zeff (effective nuclear charge), Mg =
2, Na = 1. The higher Zeff means more pull on the outer e- from the nucleus.
#21) Both end in 4s. The 2nd I.E. is the removal of a 2nd e-. This removal
of 2nd e- in Ca results in a full shell, low energy. Removal of the 2nd
e- in K takes an e- from a full shell that is closer to the nucleas (3rd shell) so it takes much higher energy.
#22) 12 sigma and 3 pi
#23) The bonded orbitals overlap. (a drawing helps)
#24) Co
#25) as wavelength increases, frequency and energy decrease
#26) v = 4.76x10(14) Hz or /s, and E = 3.16x10(-19) J
(***Technically it should be - energy because it is exothermic, but that
isn't our concern right now***)
--------------------------------------------------------------------------
Chapter 5 & 9
Gas Law Review WS
#0) a. 423.0torr, b. 296K, c. 1.11atm, d. -100.�C
#1) 1.039 atm
#2) 260. °C
#3) 1.75 L O2
#4a) .223 mol H2
#4b) 8.94 g Ca
#5) 44.6 g / mol
#6) mass is 4.003 so He
#7) 1.70 g / L
#8) 19 °C
Page 125
#5) 0.1 mol O2 lost
#11) 2.89 atm
#17) 4.52 atm
#33a) .123 mol H+
#33b) .281 M HNO3
#37) P of CO2 = 630mmHg, NO = 36mmHg, SO2 = 5.7mmHg, H2O = 69mmHg
#39) .083 atm
#43a) 2C2H2 + 5O2 --> 4CO2 + 2H2O
#43b) 2.24 atm
#43c) P of CO2 = 1.49 atm, H2O = .747 atm
#61) O2 is limiting
#72) a. C, b. 1.00atm, c. 4.50atm
#72d) bulb A (7 particles), bulb B (7 particles), bulb C (4 particles),
pA+pB= 3.50atm, pA+pB+pC= 4.50atm
#72e) bulb A (6 particles), bulb B (6 particles), bulb C (6 particles),
pA+pB= 3.00atm, pA+pB+pC= 4.50atm
Page 127
#35) 14.4 L
#45a) lighter
#45b) 18.4 g/mol
#47a) N2O5 < Ar < NO < N2
#47b) N2 < NO < Ar < N2O5
#49) .0226 mols NO
#53a) CO
#53b) increase T and decrease P
#59) a. same, b. He, c. He, d. He
#65) a. True b. False c. False d. True e. True
#71a) Pressure of B is 2.0atm, and Pressure of A + B is 4.0atm
#71b) you should have drawn 4 ◻◯◻ and 4 ◯ (most
likely 2 of each in each flask)
#71c) Pressure of A is 1.0atm, and Pressure of B is 1.0atm, and Pressure
of A + B is 2.0atm
Page 253
#21) I2 > Br2 > Cl2 > F2
#31a) PCl3 and explain
#31b) AsH3 and explain
#31c) C2H5OCH3 and explain
#31d) HCl and explain
#33a) LDF
#33b) ionic bonds
#33c) LDF
#33d) LDF
#35a) cov. net.
#35b) cov. mol.
#35c) ionic
#39a) metallic
#39b) ionic
#39c) cov. mol.
#39d) cov. net.
#39e) ionic
#64a) A
#64b) A
#64c) about 33 °C
#64d) Gas
#64e) about 200 mmHg
Ch 5&9 WS #1
#1a) .00349 mol H2
#1b) 6.94x10(19) molecules H2O
#1c) 2.99 (which means the H2 is moving about 3 times faster than the
H2O)
#1d) H2O, we assume ideal gases don't take up any volume, real gases actually
take up some volume and H2O is bigger so it takes up more volume, also
we assume real gases travel in straight paths and don't affect each other
as they pass, real gases travel in curved paths due to attraction caused
by IMF's, water has more IMF's (LDF, DD, HB) while H2 only has LDF, so
more attraction between H2O molecules
#2) 1.52 g/L
#8) 15.0 L NO
#9) 215 mmHg
#10) Low pressure because the volume of each gas particle is less significant,
and High temperature because high temp. means faster movement so the particles
have less time to attract each other.
#11a) chloroethane has stronger IMF's (LDF and DD while butane has only
LDF) so higher B.P.
#11b) 1-propanol has stronger IMF's (LDF, DD, and HB while acetone has
only LDF and DD) so higher B.P.
#12a) 3C + 2.5O2 --> 2CO2 + 1CO
#12b) .09999 mol O2
#12c) 72.05 % C
Ch 5&9 WS #2
#1a) 8.56 atm
#1bi) .874 mol N2 / 1.749 mol total
#1bii) 4.0 atm
#1c) ratio would decrease, because N2 is less massive and would leak out
faster than O2
#1d) 2NO + O2 --> 2NO2
#1e) 1.29 atm
#2) 5.6 L O2
#3) 80.0 (Note: either put no units or technically you could say AMU for
"atomic mass units." We don't put g / mol unless it is called
"molar mass.")
#4a) CO is a triple bond and an unshared pair on each atom, CO2 is C in
middle with two double bonds, each to an O and each O has two unshared
pairs
#4b) both are linear
#4c) it should be a straight line with positive slope
#4di) same Kinetic Energy because same temperature
#4dii) speed of CO2 is less than speed of CO because CO2 is bigger so
at equal temps. it moves slower
#4diii) number of CO2 molecules is less than number of CO molecules because
the pressure of CO2 container is less (if you want to be more exact you
can say about 2 times less, and if you want to be even more exact you
can do PV=nRT to prove it)
#5) SO2 because it is the largest molecule (it takes up the most space)
and it has the strongest IMF's (LDF & DD)
#6) 308 s
#7) B
#8) most dense is liquid, and least dense is gas
#9) C
--------------------------------------------------------------------------
Chapter 8 & 17
deltaH, deltaS, deltaG Review (first year) worksheet
#1) a.) gas b) 10 liter c) 3 moles
#2) +
#3) sign gets flipped as well
#4) C4H10(g) + 6.5O2(g) --> 4CO2(g) + 5H2O(g) and deltaH = -875kJ
#5) -15.0 kJ
#6) 10.7 g O2
#7) Ca(s) + S(s) + 2O2(g) --> CaSO4(s) + 1433kJ
#8) reactants
#9) 2Fe(s) + 1.5O2(g) --> Fe2O3(s) and deltaH = -1118.4kJ
#10) -174 kJ/mol so YES
#11) negative enthalpies and positive entropies (think about why the universe
wants these and we'll discuss in class)
#12) negative
#13) -1.19 x10(5) J
#14) 139 kJ
#15) +2219.9 kJ
#16) -2056.6 kJ
Page 219 (& notes page 30)
Notes page 30) -154 kJ
#5a) MgSO4(s) --> Mg+2(aq) + SO4(-2)(aq)
#5b) Yes, exothermic
#5c) 1.51 kJ
#5d) 49 °C
#7a) 89.3 kJ
#7b) -89.3 kJ
#7c) -3.58x10(3) kJ
#17a) -159 kJ
#17b) 374 kJ
#19a) Sr + C + 1.5O2 --> SrCO3 dH= -1.220x10(3) kJ
#19b) -333 kJ
#25a) .173 kJ
#25b) 1.85 kJ
#25c) 2.02 kJ
#27) -306.9 kJ / mol(rxn)
#29) -226.7 kJ / mol(rxn)
Page 221
#9) 7.62 kJ / °C
#33a) -1675.7 kJ / mol(rxn)
#33b) -205.4 kJ
#37a) 58.9 kJ / mol(rxn)
#37b) 108.5 kJ / mol(rxn)
#37c) -179.9 kJ / mol(rxn)
#43a) 2ClF3(g) + 2NH3(g) --> N2(g) + Cl2(g) + 6HF(g) and dH°= -1196kJ
#43b)-169 kJ / mol(rxn)
#77) 76.0 % sucrose
Hot Stuff WS
#1 & 2) 8 KClO3 + C12H22O11 --> 8 KCl + 12 CO2 + 11 H2O
#3) You should get 0.367mol KClO3 and 0.0438mol C12H22O11, then
then sucrose produces less of every product, so sucrose is limiting.
#4) -5518.1 kJ/mol(rxn) (if using the chart in back of our lab
manual) or -5467.7 kJ/mol(rxn) (if using the chart in the book)
#5) -242kJ (lab manual) or -239kJ (book)
Page 468
#3) a)Yes b)No c)Yes
#4) a)Yes b)No c)No d)No
#7) a) negative b) negative
# 15a) -152.8 J/K mol(rxn) or -.1528 kJ/K mol(rxn) (*note, there is a typo in the reaction,
H2S should be a gas)
#15b) +193.6 J/K mol(rxn) or +.1936 kJ/K mol(rxn)
#17a) -368 kJ / mol(rxn)
#17b) +709 kJ / mol(rxn)
#21a) +159.2 kJ / mol(rxn)
#21b) -107.7 kJ / mol(rxn)
#25) (+231.2 kJ / 2 mol methanol or +115.6 kJ / mol methanol, it depends on how you balance the reaction)
so NO
#29a) +.461 kJ/K mol(rxn) and YES
#29b) +.288 kJ/mol K
#29c) -1260.0 kJ/ mol(rxn)
Page 469
#33a) an increase in Temp. improves favoribility
#33b) a decrease in Temp. improves favoribility
#33c) reaction will never be favorable
#35a) 4950 K and explain
#35b) 5.20x10(3) K and explain
#35c) -1240 K and explain
#49a) +30.7 kJ / mol(rxn) and NO
#49b) -55 kJ / mol(rxn) and YES
#53a) -141.8 kJ / mol(rxn)
#53b) 9.09x10(-26) atm
#55) -189.5 kJ / mol(rxn)
#58a) +201.3 kJ / mol(rxn), so Not feasable
#58b) -98.9 kJ / mol(rxn), so yes feasable
#59) 12 mols ATP
#65) -103 (2 sig. figs.) kJ / mol(rxn) and -88 kJ / mol(rxn)
#80a) as Temp. goes up, spontaneity also goes up
#80b) No, endothermic
#80c) Yes
#80d) about 330K
#80e) .02
Ch 8 & 17 Review WS
#1a) -dS because 3 moles gas turning into 2 moles gas and 3 moles solid
#1b) If T increases then dS in master equation becomes more significant and dG becomes less -
#1c) If dG becomes less -, the forward rxn is not as favorable,
so less products get made and more reactants remain. Since K is products
over reactants the value of K goes down.
#1d) Weird question. T = -145kJ / dS
#2) -dG means forward rxn is favored, so more product, Cu2S
#3a) -16.9 kJ / mol(rxn)
#3b) -17.7 J/K mol(rxn)
#3c) -11.6kJ / mol(rxn)
#3d) 108
#4) 621 K or 348 °C
#5a) CH3OH + 1.5O2 --> CO2 + 2H2O
#5b) -167.8 Kcal / mol(rxn)
#5c) 48.9 cal/mol K
#6) 1.00atm
#7) NH3
#8) Na(s)
#9) sort of correct answer -9.4 kJ / mol(rxn)
best answer (adds in mass of sodium hydroxide when calculating q) -10.
kJ / mol(rxn)
Page 223 #76
a) -851.5 kJ / mol(rxn)
b) 6600 °C
c) Yes
Ch 8 & 17 Practice Test
#1) -202.6 kJ / mol(rxn)
#2) C
#4) E
#5) B (and C)
#7a) -151 kJ / mol(rxn)
#7b) -115.1 kJ / mol(rxn)
#7c) -.12 kJ/K mol(rxn)
#8a) C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g)
#8b) 2.07 K or 2.07 °C
#9a) 9470 J or 9.47 kJ
#9b) -40.1 kJ/mol rxn
--------------------------------------------------------------------------
Chapter 10
Dissociation Equations Review Worksheet
remember that Ionic compounds and Acids split into Ionis, but
Covalent compounds stay intact (except acids)
(and put states on these equations)
#1) CaCl2(s) --> Ca2+(aq) + 2Cl -(aq)
#4) Cu(NO3)2(s) --> Cu2+(aq)
+ 2NO3-1(aq)
#5) HCl(aq) --> H +(aq) + Cl
-(aq)
#6) Fe2(SO4)3(s) --> 2Fe3+(aq)
+ 3SO4-2(aq)
#7) C12H22O11(s) --> C12H22O11(aq)
(*Notice, this one didn't split up)
#8) Al(OH)3(s) --> Al3+(aq) + 3OH
-(aq)
#10) Mn(NO2)5(s) --> Mn5+(aq)
+ 5NO2-1(aq)
#12) HF(aq) <----> H +(aq) + F
-(aq) (*Notice the double arrow, why?)
#13) HNO3(aq) --> H +(aq) + NO3-1(aq)
(*But this one only has a single arrow, why?)
#14) Mg(CH3COO)2(s) --> Mg2+(aq)
+ 2CH3COO-(aq)
#16) CH3CH2OH(l) --> CH3CH2OH(aq)
#19) K2Cr2O7(s) --> 2K+(aq)
+ Cr2O7-2(aq)
Page 276
#3) .838 M CH3COOH
#15) 14.6 M H3PO4 and 57.8 m H3PO4 and .510 (mole fraction)
#17a) 1.14 m KOH and 6.04 % KOH
#17b) 6.90 M KOH and 7.64 m KOH
9c) 1.45x10(5) ppm and .985 (mole fraction) and .874 m caffeine
Page 277
#11a) weigh 48.5 g K2Cr2O7 then dissolve in enough water to make 465ml
solution
#11b) measure 220. ml of .750M K2Cr2O7 and add water to make 465ml solution
#13a) .0424 mol Al2(SO4)3
#13b) .0293 M Al2(SO4)3 and .0586 M Al(+3) and .0879 M SO4(-2)
#19a) hydrogen peroxide
#19b) NaOH
#19c) HCl
#19d) CH3OH
#21a) +23.3 kJ
#21b) yes, because endothermic
#23a) 5.0 x 10(-7) M / mmHg
#23b) 1.5 x 10(-4) M He
#23c) .095 L He
Solution Formation WS
#1) +31 kJ/mol
#2a) Mg+2
#2b) Be+2
#2c) Fe+3
#3) a)H2O b)CCl4 c)either d) H2O e) H2O
Page 278
#27a) 14.1 mmHg
#27b) 16.7 mmHg
#27c) 19.00 mmHg
#29) .9 g C10H8
#61) 6
boxes and 3 circles, Na+ is 6M, S-2 is 3M
Page 278 again
#37) -4.5 °C and 101.2 °C
#43) 94.6 % benzene
#45) .16 M NaCl
#49c) -2.3 °C and 100.65 °C
#59) shrink
#60) picture should have 4 squares and 8 circles, Ca+2 is 4M and Cl- is
8M
#62a) solution #2
#62b) neither
#62c) solution #2
#62d) neither
#62e) solution #1
#67a) CaCl2 dissociates into 3 ions while CaSO4 dissociates into only
2 ions. More ions means more particles and freezing point is a colligative
property.
#67b) Increase in temp. helps the solute break apart and also the solvent
break apart. These two processes are always endothermic.
#67c) Nature wants osmosis to occur and "pushes" with a certain
osmotic pressure. For reverse osmosis to occur we need to overcome than
pressure.
#67d) BaCl2 dissociates into 3 ions while glucose molecules remain intact.
So, BaCl2 leads to more particles and osmotic pressure is a colligative
property.
#67e) The volume of solution becomes closer and closer to the amount of
solvent as a solution is diluted.
#70a) Methed 1: Dissolve it in distilled water and see if it conducts
electricity. If it conducts it is an electrolyte.
Method 2: Dissolve it in distilled water (or whatever solvent works) and
see how it changes the boiling or freezing point. From this determine
the "i" value. If the value is 1 it is nonelectrolyte.
#70b) The carbon dioxide gas is less soluble at higher temps.
#70c) Molality is per just solvent, mole fraction is per entire solution.
#70d) There is an interaction between the solute and solvent (for example
"ion-dipole" if ionic stuff in water). This interaction holds
onto the solvent stronger than solvent to solvent interaction. This lowers
the vapor pressure making it more difficult for the solvent to become
a gas, thus requiring a higher boiling point.
Chapter 10 Review WS
#1a) 11.2 % C4H8O2
#1b) 87.72 % H2O
#1c) .0251
#2) 341
#3) 8.58 % KCl
#4) the solvent has a more difficult time escaping, for two reasons, (1)
solute particles block the solvent, (2) attractive forces between the
solute and solvent (called "Ion Dipole") are stronger than the
IMF's the solvent normally has to overcome
#5) 6.2 x 10(4)
#6) 1.3 x 10(-5) M CO2
#7) most likely reason is a "volitile" solute, in this case
the interactions between solute and solvent are weaker than normal (example
would be alcohol added to water)
#8) an increase in temp. increases solubility for solids because it is
an endothermic process,
an increase in temp. decreases solubility for gases because it is an exothermic
process
#9) the answer depends on how you approach the question, if we base it
on things dissolving in water then NO, it is not colligative because it
does depend of the TYPE of particle as to whether a solution will conduct
electricity
#10a) CH2
#10b) 24.0 g O2
#10c) 85 (no units)
#10d) C6H12
#11a) both should be 2
#11b) both i values are lower because of incomplete dissociation, the
MgSO4 is lower than NaCl because MgSO4 ions have a higher charge and thus
more "ion pairs" exist and less "ions"
#11c) as concentration goes up there is less water availalble to "hydrate"
the ions so i value goes down
#12a) .94 m (NH4)2SO4
#12b) 11.0 % (NH4)2SO4
#13) 32 kJ/mol
#14) +4.9 kJ/mol and because enthalpy is positive (endothermic), heating
the solution increases solubility
#15) HCl is 2, (NH4)2SO4 is 3, H2SO4 is >2 but <3, HF is >1 but
<2
#16) pressure applied to the solution must be higher than the osmotic
pressure (pi)
Chapter 10 WS #2 is near the end of Chapter 11 stuff
--------------------------------------------------------------------------
Chapter 11
Chapter 11 1st Year Review WS
#1) Concentration, Temperature, Reactants, Catalyst, Surface Area, Agitation,
Pressure(gases only)
#2) More collisions and collisions have higher energy. Both happen because
particles are moving faster.
#3) Catalyst provides an alternate pathway with a lower Activation Energy
#4) Ea=160kJ, deltaH=80kJ, Ea'=80kJ, deltaHreverse= -80kJ, forward direction
is endo
#5) for uncatalyzed reaction Ea=25kJ, deltaH= -15kJ, Ea'=40kJ, deltaHreverse=15kJ,
forward direction is exo
for catalyzed reaction Ea=10kJ, deltaH= -15kJ, Ea'=25kJ, deltaHreverse=15kJ,
forward direction is exo
#6) The two hills represent 2 different steps in the reaction. The 1st
step is the rate determining step because it is higher.
2nd#5) They are about the same and probably both small
2nd#6) Slowest step
Page 311
#9a) chemical A is 3rd order, chemical B is zero order, and overall is
3rd
#9b) A is 1st, B is 1st, overall is 2nd
#9c) A is 1st, B is 2nd, overall is 3rd
#9d) A is 0, B is 1st, overall is 1st
#11a) / M(2) • min (that means nothing over molarity squared times
minutes, another way of writing it is M(-2) • min(-1), and yet another
way of writing it is L(2) • mol(-2) • min(-1)
#11b) / M • min
#11c) / M(2) • min
#11d) / min
#21a) S2O8(-2) is 1st order, I(-1) is 1st order, and overall is 2nd order
#21b) rate = k[S2O8(-2)](1) [I(-1)](1)
#21c) .371 / M • min
#21d) 3.41x10(-3) M / min
#23a) H2 is 1st order, Br2 is 1/2 order, and overall is 1and1/2 order
#23b) rate =k[H2](1) [Br2](1/2)
#23c) .150 / M(1/2) • sec
#23d) .0316 M / sec
#26a) m = 1, p = 1, n = 0
#26b) rate=k[CH3COCH3](1) [I2](0) [H+](1)
#26c) 2.6x10(-5) / M • sec
#26d) 6.8x10(-5) M / sec
Page 314
#33a) .0196 M
#33b) 5100 (3 sig figs) days
#33c) 2030 days
#39) 3.23 mg
#43a) .050 atm/sec
#43b) 2.1 sec
#43c) 5.0 sec
#63a) No and show why
#63b) No and show why
#63c) Yes and show why
#63d) Yes and show why
#62) Yes and show why
Chapter 11 Worksheet #1
#1a) rate=k[A](2)[B](1)
#1b) 289 / M(2) • hour
#1c) .0765 M / hour
#1d) .180 M chemical C
#2a) rate=k[X](0)[Y](1) or rate=k[Y](1)
#2b) .0070 / sec
#2c) 58 sec
#2d) mechanism #3 is the best, and show why
#3a) increase rate because more collisions
#3b) decrease rate because less collisions
#3c) increase rate because a higher percentage of collisions
will meet the energy requirements (Ea) due to the new path with a lower
activation energy. We'll talk about graph in class.
#3d) increase rate because at higher temp. more molecules have enough energy (Ea) to react
Chapter 10 WS #2
#1) 1.86 M Mg(NO3)2
#2a) .210
#2b) 19.6 torr
#3) 9.03 x 10(23) ammonium ions
#4) 6.2 x 10(4)
#5) (Relationship) as elec. cond. goes up the B.P. also goes up
(To account for diff. in B.P.) more particles cause higher B.P. and this
can be expressed with the "i" value
sucrose must be covalent (i of 1), formic acid must be weak acid (i between
1 - 2), sodium formate must be ionic (i of 2+)
#6) measure 11.3ml of the .200M solution, put into a 50.0ml volumetric
flask, add distilled water up to the 50.0ml and mix
Chapter 11 Worksheet #2
#16.12a) 1st order
#16.12b) 1st order
#16.12c) 2.5x10(-5) M/sec
#16.26) rate=k2kc1/2[Br2]1/2[H2]1
#16.27) c and d
#16.28) step #2
#16.30a) lower temp. slows particles so fewer collisions and collisions
have less energy
#16.30b) reactant order must be something other than 1
#16.30c) heat from the exothermic reaction provides enough energy to remaining
reactants to reach activation energy (this is typical of most combustion
reactions)
#16.30d) catalyst provides a new path (with a lower Ea) for both directions
so both directions speed up (Note: it gets the reactions to equilibrium
faster but doesn't change where the equilibrium is)
#16.42a) X is 1st order, Z is 2nd order, and overall order is 3
#16.42b) rate=k[X](1)[Z](2) and k=.46 / M(2) • hour
#16.42c) .0031 M / hour
Page 311
#3) .80 M/s for CO2 and 1.2 M/s for H2O
#31a) straight line for ln [(CH3)2N2] vs. time so 1st order
#31b) about 0.20 / hour
#31c) about 12.0 hr
#31d) about .083 M / hr
#77) please label both axis, and draw a curve for both the regular and
catalyzed reactions, the Ea' is 90. kJ
-------------------------------------------------------------------------
Chapter 12
Chapter 12 1st year Review WS
#0) solids and liquids (we only put gas and aqueous)
#1) 0.153
#2) 2.2
#3) 2.37
#4) a. left, b. left, c. left, d. left, e. right
#5) a. right, b. right, c. no change, d. right, e. no change
#6) No change (the solid would just settle to the bottom, the amount of
dissolved solute would remain the same)
#7) a. left, b. left, c. right, d. left, e. right
#8) the RATES of forward and reverse reactions
#9) it gets to equilibrium faster but does not change the position of
equilibrium (equilibrium constant stays the same)
#10) temperature (so if the temp. is changed then the equilibrium constant
will change, think of Le Chatelier)
#11) Yes, equilibrium constants are written for specific coefficients
#12)
a1) .0200 M SbCl5
a2) .747 atm
b) .00240
Page 344
#3)
@10s 2.000, 1.200, 0.150
@20s 1.625, 0.950, 0.275
@30s 1.325, 0.750, 0.375
@40s 1.100, 0.600, 0.450
@50s 0.950, 0.500, 0.500
@60s 0.950, 0.500, 0.500
#15) 2.4
#17) 1.1x10(3)
#21a) NH4CO2NH2(s) <----> 2NH3(g) + CO2(g)
#21b) 4.17x10(-4)
#27a) Q = 0 so forward
#27b) Q = 4.0 so reverse
#37a) CO and H2O are .301atm, H2 and CO2 are .343atm
#37b) total Pressure initially is 1.288atm and total Pressure at equilibrium
is 1.286atm, so the same, and NO it is only true if there are equal moles
of gas on reactant side versus product side
#41) HCN is .60atm, C2N2 and H2 are .24atm
#43a) 1.)Reverse 2.)Reverse 3.)No Change 4.)Forward 5.)Reverse
#43b) only temp. affects the value of K, since it is endo. and increase
in T will increase K and a decrease in T will decrease K
#55) 61.9% remains
#63) No, it's endothermic
Chapter 12 WS #1
#1a) same b) increased c) same d) decreased
#2a) .34
#2b) 5.0
#2c) 5.0
#2d) .87
#3) E
#4) C
#5) C
#6a) 45.7
#6b) .0763
Chapter 12 WS #2
#1a) left b) left c) right d) left e) left
#1f) up, up down
#1g) down, up, up
#2) N2 and O2 are .00371M, NO is .00458M
Chapter 12 WS#3
#1) H2 is .304atm, Cl2 is .404atm, HCl is .492atm
#2) Kp = .0107
#3. 1st question) H2O and CO are .041M, CO2 and H2 are .109M
#3. 2nd question) Q=8.07 so Reverse
#3. 3rd question) H2O and CO are .043M, CO2 and H2 are .115M
Chapter 12 WS#4
#1968a) Kc = 1.12x10(-6)
#1968b) NO2 is 1.50x10(-4)mol and N2O4 is .100mol
#1969a) 5.1x10(-7)
#1969b) .05M
#1981a) .109
#1981b) H2S is .233atm and NH3 is .466atm
#1981c) 6.97x10(-3)mol
#4a) .849atm
#4b) .0394
#4c) .304 mol
#5a) Hb(O2)4
#5b) HbH4+4
#5c) Hb(O2)4
#5d) the increase of CO2 causes an increase of H+ which causes a shift
to the left, so it lowers Hb(O2)4
#5e) NaHCO3 causes a decrease of H+ which causes a shift to the right,
so it raises the Hb(O2)4
-------------------------------------------------------------------------
Chapter 4.2 & 16
WS Ksp 1st Year Review
#1) 1.95x10-11
#2) 1.12x10-8
#3) 2.6x10-24
#4) .069 M
#5) 1.44x10-8 M
#6) 2.61x10-9 M
Page 96 and 439
Page 96
#7a) Na2SO4 (soluble)
#7b) Fe(NO3)3 (soluble)
#7c) AgCl (not soluble)
#7d) Cr(OH)3 (not soluble
#11a) Mg+2(aq) + 2OH-1(aq) --> Mg(OH)2(s)
#11b) Ag+1(aq) + Cl-1(aq) --> AgCl(s) and Ba+2(aq) + SO4-2(aq) -->
BaSO4(s)
#13a) no ppt.
#13b) 2Ag+1(aq) + CO3-2(aq) --> Ag2CO3(s)
#13c) 2Co+3(aq) + 3CO3-2(aq) --> Co2(CO3)3(s)
#13d) 3Ba+2(aq) + 2PO4-3(aq) --> Ba3(PO4)2(s)
#13e) no ppt.
#20a) 2Al+3(aq) + 3CO3-2(aq) --> Al2(CO3)3(s)
#20b) .108 M AlCl3
#20c) .379 g Al2(CO3)3
#76) a with 1, b with 3, c with 2
#78) a is weak, b in non, c is strong, d is weak
Page 439
#5a) 1.3x10(-4) M
#5b) 5.1x10(-10) M
#5c) 3.1x10(-6) M
#7a) .018 M
#7b) 2x10(-19) M
#7c) 3x10(-11) M
Page 439 (other one)
#9a) 2.9x10(-4) M
#9b) 75% F- precipitated
#11) I.P. = 2.6x10(-10) so No PPT,then for second question sulfate needs
to be .95M
#13a) I.P = 7.1x10(-8) so Yes PPT
#15) 2.5x10(-9)
#17a) 1.9x10(-3) g/L
Note about #17b & c) these have something new, see if you can figure
it out, we will discuss in class
#17b) 5.2x10(-7) g / L
#17c) 1.5x10(-7) g / L
Page 440
#17b) 5.2x10(-7) g / L
#17c) 1.5x10(-7) g / L
#28a) HEY, see your notes!
#29a) HEY, see your notes!
#30a) Cu(OH)2(s) + 4NH3(aq) --> Cu(NH3)4+2(aq) + 2OH-(aq)
#38a) 4x10(-7)
#38b) 3x10(-2) M
#44) a)reverse b)forward c)forward d)reverse
#50a) 3boxes and 6circles (unsaturated)
#50b) 4boxes and 8circles (saturated)
#50c) 4boxes and 8circles and 2(box w/ 2 circles combined)
(saturated)
#54) No, it is ENDO. As temp. goes up we get more ions dissovled.
#55)
a)T b)T c)F d)T e)T
WS Ch. 4.2 / 16 WS #2
#1) 1x10(-4)M
#2) 1x10(-6)M
#3) NaCl because it has a higher ksp
#4) we can when "same ion ratio" but we can't when "different
ion ratio"
#7a) Ag+(aq) + Cl -(aq) -->
AgCl(s)
#7b) Ca+2(aq) + SO4-2(aq)
--> CaSO4(s)
#7c) Cu+2(aq) + 2OH -(aq)
--> Cu(OH)2(s)
#7d) Ba+2(aq) + CO3-2(aq)
--> BaCO3(s)
#8) .0800 L AgNO3
#10) .0029 g / 100ml
-------------------------------------------------------------------------
Chapter 13 & 14
Acid / Base 1st Year Review WS
#9) 2.757
#10a) Acid + Base --> Conjugate Acid + Conjugate Base
#10b) Acid + Base --> Conjugate Base + Conjugate Acid
#11) 2.00x10(-11) M H+
#12) 1.4
#13) 11.11
#14) 5.6x10(-12) M OH-
#15a) OH-(aq) + HF(aq) --> HOH + F-(aq)
#15b) HCN(aq) + OH-(aq) --> HOH + CN-(aq)
#15c) H+(aq) + OH-(aq) --> HOH
#15d) HS-(aq) + H+(aq) --> H2S(aq)
#16) .0220 M NaOH
#17) 104
#18a) H3O+
#18b) acidic (try to figure out why)
#18c) No
#19a) strong acid with a weak base
#19b) near a pH of 3
#19c) 40 ml
#20) .0032 M H+
Page 374, 396, ws #1A
Page 374
#1a) acid , base , conjugate acid , conjugate base
#1b) acid , base , conjugate base, conjugate acid
#1c) acid , base , conjugate base, conjugate acid
#25b) Al(H2O)6+3(aq) <----> Al(H2O)5OH+2(aq) + H+(aq)
#25c) H2S(aq) <----> H+(aq) + HS-(aq)
#25d) HPO4-2(aq) <----> H+(aq) + PO4-3(aq)
#25e) HClO2(aq) <----> H+(aq) + ClO2-1(aq)
#35) 7.6x10(-5)
#49a) NH3(aq) + HOH <----> NH4+1(aq) + OH-(aq)
#49b) NO2-1(aq) + HOH <----> HNO2(aq) + OH-(aq)
#59a) basic
#59b) slightly acidic (Note: have to lookup Ka and Kb values)
#59c) acidic
#59d) basic
#59e) neutral
#61a) CO3-2(aq) + HOH <----> HCO3-1(aq) + OH-(aq)
#61b) Ka wins so we just show how it is acidic: NH4+1(aq) <---->
H+(aq) + NH3(aq)
#61c) H2PO4-1(aq) <----> H+(aq) + HPO4-2(aq)
#61d) NO2-1(aq) + HOH <----> HNO2(aq) + OH-(aq)
#61e) No Rxn.
#67) 5.3x10(-4)
Page 396
#1a) NH3(aq) + HF(aq) --> NH4+1(aq) + F-(aq)
#1b) H+(aq) + OH-(aq) --> HOH
#1c) SO3-2(aq) + H+(aq) --> HSO3-1(aq)
#1d) H+(aq) + OH-(aq) --> HOH
#3a) H+(aq) + F-(aq) --> HF(aq)
#3b) H+(aq) + OH-(aq) --> HOH
#3c) H+(aq) + H2PO4-1(aq) --> H3PO4(aq)
WS #1A
#1) acidic
#2) basic
#3) more info needed
#4) acidic
#5) neutral
#6) NH4+1(aq) + OH-(aq) --> H2O + NH3(aq)
#7) H+(aq) + NO2-1(aq) --> HNO2(aq)
#8) H+(aq) + OH-(aq) --> HOH
#9) CH3CH2NH2(aq) + H+(aq) --> CH3CH2NH3+(aq)
#10) H2PO4-1(aq) + S-2(aq) --> HS-(aq) + HPO4-2(aq)
Page 374 (more of it)
#20) 0.7900
#23) 13.91
#37a) .00180 M H+
#55a) Cod(aq) + H2O <----> CodH+(aq) + OH-(aq)
#55b) 8.3x10(-7)
#55c) 9.61
#57) 27g ammonia
#71a) No, Kw will be larger than what we are used to
#71b) 6.81
Worksheet #1B
#1a) Neutral, b) Acidic, c) Acidic, d) Neutral
#2) best way to do this is with three equations (1st showing the dissociation
of FeCl3,
2nd showing Fe+3 adding waters, 3rd showing the release of an H+)
# 3a) .0215 L calcium hydroxide
#3b) Thymol Blue because it is a Strong Base and Weak Acid titration so
endpoint will be above 7
#4a) NH3 + H+ --> NH4+
#4b) H+ + CH3COO-1 --> CH3COOH
#4c) H+ + OH-1 --> H2O
#5a) CH3NH2 + H2O --> CH3NH3+1 + OH-1
#6) H2O and HCO3-1 are the Bronsted Acids
#5b) 11.0 or 11.00
#7) 1.548
#8) 12.85 or 12.854
#9) 11.2 or 11.20
#10) 257
#13) 10.54 or 10.544
Worksheet #2
#1a) 64.7 ml NaOH
#1b) 2.53
#1c) HA + OH- --> HOH + A-
#1d) 3.60
#1e) 8.43
#1f) 10.111
#2a) 176.1
#2b) 7.7x10(-5)
#2c) 1.3x10(-10)
#2d) 8.54
Worksheet #3
#1a) 1.71
#1b) 2.69
#1c) 3.15
#1d) 8.32
#1e) 12.3 or 12.33
#2a) 5.3x10(-6) M
#2b) NH4(+) + OH(-) --> HOH + NH3
#2c) 8.4x10(-10) M
#2d) 9.25
#2e) 10.74 or 10.7
#2f) 11.194 or 11.19
#2g) The point is to show where the pH "jump" is.
pH should be the y-axis and ml KOH should be the x-axis.
See the Hints if you need help.
Page 397 & 377 & 399
page 397
#9a) 3.85
#9b) 3.55
#11a) 0.250M OH- and 0.250M H+
#11b) 0.250M OH- and 0.125M H+
#15a) HF and F-
#15b) H2CO3 and HCO3-1
#15c) HPO4-2 and PO4-3
#13) 3.61 (you will need to lookup ka)
#19a) 4.69 (you will need to lookup ka)
#19b) 4.69
#25a) 7.01 (you will need to lookup ka)
#25b) 6.56
#25c) 7.37
#31a) No
#31b) Yes
#31c) Yes
#31d) No
#31e) Yes
page 377
#77a) figure 1
#77b) figure 2
#77c) figure 2
page 399
#55a) weak
#55b) about 5x10(-4) this may vary based on how you read the graph
#55c) about 8.5
Worksheet #4
*****A lot of help on the hints page*****
#1a) 2.22
#1b) 1.8x10(-5) M
#1c) .21 M
#2a) 27.6 % K2CO3
#2b) 56 % KOH and 16 % KCl
#3a) 4.0x10(-5) M
#3b) 8.0x10(-8) M
#3c) 2.40
#4a) OCl- + HOH <----> HOCl + OH-
#4b) 52.0 ml HCl
#4c) 10.43
#4d) OCl- + H+ --> HOCl
#4e) 7.95
#4f) 7.56
#4g) 7.03
#4h) HOCl <----> H+ + OCl-
#4i) 4.36
#4j) 2.90
Ch 13/14 Practice Test
#1) A
#2) D
#3) C
#4) 2.8x10(-6)
#5a) 1.71
#5b) 3.72
#5c) 2.94
#6a) .0133 L Ca(OH)2
#6b) .0750 M C2H3O2(-1)
#6c) 8.81
#7a) 184
#7b) .00227 mol HA unreacted
#7c) 2.2x10(-6) M
#7d) 1.4x10(-6)
#8a) acidic
#8b) basic
#8c) neutral
#9a) 9.46 M HCl
#9b) .11 L
#9c) .2084 M HCl
#10a) 3.98x10(-6) M
#10b) 1.6x10(-10)
Practice Acid / Base Net equations
#1) H+ + C2H5NH2 --> C2H5NH3+1
#2) NH4+1 + NO2-1 --> HNO2 + NH3
#3) Cu(H2O)4+2 + CN-1 --> Cu(H2O)3OH+ + HCN
#4) HF + H2PO4-1 --> F-1 + H3PO4
#5) HF + OH-1 --> HOH + F-1
-------------------------------------------------------------------------
Chapter 4.4 & 18
Page 98
#43a) O = -2 and P = +5
#43b) H = +1 and N = -3
#43c) O = -2 and C = +4
#43d) O = -2 and S = +2
#43e) H = +1 and N = -2
#45a) Reduction
#45b) Reduction
#45c) Reduction
#45d) Oxidation
#49 We are only doing a half-reaction so you need to keep the electrons
in the equation
#49a) Oxidation and 4H2O + Mn+2(aq) --> MNO4-1(aq) + 8H+(aq) + 5e-
#49b) Reduction and 3e- + 4H2O + CrO4-2(aq) --> Cr+3 (aq) + 8OH-(aq)
#49c) Reduction and 2e- + 2H2O + PbO2(s) --> Pb+2(aq) + 4OH-(aq)
#49d) Reduction and 2e- + 2H+(aq) + ClO2-1(aq) --> ClO-(aq) + H2O
#70) 0.379
% Fe
#73a) Cr2O7-2
#73b) Sn+2
#73c) Cr2O7-2
#73d) OH-
Redox Balancing WS
I expect you to do this using the 5 step process for reaction 1-10.
*I didn't put states to keep the clutter down, but you should put states.
If you get stuck on how to write the half reactions, see the Hints Page
for some help.
#1) 2Fe + 3Cu+2 <----> 2Fe+3 + 3Cu
#2) Zn + Cu+2 <----> Zn+2 + Cu
#3) 4Cr+3 + 6H2O --> 4Cr + 3O2 + 12H+
#4) Mg + 2H+ --> Mg+2 + H2
#5) 5Fe+2 + 8H+ + MnO4-1 <----> Mn+2 + 4H2O + 5Fe+3
#6) 5H2O + NO2-1 + OH- + 2Al <----> 2Al(OH)4-1 + NH3
#7) Cu + 2Ag+ <----> 2Ag + Cu+2
#8) 6H+ + ClO3-1 + 5Cl-1 <----> 3Cl2 + 3H2O
#9) 2OH- + Zn + 2H2O <----> Zn(OH)4-2 + H2
*Note: for #10-15 I'm giving the total number of e- gained and lost according to the balanced reaction*
#10) CH4 + 2O2 --> CO2 + 2H2O (C losing 8e- and four O's gaining 8e-)
#11) (NH4)2Cr2O7 --> Cr2O3 + N2 + 4H2O (two Cr's gaining 6e- and two N's losing 6e-)
#12) 2H2O --> 2H2 + O2 (two O's losing 4e- and four H's gaining 4e-)
#13) 3H2 + N2 <----> 2NH3 (six H's losing 6 e- and two N's gaining 6e-)
#14) 2H2O2 --> O2 + 2H2O (two O's losing 2e- and two O's gaining 2e-)
#15) C12H22O11 + H2O --> 4C2H5OH + 4CO2 (four C's losing 16e- and eight C's gaining
16e-)
Page 500
#1a) 3Mg(s) + 2Sc+3(aq) --> 3Mg+2(aq) + 2Sc(s)
#1b) Sn(s) + Pb+2(aq) --> Pb(s) + Sn+2(aq)
#1c) 6Cl-(aq) + 8H+(aq) + 2NO3-1(aq) --> 3Cl2(g) + 2NO(g) + 4H2O
#5) (we will draw it in class)
Half equations at Anode: Mn(s) --> Mn+2(aq) + 2e- and at Cathode: Cr+3(aq) + 3e- --> Cr(s)
Overall reaction: 3Mn(s) + 2Cr+3(aq) --> 3Mn+2(aq) + 2Cr(s)
Abbreviated notation: Mn | Mn+2 || Cr+3 | Cr
Page 503
#61a) 31.7 g Pb
#57a) 1112 mol e-
#57b) 1242 amps
#57c) 278.0 mol O2
#59a) 16.0 g Ag
#59b) .244 cm
#63) .550 g Cd and .810 g Ni2O3
#66) 98,500 C/mol e-
#69) 108 amu
Page 501
#11) Reducting agents (low to high) Cr+3 < Hg < Sn+2 < H2
Oxidizing agents (low to high) Cr+3 < Sn+2 < Br2
#15a) +.695 V
#15b) +.383 V
#15c) +.202 V
#17a) +.063 V
#17b) +1.178 V
#17c) +.948 V
#19a) +1.182 V
#19b) +.952 V
#19c) +.305 V
#72) C
Page 502
#34a) E° = -.0414 V and deltaG° = 1.60x10(4)J
#34b) deltaG° = -4.52x10(4)J and K = 8.1x10(7)
#34c) E° = .015 V and K = 10
#38) E° = +.305 V and deltaG° = -1.77x10(5) J and k = 8.4x10(30)
*Note: the "8.4" may vary depending on how the math is done
#43a) +.265V
#43b) E = +.265V - (.0257V/6) ln[(Au+3)2 / (H+)6 (H2O2)3]
#43c) +.288V
#49) .15 M
#53a) E° = +.435V so YES
#53b) E = +.186V so YES
#53c) E = -.098V so NO
#53d) pH = 3.96
#74) only b
#77a) EQ
#77b) LT
#77c) GT
#77d) EQ
#77e) MI
Ch 18 WS Review #1
#1a) H=+1 O=-2 C=+4
#1b) F=-1 Te=+6
#1c) Na=+1 O=-2 Fe=+3
#2) Oxidation half = Sn+2(aq) --> Sn+4(aq) + 2e-
Reduction half = 1e- + 2H+(aq) + NO3-1(aq) --> NO2(g) + H2O
Overall = 4H+(aq) + 2NO3-1(aq) + Sn+2(aq) --> 2NO2(g) + 2H2O + Sn+4(aq)
#5) .945 grams Cu
#6) -.916 V
#9) +1.126 V
#10) 15.0 ml
#13) In acidic medium Mn = +2 because 6H+1 + 5SO3-2 + 2MnO4-1 --> 5SO4-2
+ 2Mn+2 + 3H2O
In basic medium Mn = +6 because 7H2O + SO3-2 + 2MnO4-1 --> SO4_2 +
2Mn+6 + 14OH-1
#14a) 6H+ + 2MnO4-1 + 5H2C2O4 --> 10CO2 + 2Mn+2 + 8H2O
#14b) .009724 mol H2C2O4
#14c) .009724 mol CaCO3
#14d) 77.76 % CaCO3
Ch 18 WS Review #2A
#1a) -105.4 cal/K
#1b) -58 kJ
#1c) -189 kJ
#2a) -.127 V
#2b) No, because -E°
#3a) K+(aq) cannot be reduced in water. Because it would be come K(s)
and immediately turn back into K+(aq). Instead, H2O gets reduced.
#4a) .898 grams Cr
#4b) 1850 seconds
Ch 18 WS Reviw #2B
#1) E° is +1.09V so YES
#2) Q = Ksp = 2x10(-10)
#3) anything from Pb+2 down to I2 on the chart
#4a) +1.101V
#4b) +1.22V
#5) -.0437V
#6) 72.6 % Fe
#7) 1.57 grams Cd
Ch 18 Redox Practice Test
#1) 3H2S(aq) + 2HNO3(aq) --> 3S(s) + 2NO(g) + 4H2O
#2a) H2O --> .5O2(g) + H2(g)
#2b) .0801 grams H2
#3) .936V
#4) E = -.590V, deltaG°= +342kJ, K= 1.5x10(-60)
#5a) 15.7 L Cl2
#5b) 12.146
#6a) +2.357V
#6b) anode: Mg(s) --> Mg+2(aq) + 2e-
cathode: 2e- + 2H+(aq) --> H2(g)
#6f) smaller
-------------------------------------------------------------------------
Chapter 19 - 22
Page 605, 421, 548, 572
page 605 #1a) alkane
#1b) alkyne
#1c) alkene
#5a) 2-methylbutane
#5b) 3-methylpentane
#5c) 2,3-dimethylpentane
#5d) 2,3,5-trimethylhexane
#11a) 2-methyl-1-propene
#11b) 2,3-dimethyl-2-butene
#11c) 2-pentene
#11d) 2-methyl-1-butene
#15a) alcohol
#15b) ester and carboxylic acid
#15c) alcohol and carboxylic acid
#17a) alcohol
#17b) carboxylic acid
#17c) ketone
#17d) aldyhyde
#17e) amine
#21a) CH3-O-C(w/ double bond to O)-H
#21b) CH3-O-C(w/ double bond to O)-CH3
#21c) CH3-O-C(w/ double bond to O)-(CH2)6-CH3
#30) there are 3 isomers total (by name they are 1-butene, 2-butene, 2-methyl-1-propene)
#33) by name they are 1-bromo-2-chlorobenzene, 1-bromo-3-chlorobenzene,
1-bromo-4-chlorobenzene) and the numbers can be switched for the halogens
but still make the same molecule
page 421 #1a) H2O (no charge), Cl- (-1 charge), OH- (-1 charge)
#1b) +2
#1c) Na2[Ni(H2O)2Cl2(OH)2]
#11a) [Ag(H2O)2]+1
#11b) [Pt(H2O)4]+2
#11c) [Pd(Br)4]-2
#11d) [Fe(C2O4)3]-3 (look in book to see why it is only 3 C2O4 and not
6)
page 548 #15a) Mg + Cl2 --> MgCl2 (Magnesium Chloride)
#14a) potassium nitride, K3N
#14b) potassium iodide, KI
#14d) potassium hydride, KH
#14e) potassium sulfide, K2S
#15c) 2Li + S --> Li2S (Lithium Sulfide)
#15d) 2Na + 2H2O --> 2Na+ + H2 + 2OH- (Sodium Ion and Hydrogen Gas
and Hydroxide)
page 572 #14a) NH3
#14b) HNO3
#14c) HNO2 (also NH4+1)
#14d) HNO3
#23a) Cu+2 + 4NH3 --> [Cu(NH3)4]+2 (Note: I am assuming excess ammonia
on this one)
#23b) H+ + NH3 --> NH4+1
#23c) If we assume a regular amount of ammonia then Al+3 + 3NH3 + 3H2O
--> Al(OH)3 + 3NH4+1
or the regular amount of ammonia can also be written Al+3 + 3NH4OH -->
Al(OH)3 + 3NH4+1
{however if we assume excess ammonia then Al+3 + 6NH3 --> [Al(NH3)6]+3}
Page 525, 572, 548
page 525 *I'll do the nuclear symbols by showing the top # / bottom
#, then element symbol
#3a) 237/93Np --> 4/2He + 233/91Pa
#3b) 85/39Y --> 0/1e + 85/38 Sr
#3c) 2 12/6C --> 23/11Na + 1/1H
#3d) 239/94Pu + 1/0n --> 130/50Sn + 107/44Ru + 3 1/0n
#4a) 230/90Th --> 4/2He + 226/88Ra
#4b) 210/82Pb --> 0/-1e + 210/83Bi
#4c) 235/92U + 1/0n --> 140/56Ba + 93/36Kr + 3 1/0n
#4d) 37/18Ar + 0/-1e --> 37/17Cl
#11a) Xe + 97/38Sr
#11b) C --> 4
#11c) Tc + 0/1e
#11d) 209/83Bi and Po
#33a) .06980 grams / mol
#33b) 6.28x10(9) kJ / mol
#49) about 15 atm (sig. figs. are unknown)
page 572 #3a) HClO3 #3b) HIO4 #3c) HBRO #3d) HI
#4a) KBrO2 #4b) CaBr2 #4c) NaIO4 #4d) Mg(ClO)2
#9a) NH3 #9b) N2O #9c) H2O2 #9d) SO3
#13a) S-2 #13b) SO3-2 and HSO3-1 #13c) H2SO4 and H2SO3
#18a) 3HClO(aq) --> Cl2(g) + HClO2(aq) + H2O
#18b) 2ClO3-1(aq) --> ClO4-1(aq) + ClO2-1(aq)
page 548 #20a) Co(s) + 2H+(aq) --> H2(g) + Co+2(aq)
#20c) 14H+(aq) + Cr2O7-2(aq) + 6e- --> 2Cr+3(aq) + 7H2O
CH 19-22 Review WS
#1) decomposition (and redox)
#2) green
#3) brown-red
#4) yellow
#5) CH3CH2COOH or C3H6O2 (I like the first version better 'cuz it suggests
Lewis structure)
#6) C4H8
#9) 1.62 grams Rn-222
#11) alpha
#12a) C4H8O or CH3COCH2CH3
#12b) C3H4 or CH3CCH
#12c) C4H10O or CH3CH2CH2CH2OH
#12d) C8H16
#13) 17.2 amu
#14) nitrogen-14
#15) 0.15017g/mol and 1.35x10(10) kJ/mol
Isomer WS #6
#1) c, d, e
#2) Id=identical, Iso=isomers, DC=different compounds
a)Id b)Id c)DC d)Iso e)Id f)ID g)Iso h)Id i)Iso j)Iso k)Id l)Id m)Iso
n)DC o)Id p)Id
WS "Writing equations by type"
#3) Ba+2 + Cr2O7-2 --> BaCr2O7
#4) Fe(CH3COO)2 + 2HCO3-1 --> Fe(HCO3)2 + 2CH3COO-1
#5) SO3 + H2O --> 2H+ + SO4-2
#6) Al2O3 + 3H2O --> 2Al(OH)3
#7) 2Fe + 3Cl2 --> 2FeCl3 (Note: forming FeCl2 is OK, FeCl3 is better)
#8) C2H4 + Br2 -->(35° above the arrow) C2H4Br2
#9) 2NaCl -->(and "electricity" over the arrow) 2Na + Cl2
#10) 2H2O2 -->(MnO2 above the arrow) O2 + 2H2O
#11) MgCO3 --> MgO + CO2 (and either a triangle or the "heat"
over the arrow)
#12) Ag+1 + Br-1 --> AgBr
#13) everything reacts on this one: H2SO4 + Ba(CH3COO)2 --> BaSO4 +
2CH3COOH
#15) Zn(OH)2 + 2OH-1 --> [Zn(OH)4]-2
#16) AgCl + 2NH3 --> [Ag(NH3)2]+ + Cl-
#18) 4H+ + [Cd(NH3)4]+2 --> 4NH4+1 + Cd+2
#19) NaHCO3 + CH3COOH --> CO2 + H2O + Na+1 + CH3COO-1
#20) Cl2 + 2I- --> I2 + 2Cl-
#21) NaH + H2O --> H2 + Na+1 + OH-1
#23) CH3COOH + CH3OH --> H2O + CH3COOCH3
WS "Writing Reactions w/ extra questions"
Set A #2i) 2KClO3 -->(and "heat" over the arrow) 2KCl + 3O2
Set A #3i) CH3COOH + OH- --> HOH + CH3COO-
Set A #3ii) because the CH3COO- produced is a weak base
Set B #1i) AgCl + Cl-1 --> [AgCl2]-1
Set B #2ii) HF & F-1 or NH3 & NH4+1
Set B #3i) Zn + Cu+2 --> Zn+2 + Cu
Set C #1i) H3P + BCl3 --> H3P : BCl3
Set C #2i) Ni+2 + 2OH- --> Ni(OH)2(s)
Set C #2ii) Br- & K+
Set C #3i) 2C6H14 + 19O2 --> 12CO2 + 14H2O
Set C #3ii) 13
Set D #1ii) oxidized
Set D #2i) 2H2O2 --> (with MnO2 over the arrow) forming O2 + 2H2O
Set D #3i) C4H8 + Cl2 --> C4H8Cl2 (what is happening here is the double
bond is "opening up" and the chlorines are attaching)
Set D #3ii) synthesis (or combination or addition)
Set E #1i) C6H6 + F2 --> C6H5F + HF
Set E #2i) ZnO + 2H+ --> H2O + Zn+2
Set E #3i) 2MnO4-1 + 16H+ + 10Cl- --> 2Mn+2 + 8H2O + 5Cl2
Set E #3ii) Cl-
Set F #1i) 6H+ + 3SO4-2 + Ca3(PO4)2 --> 2H3PO4 + 3CaSO4
Set F #2ii) Bases
Set F #3i) BaH2 + 2H2O --> 2H2 + Ba+2 + 2OH-
Set G #1i) K2Cr2O7 + 14H+ + 6Br-1 --> 2Cr+3 + 7H2O + 2K+1 + 3Br2 (I
assumed the potassium chromate started as a solid. If it had been a solution
of potassium chromate K+ would have not been written at all)
Set G #2i) Cu + 4H+1 + 2NO3-1 --> Cu+2 + 2H2O + 2NO2
Set G #3i) 2Cl- + Pb+2 --> PbCl2(s)
Set H #1i) Cu + 4H+1 + SO4-2 --> Cu+2 + 2H2O + SO2
Set H #2i) I-131 --> electron + Xe-131
Set H #3i) Cf-249 + O-18 --> 4 neutron + Sg-263
Set I #1i) 2NH3 + AgCl --> [Ag(NH3)2]+1 + Cl-
Set I #2ii) flame test the solution for purple color
Set I #3i) 2H2O + 2Cl- --> Cl2 + H2 + 2OH-
Set I #3ii) H2O
Set J #1i) C6H6 + Br2 --> (with "heated" above the arrow) C6H5Br + HBr
Set J #1ii) substitution (or replacement)
Set J #2i) CaH2 + 2H2O --> Ca+2 + 2H2 + 2OH-
Set J #3i) 2C5H10 + 15O2 --> 10CO2 + 10H2O
Holy Smokes, that was difficult to type!
WS "AP Chem. Equations Worksheet"
a) NH3 + CH3COOH --> NH4+1 + CH3COO-
A/B
NH3 and NH4+1 is a pair and CH3COOH and CH3COO- is an additional pair
b) OH- + CH3COOH --> HOH + CH3COO-
A/B
>7 'cuz it made a weak base (CH3COO-)
c) 2Li + 2H2O --> H2 + 2OH- + 2Li+
Group 1&2 in water, Redox
Basic
d) BaO + H2O --> Ba+2 + 2OH-
Metal Oxide in water
Basic, metal oxides and water form hydroxides
e) CaO + CO2 --> CaCO3
Synthesis
Synthesis
f) N2O5 + H2O --> 2H+1 + 2NO3-1
Non-metal Oxid in water
Acidic, nonmetal oxides form acids
g) 2H2O --> 2H2 + O2
Decomp. and Redox
Electrolysis of water
h) 4Li + O2 --> 2Li2O
Synthesis, Redox, Formation, Combustion
Group 1
i) 2Br2 + CH3CCCH2CH3 --> CH3C(Br)2C(Br)2CH2CH3
Organic Addition
Beer
j) Zn(OH)2 + 4NH3 --> [Zn(NH3)4]+2 + 2OH-
Complex Ion
It's a precipitate
k) 3OH- + Al(OH)3 --> [Al(OH)6]-3
Complex Ion, Synthesis
Hexahydroxoaluminate (you don't
really need to know this name, but it is interesting)
l) NH4OH --> NH3 + H2O
Decomp.
Decomp.
m) 2H2O2 --> O2 + 2H2O (and write cat. above arrow)
Decomp., Redox
Oxygen is both reduced and oxidized (Cool!)
n) 2H2O + 2I- --> I2 + H2 + 2OH-
Redox
Cathode, 'cuz the H+ is attracted
to the cathode then makes H2 gas
o) 2H+ + Zn --> H2 + Zn+2
Redox
See if it is flammable
p) C6H6 + Br2 --> C6H5Br + HBr (and write cat. above arrow)
Organic substitution
Water is polar, benzene is nonpolar
q) Ag+ + Cl- --> AgCl
Precipitate
Nitrate and Sodium
r) CH3COOH + CH3OH --> H2O + CH3COOCH3
Esterification
Vinegar
s) 2Ag+1 + CrO4-2 --> Ag2CrO4
Precipitate
Ox. number is +6
t) Li2O + H2O --> 2Li+ + 2OH-
Metal Oxide in water
Basic
u) 2KClO3 -->("MnO2" written above the arrow) 3O2 + 2KCl
Decomp., Redox
New path (mechanism) with lower activation energy
v) 3Mg + N2 --> Mg3N2
Syn., Redox, Formation
Ox. number goes from zero to +2
w) 2Al + 3Cu+2 --> 2Al+3 + 3Cu
Redox
Blue color fades, solid copper forms, solid aluminum dissolves, + voltage
x) LiH + H2O --> H2 + OH- + Li+
Hydride in water
Hydrogen has an ox. number of -1
y) 2H+ + SO4-2 + Ba+2 + 2OH- --> 2H2O + BaSO4
A/B, Precipitate
Neutral, the strong acid and strong base amounts were equal so they neutralize each other
z) 3H2 + N2 --> 2NH3 (and iron cat. written above the arrow)
Syn., Redox
Haber-Bosch
That was even more difficult to type!
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AP Lab Review
Lab Review Set #1
Hydrate Lab BaCl2 • 2H2O
Conc. of Unknown Sol'n. Lab 0.735 M NaCl
Titration Pt1 0.0726 M NaOH
Titration Pt2 245 g/mol
Heat of Comb. Lab -222 kJ/mol
Lab Review Set #2
Simplest Formula Lab MgO
Spectro. Lab about .18 M Fe+3
Hydrogen Lab 1 mol H2 to 1 mol Mg, and it matches
the balanced reaction because reaction says 1 mol Mg will produce 1 mol
H2
Rate Exp. Lab Rate = k [BrO3-1]1
[Br-]1 [H+]2
Graham's Law Lab 47.3 g/mol
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