, SOLUTION MANUAL FOR CHEMISTRY IN
CONTEXT 10TH EDITION AMERICAN CHEMICAL
SOCIETY
ANSWERS AT THE
END OF EACH CHAPTER
CHAPTER 1: Portable Electronics: The Periodic Table in the Palm of Your
Hand
Emphasizing Essentials
1. Answer:
a. Compound (2 molecules of one compound made up of two different elements).
b. Mixture (2 atoms of one element plus 2 atoms of another).
c. Mixture (3 different substances, two elements and one compound).
d. Element (4 atoms of the same element).
2. Answer:
a. element (4 molecules of one diatomic element)
b. Compound (2 molecules of one compound made up of two different elements).
c. Mixture (2 atoms of one element plus 2 atoms of another).
d. Compound (4 molecules of one compound made up of two different elements).
3. Answer: Exact answer will vary depending on viewing size of text. An approximate
measurement for the period could be 0.25 mm. Converting this to nanometers:
-3
m
0.25 mm × 10 × 1 nm = 2.5×105 nm or 250,000 nm.
1 mm 10-9 m
4. Answer:
1m 1000 mm
Converting 10.0 nanometers to millimeters: 10.0 nm × × =1×
1×109 nm 1m
10−5 mm or 0.00001 mm.
5. Answer:
a. 1.5 103 m
b. 9.58 10–11 m
c. 7.5 10–6 m
6. Answer: 1×102cm, 1×106µm, 1×109nm.
,7. Answer:
a. Group 1 and Group 17
b. 1A: lithium, sodium, potassium, rubidium, cesium, francium
7A: fluorine, chlorine, bromine, iodine, astatine
c. 1A: All but hydrogen (a nonmetal) are reactive metals.
7A: All but astatine (a metalloid) are reactive nonmetals.
8. Answer:
a.
b. iron, Fe; magnesium, Mg; aluminum, Al; sodium, Na; potassium, K; silver, Ag.
c.
d. sulfur, S; oxygen, O, carbon, C, chlorine, Cl, fluorine, F (and others).
9. Answer:
a. compound b. compound c. mixture
d. element e. mixture f. element
10. Answer:
a. heterogenous mixture b. heterogenous mixture c. heterogenous mixture
d. homogenous mixture e. heterogenous mixture
11. Answer:
, Grey spheres: electrons
Black spheres: protons
White spheres: neutrons
12. Answer: There are several allotropes of sulfur. The most stable and common allotrope
consists of 8 atoms in a ring. Other common allotropes include rings of 5, 6, 7, 10 and larger
number of atoms. Most allotropes are yellow solids, although they can be found in liquid or
gaseous forms at appropriate temperatures. Most of these allotropes are created by heating sulfur
of the 8-membered ring form.
13. Answer:
An aluminum atom with mass number 27 has 13 protons, 14 neutrons, and 13 electrons.
14. Answer:
A copper atom with a mass number 64 has 29 protons, 35 neutrons, and 29 electrons.
15. Answer:
Electrical conductivity is the transport of electrons from one location to another. Thermal
conductivity is the transfer of heat from one location to another. Their difference is in
what is being transferred.
16. Answer:
mass of Hf 178.5 g/mol
a. total mass of Hf+O's = (178.5 g/mol)+(16.00 g/mol)(2)
× 100% = 84.80% Hf
mass of O's (16.00 g/mol)(2)
total mass of Hf+O's = × 100% = 15.20% O
(178.5 g/mol)+(16.00 g/mol)(2)
mass of Be 9.012 g/mol
b. total mass of Be+Cl's = × 100% = 11.28% Be
(9.012 g/mol)+(35.45 g/mol)(2)
mass of Cl's
(35.45 g/mol)(2)
total mass of Be+Cl's = (9.012 g/mol)+(35.45 g/mol)(2) × 100% = 88.72% Cl
CONTEXT 10TH EDITION AMERICAN CHEMICAL
SOCIETY
ANSWERS AT THE
END OF EACH CHAPTER
CHAPTER 1: Portable Electronics: The Periodic Table in the Palm of Your
Hand
Emphasizing Essentials
1. Answer:
a. Compound (2 molecules of one compound made up of two different elements).
b. Mixture (2 atoms of one element plus 2 atoms of another).
c. Mixture (3 different substances, two elements and one compound).
d. Element (4 atoms of the same element).
2. Answer:
a. element (4 molecules of one diatomic element)
b. Compound (2 molecules of one compound made up of two different elements).
c. Mixture (2 atoms of one element plus 2 atoms of another).
d. Compound (4 molecules of one compound made up of two different elements).
3. Answer: Exact answer will vary depending on viewing size of text. An approximate
measurement for the period could be 0.25 mm. Converting this to nanometers:
-3
m
0.25 mm × 10 × 1 nm = 2.5×105 nm or 250,000 nm.
1 mm 10-9 m
4. Answer:
1m 1000 mm
Converting 10.0 nanometers to millimeters: 10.0 nm × × =1×
1×109 nm 1m
10−5 mm or 0.00001 mm.
5. Answer:
a. 1.5 103 m
b. 9.58 10–11 m
c. 7.5 10–6 m
6. Answer: 1×102cm, 1×106µm, 1×109nm.
,7. Answer:
a. Group 1 and Group 17
b. 1A: lithium, sodium, potassium, rubidium, cesium, francium
7A: fluorine, chlorine, bromine, iodine, astatine
c. 1A: All but hydrogen (a nonmetal) are reactive metals.
7A: All but astatine (a metalloid) are reactive nonmetals.
8. Answer:
a.
b. iron, Fe; magnesium, Mg; aluminum, Al; sodium, Na; potassium, K; silver, Ag.
c.
d. sulfur, S; oxygen, O, carbon, C, chlorine, Cl, fluorine, F (and others).
9. Answer:
a. compound b. compound c. mixture
d. element e. mixture f. element
10. Answer:
a. heterogenous mixture b. heterogenous mixture c. heterogenous mixture
d. homogenous mixture e. heterogenous mixture
11. Answer:
, Grey spheres: electrons
Black spheres: protons
White spheres: neutrons
12. Answer: There are several allotropes of sulfur. The most stable and common allotrope
consists of 8 atoms in a ring. Other common allotropes include rings of 5, 6, 7, 10 and larger
number of atoms. Most allotropes are yellow solids, although they can be found in liquid or
gaseous forms at appropriate temperatures. Most of these allotropes are created by heating sulfur
of the 8-membered ring form.
13. Answer:
An aluminum atom with mass number 27 has 13 protons, 14 neutrons, and 13 electrons.
14. Answer:
A copper atom with a mass number 64 has 29 protons, 35 neutrons, and 29 electrons.
15. Answer:
Electrical conductivity is the transport of electrons from one location to another. Thermal
conductivity is the transfer of heat from one location to another. Their difference is in
what is being transferred.
16. Answer:
mass of Hf 178.5 g/mol
a. total mass of Hf+O's = (178.5 g/mol)+(16.00 g/mol)(2)
× 100% = 84.80% Hf
mass of O's (16.00 g/mol)(2)
total mass of Hf+O's = × 100% = 15.20% O
(178.5 g/mol)+(16.00 g/mol)(2)
mass of Be 9.012 g/mol
b. total mass of Be+Cl's = × 100% = 11.28% Be
(9.012 g/mol)+(35.45 g/mol)(2)
mass of Cl's
(35.45 g/mol)(2)
total mass of Be+Cl's = (9.012 g/mol)+(35.45 g/mol)(2) × 100% = 88.72% Cl
