Download Chapter 5HW_Ans

Chapter 5 Homework: 5.4, 5.8, 5.10, 5.24, 5.26, 5.30, 5.34, 5.36, 5.40, 5.44, 5.52, 5.60,
5.66, 5.70.
5.4
6.022 x 1023 atoms or molecules
?
6.022 x10 23
atoms or molecules
Conversion factor =
1
6.022 x 1023 atoms or molecules
≡
1 mole
x atoms or molecules ≡
?
1
Conversion factor =
moles
6.022 x10 23
a) 2.709 x 1023 Li atoms
b) 1.084 x1023 CO2 molecules
c) 0.129 mole of copper
d) 0.623 moles of ethane molecules
5.8
One molecules of Al2(SO4)3 has 2 aluminum ions , 3 sulfur atoms, 12 oxygen
atoms, and it has 3 sulfate ions.
Therefore one mole of Al2(SO4)3 will have as 2 moles of aluminum ions, 3 moles
of sulfur, 12 moles of oxygen, and 3 moles of SO42- ions.
1 mole
x moles
≡
≡
b) 0.8 moles of Al3+ ions.
a) 9 moles of sulfur
c) 4.5 moles of SO42- ions
5.10
For molar mass add masses of all the atoms present using periodic table.
a) 151.92 grams/mole
c) 183.19 grams/mole
e) 84.124 grams/mole
5.24
b) 101.96 grams/mole
c) 60.094 grams/mole
f) 183.496 grams/mole
Chemical change is due to change in composition. In physical change the
chemical does not change.
a) physical
c) physical
e) chemical
5.26
5.30
b) chemical
c) biochemical healing
f) physical
a) balanced
b) balanced
c) not balanced
balanced equation: 2KClO3 (s)
→
d) not balanced
balanced equation: 3Mg (s) + N2 (g) →
a) already balanced
b) 2Al (s) + 3H2SO4 (aq)
c) K2SO4 (aq) + BaCl2 (aq)
d) already balanced
→
→
2KCl (s) + 3O2 (g)
2Mg3N2 (s)
Al2(SO4)3 (aq) + 3H2 (g)
BaSO4 (s) + 2KCl (aq)
5.34
a) double replacement
c) double replacement
e) single replacement
g) double replacement
b) combination
d) combination, combustion, oxidation
f) decomposition
5.36
a) CaO (s)
c) KCl and I2
b) PbO and O2
d) CuS (s) and NaCl
5.40
Oxidation: loss of electrons and reduction: gain of electrons
a) Li0 oxidized to Li 1+ and F0 reduced to F1b) Cl0 reduced to Cl 1- and I1- oxidized to I0
c) Zn0 oxidized to Zn2+ and Cu2+ reduced to Cu0
d Fe0 oxidized to Fe2+ and Cu2+ reduced to Cu0
5.44
a) Oxidized: gain of protons (hydrogen is also reduction)
b) reduced
c) redox reactions go together as one species losses (electrons or hydrogen in
oxidation or gains oxygen) and concomitantly the other species gains
(electrons or hydrogen) in reduction.
5.52
The balanced equation:
2C2H2 (g) + 5O2 (g) →
4CO2 (g) + 2H2O (g)
Indicates that 2 molecules of acetylene react with 5 oxygen molecules and give 4
carbon dioxide molecules and 2 water molecules
In other words: 2 moles of C2H2 react with 5 moles of O2 to give 4 moles of CO2
and 2 moles of H2O
Or
2 x 26 grams of gaseous acetylene react with 5 x 32 grams of O2 gas to give 4 x
44 grams of CO2 gas and 2 x 18 grams of H2O vapor.
5molesO2
x X moles of C2H2; therefore 5 moles of O2
2molesC 2 H 2
are produced as moles of C2H2 cancel as they are in both numerator and
denominator.
4molesCO2
b) conversion factor:
x X moles of C2H2; therefore 7 moles of O2
2molesC 2 H 2
as moles of C2H2 cancel as they are in both numerator and denominator
2molesC 2 H 2
c) conversion factor:
x X moles of H2O; therefore 0.5 moles of
2molesH 2 O
C2H2 as moles of H2O cancel as they are in both numerator and denominator
a) conversion factor:
4molesCO2
x X moles of O2; therefore 0.08 moles of CO2
5molesO2
as moles of O2 cancel as they are in both numerator and denominator.
d) conversion factor:
5.60
Balanced equation:
2H2S (g) + 3O2 (g) →
2SO2 (g) + 2 H2O (g).
or
2 x 34 gm of H2S + 3 x 32 gm of O2 → 2 x 64 gm SO2 + 2 x 18 gm H2O
96 gO2
× 2.5 gH 2 S = 3.529 gm of O2
68 gH 2 S
128 gSO2
c)
× 38.5 gO2 = 51.33 gm of SO2
96 gO2
96 gO2
c)
× 55.8 gH 2 O = 148.8 gm of O2
36 gH 2 O
b)
5.66
a) exothermic: energy given out
b) exothermic
c) endothermic: energy must be supplied to carry out the decomposition.
5.70
a) adding NO will drive the reaction forward and the rate will increase
b) lowering the temperature would slow the process and the reaction rate will
decrease
c) removing hydrogen will reduce the reaction rate and when all the hydrogen is
consumed the reaction will stop.
d) adding catalyst will lower the activation energy and the reaction rate will
increase.