Below are answer explanations to the Science Test of the 3rd released full-length Digital ACT practice test.

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Answer Explanations to the 3rd Digital ACT Practice Science Test

Passage I – “β-Agarase Activity”

Question 1, “Based on Table 1, what was the β-agarase activity for an aqueous solution having a sodium ion concentration of 5 mM?” The correct answer is D: “1.96 nmol/min.”

  • This question requires students to locate a specific value in Table 1. The table displays β-agarase activity for four different metal ions at five concentrations. To find the correct value, locate the row for sodium (Na⁺) and the column corresponding to 5 mM. Their intersection gives an activity of 1.96 nmol/min, making Choice D correct. Choices A (1.06 nmol/min), B (1.82 nmol/min), and C (1.89 nmol/min) are the activities for calcium, potassium, and magnesium, respectively, at the same concentration. These choices are incorrect because they correspond to different metal ions.

Question 2, “Based on Table 2, for an aqueous solution containing 1 mM of Na⁺ at a pH of 7.0, approximately how many nanomoles of agarose would be broken down by β-agarase in 5 min at a temperature of 20°C?” The correct answer is H: “6.20 nmol.”

  • This question requires using the enzyme activity rate from Table 2 to calculate the amount of agarose broken down over a specified period. At 20°C, β-agarase breaks down 1.24 nanomoles of agarose per minute. Multiplying this rate by 5 minutes gives 1.24 × 5 = 6.20 nanomoles, making Choice H correct. Choices F (0.25 nmol) and G (0.34 nmol) are much too small and do not account for the full five-minute interval. Choice J (8.45 nmol) results from multiplying the activity at 30°C, 1.69 nmol/min, by five rather than using the specified temperature of 20°C.

Question 3, “Based on Table 1, the β-agarase activity for an aqueous solution containing 25 mM of Ca²⁺ at 40°C and a pH of 7.0 would most likely be:” The correct answer is C: “between 0.51 nmol/min and 1.01 nmol/min.”

  • This question tests the ability to estimate a value using a trend in experimental data. Table 1 shows that β-agarase activity decreases as the concentration of calcium increases. At 10 mM, the activity is 1.01 nmol/min, and at 50 mM, it is 0.51 nmol/min. Since 25 mM falls between these concentrations, the expected activity should fall between their corresponding values, making Choice C correct. Choice A places the activity below the value recorded even at 100 mM, while Choice B predicts a value lower than the activity at 50 mM. Choice D incorrectly predicts an activity greater than that recorded at 10 mM, contradicting the decreasing trend.

Question 4, “Consider the claim ‘For any given metal ion concentration, the β-agarase activity in the presence of Mg²⁺ is always less than the β-agarase activity in the presence of Na⁺.’ For how many of the concentrations listed in Table 1 are the β-agarase activities inconsistent with this claim?” The correct answer is F: “1.”

  • This question requires comparing the magnesium and sodium rows in Table 1 at each of the five concentrations. At 1 mM, magnesium produces an activity of 2.51 nmol/min, while sodium produces only 1.77 nmol/min. This contradicts the claim that magnesium always produces less activity than sodium. However, at 5, 10, 50, and 100 mM, magnesium activity is lower than sodium activity, consistent with the claim. Therefore, only one concentration contradicts the statement, making Choice F correct. Choices G (2), H (4), and J (5) incorrectly count additional concentrations at which the claim actually holds true.

Question 5, “How many of the pH values listed in Table 3 are classified as basic?” The correct answer is B: “2.”

  • This question tests knowledge of the pH scale. A solution is classified as basic when its pH is greater than 7. Table 3 lists pH values of 4.0, 5.0, 6.0, 7.0, 8.0, and 9.0. Only 8.0 and 9.0 are greater than 7, so there are two basic pH values, making Choice B correct. Choice A (1) fails to count both basic values. Choices C (3) and D (4) incorrectly include additional values that are neutral or acidic. In particular, a pH of exactly 7.0 is neutral rather than basic.

Passage II – “Vermicompost and Tomato Plant Growth”

Question 6, “Based on Tables 1 and 2, what was the average plant height for the tomato plants grown in the mixture with 35% by volume of vermicompost?” The correct answer is G: “50 cm.”

  • This question requires using information from two tables. Table 1 identifies Mixture O as containing 35% vermicompost and 65% potting soil. Once this mixture is identified, Table 2 shows that tomato plants grown in Mixture O had an average height of 50 cm, making Choice G correct. Choice F (48 cm) corresponds to Mixture P, which contained 50% vermicompost. Choice H (53 cm) corresponds to Mixture N, which contained 25% vermicompost. Choice J (55 cm) corresponds to Mixture M, which contained 20% vermicompost. The correct approach is to identify the mixture using Table 1 before locating its average height in Table 2.

Question 7, “The average number of leaves for the tomato plants changed the most between which of the following mixture pairings listed in Table 2?” The correct answer is B: “Mixture M and Mixture N.”

  • This question requires calculating the absolute difference in average leaf count for each listed pair of mixtures. Table 2 shows that Mixture L had 105 leaves and Mixture M had 108, a difference of 3. Mixture M had 108 leaves and Mixture N had 102, a difference of 6. Mixture N had 102 leaves and Mixture O had 100, a difference of 2. Finally, Mixture O had 100 leaves and Mixture P had 97, a difference of 3. The greatest change is 6 leaves between Mixtures M and N, making Choice B correct. Choices A, C, and D represent smaller differences.

Question 8, “Based on Tables 1 and 2, as the percent by volume of potting soil in a mixture was increased from 50% through 100%, which of average plant height and average number of leaves, if either, always increased?” The correct answer is J: “Neither average plant height nor average number of leaves.”

  • This question requires tracking two variables as potting soil concentration increases. Beginning with Mixture P at 50% potting soil, both average plant height and average leaf count increase through Mixture M at 80% potting soil. However, when the potting soil concentration increases to 85% in Mixture L, average height decreases from 55 cm to 51 cm, and average leaf count decreases from 108 to 105. Both decrease again at 100% potting soil in Mixture K. Therefore, neither variable always increases, making Choice J correct. Choices F and G incorrectly identify one variable as consistently increasing, while Choice H incorrectly claims that both increase throughout the entire range.

Question 9, “Based on Table 2, for the tomato plants grown in Mixture M, what was the average yield in kilograms?” The correct answer is A: “0.114 kg.”

  • This question tests unit conversion. According to Table 2, tomato plants grown in Mixture M produced an average yield of 114 grams. Since 1 kilogram equals 1,000 grams, converting grams to kilograms requires dividing by 1,000. Therefore, 114 ÷ 1,000 = 0.114 kg, making Choice A correct. Choice B (1.14 kg) incorrectly moves the decimal point only one place too far relative to the correct value. Choices C (11,400 kg) and D (114,000 kg) are much too large and result from incorrectly increasing the numerical value rather than decreasing it when converting grams to kilograms.

Question 10, “Assume that for each mixture tested in the study, 5 one-day-old tomato plants were grown in the mixture. The following table shows the plant height at 85 days for each individual tomato plant from one of the mixtures tested. Based on this table and the data listed in Table 2, what was the identity of this mixture?” The correct answer is J: “Mixture P.”

  • This question requires calculating the average height of the five plants and matching it to the corresponding mixture in Table 2. The individual plant heights are 46, 44, 51, 47, and 52 cm. Adding these values gives 240 cm, and dividing by five plants produces an average height of 48 cm. Table 2 identifies Mixture P as having an average plant height of 48 cm, making Choice J correct. Choice F (Mixture K) has an average height of 44 cm, Choice G (Mixture L) has an average of 51 cm, and Choice H (Mixture O) has an average of 50 cm. None matches the calculated average.

Passage III – “Pollination of Heliconia metallica“

Question 11, “The data that were averaged to produce the results shown in Figure 2 were most likely collected using which of the following pieces of equipment?” The correct answer is D: “Balance.”

  • This question tests knowledge of scientific measurement tools. Figure 2 displays the average mass per seed, measured in milligrams. A balance is used to measure mass, making Choice D correct. Choice A (thermometer) measures temperature, which is not the variable being reported in Figure 2. Choice B (pH meter) measures the acidity or basicity of a solution, while Choice C (telescope) is used to observe distant objects. Neither is appropriate for measuring seed mass. Recognizing the units on the graph’s vertical axis is the key to identifying the correct equipment.

Question 12, “One of the questions about H. metallica plants that the study was designed to answer was which of the following?” The correct answer is G: “Are the percent of flowers that produce fruit and the average mass per seed different when flowers are self-pollinated than when flowers are cross-pollinated?”

  • This question tests understanding of an experiment’s purpose. The researchers applied different pollination treatments, including self-pollination and cross-pollination, and measured two outcomes: the percentage of flowers producing fruit and the average mass per seed. Choice G accurately describes the relationship being investigated. Choice F concerns the plants’ location within the rainforest understory, which was not manipulated. Choice H asks about fruit ripening time, which was not measured. Choice J asks whether hummingbirds normally pollinate the flowers, a fact already established in the passage rather than a question tested by the study.

Question 13, “The pollination treatments received by Groups 1 and 2 differed in which of the following ways? The pollen received by each Group 1 flower was:” The correct answer is A: “from the same plant as the flower, whereas the pollen received by each Group 2 flower was from a different plant than the flower.”

  • This question requires interpreting the experimental treatments in Table 1. Group 1 underwent self-pollination, meaning each flower received pollen from the same plant. Group 2 underwent cross-pollination using pollen collected from a separate donor plant. Choice A correctly identifies this distinction. Choice B reverses the two treatments. Choices C and D incorrectly distinguish the groups based on whether their pollen came from one or six plants. The six-donor treatment applies to Group 3, not Group 2. The important difference between Groups 1 and 2 is whether the pollen originated from the same plant or a different plant.

Question 14, “For any group, the value shown in Figure 1 was most likely calculated using which of the following expressions?” The correct answer is H: “(number of flowers producing fruit ÷ total number of flowers) × 100.”

  • This question tests the calculation of percentages. Figure 1 reports the percentage of flowers that produced fruit, so the correct calculation is the number of flowers producing fruit divided by the total number of flowers, multiplied by 100. Choice H correctly represents this calculation. Choices F and G use seed count and seed mass, which are unrelated to the percentage of flowers producing fruit. Choice J reverses the numerator and denominator, dividing the total number of flowers by the number producing fruit. That would not produce the desired percentage and could result in values greater than 100%. Choice H correctly calculates the proportion of flowers that successfully produced fruit.

Question 15, “Do the results of the study indicate that the nylon bags successfully prevented the normal pollinators from pollinating the H. metallica flowers?” The correct answer is B: “Yes; none of the flowers receiving the no pollination treatment produced fruit.”

  • This question requires evaluating whether the experimental setup successfully prevented unintended pollination. Group 4 received no pollination treatment, and Figure 1 shows that none of its flowers produced fruit. This supports the conclusion that the nylon bags prevented normal pollinators from reaching the flowers, making Choice B correct. Choice A also answers “Yes” but incorrectly uses the 20% fruit-production rate in the self-pollination group as evidence. Choice C makes the same mistake while concluding “No.” Choice D correctly identifies the absence of fruit in Group 4 but draws the opposite conclusion. The results of the untreated group provide the most direct evidence that the bags worked.

Question 16, “What was the total mass of the seeds produced by the Group 3 flowers?” The correct answer is J: “Cannot be determined from the given information.”

  • This question tests whether the information provided is sufficient to calculate a requested quantity. Figure 2 gives the average mass per seed for Group 3, while Figure 1 provides the percentage of flowers that produced fruit. However, neither figure identifies the total number of seeds produced. Since calculating total seed mass requires multiplying average seed mass by the number of seeds, the total cannot be determined. Choice J is therefore correct. Choice F (0 mg) is incorrect because Group 3 produced seeds. Choices G (45 mg) and H (85 mg) confuse individual values reported in the figures with the total mass of all seeds.

Passage IV – “Arsenic and Antimony in Groundwater”

Question 17, “Based on the procedures of the study, at which of these 2 times—before sample collection or after sample collection but before filtering—if either, were the collection containers cleaned?” The correct answer is A: “Before sample collection only.”

  • This question tests the ability to identify the sequence of experimental procedures. The passage explicitly states that all collection containers were rigorously cleaned before sample collection. It then explains that groundwater samples were collected and immediately passed through filters into sample bottles. No additional cleaning step is described between collection and filtering. Choice A correctly identifies when cleaning occurred. Choice B incorrectly places cleaning after sample collection. Choice C suggests that the containers were cleaned at both stages, while Choice D incorrectly claims that no cleaning occurred. The correct answer follows directly from the procedure described in the passage.

Question 18, “Based on Figures 2 and 4, the difference between the greatest Sb content from a well in Aquifer 1 and the greatest Sb content from a well in Aquifer 2 was approximately:” The correct answer is G: “0.95 nmol/kg.”

  • This question requires identifying the maximum antimony (Sb) content in two graphs and subtracting the values. In Figure 2, the highest Sb content for Aquifer 1 is approximately 0.50 nmol/kg. In Figure 4, the highest Sb content for Aquifer 2 is approximately 1.45 nmol/kg. Subtracting gives 1.45 − 0.50 = 0.95 nmol/kg, making Choice G correct. Choice F (0.50 nmol/kg) reports the maximum from Aquifer 1 rather than the difference. Choice H (1.45 nmol/kg) reports the maximum from Aquifer 2. Choice J (1.95 nmol/kg) results from adding the two maximum values instead of subtracting them.

Question 19, “Consider the As content from wells located at distances 55 km to 75 km from the recharge zone of Aquifer 1. What is the order of these wells from greatest to least As content?” The correct answer is B: “55 km, 64 km, 75 km, 60 km.”

  • This question requires comparing the arsenic (As) contents of four wells using Figure 1. The well at 55 km has the greatest As content of the four, at approximately 380 nmol/kg. The well at 64 km follows with approximately 210 nmol/kg, while the well at 75 km has approximately 70 nmol/kg. The well at 60 km has the lowest content, approximately 30 nmol/kg. Choice B correctly orders these locations from greatest to least. Choice A incorrectly places the 60 km well ahead of wells with higher As contents. Choices C and D also fail to follow the descending order shown by the graph.

Question 20, “The use of which of the following measurement tools would have most likely resulted in the most accurate measurement of the volume of the portion removed from each sample?” The correct answer is H: “100 mL graduated cylinder.”

  • This question tests knowledge of laboratory equipment. The passage states that a 60 mL portion was removed from each groundwater sample for analysis. A 100 mL graduated cylinder is designed to measure liquid volumes accurately and has sufficient capacity to measure 60 mL in one step, making Choice H correct. Choice F (10 mL test tube) and Choice G (25 mL flask) cannot hold the entire 60 mL portion at once and are not the best instruments for this measurement. Choice J (250 mL beaker) can hold the necessary volume but is generally less precise than a graduated cylinder. Choice H provides the most appropriate combination of capacity and measurement accuracy.

Question 21, “Based on the results of the study, for Aquifer 2, the As content and the Sb content each showed an increase over which of the following ranges of distance from the recharge zone?” The correct answer is C: “From 79 km to 91 km.”

  • This question requires comparing trends in Figures 3 and 4, which show As and Sb contents in Aquifer 2. Between 79 km and 91 km, both graphs show increases: As content rises from approximately 0.5 to 1.0 nmol/kg, while Sb content also increases slightly. Choice C is therefore correct. Choice A (32 km to 46 km) is incorrect because both contents decrease. Choice B (46 km to 79 km) also shows a decrease in As content rather than an increase. Choice D (91 km to 100 km) is incorrect because Sb content decreases over that interval, even though As content increases.

Question 22, “Based on the procedures of the study, what was the total volume of water collected from Aquifer 2?” The correct answer is H: “3,000 mL.”

  • This question requires multiplying the number of samples by the volume collected per sample. The passage states that scientists sampled six wells in Aquifer 2, collecting 500 mL of groundwater from each well. Therefore, the total volume collected was 6 × 500 = 3,000 mL, making Choice H correct. Choice F (360 mL) results from multiplying six wells by the 60 mL portion used for analysis rather than the original sample volume. Choice G (500 mL) represents the volume collected from a single well. Choice J (5,500 mL) corresponds to eleven 500 mL samples, the number of wells sampled in Aquifer 1 rather than Aquifer 2.

Passage V – “Nitrogen Gas Experiments”

Question 23, “Which of the following statements describes how the syringe was most likely manipulated in order to obtain the pressure measurement of 68 kPa in Experiment 1?” The correct answer is B: “The syringe plunger was moved to set the volume at 15 mL.”

  • This question requires matching a pressure measurement with the corresponding experimental condition. Table 1 shows that when the nitrogen gas occupied a volume of 15 mL, its pressure was 68 kPa. Choice B correctly describes adjusting the syringe plunger to this volume. Choice A incorrectly selects 5 mL, which corresponds to a much higher pressure of 204 kPa. Choices C and D describe placing the syringe in water baths at different temperatures, but Experiment 1 was conducted at a constant temperature of 298 K. The variable adjusted during this experiment was gas volume, not temperature.

Question 24, “The statement ‘For a gas sample at constant pressure, volume increases as temperature increases’ is supported by the results of which of the experiments?” The correct answer is G: “Experiment 3 only.”

  • This question tests the ability to distinguish the relationships investigated in three experiments. Experiment 3 maintained a constant pressure of 102 kPa while varying temperature. Table 3 shows that as temperature increased from 279 K to 333 K, gas volume increased from 201 mL to 240 mL, supporting the statement. Experiment 1 held temperature constant while changing volume and measuring pressure, so it cannot establish the effect of temperature on volume. Experiment 2 used a rigid flask, keeping volume constant while examining the relationship between temperature and pressure. Therefore, only Experiment 3 provides the relevant evidence, making Choice G correct. Choices F, H, and J incorrectly include experiments that did not test the specified relationship.

Question 25, “Suppose that in Experiment 1, after the pressure had been measured at a volume of 20 mL, the temperature of the syringe was then changed to 320 K while holding the volume constant. Based on the results of Experiment 2, would the pressure measured at 320 K more likely have been 48 kPa or 55 kPa?” The correct answer is C: “55 kPa; the increase in temperature would increase the pressure.”

  • This question requires applying the temperature-pressure relationship demonstrated in Experiment 2. Table 2 shows that as temperature increases, pressure also increases when volume is held constant. In Experiment 1, the nitrogen gas had a pressure of 51 kPa at a volume of 20 mL and a temperature of 298 K. Increasing the temperature to 320 K while maintaining the same volume would therefore increase the pressure above 51 kPa. Of the two proposed values, 55 kPa is the reasonable prediction, making Choice C correct. Choices A and B incorrectly predict a decrease in pressure, while Choice D predicts the correct pressure but incorrectly describes the temperature as decreasing.

Question 26, “In Experiment 2, how many trials, if any, were done at a temperature greater than 0° Celsius?” The correct answer is J: “4.”

  • This question tests temperature conversion and comparison. Table 2 lists experimental temperatures of 279 K, 298 K, 316 K, and 333 K. Since 0°C is equivalent to approximately 273 K, every temperature listed is greater than 0°C. Therefore, all four trials were conducted above the freezing point of water, making Choice J correct. Choice F (0) incorrectly treats all temperatures as below 0°C, while Choices G (2) and H (3) fail to count all four temperatures. A full conversion of each value is unnecessary because each is clearly greater than 273 K.

Question 27, “Were the experiments performed with the gas that is most abundant in air?” The correct answer is A: “Yes; nitrogen was tested in the experiments and makes up the greatest percentage of air.”

  • This question tests basic scientific knowledge about atmospheric composition. The passage states that all three experiments used nitrogen gas (N₂). Nitrogen accounts for approximately 78% of Earth’s atmosphere, making it the most abundant gas in air. Choice A correctly identifies both the gas used and its abundance. Choice B incorrectly claims that water vapor was tested and that it is the most abundant atmospheric gas. Choice C correctly identifies nitrogen as the experimental gas but incorrectly states that oxygen is more abundant. Choice D incorrectly identifies the experimental gas as water vapor and also incorrectly claims that oxygen is the most abundant atmospheric gas.

Question 28, “Based on the procedures and results of Experiment 3, at a pressure of 102 kPa, the boiling point of N₂ must be:” The correct answer is F: “less than 279 K.”

  • This question requires interpreting the physical state of nitrogen during Experiment 3. The experiment used nitrogen gas at a constant pressure of 102 kPa, with temperatures ranging from 279 K to 333 K. Because the nitrogen remained a gas even at the lowest tested temperature of 279 K, its boiling point at this pressure must be below that temperature. Choice F is therefore correct. Choice G places the boiling point between 279 K and 298 K, while Choice H places it between 298 K and 333 K; both would imply that nitrogen was not gaseous at all tested temperatures. Choice J incorrectly places the boiling point above 333 K, contradicting the use of nitrogen gas throughout the experiment.

Passage VI – “Voltmeter Resistance and Circuit Voltage”

Question 29, “Based on the results of Study 2, for which trial was the difference between the average voltage and the calculated voltage greatest, and what was that difference?” The correct answer is D: “Trial 10; 0.214 V.”

  • This question requires comparing the absolute differences between the average and calculated voltages listed in Table 2. For Trial 1, the difference is 4.500 − 4.498 = 0.002 V. For Trial 4, it is 8.181 − 8.174 = 0.007 V. For Trial 8, it is 4.500 − 4.495 = 0.005 V. Finally, Trial 10 has a difference of 4.500 − 4.286 = 0.214 V, the largest of the four. Choice D is therefore correct. Choices A and B incorrectly identify Trial 8 as having the greatest difference, while Choice C identifies Trial 10 but provides an incorrect numerical difference.

Question 30, “Consider the circuit that had a voltage of 1.498 V across R₂. Based on the results of Study 1, what was the resistance of R₂ in this circuit?” The correct answer is F: “1 kΩ.”

  • This question requires locating a particular trial in Table 1 and identifying its corresponding resistance value. The voltage of 1.498 V appears in Trial 6. Reading across that row, the resistance of R₂ is 1 kΩ, making Choice F correct. Choice G (2 kΩ) corresponds to the R₂ resistance in Trial 2, while Choice H (5 kΩ) corresponds to Trial 3. Choice J (1,000 kΩ) is the internal resistance of the voltmeter, Rᵥ, in Trial 6 rather than the resistance of R₂. The key is to identify both the correct row and the correct resistance column.

Question 31, “Based on the results of Study 1, in which of the following trials, if any, did the student use 2 resistors having the same resistance?” The correct answer is B: “Trials 1, 8, 9, and 10 only.”

  • This question requires comparing the three resistance values listed in each row of Table 1. In Trials 1, 8, 9, and 10, both R₁ and R₂ have resistances of 1 kΩ. In the remaining trials, all three listed resistance values are different from one another. Choice B therefore correctly identifies the trials in which two resistors shared the same resistance. Choice A incorrectly includes Trials 2, 3, and 4, where R₁ and R₂ have different values. Choice C omits Trial 1, which also qualifies. Choice D incorrectly claims that no trials contained two resistors with equal resistance.

Question 32, “Based on the results of Study 1, in how many trials was the internal resistance of the voltmeter less than the resistance of R₂?” The correct answer is F: “0.”

  • This question requires comparing the resistance of the voltmeter, Rᵥ, with the resistance of R₂ across all ten trials. Table 1 shows that the smallest internal resistance of the voltmeter was 10 kΩ, occurring in Trial 10. The largest resistance of R₂ was also 10 kΩ, occurring in Trial 4. However, these values occurred in different trials, and the voltmeter’s internal resistance was greater than R₂ in every individual trial. Therefore, there were zero trials in which Rᵥ was less than R₂, making Choice F correct. Choices G (3), H (4), and J (10) incorrectly count trials that do not satisfy the specified condition.

Question 33, “Suppose another circuit had been tested in Study 1 and the voltage across R₂ had been 4.354 V. If the resistances of R₁ and R₂ were each 1 kΩ, the resistance of Rᵥ in this circuit would most likely have been:” The correct answer is B: “between 10 kΩ and 100 kΩ.”

  • This question requires identifying a trend in Table 1 and estimating a resistance from a voltage measurement. Trials 8, 9, and 10 all use R₁ and R₂ values of 1 kΩ while changing the internal resistance of the voltmeter. At Rᵥ = 10 kΩ, the measured voltage is 4.282 V, while at Rᵥ = 100 kΩ, the voltage is 4.477 V. Since 4.354 V falls between these measurements, the corresponding internal resistance should fall between 10 kΩ and 100 kΩ, making Choice B correct. Choice A would generally produce a voltage below 4.282 V, while Choice C would produce a voltage above 4.477 V. Choice D is incorrect because an internal resistance of 1,000 kΩ produces a voltage near 4.495 V.

Question 34, “The power dissipated by a resistor is equal to the square of the voltage across the resistor divided by the resistance. Based on Table 1, in which trial was the power dissipated by R₂ the greatest?” The correct answer is J: “Trial 8.”

  • This question requires applying the given power formula, P = V²/R, to the listed trials. Since the resistances are given in kilohms, the calculated powers can be expressed in milliwatts. Trial 2 gives approximately 5.996²/2 = 17.98 mW; Trial 3 gives 7.493²/5 = 11.23 mW; Trial 4 gives 8.173²/10 = 6.68 mW; and Trial 8 gives 4.496²/1 = 20.21 mW. Trial 8 has the greatest power dissipation, making Choice J correct. Choices F, G, and H correspond to the lower power values calculated for Trials 2, 3, and 4. The highest voltage does not necessarily produce the greatest power because resistance also affects the result.

Passage VII – “Distillation of Ethanol and Water”

Question 35, “Based on Figure 1, the most likely purpose of the thermometer was to measure the temperature of the:” The correct answer is D: “vapor in the DH.”

  • This question requires interpreting the placement of laboratory equipment in Figure 1. The thermometer is positioned in the distilling head (DH), where vapor passes from the distilling column toward the condenser. Its purpose is therefore to measure the temperature of the vapor at this location, making Choice D correct. Choice A refers to the liquid in the collection flask, which is located farther along the apparatus. Choice B refers to the liquid in the distilling flask, which is below the thermometer. Choice C refers to vapor in the condenser, but the thermometer is located in the distilling head rather than inside the condenser.

Question 36, “Consider the opinion given by Student 4 concerning the vapor that reached the condenser first. Suppose that enough of that vapor had condensed to form 200.0 mL of liquid. The volume of ethanol, if any, that the liquid contained would have been closest to which of the following?” The correct answer is H: “192 mL.”

  • This question requires applying the composition given in Student 4’s explanation. Student 4 states that the first vapor reaching the condenser would contain 96% ethanol and 4% water by volume. If 200.0 mL of this vapor condensed into liquid, the ethanol portion would be 0.96 × 200.0 = 192 mL, making Choice H correct. Choice F (0 mL) incorrectly assumes that the condensed liquid contains no ethanol. Choice G (96 mL) confuses the 96% concentration with a volume based on 100 mL rather than 200 mL. Choice J (200 mL) incorrectly assumes that the liquid is pure ethanol, which contradicts Student 4’s description of the mixture.

Question 37, “Which of the students indicated that the composition of the vapor formed when the ethanol/water mixture is boiled is different than the composition of the liquid mixture in the DF?” The correct answer is C: “Students 1, 3, and 4 only.”

  • This question requires comparing the students’ predictions about vapor composition. Student 1 claims that pure ethanol vapor forms initially, which differs from the original ethanol-water mixture. Student 3 argues that the vapor contains a greater proportion of ethanol than the liquid, while Student 4 predicts a vapor composition of 96% ethanol and 4% water. All three therefore describe vapor compositions that differ from the original liquid mixture. Student 2, however, explicitly states that the vapor has the same composition as the liquid. Choice C correctly includes Students 1, 3, and 4 while excluding Student 2. Choices A, B, and D incorrectly omit one or more qualifying students or include Student 2.

Question 38, “Which of the following statements describes a difference between the opinion given by Student 2 and the opinion given by Student 3?” The correct answer is J: “a 1:1 mixture by volume of ethanol and H₂O will be the only thing to collect in the CF, whereas Student 3 stated that pure ethanol will collect in the CF first, followed by a mixture of ethanol and H₂O.”

  • This question tests the ability to compare two scientific explanations. Student 2 argues that ethanol and water cannot be separated by distillation and predicts that the collection flask will contain the same 1:1 mixture throughout the process. Student 3, in contrast, predicts that pure ethanol will collect first, followed by liquid containing an increasing proportion of water. Choice J accurately describes this disagreement. Choices F and G incorrectly identify the temperatures predicted by the students; Student 2 predicts 89.2°C, while Student 3 predicts that the first vapor reaching the condenser will be at 78.4°C. Choice H reverses the students’ predictions about the composition of the collected liquid.

Question 39, “Atmospheric pressure is 1 atm at an altitude of 0 m. Suppose that the distillation described in the passage were carried out at an altitude of 1,000 m. For a distillation at that altitude, which of the following statements would be most consistent with the opinion given by Student 1? The mixture will initially boil:” The correct answer is B: “at a temperature below the BP given for ethanol at 1 atm.”

  • This question requires applying the relationship between atmospheric pressure and boiling point to Student 1’s explanation. Atmospheric pressure decreases as altitude increases, and liquids boil at lower temperatures when external pressure is reduced. Student 1 predicts that the original mixture initially boils at ethanol’s boiling point of 78.4°C under a pressure of 1 atm. At an altitude of 1,000 m, the reduced atmospheric pressure would lower this boiling point, making Choice B correct. Choice A incorrectly assumes that the boiling point remains unchanged. Choice C uses the boiling point of water rather than ethanol, while Choice D predicts a temperature above water’s boiling point at 1 atm, contradicting the effect of reduced atmospheric pressure.

Question 40, “Benzene, which has a BP of 80.2°C at 1 atm of pressure, forms an azeotrope with H₂O. Based on information provided in the passage, the BP of this azeotrope is most likely:” The correct answer is F: “less than 80.2°C.”

  • This question tests understanding of the definition of an azeotrope provided in Student 4’s explanation. The passage defines an azeotrope as a mixture of two liquids that boils at a lower temperature than either pure liquid. Benzene boils at 80.2°C, while water boils at 100.0°C at 1 atm. Therefore, an azeotrope formed from these two substances would have a boiling point lower than 80.2°C, making Choice F correct. Choices G and H place the boiling point between the boiling points of the two pure liquids, which contradicts the given definition. Choice J places the boiling point above 100°C, which is also inconsistent with the behavior described for an azeotrope.

Additional ACT practice test answer explanations