Below are answer explanations to the 2nd full-length Science test of the released ACT practice test for the updated 2025 test. 

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Answer Explanations to the ACT 2025 Practice Science Test

Passage I

Question 1, “According to Table 2, at 1 atm, what is the BP of the alkane with the chemical formula C₅H₁₀?” The answer is D: “49°C.”

  • To answer this question, the student must correctly identify that C₅H₁₀ refers to cyclopentane, not an n-alkane, and then read its boiling point directly from Table 2, which lists cycloalkanes. Table 2 shows that cyclopentane (C₅H₁₀) has a BP of 49°C at 1 atm. Choice A (−130°C) and B (−93°C) are melting points, not boiling points, and C (36°C) is the boiling point of pentane (C₅H₁₂) from Table 1, a different compound entirely. The key skill is matching the correct formula to the correct table before reading the value.

Question 2, “For the n-alkanes listed in Table 1, as the number of carbon atoms per molecule increases, the BP at 1 atm:” The answer is F: “increases only.”

  • This question requires identifying a trend in Table 1. Reading down the table from propane (C₃H₈) through nonane (C₉H₂₀), the boiling point consistently rises, from −42°C up to 151°C. There are no reversals or decreases at any point, so the trend is a steady increase. Choices G, H, and J all incorrectly suggest either a decrease or a mixed trend, which is not supported by the data. The correct approach is to ignore individual values and focus on the overall pattern across increasing carbon number.

Question 3, “At 1 atm, how many of the cycloalkanes listed in Table 2 have an MP above the MP of ice?” The answer is B: “3.”

  • Ice melts at 0°C, so the student must count how many cycloalkanes in Table 2 have melting points greater than 0°C. Cyclohexane (7°C), cyclooctane (15°C), and cyclononane (11°C) meet this condition, for a total of three. The remaining cycloalkanes all have melting points below 0°C. Answer choices A, C, and D miscount the qualifying compounds. The key here is comparing each value to a known reference point rather than to each other.

Question 4, “Based on Tables 1 and 2, what is the name of the n-alkane shown in Figure 1, and what is the name of the cycloalkane shown in Figure 1?” The answer is F: “n-alkane: propane; Cycloalkane: cyclopropane.”

  • The structural drawings in Figure 1 must be interpreted. The n-alkane shows a straight chain of three carbon atoms, which corresponds to propane (C₃H₈) in Table 1. The cycloalkane shows a three-carbon ring, which corresponds to cyclopropane (C₃H₆) in Table 2. Choices G, H, and J all incorrectly assign four-carbon names (butane or cyclobutane), which do not match the number of carbons shown in the figure. This question tests structural interpretation rather than numerical reading.

Question 5, “Based on Table 2, in a molecule of any given cycloalkane, the number of hydrogen atoms is always equal to:” The answer is C: “twice the number of carbon atoms.”

  • Looking across Table 2, every cycloalkane follows the pattern CₙH₂ₙ: for example, C₃H₆, C₄H₈, C₅H₁₀, and so on. This shows that the number of hydrogen atoms is always two times the number of carbon atoms. Choices A and B underestimate the hydrogen count, while D exaggerates it. The correct strategy is to identify the consistent formula pattern rather than checking just one example.

Passage II

Question 6, “Relative to the center of the protostar, does gravity more likely accelerate gas particles inward or outward, and does RP more likely accelerate gas particles inward or outward?” The answer is G: “Gravity: inward; RP: outward.”

  • This question tests basic physical interpretation of forces described directly in the passage. Gravity is described as attracting gas toward the protostar, meaning it accelerates particles inward, toward the center. Radiation pressure (RP), by contrast, is explicitly stated to push gas away from the protostar, meaning it accelerates particles outward. Choice G correctly pairs these directions. Choices F and H incorrectly reverse one of the forces, while J incorrectly claims both forces push outward, contradicting the description of gravitational attraction. Students should anchor their reasoning strictly to the passage’s force descriptions rather than outside assumptions.

Question 7, “Based on Scientist 2’s argument, do gas particles more likely accrete near the equator or near the poles of a protostar with a disk?” The answer is B: “Near the equator, because the effect of RP is reduced there.”

  • Scientist 2 explains that because the protostar rotates, a disk forms in the equatorial plane, and this disk reduces the effect of radiation pressure in that region. Since radiation pressure inhibits accretion, reducing RP allows gas to fall inward more easily. Therefore, gas particles are more likely to accrete near the equator, not the poles. Choice B correctly identifies both the location and the reason. Choice A gets the location right but gives the opposite reasoning, while C and D incorrectly place accretion near the poles, contradicting Scientist 2’s model.

Question 8, “One of the most massive stars known is Eta Carinae, which has an approximate mass of 120 Mₛ. Based on the arguments of Scientists 1, 2, and 3, respectively, what is the minimum number of stars, each formed entirely by accretion, that would have been required to form Eta Carinae?” The answer is H: “Scientist 1: 6; Scientist 2: 3; Scientist 3: 1.”

  • This question requires translating each scientist’s maximum accretion limit into a numerical calculation. Scientist 1 claims accretion alone can produce stars up to 20 Mₛ, so reaching 120 Mₛ would require 120 ÷ 20 = 6 stars. Scientist 2 allows accretion up to 40 Mₛ, so 120 ÷ 40 = 3 stars. Scientist 3 argues accretion is limited only by available gas, meaning a single star could reach 120 Mₛ entirely by accretion, so 1 star is sufficient. Choices F, G, and J contain at least one incorrect calculation or misinterpretation of a scientist’s claim.

Question 9, “An observation of which of the following stars emerging from its envelope would support Scientist 2’s argument but weaken Scientist 1’s argument?” The answer is C: “A 30 Mₛ star.”

  • Scientist 1 claims accretion alone is limited to 20 Mₛ, while Scientist 2 claims accretion can reach 40 Mₛ. A star emerging from its envelope at 30 Mₛ exceeds Scientist 1’s maximum but falls within Scientist 2’s range, directly supporting Scientist 2 while contradicting Scientist 1. A 15 Mₛ or 20 Mₛ star would support both scientists, and a 50 Mₛ star would exceed even Scientist 2’s limit. The key is identifying a mass that lies between the two proposed thresholds.

Question 10, “Scientists 2 and 3 agree that a disk forms around a protostar as a result of the protostar’s:” The answer is F: “motion.”

  • Both Scientists 2 and 3 attribute disk formation to the protostar’s rotation, which is a form of motion. This rotation causes material to spread into a flattened disk in the equatorial plane. Choice F correctly captures this shared assumption. Choices G, H, and J reference radiation, location in a cluster, or mergers—none of which are cited by either scientist as the cause of disk formation. This question rewards careful comparison of overlapping claims.

Question 11, “Which of the scientists, if any, would be likely to agree that the Sun could have formed entirely by accretion?” The answer is C: “Scientists 1, 2, and 3.”

  • The Sun has a mass of 1 Mₛ, which is far below the accretion limits proposed by all three scientists. Scientist 1 allows accretion up to 20 Mₛ, Scientist 2 up to 40 Mₛ, and Scientist 3 allows accretion until gas is depleted. Since 1 Mₛ is well within all these limits, all three scientists would agree that the Sun could have formed entirely by accretion. The other answer choices incorrectly exclude scientists whose models clearly allow low-mass star formation.

Passage III

Question 12, “Which of the following pie charts best represents the seedling mortality results for Group 4?” The answer is F.

  • From Table 2, Group 4 has a seedling mortality of 6%, meaning 94% survived. The correct pie chart must therefore show a very small mortality slice and a much larger survival slice. Choice F matches this proportion. The other choices misrepresent the data by either exaggerating mortality or showing equal or reversed proportions. This question is not about exact drawing precision but about recognizing which visual best reflects a small percentage versus a large remainder.

Question 13, “The pots in which group were most likely covered by the greatest number of layers of the plastic mesh?” The answer is A: “Group 1.”

  • The passage states that plastic mesh was used to reduce sunlight intensity, and Table 1 shows that Group 1 received the lowest sunlight intensity (10%). More mesh layers would block more light, so the group with the lowest intensity must have had the greatest number of layers. Groups 2–4 all received higher sunlight percentages, meaning fewer or no layers. This is a cause-and-effect reasoning question rather than a data-lookup question.

Question 14, “Based on the results of the study, approximately what percent of the seedlings that received a sunlight intensity of 10% survived to 10 weeks?” The answer is H: “92%.”

  • Group 1 received 10% sunlight and had a mortality of 8% (Table 2). Survival is the complement of mortality, so 100% − 8% = 92% survived. Choice F incorrectly restates mortality, G incorrectly uses dry mass, and J incorrectly claims the value cannot be determined even though the necessary data are explicitly given. This question tests whether students understand that survival and mortality are inverse quantities.

Question 15, “Suppose that as the plant dry mass increases, the leaf area ratio decreases. Based on the results of the study, the plants in which group most likely had the lowest leaf area ratio?” The answer is C: “Group 3.”

  • The question tells students that higher dry mass corresponds to a lower leaf area ratio. Looking at Table 2, Group 3 has the highest average dry mass (0.80 g). Therefore, Group 3 would have the lowest leaf area ratio. Groups 1, 2, and 4 all have lower dry masses. This is a classic ACT logic question where a relationship is given and must be applied to identify the correct group.

Question 16, “The effect of what abiotic factor was examined in the study?” The answer is J: “Sunlight intensity.”

  • An abiotic factor is a nonliving environmental variable. The study explicitly manipulates sunlight intensity by using plastic mesh. Average dry mass and seedling mortality are outcomes, not factors, and precipitation was held constant by watering all pots equally. Students must distinguish between what is controlled or measured and what is intentionally changed.

Question 17, “Based on the results of the study, what was the dry mass of an individual seedling in Group 2?” The answer is D: “Cannot be determined from the given information.”

  • Table 2 reports the average dry mass of surviving plants, not the dry mass of an individual seedling. The total number of surviving seedlings is not given, nor is the distribution of masses. Therefore, it is impossible to determine the dry mass of a single seedling. Choices A and B incorrectly lift values from the table without understanding what those values represent, and C invents a calculation with no supporting data.

Passage IV

Question 18, “In each trial, the flasks were most likely shaken to:” The answer is H: “maximize the contact between the Congo red and the particles of B and of MB.”

  • Shaking the flasks increases mixing, which ensures that Congo red molecules come into frequent contact with the bentonite (B) or modified bentonite (MB) particles. Since the goal of the experiments is to measure how much Congo red is removed by binding to these particles, maximizing contact is essential. Choices F and G incorrectly focus on changing concentration, which shaking does not do, and J directly contradicts the experimental goal. This question tests understanding of why a procedural step is included, not just what happens during the step.

Question 19, “Suppose that, in an additional trial of Experiment 3, a shaking time of 100 min had been tested. The % CR removed by MB in this trial would most likely have been between:” The answer is D: “80% and 90%.”

  • In Experiment 3, Figure 3 shows that % Congo red removed by MB increases rapidly with shaking time and then begins to level off. At 90 minutes, the value is just under 80%, and by 120 minutes it is near or slightly above 90%. A shaking time of 100 minutes falls between these two data points, so the expected value would be between 80% and 90%. Choices A and B are far too low and inconsistent with the upward trend, while C underestimates the value relative to nearby points on the graph.

Question 20, “In Experiment 2, the % CR removed by B from the neutral Congo red solution was closest to which of the following?” The answer is G: “20%.”

  • A neutral solution has a pH of 7, so the student must locate pH 7 on Figure 2 and then read the value for B, not MB. At pH 7, the % Congo red removed by B is slightly below 20%, making 20% the closest option. Choices F, H, and J are either too low or far too high compared to the data point shown. This question emphasizes careful identification of the correct curve and correct pH value.

Question 21, “Based on the results of Experiments 2 and 3, the % CR removed would likely be greatest for which of the following combinations of pH and shaking time?” The answer is B: “pH: 5.0; Shaking time: 240 min.”

  • From Experiment 2, % Congo red removal is highest at lower pH values, especially near pH 5, particularly for MB. From Experiment 3, % removal increases with longer shaking times, reaching its highest values near 240 minutes. Combining these two results, the greatest removal would occur at low pH and long shaking time. The other options include either higher pH, shorter shaking time, or both, which the data show would reduce removal efficiency.

Question 22, “Based on Figure 2 and additional information in the passage, how many trials were performed in Experiment 2?” The answer is G: “Six; in each trial, the % CR removed was determined for both B and MB at 1 of 6 pH values.”

  • Experiment 2 varies pH across six values while holding mass and shaking time constant. At each pH value, both B and MB are tested simultaneously, meaning each pH corresponds to one trial. Therefore, six total trials were conducted. Choices F, H, and J incorrectly double-count measurements or introduce variables not described in the passage. This question tests the distinction between number of trials and number of measurements per trial.

Question 23, “Consider the description of Experiment 1 and the % CR removed by MB in the 0.200 g trial of Experiment 1. The concentration of Congo red that remained in the solution when the shaking ended was approximately:” The answer is A: “0 mg/L.”

  • In Experiment 1, the initial Congo red concentration is 300 mg/L. Figure 1 shows that at 0.200 g of MB, the % Congo red removed is essentially 100%. If nearly all the dye is removed, almost none remains in solution. Therefore, the remaining concentration would be approximately 0 mg/L. Choices B, C, and D would only be plausible if a significant fraction of Congo red remained, which contradicts the graph.

Passage V

Question 24, “Which student would be most likely to agree that while the rod was partially submerged, the scale was supporting the entire weight of the rod?” The answer is H: “Student 3.”

  • Student 3 argues that the buoyant force B is equal to the weight of the rod, which would mean the teacher’s hand supports 0.00 N. Student 3 also claims that depth has no effect on water pressure, concluding that the force on the scale is 15.00 N—the original 10.00 N from the beaker and water plus the rod’s full 5.00 N. This implies the scale supports the entire weight of the rod. Student 1 explicitly splits the rod’s weight between the hand and the scale, while Students 2 and 4 claim the scale remains at 10.00 N. Only Student 3’s conclusion matches the question’s condition.

Question 25, “Within a fluid, pressure increases as depth increases. This fact weakens the response(s) given by which student(s)?” The answer is D: “Students 2, 3, and 4 only.”

  • Students 2, 3, and 4 all state—directly or indirectly—that depth has no effect on water pressure, which contradicts the given physical fact that pressure increases with depth. Student 1 correctly states that increased water depth leads to increased pressure at the bottom of the beaker, producing a downward force on the scale. Therefore, the correct choice includes Students 2, 3, and 4, but not Student 1. This question tests whether students can evaluate arguments against a stated physical principle.

Question 26, “Suppose that it were determined that the magnitude of B was 1.37 N. Based on Student 2’s argument, how much weight would the teacher’s hand have been supporting?” The answer is G: “3.63 N.”

  • Student 2 states that the teacher’s hand supports (5.00 N − B). Substituting B=1.37 N gives 5.00−1.37=3.63 N. Choice G reflects this calculation exactly. The other choices either ignore the subtraction (H), incorrectly equate the answer to B (F), or add instead of subtract (J). The key skill is applying the correct student’s formula, not re-deriving the physics independently.

Question 27, “Suppose that the teacher had held the rod above the water such that no portion of the rod was ever submerged. Based on Student 1’s response, how much weight would the teacher’s hand have been supporting, 0.00 N or 5.00 N?” The answer is D: “5.00 N; the teacher’s hand would have been supporting the entire weight of the rod, because the buoyant force would have been zero.”

  • Student 1 states that buoyant force exists only when water is displaced. If the rod is not submerged at all, no water is displaced, so B=0. Student 1’s expression for the hand’s support is (5.00N−B), which becomes 5.00N. Choices A and C incorrectly assume buoyancy without displacement, and B incorrectly claims zero support despite no buoyant force. This question checks logical consistency within a student’s own argument.

Question 28, “In regard to B, which of the following statements summarizes the responses given by Students 1 and 3?” The answer is F: “Student 1 claimed that B is equal in magnitude to the weight of the displaced water, whereas Student 3 claimed that B is equal in magnitude to the weight of the rod.”

  • Student 1 explicitly states that buoyant force equals the weight of the displaced water, which is the correct physical definition of buoyancy. Student 3, however, claims that because the rod is less dense than water, B equals the weight of the rod. Choice F accurately contrasts these two claims. The remaining options either reverse the students’ positions or incorrectly say they agree.

Question 29, “Consider the ‘before’ portion of Figure 1, and assume that the scale was on a lab bench. If the scale itself had a weight of 45.80 N, what total force must the lab bench have been exerting on the underside of the scale?” The answer is D: “55.80 N.”

  • In the “before” scenario, the scale displays 10.00 N, which represents the force from the beaker and water on the scale. The lab bench must support both the weight of the scale itself (45.80 N) and the 10.00 N load on top of it. Adding these gives 45.80+10.00=55.80 N. The other options reflect incomplete force accounting, such as ignoring either the scale’s weight or the load it supports.

Passage VI

Question 30, “For the batch of jam prepared with 1.0% Pectin Z by mass, as storage time increased, the TMA concentration:” The answer is G: “decreased only.”

  • Looking at Table 1 for Pectin Z at 1.0%, the TMA concentration drops steadily across time: from 28 mg/100 g (1 day) to 25 (1 month), then 20 (3 months), and finally 15 (6 months). There is no increase at any point, so the trend is a consistent decrease. Choices F, H, and J all require at least one increase in the data, which does not occur. This question tests whether students can track a single row across multiple time points and correctly describe the trend.

Question 31, “Suppose the scientists had also prepared a batch of jam using 0.5% Pectin Y. Based on the results of the experiment, at a storage time of 3 months, the TMA concentration would most likely have been between:” The answer is C: “28 mg/100 g and 32 mg/100 g.”

  • For Pectin Y, the 3-month TMA concentration is 28 mg/100 g at 0.3% and 32 mg/100 g at 0.7%. A concentration of 0.5% falls between these two values, so the expected TMA concentration would also fall between 28 and 32 mg/100 g. The other answer choices either fall outside this range or contradict the observed pattern that higher pectin concentration corresponds to higher TMA retention. This question tests interpolation rather than direct lookup.

Question 32, “Which of the following variables was not an independent variable in the experiment?” The answer is H: “TMA concentration.”

  • Independent variables are the factors the scientists deliberately changed: pectin type, pectin concentration, and storage time. TMA concentration is what was measured as a result of those changes, making it a dependent variable. Choices F, G, and J are all explicitly manipulated in the experimental design. This is a classic variables-identification question that rewards careful reading of the experiment description.

Question 33, “Suppose that the experiment had been repeated, except that the jars had been stored at 30°C. Would the TMA concentrations in this new experiment more likely have been less than or greater than the corresponding TMA concentrations listed in Table 1?” The answer is A: “less, because more TMA would have broken down at the higher temperature.”

  • The passage states that antioxidants like TMA break down faster when exposed to added heat. Increasing storage temperature from 20°C to 30°C would therefore accelerate breakdown, leading to lower TMA concentrations at every time point. Choice A correctly connects increased temperature to increased degradation. Choices B, C, and D either reverse the temperature logic or contradict the given information.

Question 34, “Assume that, in the recipe the scientists used, 100 g of jam was produced for every 70 g of blackberries. If no TMA broke down as the jam was prepared, what mass of TMA would have been found in 100 g of jam before the jars were placed in boiling water?” The answer is G: “140 mg.”

  • Fresh blackberries contain 200 mg TMA per 100 g, which is equivalent to 2 mg per gram. If 100 g of jam is made from 70 g of blackberries, then the TMA present would be 70×2=140 mg. The other choices result from using incorrect proportions or failing to convert the given concentration correctly. This question tests proportional reasoning rather than experimental interpretation.

Question 35, “A total of how many jars were prepared in the experiment?” The answer is D: “54.”

  • The experiment includes 9 batches of jam (3 pectin types × 3 concentrations). Each batch was divided equally into 6 jars, giving 9×6=54 jars total. Choices A, B, and C reflect partial counts, such as jars per batch or batches per pectin type. The key is recognizing that every combination produced its own full set of jars.

Passage VII

Question 36, “Allele A and Allele B can best be described as:” The answer is F: “different versions of the same gene.”

  • By definition, alleles are alternative forms of the same gene found at the same genetic location. The passage explicitly refers to Allele A and Allele B of “a specific gene,” confirming this relationship. Choice F correctly matches the biological definition. Choice G incorrectly suggests different genes, while H and J incorrectly claim the alleles are identical. This question tests foundational genetics vocabulary rather than data interpretation.

Question 37, “Suppose that another population of amoebas with an initial Allele A frequency of 0.5 and an initial size of 10,000 had been included in the computer simulation. Based on Figure 1, the frequency of Allele B in G2 for that population would most likely have been closest to which of the following?” The answer is B: “0.5.”

  • Figure 1 shows populations with large initial sizes (10,000 amoebas) experiencing only small fluctuations in allele frequency due to genetic drift. If Allele A starts at 0.5, Allele B must also start at 0.5, since the two frequencies sum to 1. Because large populations resist dramatic random changes, the frequency of Allele B in generation 2 would likely remain close to 0.5. Choices A, C, and D represent extreme shifts that are inconsistent with the stable trends shown for large populations in Figure 1.

Question 38, “According to Figure 2, Allele B was absent from which of P5–P8 in G2?” The answer is F: “P5.”

  • Allele B is absent when the frequency of Allele A equals 1.0, meaning only Allele A remains. In Figure 2, population P5 reaches an Allele A frequency of 1.0 by generation 2, indicating complete loss of Allele B. The other populations still show Allele A frequencies below 1.0 at G2, meaning Allele B is still present. This question tests careful reading of a graph rather than conceptual genetics alone.

Question 39, “Based on Figures 1 and 2, is the effect of genetic drift on allele frequency greater in a relatively large population or in a relatively small population?” The answer is D: “Relatively small; each of P5–P8 experienced greater fluctuations in the frequency of Allele A than did each of P1–P4.”

  • Figures 1 and 2 contrast large populations (P1–P4, 10,000 amoebas) with small populations (P5–P8, 12 amoebas). The small populations show dramatic swings, including fixation and loss of alleles, while the large populations change only slightly. This demonstrates that genetic drift has a stronger effect in small populations, where random sampling has a larger impact. The other choices incorrectly reverse or misdescribe the observed trend.

Question 40, “Based on Figure 1, in the initial generation of each of P1–P4, how many amoebas had the genotype AA, and how many amoebas had the genotype BB?” The answer is F: “Amoebas with genotype AA: 5,000 / Amoebas with genotype BB: 5,000.”

  • Figure 1 shows that in generation G0, the frequency of Allele A is 0.5 for each of P1–P4. With 10,000 amoebas total and all individuals homozygous, half must be AA and half BB, resulting in 5,000 of each genotype. The other choices incorrectly assign unequal or excessive counts that contradict the given frequency. This question requires converting allele frequency into actual population numbers.

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