ACT Past Papers with Answers

Paper 15 · ACT June 2021 · Form D06 · Science Test

The number of circular pieces of filter paper used in Experiment 2 was:

Paper 15 · ACT June 2021 · Form D06 · Science Test

Suppose that another trial had been performed in Experiment 1 and that the RT for this trial had been 8 sec. The H2O2 concentration for this trial would most likely have been closest to which of the following?

Paper 15 · ACT June 2021 · Form D06 · Science Test

Did H2O2 decompose faster in Trial 2 or in Trial 7 ? decomposition. decomposition. decomposition. decomposition.

Paper 15 · ACT June 2021 · Form D06 · Science Test

Consider the gas bubbles that were responsible for causing the filter paper to rise to the surface. The inte- rior of these gas bubbles was most likely primarily occupied by which of the following substances?

Paper 15 · ACT June 2021 · Form D06 · Science Test

Suppose that in each trial of Experiment 1 the volume of the H2O2 solution contained in the beaker in Step 2 had been 400 mL. In Trial 5, the RT would most likely have been closest to which of the following?

Paper 15 · ACT June 2021 · Form D06 · Science Test

According to the results of Experiments 1 and 2, what was the range of RT values for the trials done at a neu- tral pH ? Table 2 Trial pH H2O2 concentration (percent by volume) RT (sec) 06 5 0.2 50 07 5 0.3 39 08 5 0.5 31 09 5 1.0 24 10 5 2.0 18 11 7 0.2 19 12 7 0.3 15 13 7 0.5 12 14 7 1.0 09 15 7 2.0 07 16 8 0.2 22 17 8 0.3 17 18 8 0.5 14 19 8 1.0 10 20 8 2.0 08 39 GO ON TO THE NEXT PAGE. ACT-D06 4 4 Passage II The number of bacteria that adhere to a surface is affected by the medium in which they are grown. Two stud- ies examined how different media affect the number of bac- teria, either Escherichia coli or Salmonella typhimurium, that adhere to a stainless steel surface. Study 1 In each of 4 trials, Steps 1−8 were performed:

Paper 15 · ACT June 2021 · Form D06 · Science Test

If an incubation time of 36 hr had been tested in Study 1, the average number of E. coli cells/cm2 in 10% soy broth would most likely have been: . and 12,000 cells/cm2 . and 49,000 cells/cm2 . .

Paper 15 · ACT June 2021 · Form D06 · Science Test

Suppose a student wants to use a medium to grow adherence of the E. coli to a stainless steel surface over a 72 hr incubation period. Based on the results of Study 1, which of the 4 media tested should the student select?

Paper 15 · ACT June 2021 · Form D06 · Science Test

In each trial of the studies, what was the purpose of the euchrysine solution? attached to each chip ated with each chip prior to staining the cells staining the cells

Paper 15 · ACT June 2021 · Form D06 · Science Test

According to the results of Studies 1 and 2, for all 4 of the incubation times tested, the average number of S. typhimurium cells/cm2 was greater than the average number of E. coli cells/cm2 for which of the media tested?

Paper 15 · ACT June 2021 · Form D06 · Science Test

After the cells attached to a stainless steel chip were stained, any excess dye that remained on the chip needed to be removed prior to analyzing the cells. In the studies, which step was most likely performed to accomplish this task?

Paper 15 · ACT June 2021 · Form D06 · Science Test

Which of the following statements about the effect of either 10% soy broth or skim milk on the number of S. typhimurium that adhere to a stainless steel surface is consistent with the results of Study 2 ? As the incu- bation time increases from 1 hr through 72 hr for S. typhimurium in: S. typhimurium cells/cm2 increases and then remains constant. S. typhimurium cells/cm2 decreases and then remains constant. cells/cm2 increases only. cells/cm2 decreases only.

Paper 15 · ACT June 2021 · Form D06 · Science Test

In the studies, the macromolecule that was stained in the cells was composed of what type of subunit? 41 GO ON TO THE NEXT PAGE. ACT-D06 4 4 Passage III The presence of earthworms in forest soil can affect the amount of phosphorus (P) in the soil. Studies were done in 2 adjacent oak forests: Forest W, an old forest, and Forest Z, a young forest. Both forests had the same type of soil. Two 100 m2 plots were selected in each forest: one that had been invaded by earthworms several years before the studies began, and one that had no earthworms (see Table 1). Study 1 Three holes were dug in each plot. In each hole, a 10 cm3 volume of soil was collected from the wall of the hole at the surface and also at 3 cm, 6 cm, 9 cm, and 12 cm depths. All the soil volumes collected from the holes in a plot were combined into a single sample, and all particles larger than 2 mm in diameter were removed. The sample from each plot was oven dried and then analyzed to deter- mine the total soil P, in mg P/cm3 (see Figure 1). Figure 1 Study 2 In each plot, samples of the soil solution (aqueous solution present between soil particles) were collected using a pair of lysimeters that had been installed 2 years before the studies began. One lysimeter collected a sample of soil solution from a depth of 12 cm (shallow), and the other collected a sample from a depth of 36 cm (deep). For each sample, the concentration of soluble reactive P (SRP; one of several different forms of P found in soil solution), in micrograms of P per liter (μg P/L), was determined (see Figure 2). Figure 2 Figures adapted from Esteban R. Suarez et al., “Effects of Exotic Earthworms on Soil Phosphorus Cycling in Two Broadleaf Temper- ate Forests.” ©2003 by Springer-Verlag.

Paper 15 · ACT June 2021 · Form D06 · Science Test

In Study 1, why were the soil samples oven dried before being analyzed?

Paper 15 · ACT June 2021 · Form D06 · Science Test

Which of the following statements gives the most likely reason the lysimeters were installed 2 years before the studies began? The 2-year period allowed time for the: Table 1 Forest Plot Earthworms W 1 present 2 absent Z 3 present 4 absent Forest Z Forest W Plot 4 Plot 3 Plot 2 Plot 1 10 8 6 4 2 0 total soil P (mg P/cm 3 ) Forest Z Forest W Plot 4 Plot 3 Plot 2 Plot 1 60 50 40 30 20 10 0 SRP concentration ( μ g P/L) Key shallow deep 42 GO ON TO THE NEXT PAGE. ACT-D06 4 4

Paper 15 · ACT June 2021 · Form D06 · Science Test

Which of the following statements gives the most likely reason that only the top 12 cm of soil was con- sidered in Study 1 ? 12 cm. 12 cm.

Paper 15 · ACT June 2021 · Form D06 · Science Test

Which 2 plots should be compared to determine the effect of a forest’s age on total soil P in the absence of earthworms?

Paper 15 · ACT June 2021 · Form D06 · Science Test

In Study 1, all the particles in the samples analyzed for total soil P were:

Paper 15 · ACT June 2021 · Form D06 · Science Test

The greatest depth at which soil was collected in Study 1 was how many times as great as the greatest depth at which soil solution samples were collected in Study 2 ?

Paper 15 · ACT June 2021 · Form D06 · Science Test

Consider the results of Study 2 for Forest W. For each of the 2 depths, was the SRP concentration in the plot with earthworms greater than or less than the SRP con- centration in the plot without earthworms? shallow deep 1 _ _ 3 1 _ _ 2 43 GO ON TO THE NEXT PAGE. ACT-D06 4 4 Passage IV Certain odorants (airborne chemical compounds) can stimulate a change in the electrical activity of the olfactory neurons in the antennae of the moth Manduca sexta. The change in the electrical activity of an antenna in the pres- ence of an odorant can be detected as an increase in the antenna’s voltage relative to its voltage in the absence of odorants. The figure below shows the average change in antennal voltage for adult female M. sexta that were exposed to each of 5 odorants at each of 4 concentrations. Figure adapted from Kevin C. Daly, Lynnsey A. Carrell, and Esther Mwilaria, “Detection Versus Perception: Physiological and Behav- ioral Analysis of Olfactory Sensitivity in the Moth (Manduca sexta).” ©2007 by the American Psychological Association.

Paper 15 · ACT June 2021 · Form D06 · Science Test

As the odorant concentration increased from 1 mg/L through 100 mg/L, the average change in antennal voltage: then increased for 1 odorant. then decreased for 1 odorant.

Paper 15 · ACT June 2021 · Form D06 · Science Test

Based on the figure, if adult female M. sexta are exposed to 5 mg/L of methyl salicylate, the average change in antennal voltage will most likely be:

Paper 15 · ACT June 2021 · Form D06 · Science Test

For which of the following combinations of odorant and concentration was the average change in antennal voltage closest to the average change in antennal volt- age for 50 mg/L of nerolidol? odorant concentration

Paper 15 · ACT June 2021 · Form D06 · Science Test

For the M. sexta exposed to an odorant concentration of 1 mg/L, what is the order of the 5 odorants, from the odorant that stimulated the least average change in antennal voltage to the odorant that stimulated the greatest average change in antennal voltage? nerolidol, propylacetone linalool, propylacetone phyllic acid, linalool salicylate, linalool 480 440 400 360 320 280 240 200 160 120 80 40 0 480 440 400 360 320 280 240 200 160 120 80 40 0 average change in antennal voltage (μV*) average change in antennal voltage (μV*) Key linalool caryophyllic acid nerolidol methyl salicylate propylacetone 1 10 50 100 concentration (mg/L) *microvolts; 1 μV = 1 × 10−6 V 44 GO ON TO THE NEXT PAGE. ACT-D06 4 4

Paper 15 · ACT June 2021 · Form D06 · Science Test

According to the passage, certain chemical compounds can stimulate a change in the electrical activity of a specific type of neuron in M. sexta. This type of neuron is found in what organ of the human body?

Paper 15 · ACT June 2021 · Form D06 · Science Test

Would the neurons that are referred to in the passage be classified as belonging to the central nervous system or to the peripheral nervous system? and process information from other neurons. stimuli from the environment and then pass this information to other neurons. receive and process information from other neurons. receive stimuli from the environment and then pass this information to other neurons. 45 GO ON TO THE NEXT PAGE. ACT-D06 4 4 Passage V The orbital energy of a satellite depends on the central body (moon or planet) around which the satellite revolves, the altitude of the satellite’s orbit, and the satellite’s mass. Table 1 lists these characteristics for each of 12 different satellites. A satellite’s orbital period is the time required for a satellite to complete 1 revolution around its central body. Figure 1 shows a satellite’s orbital period as a function of altitude above the Moon, Mars, and Earth. Figure 1 Table 1 Satellite Central body Altitude (km)* Mass (kg)† Orbital energy (MJ)‡ 01 Moon 0,400 100 0,168 02 1,400 100 0,204 03 2,400 100 0,223 04 3,400 100 0,235 05 Earth 1,400 050 1,846 06 1,400 150 5,537 07 1,400 200 7,383 08 1,400 250 9,229 09 Earth 0,400 100 3,313 10 1,400 100 3,692 11 2,400 100 3,984 12 3,400 100 4,217 *kilometers † kilograms ‡ megajoules Key Moon Mars Earth orbital period (hr) 10 8 6 4 2 0 altitude (km) 400 1,400 2,400 3,400 46 GO ON TO THE NEXT PAGE. ACT-D06 4 4

Paper 15 · ACT June 2021 · Form D06 · Science Test

Consider a satellite that completes 1 revolution around Mars in exactly 5 hr. Based on Figure 1, the altitude at which this satellite orbits is most likely closest to which of the following?

Paper 15 · ACT June 2021 · Form D06 · Science Test

Consider the hypothesis “For a given central body and a given satellite mass, the greater the altitude of the orbit, the greater the orbital energy.” Which group(s) of satellites listed in Table 1 could be cited as evidence in support of this hypothesis?

Paper 15 · ACT June 2021 · Form D06 · Science Test

Consider 3 satellites: 1 orbiting the Moon, 1 orbiting Earth, and 1 orbiting Mars. Each satellite has the same mass and orbits at the same altitude above its central body. Based on Figure 1, the satellite orbiting which central body most likely completes a revolution in the least time?

Paper 15 · ACT June 2021 · Form D06 · Science Test

Consider the data for Satellites 5−8. The trend in this data set could be more completely demonstrated if which additional satellite were grouped together with Satellites 5−8 ?

Paper 15 · ACT June 2021 · Form D06 · Science Test

The diagram below (not drawn to scale) shows the radius, R, of a central body; the center-to-center dis- tance, d, between the central body and a satellite; and the satellite’s altitude. Given the data in Table 1 regarding Satellites 2 and 10, for which satellite would d be greater? greater than the altitude for Satellite 10. R for Earth. greater than the altitude for Satellite 2. for the Moon.

Paper 15 · ACT June 2021 · Form D06 · Science Test

Which of the following questions requires additional information beyond what is available in Table 1 and Figure 1 in order to be answered? orbit around Mars at an altitude of 400 km ? if its orbital energy is 3,984 MJ ? lite complete per hour while in orbit around Mars at an altitude of 400 km ? satellite to complete 1 revolution around Earth at an altitude of 400 km ? altitude d R 47 GO ON TO THE NEXT PAGE. ACT-D06 4 4 GO ON TO THE NEXT PAGE. Passage VI Each of 3 scientists provided a model of acids and bases. Scientist 1 An acid is an uncharged substance that dissociates (breaks apart into 2 or more ions) when dissolved in H2O, producing 1 or more hydrogen ions (H+ ) and a correspond- ing negative ion. A base is an uncharged substance that dis- sociates when dissolved in H2O, producing a metal ion and 1 or more hydroxide ions (OH− ). An example of an acid is hydrochloric acid (HCl) because, in H2O, it dissociates into H+ and a chloride ion (Cl− ). An example of a base is sodium hydroxide (NaOH) because, in H2O, it dissociates into a sodium ion (Na+ ) and OH− . Scientist 2 An acid is an uncharged or charged substance that donates an H+ to another substance (the base) in a chemical reaction, forming a covalent bond. A base is an uncharged or charged substance that accepts an H+ from another sub- stance (the acid) in a chemical reaction, forming a covalent bond. In the reaction of HCl with H2O, HCl is an acid because it donates an H+ to the H2O. A covalent bond forms between the donated H+ and the H2O, producing a hydronium ion (H3O+ ) and Cl− . In this reaction, H2O is a base because it accepts the H+ from HCl. Scientist 3 An acid is an uncharged or charged substance that accepts a pair of electrons from a base (an uncharged or charged substance that donates the pair of electrons) in a chemical reaction. As the electrons are transferred, a new covalent bond forms between the atom in the acid that is accepting the electrons and the atom in the base that is donating the electrons. At the same time, an existing bond within the acid is broken. In the reaction of HCl with H2O, HCl is an acid because it accepts a pair of electrons from H2O. A covalent bond forms between the H atom of HCl and the O atom of H2O, and, at the same time, the bond between the H atom and the Cl atom in HCl is broken. This produces H3O+ and Cl− . In this reaction, H2O is a base because it donates the pair of electrons to the H atom of the HCl molecule.

Paper 15 · ACT June 2021 · Form D06 · Science Test

All 3 scientists made which of the following state- ments about H2O or HCl ?

Paper 15 · ACT June 2021 · Form D06 · Science Test

Scientist 2’s description of an acid differs from that of Scientist 3 in which of the following ways? According to Scientist 2, an acid: Scientist 3, an acid donates an H+ . Scientist 3, an acid accepts an H+ . , whereas according to Scientist 3, an acid donates a pair of electrons. , whereas according to Scientist 3, an acid accepts a pair of electrons.

Paper 15 · ACT June 2021 · Form D06 · Science Test

Consider the statement “All bases contain hydroxide ions.” Which of the scientists would be likely to agree with the statement?

Paper 15 · ACT June 2021 · Form D06 · Science Test

Consider the statement “In any chemical reaction where a substance acts as an acid, there must also be a substance that acts as a base.” Which of Scientists 2 and 3, if either, would be likely to agree with this statement?

Paper 15 · ACT June 2021 · Form D06 · Science Test

The chemical formula of hypochlorous acid is HOCl. Scientist 1 would likely assert that HOCl is an acid, and explain that assertion by stating that, in H2O, HOCl dissociates into: and ClO+ . and ClO− . and Cl+ . and Cl− . 48 ACT-D06 4 4

Paper 15 · ACT June 2021 · Form D06 · Science Test

Based on Scientist 3’s model, which of these can never act as a base in a reaction: an individual atom of H or an individual atom of O ? An atom of:

Paper 15 · ACT June 2021 · Form D06 · Science Test

Consider Scientist 2’s model. The conjugate base of an acid can be defined as the part of the acid that remains after the acid has donated the H+ in a reaction. What substance is the conjugate base in the reaction described by Scientist 2 ? 49 ACT-D06 4 4 END OF TEST 4 STOP! DO NOT RETURN TO ANY OTHER TEST. 50 Scoring Keys for Form D06 Use the scoring key for each test to score your answer document for the multiple-choice tests. Mark a “1” in the blank for each question you answered correctly. Add up the numbers in each reporting category and enter the total number correct for each reporting category in the blanks provided. Also enter the total number correct for each test in the blanks provided. The total number correct for each test is the sum of the number correct in each reporting category. Test 1: English—Scoring Key Number Correct (Raw Score) for: Production of Writing (POW) _______ (22) Knowledge of Language (KLA) _______ (12) Conventions of Standard English (CSE) _______ (41) Total Number Correct for English Test _______ (POW + KLA + CSE) (75) Key Reporting Category* POW KLA CSE