Free Standardized Achievement Science Test Question and Answer — Questions and Answers

Question 1: Which of the two scientific explanations for how craters are eliminated from Europa's surface is most accurate?

Scientist 1: Filled in by water Scientist 2: Sublimation (Correct answer)
Scientist 1:Sublimation Scientist 2: Filled in by water
Scientist 1: Worn smooth by wind Scientist 2: Sublimation
Scientist 1: Worn smooth by wind Scientist 2: Filled in by water

Correct answer: Scientist 1: Filled in by water Scientist 2: Sublimation

This question refers to a scientific passage (not provided) detailing two scientists' explanations for crater elimination on Europa. Based on the correct answer, Scientist 1 posits that craters are filled in by water, suggesting a subsurface ocean. Scientist 2 attributes crater removal to sublimation, where ice directly turns into gas, indicating a process of atmospheric or surface erosion.

Question 2: Which of the following descriptions of Europa would be agreed upon by both scientists in light of the evidence provided?

Which of the following descriptions of Europa would be agreed upon by both scientists in light of the evidence provided?
Which of the following descriptions of Europa would be agreed upon by both scientists in light of the evidence provided?
Europa is completely covered by a layer of ice. (Correct answer)
Europa has a surface temperature of 20°C.

Correct answer: Europa is completely covered by a layer of ice.

This question refers to a scientific passage (not provided) about Europa. The correct answer indicates that both scientists, despite their differing theories on crater removal, agree on the fundamental observation that Europa is completely covered by a layer of ice. This shared understanding forms the basis for their respective hypotheses about the moon's surface processes.

Question 3: Which of the following descriptions of Europa would be agreed upon by both scientists in light of the evidence provided?

has remained unchanged for millions of years.
is covered with millions of meteorite craters.
is being shaped by the movement of ice. (Correct answer)
has remained unchanged for millions of years.

Correct answer: is being shaped by the movement of ice.

This question refers to a scientific passage (not provided) about Europa. The correct answer indicates that both scientists agree Europa's surface is being actively shaped by the movement of ice. This implies that the icy crust is not static but rather dynamic, undergoing continuous modification through processes like water filling or sublimation, which alter its features over time.

Question 4: Which of the following descriptions of the meteorite craters on Europa best reflects the opinions of both scientists?

Meteorites frequently strike Europa's surface but do not leave any craters.
Meteorite craters, once formed on Europa, remain unchanged for billions of years.
Meteorite craters, once formed, are then smoothed or removed by Europa's surface processes. (Correct answer)
No meteorites have struck Europa for millions of years.

Correct answer: Meteorite craters, once formed, are then smoothed or removed by Europa's surface processes.

This question refers to a scientific passage (not provided) about Europa. Both scientists, despite proposing different mechanisms for crater removal (water filling or sublimation), concur that meteorite craters on Europa do not remain permanently preserved. Instead, they are actively smoothed, eroded, or entirely removed by ongoing surface processes on the icy moon, indicating a geologically active environment.

Question 5: Ice sublimates to water vapor and enters Europa's atmosphere, according to Scientist 2. Which of the following gases would one most likely anticipate to find in Europa's atmosphere as a result of this process, if ultraviolet light then split those water vapor molecules apart?

Chlorine
Nitrogen
Oxygen (Correct answer)
Methane

Correct answer: Oxygen

Water vapor (H₂O) is composed of hydrogen and oxygen atoms. When ultraviolet light splits these molecules, a process called photodissociation occurs, breaking the chemical bonds. This releases free hydrogen and oxygen atoms, which can then recombine to form H₂ and O₂. Therefore, oxygen (O₂) would be a primary gas anticipated in Europa's atmosphere from this process.

Question 6: Which of the following materials must be present on Europa in order for a magnetic field to be formed there, according to the information in Scientist 1's point of view?

Dissolved salts (Correct answer)
Frozen nitrogen
Molten magma
Water ice

Correct answer: Dissolved salts

A magnetic field in a celestial body is typically generated by a dynamo mechanism, which requires the movement of an electrically conductive fluid. Scientist 1's viewpoint likely suggests that Europa's subsurface ocean, if it contains dissolved salts, would be electrically conductive. The convection of this salty water, driven by tidal forces, could then create the observed magnetic field.

Question 7: Assume that Scientist 2's assertion that the surface features of Europa are comparable to Earth's flowing glaciers is true. Which of the following characteristics would Earth's glaciers be least likely to display, given this presumption and the evidence presented?

Cracks
Meteorite craters (Correct answer)
Pressure ridges
Dark bands

Correct answer: Meteorite craters

Earth's glaciers are dynamic systems, constantly flowing, melting, and reshaping the landscape. This continuous movement and the processes of weathering and erosion would quickly erase any meteorite craters that might form on their surface. In contrast, cracks, pressure ridges, and dark bands are common features of glaciers, resulting from their internal stresses, movement, and accumulation of debris.

Question 8: Which of the following statements most accurately sums up how the methods employed in Experiments 1 and 2 differ? The in Experiment 1:

diameter of the tube was varied; in Experiment 2, the temperature was varied.
temperature was varied; in Experiment 2, the diameter of the tube was varied. (Correct answer)
temperature was varied; in Experiment 2, the distance between the swabs was varied.
distance between the swabs was varied; in Experiment 2, the temperature was varied.

Correct answer: temperature was varied; in Experiment 2, the diameter of the tube was varied.

To understand the difference in methods, one must identify the independent variable manipulated in each experiment. In Experiment 1, the temperature was systematically changed to observe its effect on gas diffusion, while other factors were kept constant. Conversely, in Experiment 2, the diameter of the tube was varied, likely to investigate its impact on diffusion rates, with temperature and other conditions held steady.

Question 9: Which of the aforementioned trial sets from Experiments 1, 2, and 3 was carried out under the same sets of circumstances?

Trials 4, 7, and 9
Trials 2, 3, and 4
Trials 1, 5, and 9 (Correct answer)
Trials 10, 11, and 12

Correct answer: Trials 1, 5, and 9

To find trials conducted under the same circumstances, one must compare all the controlled variables across the experiments. This includes factors like temperature, tube diameter, and the distance between the swabs. By examining the experimental parameters for Trials 1, 5, and 9, it would be evident that all these specific conditions were identical, making them comparable sets.

Question 10: The HCl swab would have been the furthest from the ring in Experiment 3 if an experiment had been conducted with the swabs 25 cm apart.

14 cm
12 cm
8 cm
10 cm (Correct answer)

Correct answer: 10 cm

This question applies Graham's Law of Diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Ammonia (NH₃, molar mass ~17 g/mol) is significantly lighter than hydrogen chloride (HCl, molar mass ~36.5 g/mol). Therefore, NH₃ diffuses faster, meaning the reaction ring will form closer to the HCl swab. With a total distance of 25 cm, the ring would form approximately 10 cm from the HCl swab and 15 cm from the NH₃ swab, as NH₃ diffuses roughly 1.46 times faster than HCl.

Question 11: Which of the following should a different student perform if he wished to test a factor that was not examined in Experiments 1-3? He ought to investigate the following factors' effects on gas diffusion rates:

tube length.
atmospheric pressure. (Correct answer)
tube diameter.
temperature.

Correct answer: atmospheric pressure.

Experiments 1-3 likely explored common factors influencing gas diffusion, such as temperature, tube diameter, and potentially the types of gases or their initial separation. Atmospheric pressure, however, directly affects the mean free path of gas molecules and thus their diffusion rate, but it is a distinct environmental factor not typically varied in the other parameters mentioned. Therefore, investigating atmospheric pressure would introduce a new, unexamined variable to the study.

Question 12: According to Study 1's general findings, peony seeds are most likely to germinate when kept at which of the following temperatures when they are placed in a petri dish with damp paper?

18° C (Correct answer)
23°C
28°C
13° C

Correct answer: 18° C

To determine the most likely optimal germination temperature, one would refer to the data presented in Study 1. The study would show that among the tested temperatures, 18°C resulted in the highest percentage or number of germinated peony seeds when placed in a petri dish with damp paper. This indicates that 18°C provided the most favorable conditions for germination under the specified circumstances.

Question 13: Imagine that Study 2 had also contained a second batch of 25 peony seeds with storage temperatures of 25°C and germination temperatures of 18°C. Considering the data available, it is most likely that the following range of seeds would have borne fruit after being kept for 30 days:

16
24
8
0 (Correct answer)

Correct answer: 0

This question requires extrapolating from the data in Study 2 regarding storage temperature. If Study 2 indicated that peony seeds stored at 25°C consistently failed to germinate, regardless of the subsequent germination temperature, then a new batch stored at 25°C would also be expected to yield zero germinated seeds. This suggests 25°C is an unsuitable storage temperature for these seeds.

Question 14: In Study 2, the number of seeds that germinated increased when the germination temperature rose from 13°C to 28°C at the storage temperature of 5°C.

increased, then decreased. (Correct answer)
decreased, then increased.
decreased only.
increased only.

Correct answer: increased, then decreased.

To answer this, one must analyze the data for seeds stored at 5°C in Study 2 and observe the trend as the germination temperature increases from 13°C to 28°C. The data would show that the number of germinated seeds initially rises as the temperature increases from 13°C, reaching an optimal point, and then subsequently decreases as the temperature continues to rise towards 28°C. This pattern indicates an optimal temperature range, beyond which higher temperatures become detrimental.

Question 15: Which of the following collections of seeds had the same treatment before being put into the petri dishes?

The seeds from Study 1 that were stored for 8 weeks and the seeds from Study 2 that were stored at 15°C
The seeds from Study 1 that were stored for 10 weeks and the seeds from Study 2 that were stored at 15°C
The seeds from Study 1 that were stored for 8 weeks and the seeds from Study 2 that were stored at 5°C
The seeds from Study 1 that were stored for 10 weeks and the seeds from Study 2 that were stored at 5°C (Correct answer)

Correct answer: The seeds from Study 1 that were stored for 10 weeks and the seeds from Study 2 that were stored at 5°C

To identify seeds with the same pre-treatment, one must compare the storage conditions (temperature and duration) across both studies. The correct answer indicates that the seeds from Study 1 stored for 10 weeks had the same storage conditions as the seeds from Study 2 stored at 5°C. This implies that the 10-week storage period in Study 1 was also conducted at a temperature of 5°C, making their pre-germination treatment identical.

Question 16: Study 2's experimental methodology changed from Study 1's in that Study 2:

germination time was varied.
germination temperature was varied.
storage time was held constant. (Correct answer)
storage temperature was held constant.

Correct answer: storage time was held constant.

To identify the methodological difference, one must compare the variables that were manipulated and those that were controlled in each study. Study 1 likely varied the storage time to investigate its impact on germination. In contrast, Study 2's methodology changed by holding the storage time constant (e.g., at a fixed duration like 30 days) to focus on the effects of other variables, such as storage temperature or germination temperature.

Free Standardized Achievement Science Test Question and Answer

Which of the two scientific explanations for how craters are eliminated from Europa's surface is most accurate?