ACT Sample Questions Science

ACT Sample Questions: Science The ACT science sample questions contain the question types like: Data Representation Conflicting Viewpoint Research Summary Passage 1 Abandoned cornfields have been the sites of investigations concerning ecological succession, the orderly progression of changes in the plant and/or animal life of an area over time (see Figure 1). During the early stages of succession, the principal community (living unit) that dominates is the pioneer community. Pioneer plants are depicted in Figure 2. The final stage of ecological succession is characterized by the presence of the climax community, the oak-hickory forest. Figure 3 depicts the gradual change from pine to hardwoods. Following are some act sample questions on the above passage: On the basis of the data presented in Figure 1, approximately 80 years after the abandonment of cropland, the land would contain: pine seedlings only. oak-hickory hardwood forests only. early invading species like horseweed, aster, and broomsedge. large pine trees with an understory of hardwood trees. Answer: D According to the information in Figure 3, a 150-year-old climax community would contain oak and hickory trees with a density of approximately: 3,000 trees per unit area. 5,000 trees per unit area. 15,000 trees per unit area. 20,000 trees per unit area. Answer: C On the basis of the data depicting the gradual change from pine forest to an oak-hickory forest, after 100 years, as the density of the pine trees: increases, the density of the oak-hickory trees increases. increases, the density of the oak-hickory trees decreases. decreases, the density of the oak-hickory trees increases. decreases, the density of the oak-hickory trees decreases. Answer: C Given the information in Figure 1, which of the following conclusions concerning ecological succession in an abandoned cornfield is most correct? Succession in an abandoned cornfield begins on bare rock. Succession is characterized by the replacement of one plant community by another until a climax community has been achieved. The height of the plants in the communities decreases as succession progresses to the climax stage. The plant species change continuously during succession, but the change is more rapid in the later stages than in the earlier stages. Answer: B According to Figure 2, pioneer plant(s) showing a progressive increase in summed diameter of stems per unit area over the course of several years of succession is(are): horseweed only. broomsedge only. aster and broomsedge only. horseweed, aster, and broomsedge. Answer: B Passage 2 The following table represents the concentration of ions and dissolved gases in the sediment at the bottom of an ocean. A depth of 0 centimeters (cm) represents the top of the sediment. The concentrations are expressed in parts per million (ppm). The acidity of a solution is represented on a scale known as pH. A pH of 1 is very acidic, a pH of 7 is neutral, and a pH of 14 is very basic. Depth(cm) Temperature(oC) pH Concentration in sediment (ppm) SO42- S2- CO2 Fe3+ Fe2+ O2 0 4 7.0 7.0 0.0 1.0 4.0 0.5 2.0 5 5 6.5 5.0 2.0 1.5 3.0 1.5 1.0 10 7 6.0 3.5 3.5 2.0 2.0 2.0 0.0 15 9 5.5 3.3 3.8 3.0 0.8 3.8 0.0 20 10 5.0 3.0 4.0 1.0 0.5 4.0 0.0 Following are some act sample questions on the above passage: According to the information provided in the table, the concentration of which of the following ions and dissolved gases is constant for sediment depths of 10 cm or more? Sulfide (S2-) Carbon dioxide (CO2) Ferric iron (Fe3+) Oxygen (O2) Answer: D The graph below best represents the relationship between concentration and sediment depth for which of the following ions and dissolved gases? Ferrous iron (Fe2+) Oxygen (O2) Carbon dioxide (CO2) Sulfate (SO42-) Answer: C If the trends indicated in the table were to continue, one would predict the pH of the sediments at a depth of 35 cm to be: 1.5. 3.5. 4.5. 6.0. Answer: B A certain type of bottom-dwelling microorganism thrives under the following environmental conditions: low concentrations of Fe2+, high concentrations of O2, and a neutral pH. Based on the table, at which of the following sediment depths would one most likely find this microorganism? 0 cm 5 cm 10 cm 15 cm Answer: A A researcher wants to determine whether an unidentified sediment sample was drawn from a depth of 15 cm or 20 cm. Based on the information in the table, which of the following would NOT confirm the depth of the sample? O2 concentration Fe3+ concentration S2- concentration pH Answer: A Passage 3 The Sun's path from sunrise to sunset varies with the time of year. A student performed the following experiments on three clear, sunny days at three- or four-month intervals throughout the course of a year to study the path of the Sun through the sky. Experiment 1 At a chosen Northern Hemisphere location, the student placed a stick vertically into the ground so that 1 meter of its length was left above ground. The student knew that the length of the shadow was related to the height of the Sun above the horizon and that the shadow would point away from the direction of the Sun. The length in meters (m) and direction of the shadow cast by the stick were measured one hour after sunrise (Shadow A), at mid-morning (B), at noon (C), at mid-afternoon (D), and one hour before sunset (E) on each of the three days. The direction of each shadow was determined by placing a magnetic compass at the base of the stick and aligning the north arrow with the north mark on the compass. The direction of each shadow was then determined by a comparison with the compass face markings. The results are recorded in Table 1. Table 1 Shadow Day 1 Day 2 Day 3 Length(m) Shadow direction Length (m) Shadow direction Length (m) Shadow direction A 5.0 SW 8.6 NW 6.8 W B 1.2 W 2.9 NNW 1.7 NW C 0.3 N 2.3 N 0.9 N D 1.2 E 3.0 NNE 1.8 NE E 5.0 SE 8.6 NE 6.9 E Experiment 2 The following year, the student repeated Experiment 1 at a chosen location in the Southern Hemisphere. The results are in Table 2. Table 2 Shadow Day 1 Day 2 Day 3 Length(m) Shadow direction Length (m) Shadow direction Length (m) Shadow direction A 9.0 SW 5.0 NW 6.9 W B 3.2 SSW 1.1 W 1.8 SW C 2.5 S 0.3 S 1.0 S D 3.2 SSE 1.1 E 1.8 SE E 9.1 SE 5.0 NE 6.9 E Following are some act sample questions on the above passage: Which of the following was a constant in both experiments? Length of vertical stick exposed Shadow direction Day of the year Shadow length Answer: A If the experiments were repeated after pounding the stick farther into the ground so that only 0.5 m was exposed, how would this affect the shadow lengths? They would be twice as long as those in the original experiments. They would be one-and-one-half times as long as those in the original experiments. They would be one-half as long as those in the original experiments. They would be one-fourth as long as those in the original experiments. Answer: C Which of the following graphs best represents the relationship between the length of the stick's shadow and the time of day? A B C D Answer: D When the Sun is at an altitude 45o above the horizon, a vertical object will cast a shadow with a length equal to the object's height. Which of the following days included a measurement taken when the Sun was at an altitude of 45o? Day 1 in Experiment 1 Day 1 in Experiment 2 Day 2 in Experiment 1 Day 3 in Experiment 2 Answer: D Which of the following statements is best supported by the direction of Shadow A on each of the three days in Experiment 1? The direction of sunrise along the horizon varies throughout the year. Shadows never point due south in the Northern Hemisphere. Shadows never point due north in the Northern Hemisphere. The Sun's brightness varies throughout the year. Answer: A Based on the shadow directions in these experiments, which of the following best describes the difference between the apparent path of the Sun as seen from the observation points in the Northern and Southern Hemispheres? The Sun travels a more southerly path in the Southern Hemisphere than in the Northern Hemisphere. The Sun travels a more northerly path in the Southern Hemisphere than in the Northern Hemisphere. The Sun rises in the East and sets in the West in the Northern Hemisphere, but the opposite is true in the Southern Hemisphere. The Sun rises in the West and sets in the East in the Northern Hemisphere, but the opposite is true in the Southern Hemisphere. Answer: B Passage 4 Measles is an extremely contagious viral infection spread by the respiratory route. Figure 1 shows the course of measles from time of exposure to recovery from the infection. After recovery from measles, the infected individual develops immunity or resistance to reinfection. Figure 1 shows the development of immunity indicated by the antibody level. Figure 1 adapted from D. M. McLean, Virology in Health Care. ©1980 by Williams and Wilkins. The number of reported cases of measles from 1950 through 1987 is depicted in Figure 2. Figure 2 adapted from MMWR, "Summary of Notifiable Diseases." ©1986 Communicable Disease Center. Following are some act sample questions on the above passage: On Day 10 after exposure to measles, one could conclude that the greatest concentration of the measles virus would most likely be found in which of the following locations? Skin Mouth Blood Throat Answer: C The introduction of the measles vaccine had which of the following effects, if any, on the occurrence of measles? It completely eradicated measles after 1963. It caused a decrease in the number of measles cases only during 1963. It caused a gradual decrease from 400,000 measles cases in 1963 to approximately 1,500 in 1983. The introduction of the measles vaccine had no effect on the occurrence of measles after 1963. Answer: C Based on the information presented in the passage and in Figure 1, would it be possible to determine that a person had immunity against the measles virus 6 months after exposure? Yes; the level of protective antibodies against measles would be elevated 6 months after exposure. Yes; the virus would still be present in the respiratory tract to protect against reinfection. No; the level of protective antibodies against measles would be undetectable 6 months after exposure. No; the virus would no longer be present in the blood to protect against reinfection. Answer: A On the basis of Figure 1, one can conclude that there is a rise in the antibody level when the: fever vanishes. rash first appears. cold symptoms are most severe. virus is present in the blood. Answer: A According to Figure 2, the highest number of measles cases between 1981 and 1987 was approximately: 10. 100. 1,000. 5,000. Answer: D Passage 5 The relationship between pressure and volume of a gas was studied while the temperature was held constant. A container was filled with helium gas at room temperature and sealed (see figure below). The pressure on the gas could be controlled by varying the force exerted on the plunger, and the volume could be determined by the height of the plunger. Experiment 1 The initial pressure in this experiment was 1 atmosphere (atm). At that pressure the gas occupied 1 liter (L). The pressure of the helium was increased very slowly, by decreasing the volume, so that the gas remained at room temperature. The heat generated was released into the surrounding atmosphere. The pressure and the volume of the helium were recorded in Table Table 1 Pressure (atm) Volume (L) 1.0 1.00 1.5 0.67 2.0 0.50 2.5 0.40 3.0 0.33 Experiment 2 Experiment 1 was repeated at room temperature with an initial pressure of 1 atm and an initial volume of 2 L. The results are shown in Table 2. Table 2 Pressure (atm) Volume (L) 1.0 2.00 1.5 1.33 2.0 1.00 2.5 0.80 3.0 0.67 Experiment 3 In this experiment the entire gas container was insulated to prevent heat loss. The procedures from Experiment 1 were repeated. It was observed that the temperature of the gas rose as the volume decreased. Table 3 shows the volume measured at each pressure during the compression. Table 3 Pressure (atm) Volume (L) 1.0 1.00 1.5 0.78 2.0 0.66 2.5 0.58 3.0 0.52 The insulation was then removed and the pressure maintained at 3 atm. As the gas cooled to room temperature, the volume of the gas slowly decreased from 0.52 to 0.33 L. Following are some act sample questions on the above passage: How is the design of Experiment 1 different from that of Experiment 2? The container is insulated in Experiment 2, but not in Experiment 1. A different gas is used in Experiment 2 than in Experiment 1. The initial volume of the gas in Experiment 1 is half that of Experiment 2. The initial volume of the gas in Experiment 1 is twice that of Experiment 2. Answer: C A seventeenth-century scientist named Robert Boyle discovered that as the pressure on any confined gas increases (with the temperature held constant), the volume decreases. The best way to verify these results would be to repeat Experiment 1 with: an identical container made of a different material than the original. an identical volume of water. several different gases. an unsealed container. Answer: C If Experiment 1 is continued and the pressure is increased to 4 atm and remains fixed at this pressure, the gas would occupy a volume of approximately: 0.25 L. 0.33 L. 0.50 L. 1.00 L. Answer: A If Experiment 1 had started with 0.5 L of gas at 1 atm, what volume would be recorded when the pressure was 2.5 atm? 0.20 L 0.30 L 0.40 L 0.44 L Answer: A Which of the following statements best explains why the temperature of the container decreased after the insulation was removed in Experiment 3? The pressure increased, causing the temperature to decrease. The volume decreased, causing the temperature to decrease. Heat flowed from the surrounding atmosphere, through the container, and into the gas. Heat flowed from the gas, through the container, and into the surrounding atmosphere. Answer: B Suppose that Experiment 3 was modified so that the insulation was removed after each pressure increase and the container was allowed to sit at that pressure until its temperature cooled to room temperature. How would the volume readings be influenced, if at all, by this modified procedure? They would be the same as those in Experiment 1. They would be the same as those in Experiment 3. They would be greater than those in Experiment 2. They would be smaller than those in Experiment 1 Answer: A Passage 6 Several methods were investigated to reduce pollution emissions from a steel mill smokestack. Steel is mostly iron (Fe), but it also contains carbon (C). Impurities containing sulfur (S), nitrogen (N), or phosphorus (P) form gaseous compounds with oxygen that become pollutants. The smoke contains these pollutants and also tiny dust particles that pollute the air, are blown around by the wind, and eventually fall to the ground. Method 1 Steel mill smoke was passed across a set of electrified plates in the smokestack. The electrified plates attracted the dust particles in the smoke and removed them from the emissions. The efficiency of dust particle removal, which is dependent on the size of the particles, is shown in Table 1. Table 1 Voltage on plates(V) Particles removed(%) 100 20 200 50 500 90 1,000 80 3,000 70 Method 2 The smoke was passed through filters of different pore size in the smokestack that trapped dust particles and also decreased the amount of smoke leaving the stack. The data are shown in Table 2. Table 2 Filter pore size(microns) Particles trapped(%) Amount of time for filtersto drop to 50% efficiency(hours) 1,000 5 2,000 500 20 1,000 200 50 200 100 80 25 50 90 1 Method 3 To reduce pollution by chemical means, the smokestack emissions were bubbled through solutions of concentrated alkali (solutions of OH– ions). Table 3 shows the percent of the 3 pollutants removed versus the alkali concentration. Table 3 Alkaliconcentration(%) Sremoved(%) Nremoved(%) Premoved(%) 1 80 60 10 3 90 65 40 5 91 60 35 10 92 50 30 15 93 20 25 Following are some act sample questions on the above passage: If the plant's smoke contained equal amounts of S, N, and P, which of the following alkali concentrations would remove the most total pollution? 1% 3% 5% 10% Answer: B Which of the following statements about the alkali concentration is consistent with the results of Method 3? A higher alkali concentration results in more P removed. A higher alkali concentration results in more S removed. A lower alkali concentration requires more filters to remove dust particles. A lower alkali concentration results in an increased time required to remove all of the chemical pollutants. Answer: B Based on the characteristics of the filters used in Method 2, which of the following best describes the practical problem involved in choosing the best type of smokestack filter? The filter that traps the highest percentage of particles: requires the highest voltage across the electrified plates. reacts with alkali solutions. needs to be replaced least often. needs to be replaced most often. Answer: D If the smokestack of the steel mill was doubled in height, what effect, if any, would this be expected to have on the removal of pollutants? Filter efficiency would decrease because dust particles could more easily fall back down the stack. Electrostatic plate efficiency would increase because voltage increases with height. Dust particles would accumulate into larger pieces and would be more easily removed from the smoke. It cannot be determined from the given information. Answer: D Which of the following assumptions about reducing pollution emissions is common to both Methods 1 and 2? Emissions can only be removed by filters. Emissions can only be removed by electrified plates. Emissions must be captured in the smokestack to be removed. Emissions must be captured after they leave the smokestack. Answer: C To further investigate the effects of voltage on the removal of dust particles from steel mill smoke; the scientists could use which of the following procedures? Determining where the particulate matter falls to the ground after leaving the smokestack Determining what sizes of particles are removed from the smoke at different voltages Determining how the filters react when an alkali solution is passed through them Determining how the filters react when an acid solution is passed through them Answer: B Passage 7 During the development of chemistry, many chemists attempted to explain the changes that occur when combustible (capable of burning) materials burn and metals corrode or rust. The following are two proposed theories. Phlogiston Theory According to this theory, combustible materials, such as wood, coal, or metal contain a massless "essence" or presence called phlogiston. When combustion occurs, the phlogiston is released from the combusting object and is absorbed by the air. For example, when a piece of wood is burned, phlogiston is released to the air and the wood is converted to ash. The ash is free of phlogiston and can no longer support combustion. Similarly, if a metal is heated, the phlogiston is lost to the air and the metal is converted into a nonmetallic, powdery substance called ash, or calx. The corrosion (changing of a substance by a chemical reaction) of metals, such as the rusting of iron (Fe), also involves the loss of phlogiston from the metal, but at a slower rate than burning. Rust can be turned back into metal by heating it in air with a substance rich in phlogiston, such as charcoal. A transfer of phlogiston from the charcoal to the rust converts the rust back to metal. Oxygen Theory According to this theory, burning and rusting involve an element called oxygen, which is found in the air. The complete combustion of a piece of wood involves the rapid reaction of the wood with oxygen gas (O2) to produce carbon dioxide (CO2), which is a nonflammable gas, and water (H20). The rusting of iron involves the slow reaction of iron with oxygen to produce iron oxides such as Fe203. These iron oxides are known as rust. Heating rust with charcoal produces iron because the charcoal combines with the oxygen in the rust. In these transformations, there is a conservation of mass (the total mass of the reactants must equal the total mass of the products in a chemical reaction). In these reactions matter is neither created nor destroyed, but merely transformed. Following are some act sample questions on the above passage: Which of the following assumptions is implicit in the Phlogiston Theory? All combustible substances combine with phlogiston as they burn. All substances that burn contain phlogiston. Metals cannot be broken down by chemical means. The loss of phlogiston by a substance will always result in the production of heat and light. Answer: B The Phlogiston Theory could best be tested by measuring the: amount of light produced in the burning of a variety of combustible substances. amount of heat produced in the burning of a variety of combustible substances. masses of all the reactants and products before and after the reaction. amount of water produced when a substance burns. Answer: C According to the Oxygen Theory, the gases produced from the complete combustion of a candle: can support the breathing of a mouse. are nonflammable. are toxic to growing plants. are rich in hydrogen gas. Answer: B According to the Phlogiston Theory, the complete corrosion of zinc metal in air will yield a powdery substance that: cannot be converted back to zinc metal. contains pure phlogiston. contains no phlogiston. is a combination of the zinc metal and phlogiston. Answer: C According to the Oxygen Theory, both the burning of a material and the rusting of a metal involve: converting the elements of the material into gaseous compounds. forming oxygen-containing compounds from the elements in the material. removing oxygen from the material and releasing it into the air. producing high temperatures as a result of the chemical reactions. Answer: B According to the Phlogiston Theory, the gases collected from the complete burning of a piece of charcoal in air would be capable of: converting the ash from corroded tin back to tin metal. supporting combustion of another piece of charcoal. rusting iron. converting wood ash into rust. Answer: A A chemist heated a sample of mercury for several days in the apparatus shown below. As the experiment proceeded, the mercury in the retort became covered with a red powder, and the volume of mercury increased in the air reservoir. The remaining material in the reservoir would not support combustion. Which of the following theories is supported by the results of this experiment? The Phlogiston Theory, because the red powder resembled an ash The Phlogiston Theory, because the air in the reservoir could not support combustion and therefore did not contain oxygen The Oxygen Theory, because the mercury level dropped in the air reservoir indicating increased oxygen content The Oxygen Theory, because the mercury level rose in the air reservoir indicating decreased oxygen content Answer: D