answers to chem 20 final review 2011/2012
Multiple Choice:
1. A the measurements closely agree, but are not close to the accepted value
2. A melting is a physical change: no new substance is produced, only a different state of matter
3. B 81 is the mass number of the isotope; Bromine only has 35 protons
4. B this element has 35 electrons….bromine in its ground state
5. A isotopes have the same atomic number (number of protons)
6. C. atomic mass = number of protons + number of neutrons
7. B E = mc2 m = 1.156 x 10-3 g X 1 kg / 1 000 g = 1.156 x 10-6 kg
E = (1.156 x 10-6 kg)(3.0 x 108 m s-1)2 = 1.04 x 1011 J
8. C mass number of 3: 1 proton + 2 neutrons
9. A metals have very low electron affinities
10. C Sc3+ has the highest positive charge….smallest radius
11. C Oxygen will have a higher IE1 than Sulfur
12. A Thallium has the most electrons in the family
13. D Nitrogen would like to gain three electrons to become stable
14. C the clues identify hydrogen
15. B
16. A
17. B
18. C NH3 is 17.04 g mol-1 ; H2O2 is 34.02 g mol-1; CH4 is 16.05 g mol-1 and SO3 is 80.07 g mol-1
19. A
20. C
21. D mol C = |75.0 g| 1 mol| = 6.24 mol C

6.24 = 1 mol C
|12.01g|
22. B
23. C
24. B
25. C
26. B
27. B
28. D
29. C
30. D
31. A
32.A
33. B
34. A
mol H = |12.5 g| 1 mol| = 12.38 mol H
| 1.01g|

6.24
= 2 mol H
mol O = |100.0 g| 1 mo l| = 6.25 mol O
| 16.00g|

6.24
= 1 mol C
the calcium/phosphate combination is not soluble: Ca2+ with PO43- gives Ca3(PO4)2
definition of catalyst
carbon compound reacts with oxygen to form carbon dioxide and water
one substance breaks down into two smaller ones
conservation of mass, or can do a mass to mass stoichiometry
mass of products = mass of reactants 1.000g – 0.550g = 0.450g
or: |1.000 g | 1 mol | 2 mol H2O | 18.0 g | = 0.450 g
| 80.0g | 1 mol NH4NO3|1 mol H2O|
Helium is stable with only two valence electrons
Aluminum is stable with 6 valence electrons; fluorine requires 8 valence electrons
graphite is a two dimensional network of planar carbon atoms
here the phosphorus is making only three bonds..no need to expand its valence
the higher the ∆EN, the more ionic the bond
35. C many metals have very high melting points; ex. Tungsten is close to 2 000 C0
36. C this creates covalent bonds; lowers the ability of the electrons to move around the solid
37.D
38. E
39. B,D and E if ∆EN is between 0.5 and 1.67, the bond is polar covalent
40. B
41. B germanium dioxide is a network solid
42. D K+ with Cl43. C nonpolar molecules have no IM forces except those
44. C collisions between particles are elastic…no energy loss
45. C gas C contributes 21.8% to the total pressure; 21.8% of 760 mm Hg is 165.7 mm Hg
46. C use V2 = P1V1T2 = (1.08 atm)(50.0 L)(283K) = 60 L
P2T1
(0.855 atm)(298K)
47. D use V2 = P1V1
= (160.0 atm)(600.0L)
= 32 000 L
P2
(3 atm)
48. B a hot air balloon relates volume and temperature of the gas; this is Charles Law
49. C use V = nRT
= (0.0685 mol)(0.0821 L atm mol-1 K-1)(273 K)
= 1.53 L
P
(1.00 atm)
50. D use P2 = P1V1
= (1 atm)(4.0 L)
= 4.7 atm
V2
(0.85 L)
51. E
mol CH4 = | 35.2 g | 1 mol | / = 2.19 mol
| 16.05 g |
volume at SATP = |2.19 mol|24.4 L |
= 53.5 L
| 1 mol |
52. B definition of pressure
53. D Helium has a molar mass much lighter than that of Sulfur dioxide; the helium particles will
be traveling at a much higher speed.
54. B
55. A
56. A
57. B
c = n/v = |6.84 g Al2(SO4)3| 1 mol |
/ 0.250L = 0.080M
| 342.3g|
58. A MgCl2 → Mg2+ + 2Cl- [MgCl2] = c = n/v = |28.5 g|1 mol | / 0.900L = 0.332
|95.3g |
[Cl ] = |0.332M MgCl2| 2 Cl | = 0.664M
|1 MgCl2 |
59. E Ca(CH3COO)2 → 2 CH3COO- + Ca2+
[CH3COO-] = |0.250M | 2 CH3COO- |
= 0.500M
|1 Ca(CH3COO)2|
n = cv = (0.500M)(0.046L) = 0.023 mol
60. C
M = mol solute/L solution;
mol K2CO3 = |52.0g|1 mol | = 0.376 mol
|138.2g |
c = 0.376 mol / 0.518 L = 0.726M
61. D [Li+] = |0.200 M Li2SO4| 2 Li+ |
= 0.40M
| 1 Li2SO4|
62. A C1V1 = C2V2;
V1 = (0.5 M)(5.00L) / (12.0M) = 0.208L = 208 mL
63. C the complete ionic equation shows both the precipitate formed and the spectator ions present
64. B the solution with the highest concentration of particles
65. D the two liquids are immiscible
66. D solutions are transparent/translucent to light rays
67. D C = |8.00 g NaOH| 1 mol | / 0.500 L = 0.400 M
| 40.00g|
68. D
69. A
70. B
71. D
72. A
Short Answer/Calculations
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
4 significant digits
0.966 mol-L/kPa
row 1: Tc-99; 43p; 99 mass/ row2: Z = 50;50e-, 69n, 119 mass
row3: 36e-, 49n / row4: Z = 34;36e-;79 mass / row5: Z=78;78p;78e-;193 mass
aam = (101u)(0.5925) + (103u)(0.4075) = 101.8u
Rutherford: nuclear model; electrons circle nucleus of protons, neutrons ( all the mass )
Bohr: electrons exist at discrete energy levels only
Quantum: electrons exist at specific orbitals which can be identified and predicted using
quantum numbers and described by wave equations
third energy level, p orbital, y orientation, clockwise spin: a 3py electron with clockwise spin
it will release energy to return to its ground state….can be detected as a spectral line in the ultra
violet region
255
Lr
103
199
Au
79
→
4
He
2
+
195
Ir
77
Mf = (69.0 g)(0.5)5.8245/2.33 = 12.2 g
Germanium
Argon
a) -3 b) group 15 (5A)
the extra electron added is in the same shell, but the attractive force between protons and 18e- is
more diffuse than that between protons and 17ea) Cu: [Ar] 4s23d9 or [Ar] 4s13d10
b) Mo: [Kr] 5s24d4 or [Kr] 5s14d5
c) Br: [Ar] 4s23d104p5
a) +2 and +1 are likely
b) +6 and +2 are most likely c) -1 is most likely
atomic radius: decreases from left to right; more protons in nucleus attract the electrons stronger
metallic character: an increase in valence electrons decrease metallic character
electron affinity: increases from left to right except for noble gases which have no affinity for ea) |4.44g CO| 1 mol | = 0.159 mol
| 28.01g|
b) |3.65 x 1027 molecules CO|
1 mol
|
|6.02 x 1023 molecules |
= 6 060 mol
19.
20.
a) |4.60 mol F2O| 54.00 g | = 248 g
| 1 mol |
b) |4.58 x 1024 molecules CO2 |
1 mol
| 44.01 g | = 335 g
23
|6.02 x 10 molecules | 1 mol |
SO3
mol S = 40.0 g / 32.07g = 1.25S; mol O = 60.0g/16.0g = 3.75 O
21.
empirical formula is: KS2O3 : mol K = 25.8g/39.10g = 0.660K; mol S = 42.4g/32.07g = 1.32S;
mol O = 31.7g/16.00g = 1.98 O
molecular formula is: K3S6O9: molar mass/empirical mass = 453.9g/151.1 g ≈ 3
22.
a) 2 MnO2 + 4 KOH + O2 + Cl2 → 2 KMnO4 + 2 KCl + 2 H2O
b) 2 C3H7OH(l) + 9 O2(g) → 6 CO2(g) + 8 H2O(g)
c) 2 Al2Cl3(s) → 4 Al(s) + 3 Cl2(g)
d) SO3(g) + 2 HNO3(aq) → H2SO4(aq) + N2O5(g)
23.
a) Mg(s) + Pb(NO3)2(aq) → Mg(NO3)2(aq) + Pb(s)
b) Cl2(g) + 2 NaBr(aq) → 2 NaCl(aq) + Br2(aq)
24.
Ca(NO3)2(aq) + Li2SO4(aq)
25.
a) Mg(OH)2(aq) + H2SO4(aq) → MgSO4(aq) + 2 H2O(l) neutralization
b) 3 Na2CO3(aq) + 2 H3PO4(aq) → 2 Na3PO4(aq) + 3 H2O(l) + 3 CO2(g) gas formed,
neutralization
26.
a) C2H5OH(l) + 3 O2(g)
b) CH3COCH3(l) + 4 O2(g)
27.
a) dissolved in water
b) CaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ca(NO3)2(aq)
c) Ca2+(aq) + 2 Cl-(aq) + 2 Ag+(aq) + 2 NO3-(aq) → 2 AgCl(s) + Ca2+(aq) + 2 NO3-(aq)
d) 2 Ag+(aq) + 2 Cl-(aq) → 1 AgCl(s)
Or Ag+(aq) + 2 Cl-(aq) → AgCl(s)
spectator ions are present, but are not involved in forming the precipitate
in this question, the nitrate ions and calcium ions are spectators
28.
a)
S8 + 8 O2
→ 8 SO2
b) 80.1 g of S8 were burned: |160.0g SO2|1 mol | 1 S8 | 256.8 g |
|64.1g | 8SO2| 1 mol S8|
29.
1 730 g O2 : |12.0 C3H6| 9 O2 | 32.0 g O2 |
|2 C3H6|
1 mol |
30.
11.6 g octane: |11.4 g C16H32| 1 mol | 2 C8H18 | 114 g |
| 224 g | 1 C16H32|1 mol |
31.
7.30 x 1023 atoms of oxygen:
|35.0 g Mg3(PO4)2| 1 mol | 6.02 x 1023 molecules | 8 Oxygen atoms |
|262.9 g|
1 mol
|
1 molecule |
32.
a) exothermic
b) limiting reagent is: oxygen gas; theoretical yield of carbon dioxide gas is 13.5 g
→ CaSO4(s) + 2 LiNO3(aq)
→ 2 CO2(g) + 3 H2O(g)
→ 3 CO2(g) + 3 H2O(g)
% yield is 81.5%
|16.0g O2 |1 mol | 8 CO2 | 44.0 g | = 13.5 g
|32.0 g| 13 O2 | 1 mol |
|40.0g C4H10| 1 mol | 8 CO2 | 44.0 g | = 121.4 g
| 58.0g | 2 C4H10| 1 mol |
oxygen is limiting reagent;
%yield = (11.0 g / 13.5 g) x 100% = 81.5%
33.
a) C3H8(g) + 5 O2(g)
→
3 CO2(g) + 4 H2O(g)
b) |10.0g C3H8| 1 mol | 4 H2O | 18.0 g | = 16.4 g H2O
| 44.0 g | 1 C3H8| 1 mol |
c) |20.0g C3H8| 1 mol | 3 CO2 | 44.0 g | = 60.0 g CO2
| 44.0 g | 1 C3H8| 1 mol |
d) |16.2 mol O2| 4 H2O | = 13.0 mol H2O
| 5 O2 |
e) |2.00 C3H8| 3CO2 | 44.0g | = 264 g
|1C3H8| 1mol |
34.
|15.0g N2| 1 mol | 2 NH3 | 17.0 g | = 18.2 g
| 28.0 g| 1 N2 | 1 mol |
|4.00 g H2| 1 mol | 2 NH3 | 17.0 g | = 22.7 g
| 2.00g | 3 H2 | 1 mol |
maximum amount that can be produced is 18.2 g ( limited by N2 reagent)
35.
36.
avogadro’s hypothesis: equal moles of gas occupy equal volumes at the same temp/pressure:
|4.00 L NO| 1 O2 | = 2.00 L O2
| 2 NO |
a) phosphorous pentachloride
b) sulfate ion
2-
Cl
O
Cl
P
Cl
Cl
S
O
Cl
O
trigonal bipyramidal
tetrahedral
c) ozone
O
O
O
v-shaped (bent)
O
37.
A bond dipole is a charge separation between two atoms due to difference in electronegativities;
a molecular dipole is a polarity in the whole molecule due to the sum of bond dipoles giving a
resulting molecular dipole, such as in water.
38.
a) bent
b) linear
c) trigonal planar
d) trigonal pyramidal
O
S
H
:C ≡ O:
F
N
e) irregular tetrahedron
Te
F
||
H
O
Cl
P
F
O
f) linear
g) bent
F – Xe – F
Cl
S
O
Cl
O
Cl
39.
a)
O=C=O
2 pi bonds;
b)
O
c) H – C ≡ C – F
S
2 pi bonds
O O
1 pi bond
40.
41.
42.
a) 150 C0 + 273 = 423 K
a) 457K – 273 = 184 C0
b) -30 C0 + 273 = 243 K
b) 65 K – 273 = -208 C0
Large particles have larger electron clouds and can therefore have a greater difference between the positive
and negative regions of a dipole. Smaller particles have smaller electron clouds, so their dipole forces are
weaker.
43.
Both of these forces involve a separation of charges from a positive end to a negative end. However, in
dispersion forces, this is temporary and caused by random shifts in the electron cloud distribution.
In dipole-dipole interactions, the dipole is caused by a polar covalent bond due to their unequal sharing of
bonding electrons.
44.
Vaporization is the process of a liquid becoming a gas at its boiling point. Evaporation is the process of
individual particles breaking free of the surface of a liquid and can occur at any temperature.
45.
Water has a higher boiling point than methane because water molecules have strong hydrogen bonds that
require relatively high amounts of energy to break, whereas methane molecules are nonpolar, have weak
hydrogen bonds, and require far less energy to break their bonds
46.
The molecules of cooking oil are nonpolar, and the electron clouds of nonpolar molecules repel each other
forming a weaker attraction. Table salt molecules are polar, and the negative and positive regions of these
molecules are strongly attracted to the oppositely charged regions of adjoining molecules forming strong
dipole-dipole forces.
47.
Use combined gas law.
0.58L
V2 = (110 kPa)(2.00L)(353K)
(303K)(440 kPa)
48.
1.7 L
Use Boyle’s Law:
V2 = (2.25 atm)(0.75L)
(1.00 atm)
49.
50.
0.18 moles; use the ideal gas law: PV = nRT
n = PV/RT =
(1.50 atm)(3.0 L)
(0.0821 L atm mol-1 K-1)(300 K)
= 0.18 mol
a) |7.07 L CO2| 2 C2H6 | = 3.54 L C2H6
| 4 CO2 |
b) |7.07 L CO2| 7 O2 |
|4 CO2 |
= 12.4 L O2
51.
I2 is a completely non polar solute; the best choice to dissolve would be a solvent that
is also mostly non polar; of the solvents listed, CCl4 is the most nonpolar
52.
53.
a) PbCl4(s) → Pb4+(aq) + 4 Cl-(aq)
c = n/v 2.40 mol/0.250L = 9.60M
54.
a) n = cv = (4.00 x 10-2 mol/L) (3.00 L) = 0.120 mol
b) g = (0.120 mol HCl)(36.5 g HCl/mol) = 4.38 g
55.
a) first determine the number of moles solute required:
n = cv = (0.064 mol/L)(0.2000 L) = 0.0128 mol
second, determine the mass of solute required to make the solution: g = n(MM) =
= (0.0128 mol)( 106.0 g / 1 mol ) = 1.36 g
third, weigh 1.36 g of sodium carbonate on a balance, and dissolve in about 100 ml of
distilled water in a 200 ml volumetric flask; swirl gently until solid is all dissolved
fourth: add distilled water carefully up to the calibration line on the 200 ml v flask;
swirl gently to mix, and pour solution into a clean, labeled container
b) Mg(NO3)2(s) → Mg2+(aq) + 2 NO3-(aq)
b) first, determine the volume of stock solution required to prepare the dilute solution:
V1 = C2V2 / C1 = (0.030 mol/L)(460.0 ml) / (1.0 mol/L) = 13.8 ml
second, place about 230 ml water in a 460 ml volumetric flask;
third: carefully transfer, using a pipette, 13.8 ml of stock solution to the v flask and
swirl gently;
fourth: carefully add distilled water up to the calibration line on the 460 ml v flask, swirl
gently to mix and pour solution into a clean, labeled container
56.
a) Na2SO4(s) → 2 Na+(aq) + SO42-(aq)
b) c = n/v
i) n = (15.00g )( 1 mol / 142.1 g) = 0.1056 mol
ii) c = (0.1056 mol)(0.5000 L) = 0.0528 M
c) [Na+] = (0.0528 M)( 2 Na+ / 1 Na2SO4) = 0.1056M
57.
a) i) CH3CH2CH2CH2CH3
pentane
and
ii) CH3CH(CH3)CH2CH3
2-methylbutane
b) i) CH3CH2CH2CH2OH
1-butanol
and
ii)
CH3CH2CH(OH)CH3
2-butanol
58.
6-fluoro-4,6-dimethyl-3-heptene
59.
a) CH3(CH3)C=C(CH3)CH2CH3
b) CH3C≡CCH(C2H5)CH2CH2CH2CH3
60.
a) CH3CH(CH3)CH2CH2CH3
b) CH2=C(CH3)CH(C2H5)CH(CH3)CH2CH3
c) CH3CH2CH(C2H5)CH2CH2CH2CH2CH3 d)
Br
e) CH3CH(CH3)CH2CH2CH2OH
f) HClC=CClCHBrCH3
g) CH3C≡CCH2CHFCH3
h) CH3CH=C(CH3)(CH3)C(CH3)C≡CCH2CH2CH2CH3
j)
Cl
OH
i)
Cl
Cl
Cl
k)
l) CH3OC(O)CH2CH2CH2CH3
Br
Br
Br
m) CH3CH(CH3)CH2CH2OH
o) CH3CH2O(O)CCH2CH2CH3
61.
62.
n) CH3CH2Cl
p)
Cl
|
CH3 – CH – CH2 – CH – CH2 – CH2 – CH2 – CH3
|
matching: : cyclohexane matches with B; 3-pentanol matches with C;
1,4 dimethylbenzene matches with E; 2-pentanol matches with A; 1,2 dimethylcyclohexane
matches with F
a. propyne
b. 5-ethyl-4-methyl-1-heptanol or 5-ethyl-1-hydroxy-4-methylheptane
c. 1,3 –diethylcyclopentane
d. 1,3,5 –trichlorocyclohexane