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ACT Science Homework
Science 2, Set 2
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24 Minutes — 25 Questions
DIRECTIONS: There are five passages in this test.
Each passage is followed by several questions. After
reading a passage, choose the best answer to each question
and fill in the corresponding oval on your answer document.
You may refer to the passages as often as necessary.
You are NOT permitted to use a calculator on this test.
Passage I
The state of a real gas can be approximated by using the ideal gas law. The amount of an ideal, or hypothetical,
gas is determined by its pressure, temperature and volume.
The ideal gas law formula is: PV = nRT (P = pressure, V = volume, T= temperature, n = number of moles, R =
gas constant)
When the number of moles is held constant, the equation can be written as:
PV
PV
1 1
= 2 2
T1
T1
A chemist places a balloon inside a rectangular environmental chamber. The balloon is filled with helium (He)
and tied so that no gases can enter or leave the balloon (assume the balloon membrane has no pores).
Figure 1
Helium inside balloon
He
Ambient air inside
environmental chamber
ACT HOMEWORK SCIENCE 2, SET 2
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Experiment 1
The helium in the balloon is heated up when a heating element located inside the environmental chamber heats
up the ambient air. The volume of the balloon is recorded each time the temperature is increased as shown in
Table 1.
Table 1
Temperature (Celsius)
Temperature (Kelvin)
0
273
10
283
20
293
30
303
* Pressure of ambient air is 1 atmosphere
Volume of Balloon (Liters)
0.700
0.725
0.750
0.775
Experiment 2
Using the same balloon and environmental chamber as used in experiment #1, a scientist hooks up a small
pump so that the pressure of the ambient air can be changed. As the pressure of the ambient air is incrementally
increased, the volume inside the balloon decreases as recorded in Table 2.
Table 2
Pressure (atm)
Volume (Liters)
1.00
0.700
1.02
0.686
1.08
0.648
1.12
0.625
* Temperature of ambient air is 0 Celsius
Experiment 3
The same balloon and environmental chamber from experiments #1 and #2 is used, except for one change: a
thin layer of woven nylon mesh is placed along the outside of the balloon. A scientist hooks up a small pump so
that the pressure of the ambient air can be decreased. This causes the pressure of the helium inside the balloon
to increase. Table 3 shows the various temperatures of the He in the balloon.
Table 3
Pressure (atmospheres)
Temperature (Celsius)
Temperature (Kelvin)
1.00
0
273
1.03
8
281
1.08
22
295
1.15
41
314
* Volume of the balloon is 0.7 liters
ACT HOMEWORK SCIENCE 2, SET 2
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1. Which of the following was held constant in the
balloon for all three experiments?
A. Pressure of helium inside the ballon
B. Amount of helium in the balloon
C. Volume of the balloon
D. Temperature inside the balloon
5. Why was a stiff, woven mesh placed around the
outside of the balloon in Experiment 3?
A. To prevent the pressure from increasing
B. To prevent the pressure from decreasing
C. To prevent the volume from increasing
D. To prevent the volume from decreasing
2. An isobaric process is a process where the
pressure remains the same. Which of the following experiments uses an isobaric process?
F. Experiments 1 and 2
G. Experiment 1
H. Experiment 2
J. Experiment 3
6. A motorist notices one morning that his car tires
looked slightly deflated compared to when he
got home from work the evening before. Which
of the following statements could be a correct
reason?
F. As the temperature increases, the volume of
air in the tires decreases.
G. The lower volume in the tires caused higher
pressure in the tires.
H. The pressure and temperature of the tires
never change.
J. The cooler air temperature in the morning
caused lower pressure in the tires.
3. The graph shown here represents the data from
which experiment?
A. Temperature and Volume in Experiment 3
B. Temperature and pressure in Experiment 2
C. Pressure and Volume in Experiment 1
D. Pressure and Volume in Experiment 2
4.
Charles Law states that
7. The results of which experiment(s) provide(s)the
best evidence for question #6?
A. Experiment 1
B. Experiment 2
C. Experiment 3
D. Experiment 1 and 2
V1 V2
when the pres=
T1 T2
sure is held constant. If the initial volume of the
balloon is 0.700 liters, what is the final volume
of the balloon when the ambient air is heated up
to 298 Kelvin?
F. 0.763 Liters
G. 0.700 Liters
H. 0.771 Liters
J. 0.735 Liters
ACT HOMEWORK SCIENCE 2, SET 2
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Passage II
Biologists have been studying the population of reptiles and amphibians in Honduras. They have been looking at the populations for several decades and recently have seen decline. These biologists have quantified that
decline in several different areas, giving each species a score. The higher the geographic score, the smaller area
in which the animal lives. The ecological distribution score deals with the number of types of places in which
the animal resides. The higher the ecological distribution score, the fewer the habitats a species lives in, and
the more endangered the species is. The last score deals with the relative ease of reproduction. The higher the
score, the more exact the circumstances have to be for the animal to reproduce. Reptiles have a human persecution score because they are more likely to come into contact with humans. Also, because of their size, they have
been known to be used for food and hides. The higher the score, the more exposure to humans the reptiles have.
Amphibian Species
Ecological
distribution
4
Reproduction
mode
4
Total score
Bolitoglossa mexicana
Geographic
distribution
1
Bolitoglossa rufescens
3
5
4
12
Bolitoglossa carri
5
8
4
17
Bolitoglossa celaque
4
8
4
16
Bolitoglossa conanti
3
7
4
14
Bolitoglossa decora
4
8
4
16
Bolitoglossa diaphora
4
8
4
16
Bolitoglossa dofleini
3
7
4
14
Bolitoglossa dunni
3
7
4
14
Geographic
distribution
1
Ecological
distribution
5
Human
persecution
3
Total
Norops lemurinus
1
5
3
9
Norops pentaprion
1
7
3
11
Norops petersii
2
8
3
13
Norops rodriguezii
2
5
3
10
Norops ocelloscapularis
5
7
3
15
Norops pijolensis
4
7
3
14
Norops purpurgularis
4
8
3
15
Norops roatanensis
4
8
3
15
Reptile type
Norops laeviventris
http://www.pubmedcentral.nih.gov/articlerender.fcgi?tool=pubmed&pubmedid=15029252 source material
ACT HOMEWORK SCIENCE 2, SET 2
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8. What is the most profound impact on the
amphibians and reptiles of Honduras?
F. Ecological distribution
G. Geographic distribution
H. Human persecution
J. Reproduction mode
9. Which species is in the most danger?
A. Bolitoglossa mexicana
B. Bolitoglossa carri
C. Bolitoglossa dunni
D. Bolitoglossa conanti
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11. Why is Boltiglossa decora one of the most
endangered of all the species listed?
A. It is not diverse in geographic locations or in
habitats.
B. It is diverse in geographic locations and in
habitats.
C. It is not diverse in geographic locations and
not diverse reproduction mode.
D. It is diverse in geographic locations but not
in habitats.
12. Why is Boltiglossa carri more endangered than
Boltiglossa conanti?
F. Boltiglossa conanti have more ease in
reproduction than Boltiglossa carri.
G. Boltiglossa carri have more ease in
reproduction than Boltiglossa conanti.
H. Boltiglossa carri is found in more habitats
than Boltiglossa conanti.
J. Boltiglossa carri is found in fewer habitats than Boltiglossa conanti.
10. Why is Norops ocelloscapularis more
endangered than Norops petersii?
F. Norops ocelloscapularis is found in more
places than Norops petersii.
G. The scores for Norops ocelloscapularis result in a combined score that is at a higher level.
H. Norops petersii is found in more places than
Norops ocelloscapularis.
J. The scores for Norops petersii result in a
combined score that is at a higher level.
ACT HOMEWORK SCIENCE 2, SET 2
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Passage III
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Samples of sap were collected from 4 different species of maple trees in April and November of the same year.
Each sample was then analyzed. Tables 1 and 2 show the volume, color, specific gravity, and concentration of
glucose in the April and November samples, respectively.
Specific gravity was calculated as follows:
Specific gravity = density of sample/density of water
The normal range for the specific gravity of tree sap is 2.000-4.520
Table 1
Samples collected April 10
Species
Volume(ml)
Color*
Specific gravity
Glucose (g/L)
A
360
10
4.501
102.3
B
418
2
3.215
76.4
C
432
7
3.722
87.2
D
476
3
3.198
75.2
* Note: Color was assigned using the following scale: 0 = clear; 10 = very dark
Table 2
Samples collected November 16
Species
Volume(ml)
Color*
Specific gravity
Glucose(g/L)
A
100
3
2.835
20.1
B
124
1
2.523
18.6
C
145
2
2.792
19.5
D
173
1
2.499
18.5
*Note: Color was assigned using the following scale: 0 = clear: 10 = very dark
ACT HOMEWORK SCIENCE 2, SET 2
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13. One of the four species of maple trees is prized
for being the best source of sap to make into
maple syrup, a thick, sweet liquid. Based on the
information provided, this species is most likely:
A. species A.
B. species B.
C. species C.
D. species D.
14. Based on the information provided, as the concentration of glucose in tree sap rises, the specific gravity of the sap:
F. increases, then decreases.
G. decreases, then increases.
H. decreases only.
J. increases only.
15. Based on the information presented, which of
the samples most likely has the highest water
content per milliliter?
A. The April sample from species A
B. The April sample from species D
C. The November sample from species A
D. The November species from species D
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16. One milliliter of sap from which of the following
would weigh the least?
F. The April sample from species A
G. The April sample from species C
H. The November sample from species D
J. THe November sample from species B
17. Does the data in Table 1 and Table 2 support
the conclusion that as tree sap volume increases,
sap color darkens?
A. Yes, because sap samples with the greatest
volumes had the greatest color values.
B. Yes, because sap samples with the greatest
volumes had the lowest color values.
C. No, because sap samples with the greatest
volumes had the lowest color values.
D. No, because there is no relationship between
sap sample volumes and color value.
18. If sap was collected from a tree in July and
found to have a color of 4 and a specific gravity of 3.732, from which species was it most
likely taken?
F. Species A
G. Species B
H. Species C
J. Species D
http://www.epa.gov/glnpo/glindicators/water/oxygenb.html source material
ACT HOMEWORK SCIENCE 2, SET 2
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Passage IV
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Experiment 1
Scientists were working to minimize friction. Friction is measured using a coefficient, where the closer to one,
the more the material is being held back. The coefficient is determined by dividing the amount of force it takes
to move something on the material by the amount of force it takes to move the same object in a vacuum. The
mass of the material does not affect the friction coefficient. They first looked at the friction of steel on different
materials.
Surface
Surface
Steel alloy A
Steel alloy A
Steel alloy A
Steel alloy A
Steel alloy A
Brass
Cast iron
Phosphates
Lead
Steel alloy A
Sliding friction
coefficient
.44
.23
.34
.95
.57
Experiment 2
Scientists looked at a different steel alloy (a different mixture) to determine which alloy would be the best
material to make ball bearings that would rub against each other in a machine.
Surface
Surface
Steel alloy B
Steel alloy B
Steel alloy B
Steel alloy B
Steel alloy B
Brass
Cast iron
Phosphates
Lead
Steel alloy B
Sliding friction
coefficient
.41
.18
.32
.90
.55
Experiment 3
The scientists then looked at using a type of lubricant to see if it made a difference in the use of either of the
metals.
Surface
Surface
Steel alloy A
Steel alloy A
Steel alloy A
Steel alloy A
Steel alloy A
Brass
Cast iron
Phosphates
Lead
Steel alloy A
Sliding friction
coefficient
.34
.133
.173
.3
.09
Surface
Surface
Steel alloy B
Steel alloy B
Steel alloy B
Steel alloy B
Steel alloy B
Brass
Cast iron
Phosphates
Lead
Steel alloy B
ACT HOMEWORK SCIENCE 2, SET 2
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Sliding friction
coefficient
.28
.08
.158
.2
.029
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19. Which alloy of steel is harder?
A. Steel alloy B.
B. Steel alloy A.
C. Unable to tell from information provided.
D. They are the same.
20. Does lubrication help improve the amount of
friction?
F. No, it doesn’t help the metals improve their
friction coefficient.
G. Yes, it helps all metals equally.
H. Yes, it helps the phosphates most of all.
J. Yes, to some metals more than others.
21. Which of the following will further the scientist’s goal of reducing friction?
A. Testing different lubricants to determine if
they will reduce friction.
B. Testing water to determine if it will increase
friction.
C. Making the materials move slower.
D. There is no way to reduce friction.
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23. What will NOT help reduce the friction coefficient?
A. Increasing the mass of the alloy.
B. Adding water between the two surfaces.
C. Adding a lubricant between the two surfaces.
D. Experimenting with different metals.
24. How does the use of lead compare with the use
of brass when using the steel alloy B?
F. The lead is much more efficient.
G. They are the same.
H. The brass is much more efficient.
J. There is no way to tell.
25. Why does alloy A always have a higher coefficient than alloy B?
A. Alloy B has the higher coefficient.
B. Alloy A always has a higher coefficient be
cause its make up is smoother.
C. Alloy A and alloy B have the same coeffi
cients.
D. Alloy A always has a higher coefficient be
cause its make up is rougher.
22. Which is the best combination with alloy A?
F. Cast iron
G. Lead
H. Phosphates
J. Steel alloy A
ACT HOMEWORK SCIENCE 2, SET 2
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