Download Devillez (ld2653) – Test 1 Review – Devillez – (99998)

Devillez (ld2653) – Test 1 Review – Devillez – (99998)
This print-out should have 133 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
1
3. 4
4. 3 correct
5. 1
001
10.0 points
A chemist investigates the
I) melting point
II) boiling point
III) flammability
of acetone, a component of fingernail polish
remover. Which is/are physical?
1. III only
2. I, II and III
3. II and III only
4. I and II only correct
5. I only
6. I and III only
7. II only
8. None of these
Explanation:
Flammability is a chemical change; melting and boiling (change of state) are physical
changes.
002 10.0 points
Consider the statement:
“The temperature of the land is an important factor for the ripening of oranges, because it affects the evaporation of water and
the humidity of the surrounding air.”
How many of these factors,
temperature
ripening
evaporation
humidity
are physical properties or changes?
Explanation:
The temperature and humidity are physical
properties and the evaporation of water is a
physical change.
003 10.0 points
All of the following are characteristic properties of phosphorus. Which one is a chemical
property?
1. The white form is soluble in liquid carbon
disulfide, but is insoluble in water.
2. The red form melts at about 600◦ C and
the white form melts at 44◦ C.
3. Both red phosphorus and white phosphorus exist in solid allotropic forms.
4. When exposed to air, white phosphorus
will burn spontaneously, but red phosphorus
will not. correct
5. The red form of phosphorus is insoluble
in both water and carbon disulfide.
Explanation:
Chemical properties are exhibited as matter undergoes chemical changes. The burning
of white or red phosphorus explains its interaction with chemical reactions and changes in
composition which justifies its being a chemical property.
004 10.0 points
Which of the following is an extensive physical
property?
1. reactivity
1. 2
2. color
2. None
3. density
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
2
4. mass correct
5. refractive index
Explanation:
Extensive properties are those which depend on the amount of substance present.
Mass depends on and is directly proportional
to the amount of matter present.
005 10.0 points
Which of the following is an intensive property?
1. number of moles of molecules
2. volume
I
II
III
IV
best characterize(s) a mixture?
3. weight
1. I, II and III only
4. mass
2. I and III only
5. density correct
3. III and IV only correct
4. IV only
Explanation:
Intensive properties are independent of the
amount of matter present. The density of a
substance will be the same regardless of the
size (large or small) of the sample.
006 10.0 points
Which is NOT a pure substance?
1. water
2. chlorine
3. copper
4. brass correct
Explanation:
The dark and light spheres represent atoms
of different elements. A mixture consists of
different substances combined together physically. Although some atoms are chemically
bonded in the figures, the samples do not consist of all of the same type of bonded atoms.
008 10.0 points
Which diagram(s)
best characterize(s) an element?
Explanation:
Brass is an alloy (a physical mixture of the
metals copper and zinc).
007 10.0 points
Which diagram(s)
I
II
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
3
4. I only correct
III
IV
best characterize(s) an element?
Explanation:
The dark and light spheres represent atoms
of different elements. A compound is a combination of two or more elements, chemically
bound together, with the same composition
throughout.
010 10.0 points
Which diagram(s)
1. III and IV only
2. II only correct
3. I only
4. I and II only
Explanation:
The dark and light spheres represent atoms
of different elements. An element consists of
only one kind of atom. In II the atoms happen
to form diatomic molecules, but only one type
of atom is present in II.
009 10.0 points
Which diagram(s)
best characterize(s) a compound?
I
II
III
IV
best represent(s) molecules?
1. II and IV only
2. I only
I
II
3. I and II only correct
4. I and III only
5. III only
6. II and III only
III
1. II and III only
2. II only
3. I and IV only
IV
7. Another combination
8. II only
9. IV only
Explanation:
The dark and light spheres represent atoms
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
of different elements. Molecules consist of two
or more atoms chemically bonded together.
011 10.0 points
Which diagram(s)
4
neous solution. correct
2. A pure substance cannot be heterogeneous.
3. Heterogeneous solutions cannot be pure
substances.
4. Mixtures can be homogeneous.
I
II
Explanation:
A solution, although homogeneous, cannot
be a pure substance since it consists of two
components: the solute and the solvent.
013 10.0 points
Which diagram(s)
III
IV
best represent(s) atoms?
1. I and III only
I
II
2. I and II only
3. I only
4. II only
5. IV only
III
6. III only correct
7. Another combination
8. II and IV only
9. II and III only
Explanation:
The dark and light spheres represent atoms
of different elements. Atoms are single and
unattached to each other.
012 10.0 points
Which of the following statements concerning
substances is NOT true?
1. A pure substance can only be a homoge-
best characterize(s) a pure substance that is
an element?
1. I and III only
2. I and II only
3. II and III only
4. None of these
5. II only correct
6. All of these
7. III only
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
5
8. I only
Explanation:
014 10.0 points
Which of the following is a pure substance?
1. a puddle of milk
2. a rod of steel
3. a block of wood
4. a cube of sugar correct
5. a strap of leather
Explanation:
A cube of sugar is an example of a pure
compound.
015 10.0 points
Which of these substances
I) carbon dioxide gas
II) distilled water
III) jello
is a pure substance?
1. I and II only correct
2. I only
3. II only
4. II and III only
5. All of these
6. III only
7. None of these
8. I and III only
Explanation:
016 10.0 points
What is the proper solution to the following
equation?
(150 + 2.5 + 36.75) ÷ (6.60 + 0.173) =
1. 30
2. 27.942
3. 27.94183
4. 28 correct
5. 27.9
6. 27.94
7. 27.9418
Explanation:
The last position retained for the first addition problem will be the tens position giving
it 2 sigfigs. The last position retained for
the second addition problem will be the hundredths position giving it 3 sigfigs. Therefore
the final answer 27.94 needs to be rounded to
2 sigfigs: 28.
017 10.0 points
Which of the following numbers has the greatest number of significant figures?
1. 130
2. 0.690 correct
3. 1.2
4. 33,000
5. 0.0032
6. 4.3 × 102
Explanation:
Zeros to the left of the decimal are not
significant. Zeros used to place the decimal
are not significant. 0.690 has 3 significant
figures. All the other choices have 2 significant
figures.
018 10.0 points
What is the length of the line segment indicated on the scale provided using proper
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
6
significant figures?
0
1
2
3
3. 3.4308 g
4. 3.431 g correct
1. 3.00 units correct
5. 3.43 g
2. 3.000 units
6. 3 g
Explanation:
sum of given values = ?
3. 3.0 units
4. 3 units
2.636 g + 0.7948 g = 3.431 g
Explanation:
The scale is in units of tenths, so hundreths
is the uncertain digit.
019 10.0 points
What is the proper measurement of the volume represented?
5
mL
4
3
021 10.0 points
Keeping in mind the rules for significant figures, divide 92.80 g by 3.91111 mL.
1. 23.73 g/mL correct
2. 23.7 g/mL
3. 23.727 g/mL
4. 23.7273 g/mL
2
1
5. 24 g/mL
Explanation:
1. 2.5 mL correct
92.80 g (4 sf)
= 23.73 g/mL
3.91111 mL (6 sf)
2. 2.0 mL
3. 2.50 mL
4. 2.500 mL
Explanation:
The scale is in units of 1, so tenths is the
uncertain digit.
020 10.0 points
Keeping in mind the rules for significant digits, calculate the sum of 2.636 g and 0.7948
g.
(round to 4 sig. figs)
022 10.0 points
For the conversion of 0.0003140 kilograms to
ounces, the conversion factors are 103 grams
per kilogram, 1 pound per 453.6 grams and
16 ounces per pound.
The correctly expressed answer is
1. 1.108 × 10−5 oz
2. 4.329 × 10−5 oz
3. 1.108 × 10−2 oz correct
1. 3.4 g
4. 1.108 × 102 oz
2. 3.43080 g
5. 8.902 × 10−1 oz
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
7
Explanation:
4. 2.69 miles/min2
? ounces = 0.0003140 kg
1000 g 1 pound
×
×
kg
453.58 g
16 ounces
×
1 pound
= 1.108 × 10−2 ounces
Zeroes at the end of numbers containing a decimal point are considered to be significant, so
0.0003140 has 4 significant figures. The 1 kg
to 1000 g and 1 pound to 16 ounces conversions are exact definitions, so these numbers
do not affect how many significant figures the
answer should have.
023 10.0 points
How many milliliters are in a 67.8 ft3 box?
1. 1.92×106 mL correct
2. 13300 mL
3. 0.00239 mL
4. 2070 mL
5. 34.7 miles/min2 correct
6. 1.15 miles/min2
7. 1.49 miles/min2
Explanation:
Acceleration rate = 15.5 m/s2 .
1 km
1 mile
m
×
15.5 2 ×
s
1000 m 1.609 km
miles
(60 s)2
= 34.7
×
2
(1 min)
min2
10.0 points
025
Given:
1 pound (lb) = 453.6 grams
1 pound = 16 ounces (oz)
How many ounces are equal to 0.0003140
kilograms? Taking significant figures into account, the correctly expressed answer is
1. 1.1076 × 10−2 oz
2. 1 × 10−2 oz
3. 1.108 × 10−2 oz correct
4. 1.1 × 10−2 oz
5. 7150 mL
Explanation:
(12 in)3
1 ft3
1 mL
(2.54 cm)3
×
×
3
1 in
1 cm3
6
= 1.92 × 10 mL
? mL = 67.8 ft3 ×
024 10.0 points
An Olympic hurdler accelerates at a rate of
15.5 m/s2 . What is the rate in miles/min2 ?
1. 31.8 miles/min2
2. 6.89 miles/min2
3. 0.581 miles/min2
5. 1.11 × 10−2 oz
Explanation:
? ounces = 0.0003140 kg
×
1000 g 1 pound
×
kg
453.58 g
×
16 ounces
1 pound
= 1.108 × 10−2 ounces
Zeroes at the end of numbers containing a decimal point are considered to be significant, so
0.0003140 has 4 significant figures. The 1 kg
to 1000 g and 1 pound to 16 ounces conversions are exact definitions, so these numbers
are not included when deciding how many
significant figures the answer should have.
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
026 10.0 points
An oxygen molecule at 25◦ C has a speed of
about 927 meters per second. Express this
speed in miles/hour. (1 mi = 5280 ft)
1. 13400 miles/hour
d=
8
4.2 mol CH3OH
5.67 L
×
1L
32 g CH3 OH
×
1 mol CH3 OH 1000 mL
= 0.0237 g/mL
2. 0.000482 miles/hour
028 10.0 points
Polycarbonate plastic has a density of 1.3
g/cm3 . A photo frame is constructed from
two 6.2 mm sheets of polycarbonate. Each
sheet measures 28 cm by 24 cm.
What is the mass of the photo frame?
3. 20700 miles/hour
4. 2070 miles/hour correct
5. 0.000160 miles/hour
6. 0.576 miles/hour
1. 730 g
7. 3340000 miles/hour
2. 1100 g correct
3. 1000 g
Explanation:
m
T = 25◦ C
v = 927
s
1 in
1 ft
m 100 cm
×
×
×
927
s
1m
2.54 cm 12 in
1 mi
3600 s
×
= 2073.64 mi/h
5280 ft
1 hr
027 10.0 points
Calculate the density of methanol (CH3 OH) if
you know that there are 4.2 moles of methanol
in 5.67 L.
4. 850 g
5. 1200 g
Explanation:
d = 1.3 g/cm3
w = 24 cm
ℓ = 28 cm
h = 2(6.2 mm) = 1.24 cm
V = length × width × height
m
V
m = d V = d (ℓ w h)
= (1.3 g/cm3 ) (28 cm) (24 cm) (1.24 cm)
= 1083.26 g
d=
1. 0.741 g/mL
2. 0.0237 g/mL correct
3. 1334 g/mL
4. 0.0431 g/mL
029 10.0 points
The density of gold is 19.3 g/mL. What is the
mass of a gold nugget which has a volume of
3.28 mL?
5. 0.131 g/mL
6. 11.34 g/mL
1. 5.88 g
2. 0.170 g
Explanation:
n = 4.2 mol
V = 5.67 L
3. 63.3 g correct
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
9
5. 797 ◦ F
4. 30.4 g
Explanation:
Explanation:
030 10.0 points
A solid has a mass of 95 grams. A graduated
cylinder contains 35 mL of water. When the
solid is completely submerged in the water,
the new water level reads 84 mL.
What is the density of the solid?
1. 2.7 g/mL
2. 0.80 g/mL
3. 1.9 g/mL correct
4. -1.9 g/mL
5. 1.1 g/mL
6. 190 g/mL
7. 0.52 g/mL
Explanation:
031 10.0 points
On the Kelvin scale, a temperature of 43◦ C is
1. 43 K.
2. 316 K. correct
3. 273 K.
4. 230 K.
Explanation:
032 10.0 points
Copper melts at 1083 ◦ C. What is its melting
temperature in ◦ F?
1. 583 ◦ F
2. 1981 ◦ F correct
3. 619 ◦ F
? ◦ F = 1083◦ C ×
1.8◦ F
+ 32◦ F = 1981◦ F
1.0◦ C
033 10.0 points
Identify the incorrect statement regarding
Dalton’s Atomic Theory.
1. A given compound always has the same
relative numbers and types of atoms.
2. All atoms of a given element are identical.
3. All matter is composed of atoms.
4. All matter is composed of protons, electrons, and neutrons. correct
Explanation:
In the early 1800s when Dalton formulated
his atomic theory, it was believed that all
atoms of the same element were the same.
The fundamental particles of the atom –
protons, neutrons, and electrons – were not
known at this time.
034 10.0 points
The following are all proposals of Dalton’s
atomic theory. Which part of Dalton’s atomic
theory was shown to be wrong by the discovery of natural radioactivity?
1. Atoms combine in simple numerical ratios
to form compounds.
2. Atoms are permanent, unchanging, indivisible bodies. correct
3. Two or more kinds of atoms may combine
in different ways to form more than one kind
of chemical compound.
4. Atoms of a given element all weigh the
same.
4. 1324 ◦ F
5. Compounds consist of collections of
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
molecules made up of atoms bonded together.
Explanation:
The discovery of natural radioactivity provided evidence that atoms can change. They
can undergo radioactive decay, emitting electrons, protons, or neutrons. They can also
undergo nuclear fission, in which a heavy nucleus splits into two lighter nuclei.
035 10.0 points
The discovery and characterization of cathode
rays was important in the development of the
atomic theory because
1. it led to the suggestion of the existence of
the neutron.
2. it indicated that all matter contained electrons. correct
3. it indicated that all matter contains protons.
4. None of the these is correct.
5. it indicated that all matter contains alpha
particles.
Explanation:
The cathode ray experiment results were
the same, even when different types of gas,
composition of electrodes, and power sources
were used.
036 10.0 points
Rutherford’s α-scattering experiment showed
that most α particles directed toward a thin
metallic foil passed through with only slight
deviations. From this evidence we can conclude that
1. the mass of the atom is concentrated in a
very small area. correct
2. an α particle is a type of light which
should not be deflected.
3. the foil was too thin.
10
4. α particles are uncharged.
5. α particles are too small to hit anything.
Explanation:
In Rutherford’s gold foil experiment in
1910, α (alpha) particles were fired at gold
foil, and the resulting deflection of the particles were observed. Most of the α particles
went through the sample undeflected, suggesting that much of the atom was empty
space. But of the few α particles that were
deflected, they were deflected at all angles,
including some very wide angles! The wide
deflections suggested a very hard (dense) positively charged core in the atom. However,
this core, or nucleus, must be small in relation
to the overall size of the atom, since so few
of the α particles were deflected in the first
place. It also suggests that the nucleus is surrounded by a cloud of electrons at relatively
great distances from the nuclei.
037 10.0 points
What experimental evidence led Thomson to
conclude that cathode rays were negatively
charged?
1. Cathode rays were deflected towards a
negatively charged plate.
2. Cathode rays pushed a paddle wheel
counterclockwise.
3. Cathode rays pushed a paddle wheel
clockwise.
4. False; cathode rays are positively charged,
not negatively charged.
5. Cathode rays were deflected towards a
positively charged plate. correct
Explanation:
Cathode rays are repelled by negative
plates and attracted to positive plates. Since
opposites attract, cathode rays must be negatively charged.
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
11
038 10.0 points
Identify the experimental evidence that indicates an atom has a positively charged nucleus.
about the same. correct
1. Electrons can be ejected from a metallic
surface with high energy light.
Explanation:
A proton weighs 1 amu, as does a neutron.
1
An electron weighs
amu. Hydrogen
2000
has one proton, one electron, and usually no
neutrons, so H weighs about 1 amu.
2. α particles are deflected at large angles
when projected toward a thin sheet of metal.
correct
3. Cathode rays produce diffraction patterns.
4. Line spectra are produced from elemental
gases in a gaseous discharge tube.
4. The mass of a hydrogen atom is about the
same as a proton.
040 10.0 points
How many electrons are present in one neon
atom?
1. 10 correct
2. 9
5. Cathode rays are attracted to the positive
plate of an applied electrical field.
3. 11
Explanation:
In Rutherford’s gold foil experiment in
1910, α (alpha) particles were fired at gold
foil, and the resulting deflection of the particles were observed. Most of the α particles
went through the sample undeflected, suggesting that much of the atom was empty
space. But of the few α particles that were
deflected, they were deflected at all angles,
including some very wide angles! The wide
deflections suggested a very hard (dense) positively charged core in the atom. However,
this core, or nucleus, must be small in relation
to the overall size of the atom, since so few
of the α particles were deflected in the first
place. It also suggests that the nucleus is surrounded by a cloud of electrons at relatively
great distances from the nuclei.
4. 8
039 10.0 points
Which of the following is FALSE?
1. The mass of a proton is much greater than
the mass of an electron.
2. The mass of protons and neutrons are
about the same.
3. The mass of neutrons and electrons are
5. 12
Explanation:
The atomic number of an element is equal
to the number of protons in an atom. In a
neutral atoms the number of protons is equal
to the number of electrons. Neon has an
atomic number of 10.
041 10.0 points
How many protons are present in one iron(III)
ion?
1. 25
2. 28
3. 26 correct
4. 27
5. 29
Explanation:
The atomic number of an element is equal
to the number of protons in that element’s
nucleus. The number of protons defines an element and cannot be changed without changing the element.
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
Iron has an atomic number of 26.
042 10.0 points
How many protons are present in one Ca2+
ion?
1. 19
2. 22
3. 20 correct
4. 21
5. 18
Explanation:
The atomic number of an element is equal
to the number of protons in that element’s
nucleus. The number of protons defines an element and cannot be changed without changing the element.
Calcium has an atomic number of 20.
043 10.0 points
What is the symbol for the element which
contains 5 electrons in the neutral state?
1. C
2. B correct
3. Be
4. Li
Explanation:
Boron has 5 protons and in its neutral state,
has 5 electrons as well.
044 10.0 points
A magnesium (Mg) atom with a mass number of 25 contains
1. 13 protons, 12 neutrons and 12 electrons.
2. 12 protons, 13 neutrons and 12 electrons.
correct
3. 25 protons, 25 neutrons and 25 electrons.
12
4. None of these
5. 25 protons, 13 neutrons and 25 electrons.
Explanation:
045 10.0 points
Identify the isotope that has atoms with 12
neutrons, 10 protons, and 10 electrons.
1. 22 Na
2. 10 S
3. None of these
4. 22 Ne correct
5. 10 P
6. 12 Mg
7. 12 Al
Explanation:
The element Ne is identified by the number
of protons (atomic number), and the syntax
is n+pp Nep−e .
046 10.0 points
Write the hyphen notation for the element
that contains 15 electrons and 15 neutrons.
1. copper-60
2. phosphorus-30 correct
3. silicon-30
4. fluorine-15
5. sulfur-31
6. phosphorus-31
7. oxygen-15
8. zinc-60
Explanation:
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
#e− = 15
#n = 15
hyphen notation for given element = ?
atomic number = number of protons
= number of electrons
mass number
= number of protons + number of neutrons
atomic number = 15 (element is phosphorus)
mass number
= 15 protons + 15 neutrons = 30
nuclide is phosphorus-30
047 10.0 points
Choose the correct nuclear symbol and hyphen notation for the isotope which has a
mass number of 28 and atomic number of 14.
1. 14
28 S, sulfur-14
2. 14
14 P, phosphorus-14
3. 14
28 P, phosphorus-14
4. 14
14 Si, silicon-14
5. 28
14 P, phosphorus-28
6. 28
14 Si, silicon-28 correct
7. 14
28 Si, silicon-14
8. 14
14 S, sulfur-14
9.
28
14 S,
with 57.25% abundance. The other has an
atomic weight of 122.8831 amu. What is the
atomic weight of the element?
1. 122.38 amu
2. 123.45 amu
3. 121.17 amu
4. 122.15 amu
5. 121.54 amu
6. 121.75 amu correct
Explanation:
Average Atomic Mass
= 0.5725(120.9038) + 0.4275(122.8831)
= 121.75 amu
049 10.0 points
Calculate the molar mass of the noble gas
krypton in a natural sample, which is 0.81%
78
Kr (molar mass 77.92 g/mol), 1.82% 80 Kr
(molar mass 79.91 g/mol), 10.92% 82 Kr (molar mass 81.91 g/mol), 12.42% 83 Kr (molar
mass 82.92 g/mol), 56.65% 84 Kr (molar mass
83.91 g/mol), and 17.38% 86 Kr (molar mass
85.91 g/mol).
Correct answer: 83.7949 g/mol.
Explanation:
sulfur-28
Explanation:
mass number = 28
atomic number = 14
nuclear symbol and hyphen notation
for given nuclide = ?
atomic number = number of protons
element with 14 protons: silicon (Si)
nuclear symbol −→ 28
14 Si
hyphen notation −→ silicon-28
048 10.0 points
A newly discovered element has two isotopes.
One has an atomic weight of 120.9038 amu
13
MMkrypton = 0.0081 (77.92 g/mol)
+ 0.0182 (79.91 g/mol)
+ 0.1092 (81.91 g/mol)
+ 0.1242 (82.92 g/mol)
+ 0.5665 (83.91 g/mol)
+ 0.1738 (85.91 g/mol)
= 83.7949 g/mol
050
10.0 points
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
Which of the following statements about a
nonmetal is true?
14
5. C only
6. B only
1. A nonmetal forms covalent bonds with
metals.
2. A nonmetal has metallic bonding.
3. A nonmetal tends to form cations.
7. None of these
8. A and C only
Explanation:
The transition metals lie in the shaded area
4. A nonmetal has more than 3 electrons in
its outer shell. correct
Explanation:
Nonmetals start in group VI of the periodic
table, so they have more than 3 electrons in
their outer shells.
051 10.0 points
Which characteristic describes a metal?
053 10.0 points
Classify the elements Si, Sr, Sn, Sb.
1. brittle
1. metal; nonmetal; metal; nonmetal
2. malleable correct
2. metalloid; metal; metal; nonmetal
3. good insulator
3. nonmetal; nonmetal; metal; metalloid
4. low conductivity
4. None of these
5. tend to be reduced
Explanation:
Metals are lustrous, malleable, ductile,
good conductors, and tend to be oxidized.
052
10.0 points
Identify the transition metals.
A. hafnium
B. radium
C. radon
5. metalloid; metal; metal; metalloid correct
Explanation:
Silicon (Si) is a group 14 metalloid, strontium (Sr) is a group 2 metal, tin (Sn) is a
group 14 metal and antimony (Sb) is a group
15 metalloid.
054 10.0 points
List the elements that are members of the
alkali metals.
1. B and C only
1. C, Si, Ge, Sn, Pb
2. A only correct
2. Li, Na, K, Rb, Cs, Fr correct
3. A, B, and C
3. B, Al, Ga, In, Tl
4. A and B only
4. None of these
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
5. Be, Mg, Ca, Sr, Ba, Ra
3. those on opposite sides
6. N, P, As, Sb, Bi
4. those in the same rows
Explanation:
Elements in the same group have the same
number of valance (outer) electrons and often
react in similar ways.
7. F, Cl, Br, I, At
8. He, Ne, Ar, Kr, Xe, Rn
9. O, S, Se, Te, Po
Explanation:
The name for the group I metal is alkali metals
055
15
057 10.0 points
The letters in the periodic table are used to
identify four different elements.
10.0 points
D
C
Consider the periodic table
B
A
X
Y
Z
Based on the organization within the periodic table, which element is expected to have
2+ charge when it forms an ion?
1. C
Identify the nonmetal.
1. Z correct
2. D
2. X
3. B
3. Y
4. A correct
Explanation:
The nonmetals lie in the shaded area:
X
Explanation:
Group II elements form ions with a 2+
charge.
Y
Z
058 10.0 points
Of the following elements, the one which is
NOT a metal is
1. potassium.
2. iron.
056 10.0 points
In the periodic table, which elements typically
have similar properties?
1. those in the same columns correct
2. those related diagonally
3. cobalt.
4. arsenic. correct
5. silver.
Explanation:
Iron, cobalt, silver, and potassium are all
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
metals since they are to the left of the line
dividing metals and non-metals. Arsenic is
the only element to the right of that line.
3. a transition.
059 10.0 points
Of the following elements, the one which is
NOT a nonmetal is
5. an orbital.
1. phosphorous.
2. bismuth. correct
3. sulfur.
4. silicon.
5. iodine.
Explanation:
Only bismuth is a metal in the list of elements because it is to the left of the line
separating the metals and non-metals.
060 10.0 points
Which of the following elements would be
expected to resemble strontium (Sr) most
closely in chemical properties?
1. Si
2. Ba correct
4. a lanthanide.
Explanation:
Horizontal rows on the periodic table are
called periods. The vertical columns are
groups or families.
062 10.0 points
In the periodic table, a vertical column of
elements is called
1. a lanthanide.
2. a group. correct
3. a period.
4. an orbital.
5. a transition.
Explanation:
Horizontal rows on the periodic table are
called periods. The vertical columns are
groups or families.
063 10.0 points
Arsenic (As) is classified as
3. Li
1. a transition element.
4. Cs
2. an actinide.
5. Sc
3. a lanthanide.
Explanation:
Elements with similar chemical properties
are arranged in the same group or vertical column. Thus strontium (Sr) would be expected
to have similar properties to barium (Ba).
061 10.0 points
In the periodic table, a horizontal row of elements is called
1. a group.
2. a period. correct
16
4. a noble gas.
5. a representative element. correct
Explanation:
Representative elements are found in the A
Groups of the periodic table. As is a Group
VA element.
064 10.0 points
What is the formula for the ionic compound
formed from sodium and oxide ions?
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
17
1. NaO2
7. Ca3 N
2. None of these
3. Na3 O2
4. NaO
5. Na2 O correct
6. Na2 O3
Explanation:
Ca is in Group 2 so it forms a +2 ion; N is
in Group 5 so it forms a −3 ion. To balance
charges you need three Ca2+ for each two N3− .
067 10.0 points
What is the formula of the compound made
by combining calcium (Ca) with sulfur (S)?
Explanation:
The sodium ion is Na+ and the oxide ion is
O2− , so the formula is Na2 O.
1. SCa2
065 10.0 points
What is the formula for the ionic compound
formed from lithium and sulfide ions?
3. SCa
1. LiS
2. None of these
2. CaS correct
4. CaS2
Explanation:
Ca has two valence electrons and forms a
2+ ion. S has six valence electrons and forms
a 2− ion. The ratio will be 1 : 1, so the
formula is CaS.
3. Li4 S
4. Li2 S correct
5. Li2 SO4
6. Li2 S3
Explanation:
The lithium ion is Li+ and the sulfide ion is
S2− , so the formula is Li2 S.
066 10.0 points
Write the formula of calcium nitride.
1. Ca2 N
2. Ca3 N2 correct
3. Ca2 N3
068 10.0 points
What is the formula for the salt which contains both Ca and NO3 ions? (Note that the
charges of the ions are not indicated.)
1. Ca(NO3 )2 correct
2. Ca3 (NO3 )2
3. Ca2 (NO3 )2
4. Ca2 NO3
5. CaNO3
Explanation:
The calcium ion is Ca2+ and the nitrate
−
ion is NO−
3 . Two NO3 are needed to balance
the charge on each Ca2+ . The formula is
Ca(NO3 )2 .
4. None of these
5. CaN2
6. CaN3
069 10.0 points
Which of the following ionic compounds has
the wrong chemical formula?
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
18
1. NaO2 correct
2. KBr
3. K2 S
4. CaO
072 10.0 points
An element (call it X) in Group IA of the periodic table forms a compound with an element
(call it Y) of Group VIA. What would be the
most likely formula and nature of the compound?
1. X2 Y; covalent
5. MgCl2
Explanation:
The sodium ion is Na1+ and the oxide ion
is O2− . The compound should thus be Na2 O.
070 10.0 points
Which of the following species has the greatest
number of electrons?
1. Cl−
2. F−
3. O2−
4. S2−
5. I− correct
Explanation:
The atomic number is the number of protons in an atom. In a neutral atom this also
equals the number of electrons. For ions you
must account for the lost or gained electrons
to find the total number. O2− has 10 electrons, F− has 10 electrons, S2− has 18 electrons, Cl− has 18 electrons, and I− has 54
electrons.
071 10.0 points
Which of the following species has the greatest
number of electrons?
1. Al
2. Na+
3. F−
4. Fe2+ correct
5. Ne
Explanation:
2. XY2 ; ionic
3. XY6 ; ionic
4. X6 Y; ionic
5. X2 Y; ionic correct
6. XY2 ; covalent
7. XY6 ; covalent
Explanation:
Metals reacts with non-metals to form ionic
compounds, which involve the transfer of electrons.
A metal X in group IA would have one
valence electron. It would lose one electron
to become isoelectronic with the preceding
noble gas. This metal would now have one
fewer electron than protons and would have a
+1 charge.
An element Y in group VIA would have six
valence electrons. It would gain two electrons
to become isoelectronic with the nearest noble
gas. This element would now have two more
electrons than protons and would have a −2
charge.
We need twice as many X+ ions as Y2− ions
to equalize the overall charge, so the ratio of
X to Y ions is 2:1, and the formula would be
X2 Y.
073 10.0 points
Choose that pair of names and formulas that
do not match.
1. Fe2 (SO4 )3 : iron(III) sulfate
2. SnCl4 : tin(IV) chloride
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
3. MnCO3 : manganese(II) carbonate
4. Cu2 O : copper(II) oxide correct
5. SnO : tin(II) oxide
Explanation:
The copper(II) ion is Cu2+ ; the oxide ion
is O2− . One O2− is needed to balance the
charge of each Cu2+ , so the correct formula
for copper(II) oxide is CuO.
In Cu2 O, the total charge from copper ions
must be +2 to balance the −2 charge from
the oxide ion. Each copper ion would have
a charge of +1. This compound is copper(I)
oxide.
The manganese(II) ion is Mn2+ ; the carbonate ion is CO3 2− . One CO3 2− is needed
to balance the charge of each Mn2+ , so the
formula is MnCO3 .
The tin(IV) ion is Sn4+ ; the chloride ion
is Cl− . Four Cl− are needed to balance the
charge of each Sn4+ , so the formula is SnCl4 .
The tin(II) ion is Sn2+ ; the oxide ion is O2− .
One O2− is needed to balance the charge of
each Sn2+ , so the formula is SnO.
The iron(III) ion is Fe3+ ; the sulfate ion is
SO4 2− . Three SO4 2− are needed to balance
the charge of every two Fe3+ . (This gives a
total anion charge of −6 and a total cation
charge of +6.) The formula is Fe2 (SO4 )3 .
074 10.0 points
Which of the following pairs of names and
formulas are correct?
Z1) dinitrogen tetroxide : N2 O4
Z2) calcium hydroxide : CaOH
Z3) tin : Sn
19
The calcium ion is Ca2+ ; the hydroxide ion
is OH− . Two OH− are needed to balance the
charge on each Ca2+ , so the formula should
be Ca(OH)2 .
N2 O4 is a covalent compound and is correctly named using prefixes to indicate the
number of atoms of each element in the compound.
Sn is the symbol for tin.
075 10.0 points
The binary compound PCl3 is called
1. phosphorus chloride.
2. None of these.
3. monophosphorus trichloride.
4. triphosphorus chloride.
5. phosphorus trichloride. correct
Explanation:
PCl3 is a covalent compound and is therefore named using prefixes to indicate the number of each type of atom. The prefix for three
is “tri” so this molecule is named phosphorous
trichloride. The “mono” prefix is commonly
left off of the first element name in a binary
compound.
076 10.0 points
What is the name of the compound with the
formula CCl4 ?
1. carbon tetrachloride correct
2. carbon chloride
1. Z1) and Z3) correct
3. chlorine carbonide
2. Z1) only
4. carbon(IV) chloride
3. Z1) and Z2)
4. Z1), Z2), and Z3)
5. Z3) only
Explanation:
Explanation:
This is a covalent compound and should be
named using prefixes to indicate the number
of atoms of each element in the compound.
Binary covalent compounds are named with
the name of the first element followed by the
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
name of the second element with the suffix
“-ide” added. CCl4 is carbon tetrachloride.
077 10.0 points
Write the formula of perchloric acid.
20
1. H2 SO4 : sulfurous acid correct
2. HIO (or HOI) : hypoiodous acid
3. HNO3 : nitric acid
1. HClO2
2. HClO
3. HClO5
4. HClO4 correct
5. HClO3
Explanation:
Change ‘-ic’ to ‘-ate’. The perchlorate an+
ion is ClO−
4 . Now add enough H to make
the overall species neutral.
078 10.0 points
What is the chemical formula for hypopophosphorous acid?
4. HMnO4 : permanganic acid
Explanation:
Sulfuric acid is H2 SO4 . Sulfurous acid
has one fewer oxygen atom, so its formula
is H2 SO3 .
Iodic acid is HIO3 . Hypoiodous acid has
two fewer oxygen atoms, so its formula is HIO.
MnO−
4 is the permanganate ion. HMnO4 is
permanganic acid.
NO−
3 is the nitrate ion. HNO3 is nitric acid.
080 10.0 points
Pick the name and formula that do NOT
match.
1. HCl : hydrochloric acid
1. H2 PO
2. H2 CO3 : carbonic acid
2. H2 PO4
3. H2 SO4 : sulfuric acid
3. H2 PO5
4. HClO4 : chloric acid correct
4. HPO5
5. HPO8
6. HPO
7. H2 PO2
8. H2 PO3
9. H2 PO6
10. H3 PO2 correct
Explanation:
Cloric acid is HClO3 . P erchloric acid
has one more oxygen atom, so its formula
is HClO4 .
HCl is hydrogen chloride. In aqueous solution, HCl is known as hydochloric acid.
SO2−
4 is the sulfate ion so H2 SO4 is sulfuric
acid.
CO2−
is the carbonate ion so H2 CO3 is
3
carbonic acid.
081 10.0 points
Which one has the greatest number of atoms?
Explanation:
1. 3.05 moles of CH4 correct
079 10.0 points
Pick the name and formula that do NOT
match.
2. 3.05 moles of water
3. 3.05 moles of helium
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
4. All have the same number of atoms
5. 3.05 moles of argon
How many moles of H2 O are in 1.8 g of
AuCl3 · 2 H2 O?
Correct answer: 0.0106085 mol.
Explanation:
mAuCl3 ·2 H2 O = 1.8 g
Explanation:
For 3.05 moles of water:
? atoms = 3.05 mol H2O ×
21
6.02 × 1023 molec
1 mol
3 atoms
1 molecule
= 5.51 × 1024 atoms
×
MWAuCl3 ·2 H2 O = 196.97 g/mol
+ 3 (35.45 g/mol)
+ 4 (1.00079 g/mol)
+ 2 (16.0 g/mol)
= 339.352 g/mol
For 3.05 moles of CH4 :
6.02 × 1023 molec
? atoms = 3.05 mol CH4 ×
1 mol
5 atoms
×
1 molecule
= 9.18 × 1024 atoms
nH 2 O
1.8 g AuCl3 · 2 H2 O
=
339.352 g/mol
2 mol H2 O
×
1 mol AuCl3 · 2 H2 O
= 0.0106085 mol H2 O
For 3.05 moles of helium:
6.02 × 1023 atoms
1 mol
atoms
? atoms = 3.05 mol He ×
= 1.84 × 1024
For 3.5 moles of argon:
1. 1.5 × 1023 atoms
6.02 × 1023 atoms
1 mol
atoms
? atoms = 3.05 mol Ar ×
= 1.84 × 1024
082 (part 1 of 2) 10.0 points
How many moles of Ag+ are in 3 g of
AgCl?
Correct answer: 0.0209322 mol.
Explanation:
mAgCl = 3 g
MWAgCl = 107.87 g/mol + 35.45 g/mol
= 143.32 g/mol
nAgCl =
084 10.0 points
How many fluorine atoms are in 4.0 moles of
fluorine molecules?
3g
= 0.0209322 mol
143.32 g/mol
The number of moles of Ag+ ions equals the
number of moles of AgCl.
083 (part 2 of 2) 10.0 points
2. 6.6 × 10−24 atoms
3. 2.4 × 1024 atoms
4. 4.8 × 1024 atoms correct
Explanation:
nF = 4.0 mol
Fluorine is diatomic. Each F2 molecule
contains two fluorine atoms. We can use
Avogadro’s number and the ratio of F atoms
to F2 molecules to find the number of fluorine
atoms:
? atoms F = 4.0 mol F2
6.022 × 1023 F2
×
1 mol F2
2 atoms F
×
1 molec F2
= 4.8 × 1024 atoms F
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
22
= 2.41 × 1024 H atoms
085 10.0 points
Each mole of Al(NO3 )3 contains how many
moles of oxygen atoms?
1. 12 mol
087 10.0 points
How many moles of carbon are in 8.00 moles
of C12 H22 O11 ?
2. 6 mol
1. 8.00 mol
3. 1 mol
2. 88.0 mol
4. 3 mol
3. 96.0 mol correct
5. 9 mol correct
4. 10.0 mol
Explanation:
There are 9 O atoms in each molecule of
Al(NO3)3 . This same ratio holds for the number of moles of O in one mole of Al(NO3)3 .
9O
? mol O
=
mol Al(NO3 )3
1 mol Al(NO3 )3
Therefore there are 9 mol O in one mole of
Al(NO3)3 .
086 10.0 points
How many atoms of hydrogen are contained
in 1 mole of methane (CH4 )?
1. 6.02 × 10
23
atoms
2. The correct answer is not given.
3. 2.41 × 1024 atoms correct
4. 4 atoms
5. 3.01 × 1024 atoms
Explanation:
n = 1 mol
Each methane molecule contains 4 hydrogen atoms. There are Avogadro’s number of
methane molecules in one mole of methane
molecules:
nH = 1 mol CH4
6.02 × 1023 molec CH4
×
1 mol CH4
4 H atoms
×
1 molec CH4
Explanation:
n = 8.00 mol
In 1 mole of C12 H12 O11 , there are 12 moles
of C, 22 moles of H, and 11 moles of O. If you
have 8 moles of C12 H12 O11 , then
12 mol C
x
=
1 mol C12 H22 O11
8 mol C12 H22 O11
12 mol C
x=
1 mol C12 H22 O11
× 8 mol C12 H22 O11
= 96 mol C
088 10.0 points
How many sulfur atoms are in 78.4 g of sulfur?
Correct answer: 1.47235 × 1024 atoms.
Explanation:
mSu = 78.4 g
To solve this problem, we will need both the
atomic mass of sulfur (S, 32.066 g/mol) and
Avogadro’s number (6.022×1023 atoms/mol).
First convert grams to moles of sulfur using
the atomic mass:
1 mol S
? mol S = 78.4 g S ×
32.066 g S
= 2.44496 mol S
Now convert moles to the number of atoms
using Avogadro’s number:
? atoms S = 2.44496 mol S
6.022 × 1023 S atoms
×
1 mol S
= 1.47235 × 1024 atoms S
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
23
Explanation:
089 10.0 points
How many moles are in 418.8 g of Ba(OH)2?
Correct answer: 2.44412 mol.
Explanation:
mBa(OH)2 = 418.8 g
137.33 g Ba
1 mol Ba ×
mol Ba
16.00 g O
2 mol O ×
mol O
1.01 g H
2 mol H ×
mol H
Molar mass
n=?
= 137.33 g
= 32.00 g
1. 1.68% correct
=
2.02 g
2. 0.413%
= 171.35 g /mol
3. 42.8%
418.8 g
171.35 g/mol Ba(OH)2
= 2.44412 mol Ba(OH)2
nBa(OH)2 =
090 10.0 points
How many moles are in 158.55 g of
Fe3 (PO4 )2 ?
Correct answer: 0.443447 mol.
Explanation:
mFe3 (PO4 )2 = 158.55 g
55.85 g Fe
3 mol Fe ×
mol Fe
30.97 g P
2 mol P ×
mol P
16.00 g O
8 mol O ×
mol O
Molar mass
092 10.0 points
Although coal is not a pure chemical compound, its elemental composition may be
approximated by the formula C135 H96 NO9 S.
What is the approximate percentage by mass
of sulfur in coal?
n=?
= 167.6 g
= 61.94 g
= 128.00 g
4. 3.11%
Explanation:
FWcoal = 135(12.0107 g/mol)
+ 96(1.00794 g/mol)
+ 1(14.0067 g/mol)
+ 9(15.9994 g/mol)
+ 1(32.065 g/mol)
= 1908.27 g/mol
1(32.065 g/mol)
% S in coal =
× 100%
1908.27 g/mol
= 1.68032 %
093 (part 1 of 2) 10.0 points
Zinc chloride (ZnCl2 ) is 52.02% chlorine by
mass. What mass of chlorine is contained in
61.7 g of ZnCl2 ?
= 357.54 g /mol
Correct answer: 32.0963 g.
158.55 g
357.54 g/mol Fe3 (PO4 )2
= 0.443447 mol
nFe3 (PO4 )2 =
091 10.0 points
Aspirin has a formula C9 H8 O4 . What is the
molar mass of aspirin?
1. 95 g
2. 220 g
Explanation:
% mass of chlorine in ZnCl2 = 52.02%
mCl in 61.7 g of ZnCl2 = ?
mCl = mass % × mass ZnCl2
= 0.5202 × 61.7 g
= 32.0963 g Cl
3. 325 g
094 (part 2 of 2) 10.0 points
How many moles of Cl is this?
Correct answer: 0.905397 mol.
4. 180 g correct
Explanation:
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
mass Cl
molar mass of Cl
32.0963 g Cl
=
35.45 g Cl
= 0.905397 mol Cl
nCl =
095 10.0 points
What is the percentage of hydrogen in
C7 H8 O? The molecular weight of C is
12.011 g/mol, of H 1.00794 g/mol, and of
O 15.9994 g/mol.
Correct answer: 7.45656%.
Explanation:
MWC = 12.011 g/mol
MWH = 1.00794 g/mol
MWO = 15.9994 g/mol
24
097 (part 1 of 3) 10.0 points
a) How many CaH2 formula units are present
in 6.341 g of CaH2 ?
Correct answer: 9.08837 × 1022 formula units.
Explanation:
mCaH2 = 6.341 g
MWCaH2 = 40.0 g/mol + 2 (1.0079 g/mol)
= 42.0158 g/mol
n=
6.341 g
42.0158 g/mol
× 6.022 × 1023 formula units/mol
= 9.08837 × 1022 formula units
MWtotal = #C · MWC + #H · MWH
+ #O · MWO
= 7(12.011 g/mol)
+ 8(1.00794 g/mol)
+ 1(15.9994 g/mol)
= 108.14 g/mol ,
so the percentage of hydrogen in 0-cresol is
#H · MWH
100.0%
MWtotal
8 (1.00794 g/mol)
=
100.0%
108.14 g/mol
= 7.45656%
%=
096 10.0 points
What mass of rhodium contains as many
atoms as there are iron atoms in 28 g of iron?
098 (part 2 of 3) 10.0 points
b) Find the mass of 6.22 × 1024 formula units
of NaBF4 (sodium tetrafluroborate).
Correct answer: 1134.1 g.
Explanation:
nNaBF4 = 6.22 × 1024 formula units
MWNaBF4 = 22.99 g/mol + 10.81 g/mol
+ 4 (19.0 g/mol)
= 109.8 g/mol
6.22 × 1024 formula units
units
6.022 × 1023 formula
mol
× (109.8 g/mol)
= 1134.1 g
mNaBF4 =
Correct answer: 51.596 g.
Explanation:
mFe = 28 g
mRh
28 g Fe
=
55.847 g/mol N
× (102.91 g/mol Rh)
= 51.596 g Rh
099 (part 3 of 3) 10.0 points
c) How many moles of CeI3 are in 9.66 ×
1021 formula units of CeI3 (cerium(III) iodide;
a bright, water, soluble solid).
Correct answer: 0.0160412 mol.
Explanation:
nCeI3 = 9.66 × 1021 formula units
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
9.66 × 1021 formula units
6.022 × 1023 formula units/mol
= 0.0160412 mol
nCeI3 =
100 (part 1 of 3) 10.0 points
Find the mass of 2.72 mol Cl2 .
Correct answer: 193.12 g.
Explanation:
nCl2 = 2.72 mol
71 g Cl2
m = (2.72 mol Cl2 )
mol Cl2
= 193.12 g Cl2
101 (part 2 of 3) 10.0 points
Find the mass of 2.65 × 1023 molecules H2 S.
Correct answer: 14.9618 g.
Explanation:
nH2 S = 2.65 × 1023 molec
m=?
m = 2.65 × 1023 molec
34 g H2 S
mol
·
6.022 × 1023 molec
mol
= 14.9618 g H2 S
102 (part 3 of 3) 10.0 points
Find the mass of 17 molecules SO2 .
Correct answer: 1.80671 × 10−21 g.
Explanation:
nSO2 = 17 molec
m=?
mol
m = (17 molec)
6.022 × 1023 molec
64 g SO2
·
mol
= 1.80671 × 10−21 g SO2
103 10.0 points
How many molecules are in 1.20 g of SO2 ?
Correct answer: 1.12789 × 1022 molec.
Explanation:
mSO2 = 1.20 g
32.07 g S
1 mol S ×
= 32.07 g
mol S
16.00 g O
2 mol O ×
= 32.00 g
mol O
Molar mass
= 64.07 g /mol
n=?
1.20 g
64.07 g/mol SO2
6.022 ×1023 molec
×
1 mole
= 1.12789 × 1022 molec SO2
nSO2 =
m=?
25
104 10.0 points
A can of mineral water contains 345 mL of
water. How many molecules of water does the
can contain? (water = 1.00 g/mL)
1. 57.6 molec
2. 3.74 × 1027 molec
3. 3.46 × 1025 molec
4. 1.15 × 1025 molec correct
5. 19.2 molec
Explanation:
VHCl = 345 mL
We can use the density of water (H2 O) to
convert from mL of water to grams of water:
? g H2 O = 345 mL H2 O ×
1 g H2 O
1 mL H2 O
= 345 g H2 O
Before we can use Avogadro’s number to
convert to the number of water molecules,
we must first convert from grams water to
moles of water. We do this using the formula
weight, which can be determined from the
atomic weights on the periodic table:
1.01 g H
mol H
16.00 g O
+ 1 mol O ×
mol O
? FWH2 O = 2 mol H ×
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
=
18.02 g H2 O
mol H2 O
×
1 mol H2 O
18.02 g H2 O
= 19.1 mol H2 O
Now we can use Avogadro’s number to find
the number of water molecules:
? molec H2 O = 19.1 mol H2 O
6.02 × 1023 H2 O
×
1 mol H2 O
= 1.15 × 1025 molec H2 O
1.01 g H
mol H
14.01 g N
+ 1 mol N ×
mol N
17.04 g NH3
=
mol NH3
FW = 3 mol H ×
in
Explanation:
m = 198.273 g
Each Al2 O3 molecule contains two aluminum atoms. There are Avogadro’s number of Al2 O3 molecules in one mole of Al2 O3 .
We need the formula mass of Al2 O3 so we can
convert grams of Al2 O3 to moles Al2 O3 .
Molecular mass of Al2 O3 :
? g/mol = 2(26.98 g/mol)
+ 3(16.00 g/mol)
= 101.96 g/mol
We can use this formula mass to convert g
Al2O3 to mol Al2 O3 :
? mol Al2 O3 = 198.273 g Al2 O3
×
106 10.0 points
A 46 gram sample of NH3 contains how many
grams of H?
Correct answer: 8.17958 g.
Explanation:
m = 46 g
To solve this problem we need the formula
weight of NH3 :
10.0 points
How many aluminum atoms are
198.273 g of Al2O3 ?
Correct answer: 2.34209 × 1024 atoms.
2 atoms Al
1 molec Al2 O3
= 2.34209 × 1024 atoms Al
? mol H2 O = 345 g H2 O ×
105
26
1 mol Al2 O3
101.96 g Al2 O3
This formula weight is used to convert
grams NH3 to moles NH3 :
1 mol NH3
17.04 g NH3
= 2.69953 mol NH3
? mol NH3 = 46 g NH3 ×
Each mole of NH3 contains 3 mol of H. We
can use this ratio to convert from moles NH3
to moles H:
? mol H = 2.69953 mol NH3 ×
3 mol H
1 mol NH3
= 8.09859 mol H
Finally, the atomic weight of H can be used
to convert from moles H to grams H:
= 1.94462 mol Al2 O3
We can now use Avogadro’s number and the
ratio of Al atoms to Al2 O3 molecules to find
the number of aluminum atoms:
? g H = 8.09859 mol H ×
1.01 g H
1 mol H
= 8.17958 g H
? atoms Al = 1.94462 mol Al2 O3
×
6.022 × 1023 Al2 O3
1 mol Al2 O3
107 10.0 points
Iron forms a compound called ferrocene. It
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
has the composition 64.56% C, 5.42% H, and
30.02% Fe. What is the empirical formula of
ferrocene?
1. FeC6 H5
2. Fe2 C10 H11
27
21.18 g
= 1.324 mol
16.0 g/mol
62.93 g
= 0.3309 mol
moles of Os =
190.2 g/mol
Dividing each number by 0.3309 mol gives
a ratio of 4.0 C : 4.0 O : 1.0 Os .
The empirical formula is OsC4 O4 .
moles of O =
3. FeC10 H10 correct
4. FeC11 H9
5. Fe3 C11 H12
109 (part 2 of 2) 10.0 points
From the mass spectrum of the compound,
the molecule was determined to have a molar
mass of 907 g/mol. What is its molecular
formula?
6. FeC7 H8
Explanation:
For 100 g of compound,
64.56 g
moles of C =
= 5.376 mol
12.01 g/mol
5.42 g
= 5.38 mol
moles of H =
1.0079 g/mol
30.02 g
moles of Fe =
= 0.5375 mol
55.85 g/mol
Dividing by 0.5375 mol gives 10.0 C : 10.0
H : 1.0 Fe.
The empirical formula is FeC10 H10 .
108 (part 1 of 2) 10.0 points
Osmium forms a number of molecular compounds with carbon monoxide. One lightyellow compound was analyzed to give the
following elemental composition: 15.89% C,
21.18% O, and 62.93% Os. What is the empirical formula of this compound?
1. OsCO
2. Os2 C4 O5
3. OsC3 O5
4. OsCO4
5. OsC4 O4 correct
6. Os2 CO4
Explanation:
For 100 g of osmium carbonyl compound,
15.89 g
moles of C =
= 1.323 mol
12.01 g/mol
1. Os3 C12 O12 correct
2. Os3 C3 O12
3. Os3 C9 O15
4. Os4 C8 O11
5. Os4 C2 O8
6. Os4 C8 O10
Explanation:
The formula mass of OsC4 O4 is 302.24
g/mol. The molar mass is 907 g/mol which
is 3 times the formula mass, so the molecular
formula is Os3 C12 O12 .
110 10.0 points
Paclitaxel, which is extracted from the Pacific
yew tree Taxus brevifolia, has antitumor activity for ovarian and breast cancer. It is sold
under the trade name Taxol. On analysis,
its mass percentage composition is 66.11% C,
6.02% H, and 1.64% N, with the balance being oxygen. What is the empirical formula of
paclitaxel?
1. C47 H51 NO14 correct
2. C45 H50 NO13
3. C3 H5 NO4
4. C4 H5 NO9
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
28
5. C35 H41 NO8
6. C49 H53 NO16
Explanation:
For 100 g of Paclitaxel,
66.11 g
moles of C =
= 5.504 mol
12.01 g/mol
6.02 g
moles of H =
= 5.97 mol
1.0079 g/mol
1.64 g
= 0.117 mol
moles of N =
14.01 g/mol
mass of O = 100 − 66.11 − 6.02 − 1.64
= 26.23 g
26.32 g
= 1.639 mol
16.0 g/mol
Dividing by 0.117 mol gives 47.0 C : 51.0 H
: 1.0 N : 14.0 O.
The empirical formula is C47 H51 NO14 .
112 10.0 points
An analysis of nicotine (MW = 162 g/mol)
gives 74.0% carbon, 8.65% hydrogen, and
17.3% nitrogen.
What is the TRUE
(MOLECULAR) formula for nicotine?
1. C12 H4 N
2. C4 H5 N2
3. C10 H14 N2 correct
4. C9 H12 N3
5. C6 H8 N5
moles of O =
111 10.0 points
What is the molecular formula of the molecule
that has an empirical formula of CH2 O and a
molar mass of 120.12 g/mol?
1. CH2 O
2. C3 H6 O3
3. C4 H8 O4 correct
4. C5 H10 O5
6. C5 H7 N
7. C8 H10 N4
8. C4 H5 N4
9. C11 H16 N
Explanation:
% C = 74.0%
% H = 8.65%
% N = 17.3%
MW = 162 g/mol
Assume we have 100 g of nicotine. (In fact,
we can assume any mass, but choosing 100 g
makes the math easier.) If we have 100 g of
nicotine, 74.0 g (74.0%) of it is carbon, 8.65 g
(8.65%) of it is hydrogen, and 17.3 (17.3%) of
it is nitrogen.
This means that, in this 100 g sample:
5. C2 H4 O2
6. C6 H12 O6
Explanation:
empirical formula: CH2 O
molar mass = 120.12 g/mol
molecular formula mass
x=
empirical formula mass
Empirical formula mass of CH2 O
= 12.01 amu + 2.02 + 16.00 amu
= 30.03 amu
120.12 g/mol
= 4.000 → 4
x=
30.03 g/mol
molecular formula: Cx H2x Ox = C4 H8 O4 .
? mol of C = 74.0 g C ×
1 mol C
12.011 g C
= 6.16 mol C
? mol of H = 8.65 g H ×
1 mol H
1.0079 g H
= 8.58 mol H
? mol of N = 17.3 g N ×
1 mol N
14.0067 g N
= 1.24 mol N
The smallest number of moles here is then
1.24, and we can use this number to detemine the empirical (simplest) formula for this
compound. For each element, we should now
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
divide the number of moles we just calculated
by this smallest number of moles to force the
mole ratio to integers.
So for C, we have
6.16 mol C
= 5 mol C
1.24
For H, we have
8.58 mol H
= 7 mol H
1.24
And for N, we have
1.24 mol N
= 1 mol N
1.24
This means this compound’s empirical formula is C5 H7 N1 , better written as C5 H7 N.
However, the question asked what the
compound’s TRUE molecular formula was.
For this, we need to compare the empirical
formula’s apparent molecular weight to the
known molecular weight of the compound.
The apparent molecular weight of the empirical formula is 81 g/mol, exactly half of the
known molecular weight. This means that
the TRUE formula is double the empirical
formula, i.e., C10 H14 N2 .
113 10.0 points
Calculate the nuclear binding energy of a 32
16 S
S
atom. The measured atomic mass of 32
16 is
31.972 070 amu.
1. 2.62 × 10−11 J
nuclear binding energy of 32
16 S = ?
16 protons: (16×1.007 276 amu)
= 16.116 416 amu
16 neutrons: (16×1.008 665 amu)
= 16.138 640 amu
16 electrons: (16×0.000 548 6 amu)
= 0.008 777 6 amu
total combined mass: 32.263 834 amu
mass defect
= 32.263 834 amu − 31.972 070 amu
= 0.291 764 amu
1.6605 × 10−27 kg
= (0.291 764 amu)
1 amu
−28
= 4.8447 × 10
kg
E = m c2
= (4.8447 × 10−28 kg)(3.00 × 108 m/s)2
= 4.36 × 10−11 kg·m2 /s2
= 4.36 × 10−11 J
114 10.0 points
The mass of a 73 Li atom is 7.016 00 amu.
Calculate its mass defect.
1. None of these
2. 7 amu
3. 0.294 91 amu
4. 0.042 13 amu correct
5. 3 amu
2. 4.36 × 10−11 J correct
6. 1 amu
3. None of these
7. 0.126 39amu
4. 4.36 × 1011 J
5. 2.73 × 10−12 J
6. 1.36 × 10−12 J
7. 2.62 × 1013 J
Explanation:
measured atomic mass of
29
32
16 S
= 31.972 070 amu
Explanation:
measured atomic mass of 73 Li
= 7.016 00 amu
mass defect = ?
3 protons: (3×1.007 276 amu)
= 3.021 828 amu
4 neutrons: (4×1.008 665 amu)
= 4.034 66 amu
3 electrons: (3×0.000 5486 amu)
= 0.001 646 amu
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
30
total combined mass: 7.058 134 amu
mass defect = 7.058 134 amu − 7.016 00 amu
= 0.042 13 amu
118 10.0 points
What type of particle is emitted in the transformation
201
115 10.0 points
What is the mass defect of 28
14 Si which has an
actual mass of 27.97693 g/mol?
1. 0.0102 g
2. 0.52416 g
3. 0.1004 g
4. 0.25406 g correct
Explanation:
116 10.0 points
When
I emits a β particle, what nuclide is
produced?
131
1. 131 Xe correct
2. 131 Te
3. 130 I
4. 130 Te
5. 127 Sb
Explanation:
1. positron
2. β particle correct
3. γ particle
4. No particle is emitted because electron
capture occurs.
5. α particle
Explanation:
119 10.0 points
What nuclide is formed when 24 Na undergoes
β decay?
1. 24 Ne
2. 28 P
3. 28 S
4. 24 Mg correct
Explanation:
The nuclear reaction is
24
11 Na
117 10.0 points
A nuclide undergoes proton emission to form
52
Fe. What is the nuclide?
3. 53 Co correct
4. 52 Co
5. 53 Fe
Explanation:
0
→A
Z ? + −1 e
A = 24 − 0 = 24
Z = 11 − (−1) = 12
1. 56 Ni
2. 52 Mn
Pt → 201 Au ?
24
12 ?
is magnesium:
24
11 Na
0
→ 24
12 Mg + −1 e
120 10.0 points
What is the least massive particle?
1. proton
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
2. α particle
31
123 10.0 points
For the nuclear reaction
239
235
94 Pu → 92 U + ?,
the missing particle is
3. electron correct
4. neutron
1. 1 n.
Explanation:
121 10.0 points
Uranium-238 decays by emission of an alpha
particle. The product of this decay is
2. e− .
3. 42 He. correct
4. 1 p.
1. 234
88 Ra .
Explanation:
2. 234
91 Pa .
239
94 Pu
3. 234
90 Th . correct
4
→ 235
92 U +2 He
124 10.0 points
The half-life of carbon-14 is 5,730 years. If
you started with 100.0 g of carbon-14, how
much would remain after 4 half-lives?
4. 234
92 U .
Explanation:
1. 12.5 g
122 10.0 points
Balance the nuclear equation
2. 57.3 g
? −→ 33
15 P
3. 25.0 g
32
15 P +
4. 6.25 g correct
1. None of these
Explanation:
After 4 half lives the amount of carbon has
been halved 4 times, so
2. ? = −10 β
A=
3. ? = 34
15 P
4. ? = 42 He
A0
100 g
=
= 6.25 g
n
2
24
125 10.0 points
If you have 200.0 g of radioisotope with a halflife of 5 days, how much isotope would remain
after 15 days?
5. ? = 10 n correct
6. ? = 151 P
1. 40.0 g
7. ? = 11 p
2. 13.3 g
Explanation:
mass number: 33 − 32 = 1
atomic number: 15 − 15 = 0
32
1
15 P + 0 n
−→
33
15 P
3. 12.5 g
4. 25.0 g correct
Explanation:
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
The half life is 5 days, so after 15 days,
three half lives have elapsed. The amount of
isotope remaining is
A=
A0
200.0 g
=
= 25 g
n
2
23
126 10.0 points
The half-life of P-32 is 14 days. How long after
a sample is delivered can a laboratory wait to
use a sample in an experiment if they need at
least 10 percent of the original radioactivity?
1. 42 days correct
2. 56 days
3. 14 days
4. 28 days
5. 70 days
Explanation:
After one half life 50% remains; after
two half lives 25% remains; after three
half lives 12.5% remains. So, after three
half-lives, over 10% of the 32 P remains.
3(14 days) = 42 days, so the laboratory can
wait about 42 days but shouldn’t wait longer
than that if at least 10% of the original radioactive substance 32 P is needed.
127 10.0 points
The isotope 90 Sr has a half-life of 28 years. If
you have 16 g of 90 Sr today, how much will be
left in 112 years?
1. 1 g correct
2. 8 g
3. 2 g
4 g to 2 g is the third half-life;
2 g to 1 g is the fourth half-life.
1 g of 90 Sr will be left after four half-lives
(112 years).
128 10.0 points
Calculate the energy change when one
nucleus undergoes the fission reaction
235
235
U
U + n → 142 Ba + 92 Kr + 2 n .
The masses needed are 235 U : 235.04 u;
Ba : 141.92 u; 92 Kr : 91.92 u; n : 1.0087 u.
142
1. −2.8 × 10−11 J correct
2. −1.7 × 10−10 J
3. +1.8 × 10−10 J
4. +2.9 × 10−11 J
5. −3.2 × 10−28 J
Explanation:
129 10.0 points
What nuclear reaction probably does not occur in a conventional fission nuclear reactor
designed to produce energy?
1. 235 U + n →239 Pu correct
2. 235 U + n →103 Mo + 131 Sn + 2 n
3. 235 U + n →90 Sr + 143 Xe + 3 n
4. 235 U + n →141 Ba + 92 Kr + 3 n
Explanation:
If 235 U captures a neutron, it forms an unstable species that breaks into barium and
krypton. To make plutonium, 238 U is needed
as the starting material.
4. 4 g
Explanation:
112
= 4, so 112 years is 4 half-lives.
18
16 g to 8 g is the first half-life;
8 g to 4 g is the second half-life;
32
130 10.0 points
In the fission process
239
143
86
92 U → 56 Ba + 36 Kr + X n,
what is X?
1. 5
Devillez (ld2653) – Test 1 Review – Devillez – (99998)
2. 20
3. 15
4. 10 correct
5. 25
Explanation:
239
143
86
1
92 U → 56 Ba + 36 Kr + 10 0 n
131 10.0 points
In nuclear fission, which is true?
1. One large atom splits into two or more
smaller atoms. correct
2. Two or more smaller atoms combine to
form one larger atom.
3. Individual protons split into smaller
pieces.
4. Electrons combine to form larger, negatively charged particles.
Explanation:
132 10.0 points
The nuclear equation
2
3
4
1
1 H + 1 H → 2 He + 0 n
is an example of
1. a branching chain.
2. nuclear fission.
3. nuclear fusion. correct
4. a chain reaction.
Explanation:
Two individual atoms combined to form
one atom. This is nuclear fusion.
133 10.0 points
Which reaction best describes nuclear fusion?
1. 2 C2 H6 + 3 O2 → 4 C + 6 H2 O + energy
33
90
143
2. 10 n + 235
92 U → 38 Sr + 54 Xe + energy
3. 21 H + 31 H → 42 He + 10 n + energy correct
238
4. 42 He + 235
92 U → 94 Pu + energy
Explanation:
Fusion starts with multiple atoms which
form a single atom.