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Chemistry
Answers to Chapter 2 Study Questions
1. a) An element is a fundamental substance that cannot be broken down into simpler
substances by chemical means. It is composed of atoms.
b) A compound is a pure substance made of at least two elements in a fixed ratio.
c) A pure substance can be an element or a compound. It has a constant composition.
2. a) air = mixture, solution
c) oak = mixture
e) oxygen = pure, element
g) wine = mixture, solution
b)
d)
f)
h)
titanium = pure, element
baking soda = pure, compound
7-Up = mixture, solution
carbon monoxide = pure, compound
3. a) element
b) compound
a) and b) contain molecules.
c) element
d) mixture
4. solid: particles not moving and close; liquid: particles moving and close; gas: particles
moving and far apart; aqueous (dissolved in water): particles moving and far.
5. chemical: Chlorine reacts with sodium to form NaCl. You could also list the formula of any
ionic compound chlorine forms, such as MgCl2, CaCl2, etc.
physical: Chlorine is a pale green gas at room temperature. It’s a nonmetal made up of
diatomic molecules.
6. Chemical reactions are frequently accompanied by:
a) mass changes and/or bubbles which show that a gas is involved in the reaction.
b) heat changes; heat is evolved in exothermic reactions; heat is used up in endothermic
reactions. Exothermic reactions also often result in the production of light and sound.
c) color changes which often signify a change in chemical composition.
d) the formation of a precipitate (formation of a solid from mixing solutions) which
represents the formation of an insoluble substance from soluble substances.
7. a) chemical
e) chemical
b) physical
f) chemical
c) physical
g) physical
d) chemical
8. a) physical b)chemical c) physical d)physical e) physical f)chemical g)chemical
8. a) Qualitative: This page is colorful. This page contains questions.
b) Quantitative: This page is 8.5 in x 11 in. The page contains 9 questions.
Theory: The questions on this page will be useful in studying for the test.
9. a) beaker
b) Erlenmeyer flask
c) graduated cylinder
d) pipet
Chemistry
Answers to Chapter 5 Study Questions
1. a) 6.5 x 102
b) 5 x 104
c) 2.07 x 105
d) 1.0 x 106
e) 5.0 x 104
2. a) liters (L) or cm3, graduated cylinder, buret, or volumetric flask
b) grams (g), balance
c) meters (m), ruler or meterstick
3. a) 4
b) 5
c) 2
d) 3
e) 4
4. How many significant figures are there in the following numbers or answers?
a) 3
b) 2 c) 2 d) 2 (answer 7.6)
e) 2
5. a) 1.24 x 8.2 = 10.
c) 9.999 + 0.22 = 10.22
b) 6.78 - 3.3 = 3.5
d) (5.67 x 103) x (2.1 x 102) = 1.2 x 102, or 120
0.822 g
mass

 2.35 g / cm 3
3
volume 0.350 cm
| 2.702.35 |
x 100%  13%
b) % accuracy error =
2.70
6. a) density 
7. 275 grams x 1 kg/1000g = 0.275 kg
8. 0.286m x 100cm/1m = 28.6 cm
9. 11.8 g x 1cm3/7.87g = 1.59 ml (note cm3 = ml)
10. a) Container A is the most precise because it has the most number of digits and taking into
account the fact that it is off by 2 mL, it never varies by more than 0.08 mL. You might also
say that Container C is the most precise because its volumes are reproducible, but you don’t
know whether the actual differences are more or less than Container A.
b) Container B is the most accurate, since it is consistently closest to the actual volume.
Chemistry
Answers to Chapters 3 & 4 Study Questions
1. Nuclear
Symbol
Atomic
Number
Mass
Number
Number of
Protons
Number of
Electrons
Number of
Neutrons
Charge
40
18 Ar
__18__
__40__
___18____
___18____
____22___
___0__
39 
19 K
__19__
39
19
16
__36__
___16____
36
16
S 2
18
____20___
___18____
__+1__
20
-2
2. Rutherford’s experiment supported the ideas that atoms contain a small dense center (nucleus)
and are mostly empty space.
3.
20
10 Ne
and
22
10 Ne
4. a) MG, Group 2, metal, Period 5
c) TM, Period 5
e) MG, Group 13, metalloid, Period 2
g) MG, Group 14, metal, Period 5
b)
d)
f)
h)
MG, Group 17, nonmetal, Period 4
MG, Group 15, nonmetal, Period 3
ITM, Period 7
TM, Period 6
5. Group 1 = alkali metals; Group 2 = alkaline earth metals; Group 17 = halogens; Group 18 =
noble gases.
6. a) Elements made of molecules: O2, N2, Cl2, or any other diatomic element.
b) Compounds made of molecules: CO2, H2O, NH3, or any other covalent compound.
c) Compounds made of ions: NaCl, MgSO4, or any other ionic compound.
7. a) positive, +1
b) positive, +2
c) negative, 1
d) negative, 2
e) positive, +1
8. a) covalent, dinitrogen oxide or dinitrogen monoxide
c) covalent, phosphorus trichloride
e) covalent, hydrochloric acid
g) ionic, lead(II) nitrite
b)
d)
f)
h)
9. a) calcium carbonate
d) magnesium perchlorate
c) copper(I) hydroxide
b) zinc sulfide
10. a) K3PO4
b) (NH4)2SO4
c) Co(OH)2
11. a) PI3
b) N2O5
c) HClO3
ionic, potassium oxide
ionic, aluminum phosphate
ionic, ammonium fluoride
covalent, sulfurous acid
d) FeN
12. a)covalent P2O5 b) ionic Mg(NO3)2 c) ionic AgO2
d) ionic KOH
13. a)lead II chloride b)copper I sulfate
c) carbón disulfide
d)hydrofluiric acid
e)sodium chlorate
Answers to Chapters 6 & 7 Study Questions
1. a) 3(12.0) + 8(1.01) + 3(16.0) = 92.1 g/mole
92.1 g
d) 0.217 moles 
= 20.0 g
1 mole
c) 6.02 x 1023 molecules
1 mole
e) 783 g 
= 8.50 moles
92.1 g
b) 92.1 g
2. a) 4 atoms (one N + 3 H)
6.02  1023 molecules
4 atoms
b) 1 mole NH 3 
=

1 mole NH 3
1 molecule
c) 3.40 g NH 3 
2.41 x 1024 atoms
1 mole NH 3 6.02  1023 molecules
4 atoms
=


17.0 g NH 3
1 mole NH 3
1 molecule
4.82 x 1023 atoms
3. Molar mass of NaNO2 = 23.0 + 14.0 + 2(16.0) = 69.0 g/mole
% Na = 23.0/69.0 = 33.3% Na; % N = 14.0/69.0 = 20.3% N; %O = 2(16.0)/69.0 = 46.4% O
33.3% Na, 20.3% N and 46.4% O.
(note, questions 4-5 where intentionally skipped)
6. Chemical reactions are frequently accompanied by:
a) bubbles which show that a gas is one of the products of the reaction.
b) heat changes; heat is evolved in exothermic reactions; heat is used up in endothermic
reactions. Exothermic reactions also often result in the production of light and sound.
c) color changes which often signify a change in chemical composition.
d) the formation of a precipitate which represents the formation of an insoluble ionic
compound from soluble ionic compounds.
7. a) 4 Fe(s) + 3 O2(g) 

b) B2H6(l) + 3 O2(g)
9. a) 6 Li(s) + N2(g)
b) C3H8(g)
10.
B2O3(s) + 3 H2O(l)

c) 4 PH3(g) + 8 O2(g)
8. a) PH3(g) and O2(g)
2 Fe2O3(s)
6 H2O(l) + P4O10(s)
b) H2O(l) and P4O10(s)

c) 6
2 Li3N(s)
+ 5 O2(g) 
3 CO2(g)
+ 4 H2O(l)
Chemistry
Answers to Chapters 8 & 9 Study Questions
1.
2.
4. a) 3 Na2CO3(aq) + 2 FeCl3(aq)  6 NaCl(aq) + Fe2(CO3)3(s)
5. a) Fe3+(aq) + 3 OH (aq)  Fe(OH)3(s)
b) No Reaction ((NH4)2CO3 and LiCl are both soluble)
c) Ni2+(aq) + S2-(aq)  NiS(s)
6. a) S, OR; 2 Li(s) + Cl2(g)  2 LiCl(s)
b) DD, P; Sr(NO3)2(aq) + K2SO4(aq)  2 KNO3(aq) + SrSO4(s)
c) C, OR; 2 C3H6(g) + 9 O2(g)  6 CO2(g) + 6 H2O(l)
d) DD; CaCl2(aq) + 2 NaNO3(aq)  No reaction (all products are soluble)
e) SR, OR; Fe(s) + MgSO4(aq)  No reaction (Mg is more active than Fe)
f) D, OR; 2 KI(l)  2 K(s) + I2(s)
g) SR, OR; 2 Al(s) + 6 HCl(aq)  3 H2(g) + 2 AlCl3(aq)
h) DD, AB; HNO3(aq) + KOH(aq)  H2O(l) + KNO3(aq)
6 mol HCl
= 4.80 moles HCl
2 mol CrCl3
2 mol Cr
52.0 g Cr
x
b) 0.450 mol HCl x
= 7.80 g Cr
6 mol HCl 1 mol Cr
7. a) 1.60 mol CrCl3 x
2 mol Cr 6.02 x 10 23 atoms Cr
= 4.8 x 1024 atoms
x
3 mol H 2
1 mol Cr
1 mol H 2
2 mol Cr 52.0 g Cr
d) 3.20 g H2 x
= 55.0 g Cr
x
x
2.016 g H 2 3 mol H 2 1 mol Cr
c) 12 mol H2 x
2 mol CrCl3 158.35 g CrCl3
1 mol HCl
= 12.0
x
x
36.46 g HCl 6 mol HCl
1 mol CrCl3
theoretical yield = 12.0 g CrCl3
e) 8.30 g HCl x
% Yield =
actual yield
10.2 g
x 100% =
x 100% =
theoretical yield
12.0 g
f) 6.0 moles Cr x
2 mol CrCl3
= 6.0 moles CrCl3
2 mol Cr
Cr
g CrCl3
85.0%
2 mol CrCl3
= 4.0 moles CrCl3; therefore, HCl is limiting
6 mol HCl
2 mol Cr
4.0 moles CrCl3 x
= 4.0 moles Cr used up.
2 mol CrCl3
12.0 moles HCl x
6.0 - 4.0 = 2.0
moles Cr left over.
1 mol Cr 2 mol CrCl3 158 g CrCl3
= 39.5 g CrCl3
x
x
52.0 g Cr
2 mol Cr
1 mol CrCl3
1 mol HCl 2 mol CrCl3 158 g CrCl3
43.8 g HCl x
= 63.2 g CrCl3
x
x
36.5 g HCl
6 mol HCl
1 mol CrCl3
g) 13.0 g Cr x
since 39.5 g < 63.2 g, 39.5 g CrCl3 is produced.
8. a)CH4 + 2 O2 -> CO2 + 2 H20
b) 2AL + 3Cl2 -> 2 ALCL3, synthesis
c)2NH4OH + Cu(NO3)2 -> 2 NH4NO3 + Cu(OH)2 , double replacement
d) 2 Fe + 6 HNO3 -> 2Fe(NO3)3 + 3 H2 single replacement
e) 2 H2O -> 2 H2 + O2 decomposition
9.
10.
11.
Chemistry
Answers to Chapter 11 Study Questions
1. Wavelength is inversely proportional to frequency. Energy is proportional to frequency.
2. The new idea in Bohr's model was that electrons can only exist in specific energy states.
Bohr's model included an electron orbiting the nucleus as a planet does the sun; according to
the quantum mechanical model, we can only define the probability of finding an electron at a
given location. When electrons drop from higher energy levels to lower ones, they give off
energy in the form of light. The color of light emitted depends on the energy difference
between the levels. The greater the energy difference, the shorter the wavelength of light, the
more violet the color.
3. The electron configurations of all Group 1 metals end with a single s electron. When these
metals lose this s electron, they acquire noble gas electron configurations which end in
completed energy levels. They have a strong tendency, therefore, to lose their final single s
electrons. This makes them extremely reactive and the metals with the greatest tendency to
lose electrons. Group 17 elements need only 1 p electron to complete their outermost energy
levels. They have a strong tendency to gain an electron and thus are the most reactive
nonmetals. The energy levels of noble gases are all full so these elements have no need to
gain or lose electrons and therefore don't react with anything.
4. a) 2n2, where n = Principal Energy level
b) s = 2 e-, p = 6 e-, d = 10 e-, f = 14 ec) 2 e5. Does not exist: c) 2d; Increasing energy: 1s < 2s < 3d < 4p < 4f
6. a) s
b) d
7. a) ns1
c) p
b) ns2np5
8. a) ground state
d) f
c) ns2np6
b) impossible
9. a) 1s22s22p63s23p4
10. a) [36Kr] 5s2
c) excited
d) excited
b) 1s22s22p63s23p64s23d9
b) [54Xe] 6s24f145d106p2
11. a) Ba (Period 6, Group 2)
b) He (1s2)
c) As (4s24p3 = Period 4, Group 15)
d) C (2s22p2 = Period 2, Group 14)
e) Cl (Period 3, Group 17)
f) Mn (3d5 = 5th transition metal where 3d is being filled) and also Cr (which is 4s13d5)
12.
s:
p:
As n increases, the size of an orbital increases since the probability of finding an electron
farther from the nucleus increases.
13. a) O: ()
1s
()
2s
()( )( )
2p
b) Ti: () () ()()() () ()()() () ( )( )( )( )(
1s
2s
2p
3s
3p
4s
3d
)
14. Chemical properties are shared within a Group but not within a Period. Group number is a
good predictor of chemical properties; Period number is not.
15. a) Ar; Na
16.
17.
b) Ca; S
Chemistry
Answers to Chapter 12 Study Questions
1. A chemical bond is the force holding atoms together. Atoms form chemical bonds to acquire
a stable number of electrons, which is often the same number of electrons as the nearest
noble gas. Covalent bonds involve the sharing of electrons between atoms, so that both
atoms act as if they've acquired the shared electrons. In ionic bonds, electrons are NOT
shared; a metal atom gives up one or more electrons to form a positive ion, a nonmetal gains
one or more electrons to form a negative ion, and then the two ions are attracted to each other
due to their opposite charges. Both covalent bonding and ionic bonding are strategies by
which an atom attains a stable number of electrons.
2. The number of bonds a nonmetal forms usually = 8 – valence.
3. Covalent bonds are found in molecules (covalent compounds and some nonmetal elements)
and in polyatomic ions.
4. a) As < P < N
b) Li < C < O
c) K < Mg < B
5.
6. a) Li+
b) Na+
c) F
7. a) 1s22s22p6; N3, F, Na+, Mg2+, Al3+; Ne
b) 1s22s22p63s23p6; S2-, Cl, K+, Ca2+; Ar
8. Ca(s) + I2(s)  CaI2(s)
9. a)
10. a)
e)
h)
13.
Be
b)
C
c)
b)
F
c)
f)
d)
g)
i)
Chemistry
Answers to Chapter 13 Study Questions
1. P1 = 722 mmHg 
P1
T1

P2
T2
;
1 atm
= 0.950 atm; T1 = 22 + 273 = 295 K; P2 = 1.07 atm; T2 = ?
760 mmHg
T2 = T1 
P2
1.07 atm
= 295 K 
= 332
0.950 atm
P1
K or 59°C
2. a) PT = PH2 + PH2O; Find PH2O in Table from lab report; at 19°C, PH2O = 16 mmHg
PH2 = PT - PH2O = 756 - 16 =
b) 740 mmHg 
1 atm
=
760 mmHg
740. mmHg
0.974 atm
3. V1 = 600. cm3; T1 = 25°C = 298 K; P1 = 750. mmHg
V2 = 480. cm3; T2 = 41°C = 314 K; P2 = ?
V
T
P2  P1  1  2
V2 T1
600 cm 3 314 K
 750 mmHg 

=
480 cm 3 298 K
988 mmHg
4 methane = CH4. To find grams, use PV = nRT to calculate n. molar mass (CH4) = 16.0 g/mol
n = ?; V = 28.0 L; T = 68 + 273 = 341 K; P = 2.00 atm.
16.0 g CH 4
(2 atm)( 28.0 L)
PV
n

= 2.00 mol; 2.00 mol 
= 32.0
RT (0.08206)(341 K )
1 mol CH 4
g
1 mol CO2
1 mol CH 4
= 0.341 mol CO2; 12.0 g CH4 
= 0.750 mole CH4;
44.0 g CO2
16.0 g CH 4
V V
n
0.750 mol
At constant T and P, 1  2 ; V2  V1  2  7.16 L 
= 15.7 L
n1 n2
n1
0.341 mol
5. 15.0 g CO2 
6.
7.
Chemistry
Answers to Chapter 14 Study Questions
1. In gases, the particles are far apart, move independently in the ideal case, take the shape of
their container and can have a variety of volumes. In ideal gases, the particles are not
attracted to each other at all. In liquids, the particles are close together giving them a
constant volume although liquids take the shape of their container. The particles are mobile
and attracted to each other. In solids, the particles are close together, fixed in position and
strongly attracted to each other. Solids hold their own shape.
2. The heat of vaporization is endothermic because the bonds between particles in a liquid must
be broken in order to separate the particles from each other. The breaking of bonds always
requires energy.
3. The boiling point of a liquid is the temperature at which both liquid and gas exist in
equilibrium; it is the temperature at which the equilibrium vapor pressure of the liquid is
equal to external pressure.
Chemistry
Answers to Chapter 15 Study Questions
mass KMnO4
x 100%;
mass solution
mass KMnO4 = 1.00 mole = 158 g; mass solution = 158 g KMnO4 + 158 g H2O = 316 g
158 g KMnO4
mass percent =
= 50.0 %
316 g solution
1. mass percent =
2. 335 g solution x
22.0 g KMnO4 1 mol KMnO4
=
x
100 g solution 158 g KMnO4
0.466 moles
3. 275 mL solution x
0.500 mol NaCl
=
1000 mL solution
4. 250 mL solution x
2.00 mol NaCl
58.5 g NaCl
x
=
1000 mL solution 1 mol NaCl
0.138 moles
29.2 g
5. V1 x M1 = V2 x M2; V1 x 2.00 M = 125 mL x 0.350 M
V1 = 125 mL x 0.350 M/2.00 M = 21.9 mL
6. molarity =
moles solute
;
L solution
90.0 g glu cos e 1000 mL 1 mol glu cos e
x
x
200 mL solution 1 liter
180 g glu cos e
=
2.50 M