ACT Past Papers with Answers

Paper 55 · ACT June 2011 · Form 69F · Science Test

According to the results of Study 1, the average CH4 emission rate for air-dried, sterilized beech leaves incu­ bated at 60° C was closest to which of the following?

Paper 55 · ACT June 2011 · Form 69F · Science Test

A complicating factor in interpreting the results of Study 3 was that exposing the chambers to sunlight probably also: both chambers. both chambers.

Paper 55 · ACT June 2011 · Form 69F · Science Test

Which of the following served as a control in Study 3 ? leaves leaves

Paper 55 · ACT June 2011 · Form 69F · Science Test

. Is the statement "Sterilization had no effect on the CH4 emission rate for ash leaves" supported by Figures 1 and 2 ? emission rates for both the sterilized and unsteril­ ized ash leaves were the same. emission rate for the sterilized ash leaves was less than half that for the unsterilized ash leaves. emission rates for both the sterilized and unsteril­ ized ash leaves were the same. . emission rate for the sterilized ash leaves was less than half that for the unsterilized ash leaves.

Paper 55 · ACT June 2011 · Form 69F · Science Test

In Study 1, it was unnecessary to put an identical mass of air-dried, sterilized leaves in each tube because the average CH� emissi�n rates were determined: ACT-69F 39 GO ON TO THE NEXT PAGE. 4 lO�CT 微信号 loviact 独角兽在这里 4 Passage II As fish mature, their diet often changes.Researchers investigated this phenomenon in 2groups of winter floun­ der at a location in the Hudson River estuary where there is significant mixing of river water with ocean water. All the flounder were between 2mo and 5mo old and at least 20mm long. Study 1 Each June of 1996 , 1997, and 1998, 100flounder were captured and placed in cages that were sitting on the bottom ofthe estuary .Eachcage was linec;lwith amesh bag that had square openings 3mm in diameter.At the end of 10days, the fish were removed from the cages and sorted into 4classes based on length.The stomach contents ofthe fish in each class were analyzed to determine the percent by mass of each type of prey, The results , averaged over the 3years, are shown in Figure 1 .The percent of fish in each class with empty stomachs is shown in Table I. Key Lengthof fish ■ 20mm-30mm 口 31mm-40mm 巨41mm-SO mm 口 51mm-65mm 5 p u n o g z; d n O J D J O S lU � l U O :) 号 B UJ O l S J O S S B UJ ,{ q l m 1 :, J g d 1 a p u n o g l d n O JD J O S lU a 1 u o :, t p B W O l S J O S S B W ,( q 1 u a :, 1 a d 100. 90 80 70 60 50 40 30 ＿ 』 - � Illh·:·····n:;··················-···············:·············--········寻．．．．．．．．．．．．．．．．．． -20 靡lIl||_l||Jl||.』f叮h —-才1, － ，dl 。 令了 令- ＄ ％ 令 type ofprey Figure 1 Table 1 Group 1flounder Length Number of Percent with (mm) flounder empty stomachs 20-30 45 15.5 31-40 87 32 .2 41-50 100 35.0 51-65 68 23.5 Study 2 In June of 1998, 30 more flounder were captured, removed from the river, and sorted into 4classes based on length.Immediately after thefish were sorted, their stom­ ach contents were analyzed to determine the percent° by mass of the same types of prey identified in Study 1.The results are shown in Figure 2. The percent of fish in each class with empty stomachs is shown in Table 2 . Key Lengthoffish ■ 20mm-30mm 曰31mm-40mm 巴41 mm-50mm 口 51mm-65mm 100 90

Paper 55 · ACT June 2011 · Form 69F · Science Test

For the class 51 mm-65 mm in length, were the stom­ ach contents of Group 1 flounder and of Group 2 flounder the same? in this class ate calanoids only. in this class ate ostracods only. types of prey, whereas the Group 2 flounder in this class ate malacostraca only. traca only, whereas the Group 2 flounder in this class ate many types of prey.

Paper 55 · ACT June 2011 · Form 69F · Science Test

In Study 2, which type(s) of prey was(were) NOT found in the stomachs of any of the 31 mm-40 mm long Group 2 flounder?

Paper 55 · ACT June 2011 · Form 69F · Science Test

For how many of the 4 classes was the percent of Group 1 flounder with empty stomachs higher than the percent of Group 2flounder with empty stomachs?

Paper 55 · ACT June 2011 · Form 69F · Science Test

In Study 1, what must the researchers have assumed about winter flounder feeding habits when they placed the cages in the estuary? Winter flounder feed in: estuary. . estuary.

Paper 55 · ACT June 2011 · Form 69F · Science Test

The total mass of the stomach content� of Group 2 flounder 20 mm-30 mm long was 50 mg. Based on Figure 2, the mass of bivalves in小ose stomach con­ tents was closest to which of the following?

Paper 55 · ACT June 2011 · Form 69F · Science Test

A researcher predicted that the diet of Group 1 floun­ der would change when flounder length exceeded 40 mm. Are the data in Figure 1 consistent with this prediction? length was at least 75% harpacticoids, whereas the diet of flounder greater than 40 mm in length was more varied. length was 100% harpacticoids, whereas the diet of flounder greater than 40 mm in length was more varied. length was at least 75% harpacticoids, whereas the diet of flounder more than 40 mm in length was more varied. length was 100% harpacticoids; whereas· the diet of flounder more than 40 mm in length was more varied. ＼ ， ＇ ACT七9F 41 GO ON TO THE NEXT PAGE. 4 10}\._CT 微信号loviact 独角兽在这里 4 Passage Ill As calcium oxalate hydrate (CaC20矿H20) is heated, it first dehydrates to yield calcium oxalate (CaC204). Next, the CaC204 decomposes into calcium carbonate (CaC03) and carbon monoxide (CO). The CaC03 then further decomposes into CaO and CO2 (see equations below). heat CaC20（比O(solid, s) � CaC204Cs) + H20(gas, g) CaC凸（s)上CaCOJCs) + CO(g) CaCO心）上CaO(s) + C02(g) In a thermogravimetric analysis (TGA), the mass of a sample is monitored as the sample's temperature is steadily increased. A nonreactive gas is streamed over the sample to carry away any gaseous products. The figure shows the results of a TGA of a 100 g sample of CaC204·H20. 110 I ： ： CaC20矿H20 100 4 i · ·· 4 4 4 4 4 . ! 90 + 千 . CaC凸 · . ， ••••··一·•••••••·, .．．．．．．．．．．．．．．. ••••• ••••••••••••••·,·••••••••••••••••• ; ·••••·.．．.．．．...．.i··················!··················\··················!·················· , 7 8 : 1 CaCO3 60 J. 4 • • 1 1 1, I ． ： ； ； ； ； ； 50 + ” 勺 、 、 、 、 、 ｀ ． ; : ； ； ； : · ； : ； } : 40 + 人 1 1 • 1 : 1 CaO ; ； ; ; ; ； —; - - － — · : • - ： ._ ： 30 O 100 200 300 400 500 600 7 00 800 900 1,000 temperature (0 Figure adapted from Mlle." Simonne Peltier and Clement Duval, "Sur la Thermogravimetrie des Precipites Analytiques: II. Dosage Du Cal- cium." ©1947 by Elsevier Science B.V.

Paper 55 · ACT June 2011 · Form 69F · Science Test

If the sample of CaC204·H20 had been 50 g, the mass of the sample at 900° C would.have been approximately:

Paper 55 · ACT June 2011 · Form 69F · Science Test

During the TGA, the largest change in mass resulted from the decomposition of:

Paper 55 · ACT June 2011 · Form 69F · Science Test

Calcium oxalate started to decompose at approxi­ mately what temperature? C C C C

Paper 55 · ACT June 2011 · Form 69F · Science Test

When the TGA began, approximately what percent of the mass of the sample was made up of H20 ?

Paper 55 · ACT June 2011 · Form 69F · Science Test

'fhe CO2 that wa_s part of the reactions that occurred during the TGA could best be described as a: CaC204. CaC03. CaC204. . ;r-;d��r formed directly by the decomposition.of CaC03. ACT-69F 42 GO ON TO THE NEXT PAGE. 41o�T 微信号loviact 独角兽在这里 4 Passage IV The respiratory cycle for resting humans who are breathing normally has been studied by physiologists. The figure shows 4 average measures of the cycle: lung volume, air flow (the rate at which air flows into or out of the lungs), pleural pressure (the pressure in the area between the lungs and the chest wall), and alveolar pressure (the pressure in the alveoli, small sacs in the lungs where air exchange takes place). Pleural pressure and alveolar pres­ sure are represented as values above or below atmospheric pressure (1,033 cmH20). The next inspiration begins at the end of the pause. lungvolume (L) airflow (Lisee) 。 -l.O O芒 pleuralpressure -� (cmH20 below -6 atmosphericpressure) 』 alveolar pressure 1.0 (cmH20 0.5 aboveor below 。 atmosphericpressure) 飞5 F 0 1 2 3 4 seconds Lovi *corresponds to atmosphericpressure (1,033 cmH20) Figure adapted from Robert M. Berne and Matthew N. Levy, eds., Principles ofPhysiology. ©1990 byC. V. MosbyCompany.

Paper 55 · ACT June 2011 · Form 69F · Science Test

Functional residual capacity (PRC) refers to average lung volume during the pause portion of the respira­ tory cycle. Based .on the figure, PRC is closest to which of the follow.ing values?

Paper 55 · ACT June 2011 · Form 69F · Science Test

For a resting human who is. breathing normally,.how does the duration of inspiration compare with the dura­ tion of expiration? tion of exp.iration. duration of expiration. expiration do not differ.

Paper 55 · ACT June 2011 · Form 69F · Science Test

The figure shows that when pleural pressure is lowest, air flow is closest to which of the following? 微信号： Loviact

Paper 55 · ACT June 2011 · Form 69F · Science Test

When the diaphragm contracts during the first O.5 §ec of inspiration, the chest wall expands away from the lungs, which causes the lungs to expand,. Based on the figure, what is the effect of the contraction of the diaphragm onpleuralpressure and alveolar pressure? pleural pressure alveolar pressure

Paper 55 · ACT June 2011 · Form 69F · Science Test

According to the information provided, which of the following statements best describes the relationship between alveolar pressure, atmospheric pressure, and the direction of air flow duringmost of expiration? pressure during most of expiration, air flows into the lungs. pressure during most of expiration,air flows out of the lungs. pressure during most of expiration, air flows into the lungs. pressure during most of expiration, air flowsout of the lungs. ACT歪 43 GO ON TO THE NEXT PAGE. 4 lOVACT 微信号loviact 独角兽在这里 4 Passage V A material is a superconductor (its electrical re.sis­ tance, R, is zero) if its temperature, T, is lower than its critical temperature, Tc. Students measured R and T for both a sample of Material X and a sample of Material Y while each sample was cooled. They used an ohmmeter to measure R (in ohms) and a thermocouple to measure T. The thermocouple displayed the result of each. measurement of T as a voltage (in millivolts, mV). Accordingly, the students plotted R versus voltage (see Figure 1). To convert voltage in mV to T in kelvins (K), they used Table 1. , ＇ ; … ； ； ＇ ; ' , ＇ ． : ：． , : : ， 1 , ＇ , ' ， ' ， ' 飞 ， ＇ ， ＇ … ＇ ， ＇ 主 / ． , ； ' , ＇ , ' ； ＇ ， ＇ , ＇ ; ; · ' ; ； ' ， ＇ , ' , ＇ : ： : ： : ： ， ' ， ＇ ， ＇ } } } , ＇ ， ＇ i , � � � } ， 了 ： ． … ， 1 ； ； ' , ＇ ； 7 : ,' : ： + ： ： … ， 宁 ， ： ： … ， 了 ， I 了 ： … ， ＇ 人 ， ＇ … ， ＇ ， ＇ ？ ＇ ： ： ： ， 了 ＇ ： ： ， ＇ ， 于 ' : : ,＇ ， ' } ， ＇ } + － i - + r l 60 50 40 30 20 10 ( s wq o ) N

Paper 55 · ACT June 2011 · Form 69F · Science Test

Based on Table 1, if the thermocouple were used to measure the temperature of a sample at 112K, the voltage displayed would most likely be closest to which of the following?

Paper 55 · ACT June 2011 · Form 69F · Science Test

In Table 1, as voltage decreases, temperature: J . remams constant. battery

Paper 55 · ACT June 2011 · Form 69F · Science Test

Based on Figure 1 and Table 1, the range of tempera­ tures over which Material X is a superconductor is closest to which of the following? switch Based on Figure 1 and Table 1, if the sample is kept at 81K, will the sample generate any heat as a result of the electricalcurrent flowing through the sample? zero. zero. ACT歪 45 GO ON TO THE NEXT PAGE. 41o�CT 微信号 loviact 独角兽在这里 4 Passage VI A teacher described the procedure of a study to stu­ dents in a science class: A 1 kg sphere, Sphere X, and a 2 kg sphere, Sphere Y, were released from rest, one at a time, from Point P on the right side of a frictionless, U-shaped incline. Hp was the height of Point P above Point L, the lowest point on the incline (see Figure 1). L Figure 1 Each sphere was allowed to slide as far up the left side of the incline as it could go. Next, the teacher gave the students the following definitions: l. GPEp, the gravitational potential energy of a sphere at Point P, equaled mgHp, where rri was the sphere's mass and g was the acceleration of the sphere due to Earth's gravity.

Paper 55 · ACT June 2011 · Form 69F · Science Test

Based on Figure 1 and Table 1, Tc for Material Y is most likely closest to which ofthe following?

Paper 55 · ACT June 2011 · Form 69F · Science Test

Suppose that the sample of Material Y is included in an electrical circuit, as diagrammed below.

Paper 55 · ACT June 2011 · Form 69F · Science Test

Which sphere, X or Y, was subjected to the greater amount of force from Earth's gravitational field?

Paper 55 · ACT June 2011 · Form 69F · Science Test

Suppose 2 other spheres, Sphere S and Sphere T, are released from Point P. The-m and VL for -each sphere are given in the table below. Student 1 As a sphere slid down the incline, its GPEp was con-. verted to kinetic energy. By the time it reached Point L, all of its GPEp had been converted. Based on Definition 1, because·Sphere Y had a greater m than did Sphere X, but the same g and Hp as Sphere X, Sphere Y had a greater GPEp than did Sphere X. As a result, Sphere Y had a greater KEL than did Sphere X. Thus, Sphere Y slid farther up the left side of the incline than did Sphere X. Student2 Because the 2 spheres had the same g and Hp, they had the same VL. Based on Definition 3, because Sphere Y had a greater m than did Sphere X, Sphere Y had a greater MOL than did Sphere X. Thus, Sphere Y slid farther up the left side of the incline than did Sphere X. S三Lec) |f Based on Definition 3 and Student 2's statements, which sphere will slide farther up the left side of the incline? will Sphere T: will Sphere T. will Sphere S. will Sphere S. ACT-69F 46 GO ON TO THE NEXT PAGE. 4 IOAC T 微信号loviact 独角兽在这里 4

Paper 55 · ACT June 2011 · Form 69F · Science Test

S u ppose that the study were conducted on the Moon instead of on Earth. Based on Definition 1 and S t uden t l's statements , compared to the KEL of Sphere X for the study on Earth, the KEL of Sphere X for the study on the Moon would be: the Moon is grea ter than the acceleration due to gravi ty on Earth. the Moon is less than the acceleration due to grav­ ity on Earth. , because the acceleration due to gravity on the Moon is greater than the acceleration due to grav­ ity on Earth. Moon is . less than the acceleration due to gr avity on Earth.

Paper 55 · ACT June 2011 · Form 69F · Science Test

Consider the statement "The greatest height attained by a sphere sliding up the left side of the incline does not depend on the sphere 's mass." This statement is consistent with the prediction(s) of which of the · students?

Paper 55 · ACT June 2011 · Form 69F · Science Test

Based on Student 3's statements, how dict· the amoun t of time fo r Sphere Y to slide from Point P to Point L compare to the amoun t of time for Sphere X to slide from Po int P to Point L ? The amount of time for Sphere Y to slide from Point P to Point L was: 1 F. + as g reat. 4 1 G. -¼ as g reat. 2 h e same.

Paper 55 · ACT June 2011 · Form 69F · Science Test

Suppose that a sphere is released from a new point on the incline, Point Q, that is between Point P and Point L. At Point Q, the sphere's gravitational poten­ tial energy is equal to mgHQ, where HQ is the height of Point Q relative to Point-L. Based ·on Student 1's statements about the conversion of gravitational poten­ tial energy to kinetic energy, would the sphere's KEL follo'.Ving the release from Point Q be less than or greater than the sphere's KEL following the release from Point P ? GPEp.. �- 9reater, because GPEQ would be less than GPEp. �- Less, because GPEQ would be greater than GPEp.

Paper 55 · ACT June 2011 · Form 69F · Science Test

Consider the 3 students'hypotheses concerning which sphere, if either, slid farther up the left side o f the incline. Based on the results o f the study, which o f the students'predictions, i f any, was(were) correct ? 1'sonly t 1'sand Student 2's only t l's, Student 2's, no r S�udent 3's AC T歪 47 GO ON TO THE NEXT PAGE. 4 lO�CT 微信号 loviact 独角兽在这里 4 Passage VII Students were given 12 known compounds and 4 unidentified compounds (Unknowns A-D) to analyze. Experiment 1 Students placed a solid sample of each known com­ pound in a separate capillary tube (a thin glass tube). Sam­ ples were then heated, 4 at a time, in a melting point apparatus (see Figure 1). 昔 y d :i q J d n S :l A O W l U :l A [ O S TLC plate tank solvent capillary tubes thermometer ． 。 . l O • . O , < j y b 二 e di tr th ＝ a b = distance traveled by the solvent beating block ＇ ， L ，王 王 capillary tube containing sample melting point apparatus Figure 1 TLC plate viewed under UV light Figure 2 The R1 value of each compound was then calculated by dividing distance a by distance b. The results of Experi­ ments 1 and 2 are shown in Table 1. The samples were monitored through the eyepiece. When liquid first appeared in a sample, the temperature was recorded. When no solid remained in the sample, the tern­ perature was again recorded. The compounds were placed in Groups I-IV based on similarity in melting point (MP). Experiment 2 For each compound in a group, students dissolved a sample and placed a small spot of the solution 2 cm from the bottom of a 13 cm tall and 7 cm wide thin-layer chro­ matography (TLC) plate. Each spot was circled with a pencil and allowed to dry. The plate was placed in a tank containing a small amount of a particular solvent. When the solvent had moved up to 1 cm from the top of the plate, the plate was removed and allowed to dry. Four distances were measured, in cm, under UV light: distance a for each spot and distance b (see Figure 2). Table 1 MP range Rf Group Compound (° biphenyl 69-72 0.82 I 2,5-DMP 68-71 0.24 4-NBC 70-73 0.40 PNCB 83-84 0.67 II naphthalene 80-82 0.85 vanillin 81-83 0.05 acenaphthene 93-95 0.78 III m-toluamide 94-96 0.02 methyl 4-nitrobenzoate 94-96 0.51 benzoic acid 122-123 0.19 IV traris-stilbene 122-123 0.77 succinimide 122-124 0.00 ACT-69F 48 GO ON TO THE NEXT PAGE. 4 lOVACT 微信号loviact 独角兽在这里 4 Experiment 3 The techniques from Experiments 1 and 2 were used to analyze Unknowns A-D (see Table 2). Table 2 MP range Rr Unknown (° A 93-95 0.02 B 123-125 0.74 C 68-71 0.85 D 80-82 0.06 Figures adapted from Samuel G. Levine, "Identification of Unknowns by Melting Point and Thin-Layer Chromatography in Combination." ©1990 by Division of Chemical Education, Inc., AmericanChemical Society.

Paper 55 · ACT June 2011 · Form 69F · Science Test

In Experiment 1, liquid first appeared in the biphenyl sample at which of the following temperatures? C C C C

Paper 55 · ACT June 2011 · Form 69F · Science Test

Th·e spot for which of the following samples traveled farthest up the TLC plate? END OFTEST4 STOP! DO NOT RETURN TO ANY OTHER TEST. ACT七年 49 lO�CT 微信号loviact Form 16R ACT Writing Test Prompt (June 2011) 独角兽在这里 Global Positioning System (GPS) technology can identify a person's precise location on a map. Many cell phones have GPS systems that allow parents to keep close track oftheir teenagers through the teens' cell phones. Some parents support using such technology to keep track of teens because they think it helps keep young people safe by encouraging. good behavior. Other parents do not support using such technology to keep track of teens because they think teenager·s will only resent their parents if they know their every movement is being monitored. In your opinion,·should parents use GPS systems to keep track of their teenagers? In your essay, take a position on this question. You may write about either one of the two poi11:ts of view given, or you may present a different point of view on this question. Use specific reasons and examples to support your position. 52 lO�CT 微信号 loviact 独角兽在这里 2011年6月ACT试题 (69F) 答案 1 IA L 1 C 3 1 E 1 B 1 B 2 G 。 2 F 32 K 2 F 2 G 3 A V 3 A 33 A 3 D 3 C 4 J l 4 G 34 H 4 F 4 J 5 IC 5 E 35 A 5 B 5 A 6 G 4 1 A 6 J 36 K 6 H 6 F 7 C 42 F 7 E 37 E 7 B 7 C 8 G 43 B 8 H 38 H 8 J 8 H 9 C 44 G 9 C 39 A 9 C 9 C 10 H 45 C 10 J 40 G 10 F 10 J 1 1 1 1 1 1 A 2 G 3 C 4 J 5 C 46 J 1 1 B 4 1 D 1 1 D 1 1 C 47 C 12 K 42 K 12 J 12 F 48 F 13 A 43 B 13 C 13 A 49 D 14 J 44 J 14 H 14 H 50 H 15 B 45 D 15 A 15 B 16 F 5 1 B 16 F 46 G 16 J 16 F 17 D 52 G 17 C 47 D 17 A 17 D 18 G 53 B 18 F 48 J 18 G 18 G 19 C 54 G 19 B 49 D 19 A 19 B 20 J 55 A 20 K 50 J 20 G 20 G B - H - D _ J - c 21 22 23 24 25 H - B _ J - A - F 56 57 58 59 60 c - G - c - F - B 21 22 23 24 25 E - G - c - K - A 51 52 53 54 55 c - F - D _ J - c 21 22 23 24 25 D _ J - B - F - c 21 22 23 24 25 H - A - G - A _ J 26 27 28 29 30 A - G - D - F - B 61 62 63 64 65 F - c 飞 “ - c - G 26 27 28 29 30 k - c _ J - D _ J 56 57 58 59 60 J P - G - B - G 26 27 28 29 30 J - A 飞 “ - A _ J 2 6 2 7 2 8 2 9 30 3 1 B 66 F 3 1 IB |3 32 1 尸 I H 32 J 67 D 回 · 32 H 33 IA I68 I H 33 IC I 33 ID '. 34 F 69 A 34 IG ｀ 35 C 70 J 35 A 36 H 7 1 ID • 哪＿ -- 二．尾二E,. 36 IH I 36 I H 37 D 72 IG ＿ ■尺＿＿，一蠡I 37 ID I 37 IA . - 38 F 73 IB 回工臣.-,JI'■ 一 ．r 38 IG I 38 IG 39 A 74 IF ~l.．．1-1...-- 39 ID | 3 4 9 0产 I F 40 J 75 D 40 H 微信号：loviact QQ交流群：411-279-476 在线真题：loviact.com 2016年第六版Lovi lO�CT 微信号 loviact 独角兽在这里 Explanation of Procedures U�ed to Obtain Scale Scores from Raw Scores On each of the four tests on which you rn釭ked any responses, the total number of correct responses yields a raw score. Use the table below to convert your raw scores to scale scores. For each test, locate and circle your raw score or仇e range of raw scores that includes it in the table bel叩．Then, read across to either outside column of the table and circle仇e scale score that corresponds to that raw score. As you determine your scale scores, enter them in the blanks provided on the right. The highest possible scale score for each test is

Paper 55 · ACT June 2011 · Form 69F · Science Test

One of the students calculated the Rr value for Unknown C as 1.18. There had to be an error in the student's calculation because:

Paper 55 · ACT June 2011 · Form 69F · Science Test

A chamber contains samples of each of Unknowns A-D maintained ·at 150° temperature in the chamber is slowly cooled to 40° Based on the results of Experiment 3, which of the samples would most likely start to solidify first?

Paper 55 · ACT June 2011 · Form 69F · Science Test

Based on the results of Experiments 1-3, Unknown A is most likely which of the 12 known compounds listed in Table 1 ?

Paper 55 · ACT June 2011 · Form 69F · Science Test

Which of the following diagrams best represents how the TLC plate appeared under UV light for Group II ? • I ® PNCB o naphthalene • vanillin