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Paper 50

ACT December 2012

Form 71A · Science Test · 40 questions

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Paper 50 · ACT December 2012 · Form 71A · Science Test

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AChoice A — see source imageBChoice B — see source imageCChoice C — see source imageDChoice D — see source image

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Paper 50 · ACT December 2012 · Form 71A · Science Test

Suppose that in Study 2 the steel ball had been dropped 16. Which of the following statements best explains why onto the bare concrete surface and the foam-covered the students conducted the studies at the location concrete surface from a height of 2.00 m. Approxi- referred to in the passage? The students most likely mately nom _ lower - higher — hes steel ball wanted to: ave bounce er its collision wi e bare concrete : surface than after its collision with the foam-covered * eoceeare the force exerted by gravity on the balls concrete surface? G. decrease the force exerted by air resistance on the balls tested. Passage IV At the end of the 3-week treatments, the plants were harvested, and their roots were washed. The roots were A researcher investigated the growth of 6 species of separated from the shoots, and then both were dried at floodplain plants. Each spores was Classified as cither 70°C for 48 hr. The wrongs dry shoot biomass per plant flood-sensitive (S) or flood-tolerant (T) (see Table 1). and the _— dry root biomass per plant were deter- mined for each species and treatment combination (see Figures 1 and 2, respectively). Table 1 16 — Classification ee tate ‘eo 14 ome Key = awe Achillea ptarmica Ap T 3 9223590 Dement 1 a Achillea millefolium Am S ze 10 + @ Treatment 2 [er g panapusenssemenent Festuca arundinacea T 8 4- G momma aaa SLT Festuca rubra Ss £ 6 >t seiaieanaaanaaitine Z sional Rumex palustris T 3.0 5 i ET NN 7 4g Rumex thyrsiflorus S 2 2 cccccccereveccoses i ws cecccccsce. Z ee i pcowsoee Z we a 0 a 2 Sa Z ove Z cove Z we Z ness Ap Am Fa Fr Rp Rt species | Figure | Experiment Seeds from each of the 6 species were germinated in a 5 growth chamber maintained at 25.5°C during the day and -_ cones K wosceensuennnceeveeunennsceeenseenveresenanseenecece at 10°C at night. ' wo 4-1-- ey ce: $8 Sie After germination, the seedlings of each species were B. 3, Treatment? Qo planted in separate 0.5 L pots and grown for 4 weeks. The s es Treatment 3 fo) HM seedlings were watered 3 times per week. % 9 nn | BE Treatment a ytd WE ee ; Z At the end of the 4 weeks, each plant was repotted > : a eee Z ac into its own 4 L pot containing a substrate a: srag of a * l macagepesacvonsverseneassoeooseses ena are Z nuccoceesoue==eees 4:1 mixture, by mass, of sand to compost. The Dp ants were PSG ‘| eessecccsons Z eee i Y “T $60eesees grown for 2 more weeks; during this time, they were 0 z a - a Z 2 watered 3 times per week. Ap Am Fa Fr Rp Rt Next, the potted plants were placed in plastic tubs — such that each tub contained 8 plants of each species. Each Fi - tub of plants was subjected to a different treatment (see spare Table 2 for 3 weeks. Figures 1 and 2 adapted from C. Jansen, H. M. Van De Steeg, and _ ©2005 by British Ecological Society. | Pots were watered 3 times per week. Water level was always below the substrate surface. Water level was maintained at the substrate surface. — 17. Which treatment resulted in the greatest average dry | oe Water level was maintained 2 cm shoot biomass per plant for Rp plants? above the substrate surface. i Taaeanet Water level was maintained 6 cm B. Treatment 2 above the substrate surface. C. Treatment 3 ,

A0.5 m lower balls tested.B1.0m lower H. increase the force exerted by gravity on the ballsC0.5 m higher tested.D1.0 m higher J. increase the force exerted by air resistance on theEDe Kroon, “Investigating a Trade-Off in Root Morphological6Treatment 4

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Paper 50 · ACT December 2012 · Form 71A · Science Test

When the water level was maintained 2 cm above the Pevertaeet? Ropadlees et the yaa of the substrate surface, which floodplain — the T species plants or the S species plants, had the greater A. both S and T plants produce, on average, more : lant? shoot than root biomass per plant. average dry root biomass per plant | . B. both S and T plants produce, on average, more per plant was greater for Ap plants than for C. S plants produce, on average, more shoot than root Am plants, _— for Fa ey than for Fr plants, biomass per plant, whereas T plants produce, on and greater for RP plants than for Rt plants. average, more root than shoot biomass per plant. Ap plants, greater for Fr — than for Fa plants, average, more shoot than root biomass per plant. and greater for Rt plants than for Rp plants. r plant was greater for ants than for ' ; , Oe Fi sire vente sae Fa plants een for Fr plants, 22. _ plants Pos asi in ~ experiment were from and greater for Rp plants than for Rt plant, Ow many Ciikerent genuses per a was greater for Am plants than for G. 3 Ap plants, greater for Fr oe than for Fa plants, H. 4 and greater for Rt plants than for Rp plants. Jj. 6 Passage V In om 1, the equivalence point indicates the volume of NaOH solution needed to react with all the acid Students were given unknown aqueous acid solutions in the solution in the beaker. At the halfway point, the UAX, UAY, and UAZ. Each solution had an acid concen- volume of NaOH solution added is half of the volume tration of 0.100 mole/L but contained a different acid. The needed to reach the equivalence point, and the pH of the students were also given Table 1, which gives the pK, of solution in the beaker equals the pK, of the acid in the each of 5 acids (the lower the pX,, the stronger the acid). solution. Table I Experiment 2 | Acid | pK The procedure from Experiment | was repeated with wis 2.5 UAY instead of UAX (see Figure 2). | Chicroscetic 2.9 Formic 3.7 10 Hn KAA Beles ee ee | Acetic 4.8 voveeneee @QUIVAIENCE POINE «== 4 In a titration, the titrant (a solution of known identity i i ee ae ka Vane and concentration) is slowly added to the analyte (a solu- fe Pre See DEST Sei ler tion of unknown identity or concentration). Students used 7 eS SS a titration to attempt to identify the acid present in UAX, a Sa oe _- enn oan Experiment I 0 : : . : A 0.100 mole/L aqueous sodium hydroxide (NaOH, a oO Ss 0 BL 0 2B base) solution was added to a buret (a graduated tube with volume of NaOH added (mL) a valve at the bottom that can be opened to dispense pre- cise volumes of liquid). The initial volume of solution in Fi 2 the buret was recorded. A beaker rep Syne mL of one UAX was placed under the buret. A pH probe was placed in the beaker. Then 25.00 mL of the NaOH solution was added in small increments to the acid solution while the j 3 acid solution was continuously stirred. After each addition, Experiment the pH of the solution in the beaker was recorded (sce The procedure from Experiment 1 was repeated with Figure 1). UAZ instead of UAX (see Figure 3). 12 12 ! : ; s alence > t saan eete ene. beseeserrnneaeds ans sese sanes sot teRansesene se Héeteccotes 10 equivalence poin ths-l ev nabealontia palette weneenveresapecosenvtis neiinoontevsomvafeneencenessifpeanrwrreners 0 0 . ’ ‘ ‘ 0 5 10 15 20 £25 0 5 10 IS 20 £25 volume of NaOH added (mL) volume of NaOH added (mL) Figure 1 Figure 3

AThe T species plants; the average dry root mass root than shoot biomass per plant.BThe T species plants; the average dry root mass D. S plants produce, on average, more root than shootCThe = species plants; ~ average dry ot “=DThe S species plants; the average dry root mass F. 2

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Paper 50 · ACT December 2012 · Form 71A · Science Test

A student claimed that the acid in UAX is pyruvic acid. Based on Table 1 and Figure 1, this claim is incorrect because the pK, of pyruvic acid is: ? D. greater than was the pH at the halfway point. : : 7) 28. Suppose Experiment 3 is repeated, but the concentra- aA. oe Expermeuts 1 on’ * 3" ee a differ tion of the NaOH solution FS 0.200 mole/L. Will the volume of NaOH added at the halfway point and at the F, In Experiment 1, the analyte was stirred; in Exper- equivalence point be less than, greater Sen, or equal to iment 3, the analyte was not stirred. ; the corresponding results shown in Figure 3 ? ment 3, the titrant was not stirred. halfway point equivalence point the analyte in Experiment 3. G. greater less titrant in Experiment 3. J. equal equal Passage Vi 29. Which of the following terms best describes the mix- ture of iodine and solvent in cach of the beakers at the A teacher opr ame ae he a reac — —— end of the teacher’s demonstration? beaker contained mineral oil an other contai + as «ae Solid iodine was added to each beaker. The solid iodine lodine and mineral oil iodine and H1,0 quickly dissolved in the mineral oil, which turned pink. A. heterogeneous heterogeneous The solid stayed at the bottom of the H,O, which remained B. heterogencous homogeneous colorless. The teacher asked 3 students to explain these Cc, homogencous heterogeneous results. D. homogeneous homogeneous Student I Mineral oil is made up of very large molecules, while , « H,O is made up of very small molecules. Thus, the spaces 30. Based on Student 1’s explanation, is it likely that solid between mineral oil molecules are much larger than the H,0 would be soluble in mineral oil? spaces between H,O molecules (sce figure). F. Yes, because the solute molecules and the solvent molecules are yan than mineral oil molecules. solvent molecules are sem ge aA mineral oil molecules. smaller space between HO molecules larger space between mineral oil molecules 31. How would Student 1 most likely rank the 3 sub- stances used in the demonstration, from the substance For a solid to dissolve in a solvent, its molecules must with the smallest molecular mass to the substance with suerte in between solvent molecules. The molecular mass the largest molecular mass? (the mass of 1 molecule in atomic mass units, amu) of a A. Iodine < H,O < mineral oil - gubstance is a good indicator of molecular size, so if a B. H,O < iodine < mineral oil solid has a molecular mass less than that of a solvent, it C. Mineral oil < iodine < H,O : will most likely dissolve in the solvent. Therefore, the D. Mineral oil < HO < iodine iodine dissolved in the mineral oil, but not in the H-O Student 2 _An H,0 molecule is polar because it has a region of = 32, In Student 2’s explanation, the sentence “Therefore, positive charge and a region of negative charge. Nonpolar the iodine dissolved in the mineral oil, but not in the molecules, like those that make up mineral oil, do not have H,O” implied that iodine molecules are: differently charged regions. Polar molecules are attracted F ; to each other, and nonpolar molecules are attracted to each G catty = other, but pou molecules repel nonpolar molecules. Thus, aH Snail han H.O molecul polar solids dissolve in polar solvents, and nonpolar solids . : a Aaa 3 = sal cs. dissolve in nonpolar solvents. Therefore, the iodine dis- J. larger a MOLCCUICS. solved in the mineral oii, but not in the H,O. Molecular size is unrelated to solubility. Student 3 33. Suppose age cg ee = molecules Student 2 is correct, but with one exception. Nonpolar were smaller than iodine molecules. What impact, molecules are actually strongly attracted to polar mole- if any, a this observation have on Student 3's cules. The reason that nonpolar substances don’t dissolve explanation in polar substances is that polar molecules are too strongly A. It would prove that the explanation is correct. attracted to each other to allow nonpolar molecules to B. It would support the explanation, but not prove come between them. If a — of mineral oil is placed on that the explanation is correct. the surface of a pan full of H,O, the drop will spread to C. It would weaken the explanation, but not prove form the thinnest layer possible in order to maximize con- that the explanation is incorrect. tact with H,O molecules. D. It would have no impact on the explanation.

Aless than was the pH at the equivalence point.Bgreater than was the pH at the equivalence point.Cless than was the pH at the halfway point.DIn a prs 1, the titrant was stirred; in Experi- halfw i ,EThe analyte in Experiment 1 was different from KF, less less6The titrant in Experiment 1 was different from the H. greater greater7Yes, because H,O molecules are much smaller8No, because the solute molecules are polar and the9No, because H,O molecules are much larger than

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Paper 50 · ACT December 2012 · Form 71A · Science Test

Paraffin wax is a nonpolar solid and acetic acid is a 35. The nonpolar solid 9,10-diphenylanthracene has a polar solvent. Based on Student 3’s explanation, would molecular mass of 330 amu and readily dissolves in paraffin wax be soluble in acetic acid? censae. apa is a J oes por solvent with — strongly attracted to the acetic acid molecules. tent with the explanation(s) put forth by: more attracted to each other than to the acetic acid B. Student 2 pr molecules. C. Student 1 and Student 2 only. repelled by the acetic acid molecules. more attracted to each other than to the paraffin wax molecules. Passage Vil Key Radar observations of Saturn’s moon Titan have : ani revealed landforms that resemble longitudinal sand dunes « Desert C found in some deserts on Earth. Such dunes are produced 7 Asie D when sand-size solid particles are deposited in linear, a Tn parallel piles (see Figure 1). Figure 2 shows the average © titan | dune spacing and average dune height for each of a number 200 : of areas of dunes on Titan and in 4 deserts on Earth ! ! g (Deserts A-D). Figure 3 shows plots of the surface eleva- 0 | -~ tion (elevation above a horizontal reference same across a S Ag "9 E 10 fe osu a Wi ° . fo° 0. —. 2 2 (* saa onan IE on [3 J | ——s spacin See ee” ner: (yee LMT Neat Feeieene er ee Pal al ry 100 200 500 1,000 2,000 5,000 tP Ad = average dune spacing (m) AG, Vee / dunc height Note: Each symbol on the graph represents data for a different GS, J Ws | area of dunes in a desert or on Titan. air movement perpendicular cross section through Figure 2 longitudinal sand dunes Figure 2 adapted from Nicholas Lancaster, “Linear Dunes on Titan.” ©2006 by the American Association for the Advancement of Figure 1 Science. Key 37. Assume that for the dunes represented in Figure 2, as Titan avcrage wind speed increases, average dune height os d A increases. Did Desert A or Desert B more likely have Desert the greater average wind speed? aN A. Desert A, because the areas in that desert had ee ae eee Gee esr average dune heights. a a he B. Desert A, because the areas in that desert had ee i eure \eee eee | ee | Oe | lesser average dunc heights. & oe | iE | C. Desert B, because the areas in that desert had EF ys ttf EP PL poen average dune heights. 3 : | Nf D. Desert B, because the areas in that desert had > 1 FoR BF ih ii lesser average dune heights. : 95 Bd UE. BReHHE Riedie 38. Apoording to Figure 3, across the 25 km wide cross WA i i section of dunes on Titan and in Desert A, how does 5 50 -aMei hi Sim eieiee : the surface elevation of the Titan dunes differ, if at all, ° i from the surface elevation of the Desert A dunes? The PT ee es | ae | AV | ln | | surface elevation of the Titan dunes is: ) F, the same at all distances. 0 ; ) G. pa at all distances. distance (km)

AYes, because the paraffin wax molecules would be ar mass Of /o amu. ibese ovservalions are inconsts-BYes, because the paraffin wax molecules would be A. Student | only.CNo, because the paraffin wax molecules would be D. Student 2 and Student 3 only.DNo, because the acetic acid molecules would beEgreater at some distances but less at other distances.

Form 71A · Science Test · 40 questions

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