Experimental Design and Scientific Method Practice Questions — ACT — Page 2

Question 21Question →

A student investigates how the surface roughness of a ramp affects the distance a wooden block travels after sliding down. The student proposes the following hypothesis: 'A block will travel a shorter distance on smoother surfaces because smooth surfaces offer less frictional resistance.'

The student collects the following data:

Surface	Relative Roughness	Sliding Distance (cm)
Sandpaper	High	12
Unfinished wood	Medium	28
Polished plastic	Low	65

Based on these results, which of the following represents the most accurate modification to the student's hypothesis?

The block will travel a longer distance on smoother surfaces because smooth surfaces offer less frictional resistance.

The block will travel a shorter distance on smoother surfaces because smooth surfaces offer more frictional resistance.

The block will travel a longer distance on rougher surfaces because rougher surfaces offer less frictional resistance.

The surface roughness of the ramp does not affect the sliding distance of the block.

Show answer & explanation

Answer: The block will travel a longer distance on smoother surfaces because smooth surfaces offer less frictional resistance.

Answer

The block will travel a longer distance on smoother surfaces because smooth surfaces offer less frictional resistance.

The correct answer is that the block will travel a longer distance on smoother surfaces because smooth surfaces offer less frictional resistance. According to the data, the polished plastic surface (low roughness) resulted in a sliding distance of 65 cm, which is much greater than the 12 cm distance on sandpaper (high roughness). This confirms that decreasing roughness increases the distance traveled, so the hypothesis must be modified to reflect this positive relationship between smoothness and distance.

Step-by-Step Solution

1

Analyze the student's initial hypothesis.

The student hypothesized that a block travels a shorter distance on smoother surfaces due to less friction.

To evaluate the hypothesis, we must first identify its specific claim about the relationship between surface smoothness and distance.

2

Examine the experimental data in the table.

As surface roughness decreases from sandpaper (high) to polished plastic (low), the sliding distance increases from 12 cm to 65 cm.

Comparing the relative roughness to the measured sliding distance shows how the dependent variable responds to changes in the independent variable.

3

Compare the data trend with the initial hypothesis.

The data contradicts the initial hypothesis because the block traveled further on the smoother surface, not shorter.

Determining whether the data supports or refutes the hypothesis is necessary to formulate the correct modification.

4

Modify the hypothesis to align with the data.

The modified hypothesis must state that the block travels a longer distance on smoother surfaces because less friction allows for more motion.

A valid scientific hypothesis must accurately reflect the empirical trends observed in the experimental results.

Key Concept

Formulating and Modifying Hypotheses

Question 22Question →

An investigator conducted a study to evaluate how different types of dissolved organic matter (DOM) influence the rate of photochemical degradation of a synthetic pesticide, Pesticide $X$ , in natural sunlight. Pesticide $X$ degrades when exposed to ultraviolet (UV) light, but dissolved organic matter can either shield the pesticide from light (attenuation) or sensitize it by generating reactive oxygen species (sensitization).

*Experiment 1*
Four identical quartz tubes were prepared, each containing a $10\text{ mg/L}$ aqueous solution of Pesticide $X$ . To three of the tubes, a different type of DOM (humic acid, fulvic acid, or amino acids) was added at a concentration of $5\text{ mg/L}$ . The fourth tube received no DOM. All four tubes were exposed to natural sunlight for $24\text{ hours}$ . The percentage of Pesticide $X$ degraded in each tube was measured. The results are shown in Table 1.

### Table 1

Tube	DOM Type Added	DOM Concentration ( $\text{mg/L}$ )	Percentage of Pesticide $X$ Degraded
1	Humic acid	5	42%
2	Fulvic acid	5	58%
3	Amino acids	5	74%
4	None	0	65%

*Experiment 2*
To determine whether the degradation was driven specifically by UV light rather than thermal decomposition (since the sun also heats the samples), the investigator prepared two additional quartz tubes. Each tube contained a $10\text{ mg/L}$ aqueous solution of Pesticide $X$ and $5\text{ mg/L}$ of humic acid. One tube was exposed to sunlight (Tube 5), while the other tube was wrapped in aluminum foil to block all light and placed adjacent to the first tube in the same outdoor environment (Tube 6). After $24\text{ hours}$ , the percentage of Pesticide $X$ degraded was measured. The results are shown in Table 2.

### Table 2

Tube	Wrapped in Foil?	Percentage of Pesticide $X$ Degraded
5	No	42%
6	Yes	3%

Based on the design of Experiments 1 and 2, which of the following options correctly identifies the control group for the presence of DOM in Experiment 1, and the control group for light exposure in Experiment 2, along with the correct justification for their selection?

Tube 4 in Experiment 1, because it lacks DOM, allowing the investigator to isolate the effect of DOM on Pesticide

X

degradation; and Tube 5 in Experiment 2, because it represents the standard environmental setup containing both light and heat.

Tube 1 in Experiment 1, because humic acid serves as the baseline DOM for comparison; and Tube 6 in Experiment 2, because the foil prevents any chemical interactions from occurring.

Tube 4 in Experiment 1, because it lacks DOM, allowing the investigator to isolate the effect of DOM on Pesticide

X

degradation; and Tube 6 in Experiment 2, because it lacks light exposure, isolating the effect of light from temperature changes.

Tube 4 in Experiment 1, because it acts as a baseline showing that Pesticide

X

does not degrade in the absence of organic matter; and Tube 5 in Experiment 2, because it allows the investigator to measure degradation in the presence of both heat and light.

Show answer & explanation

Answer: Tube 4 in Experiment 1, because it lacks DOM, allowing the investigator to isolate the effect of DOM on Pesticide $X$ degradation; and Tube 6 in Experiment 2, because it lacks light exposure, isolating the effect of light from temperature changes.

Answer

Tube 4 in Experiment 1, because it lacks DOM, allowing the investigator to isolate the effect of DOM on Pesticide

X

degradation; and Tube 6 in Experiment 2, because it lacks light exposure, isolating the effect of light from temperature changes.

The correct option correctly identifies that Tube 4 serves as the control group for DOM presence in Experiment 1 because it contains no DOM, providing a baseline to isolate the DOM's effect. It also correctly identifies that Tube 6 serves as the control group for light exposure in Experiment 2 because it is shielded from light by aluminum foil while experiencing the same outdoor temperature, isolating light exposure from temperature as the cause of Pesticide

X

degradation.

Step-by-Step Solution

1

Analyze Experiment 1 to identify the independent variable and the baseline/control setup.

The independent variable is the type of DOM added. To determine its effect, the investigator must compare the DOM-added trials (Tubes 1-3) to a trial with no DOM. Tube 4 contains

0\text{ mg/L}

DOM and thus serves as the control group.

A control group provides a baseline to isolate the effect of the independent variable being tested.

2

Analyze Experiment 2 to identify the independent variable and the baseline/control setup.

The independent variable is light exposure, used to distinguish photochemical degradation from thermal decomposition. Tube 6 is wrapped in foil to exclude light while keeping temperature constant relative to Tube 5. Thus, Tube 6 is the control group for light exposure.

By blocking light while keeping temperature identical, the investigator isolates light exposure as the variable driving the reaction.

3

Synthesize findings to select the correct choice.

Tube 4 is the control for Experiment 1 and Tube 6 is the control for Experiment 2, with the correct justifications regarding the isolation of DOM and light variables, respectively.

This matches the option identifying Tube 4 and Tube 6 with their respective variable isolations.

Key Concept

Identifying control groups to isolate independent variables and establish baseline conditions in multi-experiment designs.

Estimated Time:2m 0s

Question 23Question →

A student hypothesized that under constant light and temperature, the rate of transpiration in bean plants increases as the relative humidity of the surrounding air increases. To test this hypothesis, the student measured the transpiration rates of several identical bean plants at different relative humidity levels. The results are shown in the table below:

Relative Humidity (%)	Transpiration Rate ( $mg/dm^2/hr$ )
30	15.2
50	10.4
70	5.8
90	1.3

Based on these results, how should the student modify their hypothesis to accurately reflect the relationship between relative humidity and transpiration rate?

The student should hypothesize that the transpiration rate increases as relative humidity increases, because a lower relative humidity slows the rate of evaporation from the leaves.

The student should hypothesize that the transpiration rate remains constant at all relative humidity levels, because relative humidity has no effect on transpiration.

The student should hypothesize that the transpiration rate decreases as relative humidity increases, because a higher humidity decreases the water vapor concentration gradient between the leaf interior and the air.

The student should hypothesize that the transpiration rate decreases as relative humidity increases, because a higher relative humidity increases the rate of water absorption by the leaves from the air.

Show answer & explanation

Answer: The student should hypothesize that the transpiration rate decreases as relative humidity increases, because a higher humidity decreases the water vapor concentration gradient between the leaf interior and the air.

Answer

The student should hypothesize that the transpiration rate decreases as relative humidity increases, because a higher humidity decreases the water vapor concentration gradient between the leaf interior and the air.

The correct option states that the transpiration rate decreases as relative humidity increases, which matches the trend in the data. It also provides the correct scientific explanation: higher humidity reduces the water vapor concentration gradient between the leaf's wet interior and the air, reducing the rate of diffusion.

Step-by-Step Solution

1

Analyze the experimental data in the table to determine the relationship between the independent variable (relative humidity) and the dependent variable (transpiration rate).

As relative humidity increases from 30% to 90%, the transpiration rate decreases from 15.2 mg/dm^2/hr to 1.3 mg/dm^2/hr.

This establishes the empirical trend that refutes the student's initial hypothesis of a positive correlation.

2

Identify which proposed hypothesis modifications align with the observed inverse relationship.

The modifications stating that transpiration rate decreases as relative humidity increases align with the trend.

This eliminates hypotheses predicting increasing or constant trends.

3

Evaluate the scientific reasoning behind the remaining options to select the correct physical/biological mechanism.

A higher humidity reduces the concentration gradient of water vapor between the humid interior of the leaf and the outside air, which reduces the evaporation/transpiration rate.

The correct hypothesis must couple the correct trend with scientifically sound reasoning.

Key Concept

Formulating and Modifying Hypotheses

Question 24Question →

A student hypothesized that the solubility of carbon dioxide ( $CO_2$ ) in water is directly proportional to the water temperature because higher temperatures increase the kinetic energy of the gas molecules, allowing them to interact more with water molecules and remain dissolved. The student measured the solubility of $CO_2$ in water at a constant pressure of $1.0\text{ atm}$ across several temperatures and recorded the data in the table below:

Temperature ( $^\circ\text{C}$ )	$CO_2$ Solubility ( $g/kg\text{ of }H_2O$ )
10	2.4
20	1.7
30	1.3
40	1.0

Based on the results in the table, which of the following modifications to the student's hypothesis and explanation is most appropriate?

The student should retain the hypothesis because the solubility of

CO_2

increases with temperature as gas molecules gain kinetic energy and bind more tightly to water molecules.

The student should modify the hypothesis to state that

CO_2

solubility is inversely proportional to water temperature, because at higher temperatures, gas molecules lose kinetic energy and settle at the bottom of the container.

The student should modify the hypothesis to state that

CO_2

solubility is inversely proportional to water temperature, because the increased kinetic energy of the gas molecules at higher temperatures allows them to overcome intermolecular forces and escape the solution.

The student should retain the hypothesis because the decrease in solubility from

2.4\text{ g/kg}

to

1.0\text{ g/kg}

as temperature rises demonstrates a direct relationship between temperature and solubility.

Show answer & explanation

Answer: The student should modify the hypothesis to state that $CO_2$ solubility is inversely proportional to water temperature, because the increased kinetic energy of the gas molecules at higher temperatures allows them to overcome intermolecular forces and escape the solution.

Answer

The student should modify the hypothesis to state that

CO_2

solubility is inversely proportional to water temperature, because the increased kinetic energy of the gas molecules at higher temperatures allows them to overcome intermolecular forces and escape the solution.

The correct answer correctly identifies that the data shows an inverse relationship between temperature and

CO_2

solubility (solubility decreases as temperature increases). It also provides the correct scientific explanation: higher temperatures increase the kinetic energy of the dissolved gas molecules, enabling them to break intermolecular bonds with the solvent and escape as gas.

Step-by-Step Solution

1

Analyze the solubility data in the table to determine the relationship between water temperature and

CO_2

solubility.

As the water temperature increases from

10^\circ\text{C}

to

40^\circ\text{C}

, the

CO_2

solubility decreases from

2.4\text{ g/kg}

to

1.0\text{ g/kg}

. This indicates an inverse relationship, contradicting the student's hypothesis of a direct relationship.

Evaluating the data trend is necessary to see if the hypothesis is supported or refuted.

2

Assess the physical reasoning of the student's hypothesis regarding kinetic energy.

Higher temperatures do increase the kinetic energy of gas molecules. However, higher kinetic energy makes gas molecules more active, enabling them to break intermolecular bonds with the solvent and escape into the gas phase, which explains the decreased solubility.

Understanding the correct physical mechanism explains why the relationship is inverse rather than direct.

3

Identify the option that correctly describes the necessary modification (inverse relationship) and the accurate physical reasoning.

The option suggesting that solubility is inversely proportional because higher kinetic energy allows gas molecules to escape the solution is the correct choice.

Matching both the data trend and the correct scientific explanation leads to the correct answer.

Key Concept

Formulating and Modifying Hypotheses

Question 25Question →

A student hypothesized that planets located farther from the Sun have shorter orbital periods. The student gathered average orbital data for four planets:

Planet	Average Distance from Sun (AU)	Orbital Period (years)
Mercury	0.39	0.24
Venus	0.72	0.62
Earth	1.00	1.00
Mars	1.52	1.88

Based on these data, how should the student modify their hypothesis?

Modify the hypothesis to state that planets farther from the Sun have longer orbital periods, because orbital period increases as distance increases.

Keep the hypothesis as is, because Mars is farther from the Sun and has the longest orbital period.

Modify the hypothesis to state that distance from the Sun has no effect on a planet's orbital period.

Modify the hypothesis to state that as distance increases, orbital period decreases at a constant rate.

Show answer & explanation

Answer: Modify the hypothesis to state that planets farther from the Sun have longer orbital periods, because orbital period increases as distance increases.

Answer

Modify the hypothesis to state that planets farther from the Sun have longer orbital periods, because orbital period increases as distance increases.

The correct answer is correct because the data in the table shows that as distance from the Sun increases (from

0.39\text{ AU}

to

1.52\text{ AU}

), the orbital period also increases (from

0.24\text{ years}

to

1.88\text{ years}

). Therefore, the student must modify the hypothesis to reflect this direct relationship.

Step-by-Step Solution

1

Analyze the student's original hypothesis.

The original hypothesis states that planets farther from the Sun have shorter orbital periods.

To evaluate a hypothesis, we must first understand the relationship it proposes (farther distance

\rightarrow

shorter period).

2

Examine the data table to identify the actual relationship.

As distance increases from

0.39\text{ AU}

to

1.52\text{ AU}

, the orbital period increases from

0.24\text{ years}

to

1.88\text{ years}

.

This establishes the empirical relationship shown by the experimental data (larger distance

\rightarrow

longer period).

3

Compare the data trend with the original hypothesis to determine the necessary modification.

The data contradicts the original hypothesis, showing a direct relationship instead of an inverse one. Thus, the hypothesis must be modified to state that planets farther from the Sun have longer orbital periods.

Hypotheses must be updated to align with observed scientific evidence.

Key Concept

Formulating and Modifying Hypotheses

Estimated Time:45s

Question 26Question →

A student investigates how the temperature of carbonic acid ( $H_2CO_3$ ) affects the chemical weathering of basalt. The student hypothesizes that as the temperature of the acid solution increases, the rate of chemical weathering (measured as the mass of basalt dissolved over a 24-hour period) will increase. The student performs four trials, starting with a $50.0\text{ g}$ sample of basalt in each trial, while keeping the acid volume and concentration constant. The results are shown in the table below:

Trial	Temperature (°C)	Final mass of basalt (g)
1	15	48.5
2	25	46.8
3	35	45.2
4	45	43.7

Based on the student's hypothesis, if a fifth trial were conducted at $55^\circ\text{C}$ under the same conditions, what would be the predicted final mass of the basalt sample?

Between

43.7\text{ g}

and

45.2\text{ g}

Less than

43.7\text{ g}

Exactly

48.5\text{ g}

Greater than

50.0\text{ g}

Show answer & explanation

Answer: Less than $43.7\text{ g}$

Answer

Less than

43.7\text{ g}

According to the student's hypothesis, as temperature increases, the rate of weathering increases, meaning more basalt dissolves over 24 hours. Because the initial mass of basalt is constant at

50.0\text{ g}

, a higher dissolution rate results in a lower final mass of basalt. Since

55^\circ\text{C}

is higher than the maximum tested temperature of

45^\circ\text{C}

, the final mass at

55^\circ\text{C}

must be less than the final mass measured at

45^\circ\text{C}

(

43.7\text{ g}

). Therefore, the predicted final mass is less than

43.7\text{ g}

.

Step-by-Step Solution

1

Analyze the student's hypothesis.

The hypothesis states that as temperature increases, the rate of chemical weathering (mass of basalt dissolved) increases.

This establishes the direct relationship between temperature and dissolution rate.

2

Relate dissolution rate to the final mass of basalt.

Since each trial begins with

50.0\text{ g}

of basalt, a higher dissolution rate (more basalt dissolved) will result in a lower remaining final mass.

This translates the hypothesis about dissolution rate into a prediction about the measured variable (final mass).

3

Predict the final mass at

55^\circ\text{C}

relative to the existing data points.

A temperature of

55^\circ\text{C}

is higher than the highest tested temperature of

45^\circ\text{C}

(which had a final mass of

43.7\text{ g}

). According to the hypothesis, more dissolution must occur at

55^\circ\text{C}

than at

45^\circ\text{C}

, so the final mass must be less than

43.7\text{ g}

.

This applies the relationship to the target temperature value to find the correct range.

Key Concept

Formulating and testing predictions based on a scientific hypothesis.

Estimated Time:1m 0s

Question 27Question →

A student hypothesized that as the concentration of salt ( $\text{NaCl}$ ) dissolved in water increases, the boiling point of the water decreases. To test this, the student added different masses of $\text{NaCl}$ to $100\text{ mL}$ of pure water and recorded the boiling point of each solution. The results are shown in the table below:

Trial	Mass of $\text{NaCl}$ added ( $\text{g}$ )	Boiling point ( $^\circ\text{C}$ )
1	0	100.0
2	5	100.5
3	10	101.0
4	15	101.5

Based on these results, how should the student modify their hypothesis?

Modify the hypothesis to state that the boiling point of the water remains constant regardless of the concentration of

\text{NaCl}

.

Modify the hypothesis to state that as the concentration of

\text{NaCl}

increases, the boiling point of the water increases.

Modify the hypothesis to state that as the concentration of

\text{NaCl}

increases, the boiling point of the water decreases.

Modify the hypothesis to state that as the concentration of

\text{NaCl}

increases, the boiling point of the water decreases by

0.5^\circ\text{C}

for every

5\text{ g}

of

\text{NaCl}

added.

Show answer & explanation

Answer: Modify the hypothesis to state that as the concentration of $\text{NaCl}$ increases, the boiling point of the water increases.

Answer

Modify the hypothesis to state that as the concentration of

\text{NaCl}

increases, the boiling point of the water increases.

The experimental data show that as the mass of dissolved salt increases, the boiling point of the water increases from

100.0^\circ\text{C}

to

101.5^\circ\text{C}

. This contradicts the student's initial hypothesis that the boiling point would decrease. Therefore, the hypothesis must be modified to state that the boiling point increases as the salt concentration increases.

Step-by-Step Solution

1

Analyze the student's initial hypothesis.

The student proposed that increasing salt concentration decreases the boiling point of water.

Understanding the starting hypothesis is necessary to determine how it should be modified.

2

Examine the experimental data in the table.

As the mass of salt increases from

0\text{ g}

to

15\text{ g}

, the boiling point increases from

100.0^\circ\text{C}

to

101.5^\circ\text{C}

.

Analyzing the empirical data reveals the actual relationship between the variables.

3

Compare the observed data trend with the initial hypothesis and modify it accordingly.

The data shows that the boiling point increases rather than decreases. The hypothesis must be modified to reflect this direct relationship.

A scientific hypothesis must be revised when experimental evidence contradicts its initial prediction.

Key Concept

Formulating and Modifying Hypotheses

Estimated Time:45s

Question 28Question →

To investigate the factors affecting the rate of a chemical reaction, researchers conducted two experiments measuring the rate of decomposition of nitrogen dioxide ( $NO_2$ ) into nitrogen monoxide ( $NO$ ) and oxygen ( $O_2$ ):

2NO_2(g) \rightarrow 2NO(g) + O_2(g)

Experiment 1
Researchers introduced $1.0\text{ mol}$ of $NO_2$ gas into four separate rigid 10-liter containers at different temperatures. No other gases were initially present. The reaction rate was measured at the start of the reaction (initial rate). The results are shown in Table 1.

Table 1

Container	Temperature ( $^\circ\text{C}$ )	Initial Rate ( $\text{mol}/(\text{L}\cdot\text{s})$ )
1	25	$1.2 \times 10^{-5}$
2	100	$4.8 \times 10^{-5}$
3	200	$1.9 \times 10^{-4}$
4	300	$7.6 \times 10^{-4}$

Experiment 2
Using Container 1 ( $25^\circ\text{C}$ ), researchers repeated the reaction but added different amounts of helium ( $He$ ), an inert gas that does not participate in the reaction, to test whether the total pressure of the container affects the reaction rate. The initial concentration of $NO_2$ was kept at $0.10\text{ mol/L}$ ( $1.0\text{ mol}$ in the 10-liter container) in all trials. The results are shown in Table 2.

Table 2

Trial	Amount of $He$ added (mol)	Total Initial Pressure (atm)	Initial Rate ( $\text{mol}/(\text{L}\cdot\text{s})$ )
5	0.5	3.6	$1.2 \times 10^{-5}$
6	1.0	4.8	$1.2 \times 10^{-5}$
7	2.0	7.2	$1.2 \times 10^{-5}$

Based on the results of Experiments 1 and 2, which of the following setups serves as the control group to determine the effect of adding helium gas on the initial reaction rate in Experiment 2?

Container 1, because it represents the reaction under the same conditions as Experiment 2 but with no helium gas added.

Container 4, because it represents the reaction at the highest temperature, which maximizes the initial reaction rate.

Trial 5, because it has the lowest amount of helium added among the trials in Experiment 2.

Trial 7, because it has the highest initial pressure, establishing the upper limit for the pressure variable.

Show answer & explanation

Answer: Container 1, because it represents the reaction under the same conditions as Experiment 2 but with no helium gas added.

Answer

Container 1, because it represents the reaction under the same conditions as Experiment 2 but with no helium gas added.

The correct answer is the option specifying Container 1. In Experiment 2, researchers are testing the effect of adding helium gas on the initial reaction rate at a constant temperature of

25^\circ\text{C}

and initial

NO_2

concentration of

0.10\text{ mol/L}

. To determine if helium has any effect, they must compare these trials to a baseline setup where no helium was added. Container 1 in Experiment 1 provides this baseline, as it was conducted at

25^\circ\text{C}

with the same initial

NO_2

concentration but without any helium gas.

Step-by-Step Solution

1

Identify the independent variable in Experiment 2.

The independent variable is the amount of helium gas (

He

) added to the reaction mixture, which in turn changes the total initial pressure.

To determine the control group, we must first establish what variable is being manipulated in the experiment.

2

Determine the baseline condition for the independent variable.

The baseline condition is the reaction rate when zero helium (

0.0\text{ mol}

) is added, while keeping all other controlled variables constant.

A control group or baseline condition represents the state of the system before the independent variable is manipulated.

3

Locate the experimental setup that matches the baseline condition.

In Table 2, all trials (5, 6, and 7) have helium added. Looking back at Experiment 1, Container 1 represents the reaction at

25^\circ\text{C}

with

1.0\text{ mol}

of

NO_2

in a 10-L container (concentration of

0.10\text{ mol/L}

) and no other gases (helium =

0.0\text{ mol}

).

Comparing the reaction rates in Experiment 2 to Container 1 allows researchers to isolate and measure the specific effect of adding helium on the reaction rate.

Key Concept

Determining Control Groups and Baseline Conditions

Question 29Question →

A biochemist investigated the catalytic activity of amylase extracted from *Geobacillus stearothermophilus* at $70^\circ\text{C}$ in the presence of various divalent metal ions ( $\text{Mg}^{2+}$ , $\text{Ca}^{2+}$ , $\text{Zn}^{2+}$ , and $\text{Cu}^{2+}$ ). Five test tubes were prepared as shown in the table below. Each tube contained $1.0\text{ mL}$ of a $1\%$ starch solution, $1.0\text{ mL}$ of buffer solution ( $\text{pH } 7.0$ ), and $0.1\text{ mL}$ of purified amylase enzyme.

Tube	Added Solution ( $0.5\text{ mL}$ )
Tube 1	Distilled water
Tube 2	$10\text{ mM } \text{MgCl}_2$
Tube 3	$10\text{ mM } \text{CaCl}_2$
Tube 4	$10\text{ mM } \text{ZnCl}_2$
Tube 5	$10\text{ mM } \text{CuCl}_2$

All five tubes were incubated at $70^\circ\text{C}$ for 15 minutes, after which the rate of starch hydrolysis was determined for each tube.

Which of the following test tubes served as the control group to establish the baseline rate of starch hydrolysis in the absence of added metal ions?

Tube 1, because it contained all reaction components except the tested metal ions.

Tube 2, because magnesium ions are naturally present in bacterial cells and provide standard physiological conditions.

Tube 3, because calcium ions stabilize thermophilic enzymes to ensure maximum baseline activity.

Tube 5, because copper ions completely halt enzyme activity, setting the lower boundary for the experiment.

Show answer & explanation

Answer: Tube 1, because it contained all reaction components except the tested metal ions.

Answer

Tube 1 served as the control group because it contained distilled water in place of metal ion solutions, establishing a baseline rate of starch hydrolysis with no added metal ions.

The correct answer identifies Tube 1 as the control group. A control group provides a baseline measurement by keeping all experimental factors identical except for the independent variable being tested. Since Tube 1 contains distilled water in place of metal ion solutions, it establishes the baseline rate of starch hydrolysis under standard reaction conditions without metal ion intervention.

Step-by-Step Solution

1

Identify the independent variable tested across the experimental setups.

The independent variable is the type of added divalent metal ion (

\text{Mg}^{2+}

,

\text{Ca}^{2+}

,

\text{Zn}^{2+}

, or

\text{Cu}^{2+}

).

Control groups require isolating the independent variable by withholding it or substituting it with a neutral substance.

2

Examine the composition of each test tube to find the setup where the independent variable is omitted.

Tube 1 receives

0.5\text{ mL}

of distilled water instead of a metal ion solution, keeping all other reaction conditions (enzyme, substrate, buffer, volume, temperature) constant.

Distilled water acts as a neutral solvent control to measure baseline enzyme kinetics.

3

Select the option that correctly identifies Tube 1 and explains its role as a baseline control.

The option stating Tube 1 is the control group because it contains all components except the tested metal ions is correct.

A true control group isolate changes caused solely by the experimental treatments.

Key Concept

Identifying Negative Control Groups and Baseline Conditions

Estimated Time:1m 0s

Question 30Question →

### Passage
An agricultural biologist investigated the physiological stress responses of the green alga *Chlorella vulgaris* exposed to common components of agricultural runoff. The study focused on four environmental variables: nitrate ( $NO_3^-$ ) enrichment, phosphate ( $PO_4^{3-}$ ) enrichment, atrazine (a widely used herbicide) exposure, and elevated temperature.

The biologist set up 5 culture flasks with identical initial densities of *C. vulgaris*. Each flask was subjected to a specific combination of nutrient concentrations, atrazine concentration, and temperature for 7 days. The experimental conditions for each flask are detailed in Table 1.

### Table 1

Flask	Temperature ( $^\circ\text{C}$ )	Added $NO_3^-$ ( $\text{mg/L}$ )	Added $PO_4^{3-}$ ( $\text{mg/L}$ )	Atrazine ( $\text{mg/L}$ )
1	20	0.0	0.0	0.0
2	20	5.0	0.0	0.0
3	20	5.0	1.0	0.0
4	20	5.0	1.0	0.1
5	25	5.0	1.0	0.1

To evaluate the specific, independent contribution of each variable or combination of variables to algal stress, the biologist must compare the growth rates of algae in the experimental flasks against their appropriate control groups or baseline conditions.

### Matching Task
Match each of the following experimental objectives with the specific Flask that serves as its primary control group or baseline condition.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Matching pairs: (1) Baseline growth under standard conditions matches Flask 1; (2) Isolating the effect of phosphate in a nitrate-containing environment matches Flask 2; (3) Isolating the effect of atrazine matches Flask 3; (4) Isolating the effect of a temperature increase matches Flask 4.

Each experimental objective requires a control group that differs by exactly one variable to establish a proper comparison. The baseline growth rate is measured by Flask 1 because it has no experimental manipulations. The independent effect of phosphate in a nitrate environment (Flask 3) is isolated by Flask 2, which contains nitrate but no phosphate. The independent effect of atrazine (Flask 4) is isolated by Flask 3, which has the same nutrient levels but no atrazine. The independent effect of elevated temperature (Flask 5) is isolated by Flask 4, which has the same nutrients and atrazine but at the baseline temperature.

Step-by-Step Solution

1

Identify the baseline or reference state of the experiment.

Flask 1 has no added nitrate, no added phosphate, no atrazine, and is at the baseline temperature of

20^\circ\text{C}

.

This serves as the negative control or baseline condition for the entire experiment.

2

Determine the control needed to isolate the effect of added phosphate when nitrate is present.

Flask 3 introduces phosphate (

1.0\text{ mg/L}

) to a medium already containing nitrate (

5.0\text{ mg/L}

) at

20^\circ\text{C}

. Its control must keep temperature and nitrate constant but lack phosphate, which corresponds to Flask 2.

By comparing Flask 3 to Flask 2, the only difference is the presence of phosphate, thereby isolating its independent effect.

3

Determine the control needed to isolate the effect of atrazine in a nutrient-enriched medium.

Flask 4 contains nutrients (

5.0\text{ mg/L}

nitrate,

1.0\text{ mg/L}

phosphate) and introduces

0.1\text{ mg/L}

atrazine at

20^\circ\text{C}

. Its control must have the same nutrients but no atrazine, which corresponds to Flask 3.

Comparing Flask 4 to Flask 3 isolates the biological impact of the herbicide atrazine.

4

Determine the control needed to isolate the effect of temperature under nutrient-enriched and herbicide-exposed conditions.

Flask 5 is at

25^\circ\text{C}

with nutrients and atrazine. Its control must have the exact same chemical additions but be held at the standard temperature of

20^\circ\text{C}

, which corresponds to Flask 4.

Comparing Flask 5 to Flask 4 isolates the independent effect of the

5^\circ\text{C}

temperature increase.

Key Concept

A control group must be identical to the experimental group in every factor except the single independent variable being tested. This isolates the independent variable's effect on the dependent variable.

Estimated Time:3m 0s

Question 31Question →

A research team investigated the photoelectrochemical (PEC) water-splitting efficiency of a bismuth vanadate ( $BiVO_4$ ) photoanode. The experimental apparatus consisted of a three-electrode PEC cell connected to a potentiostat. The working electrode (photoanode) was illuminated by a simulated solar light source equipped with an Air Mass (AM) 1.5G filter and a water-filled optical filter. The electrochemical cell contained a $0.5\text{ M } Na_2SO_4$ aqueous electrolyte. A platinum ( $Pt$ ) wire counter electrode was used to complete the circuit, and a silver/silver chloride ( $Ag/AgCl$ ) electrode served as the reference. The gaseous products evolved at the electrodes were swept by an inert carrier gas into a gas chromatograph for quantification.

Match each component of the experimental apparatus to its primary function in this experimental setup.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The water-filled optical filter matches with absorbing infrared radiation to prevent temperature-induced changes. The platinum counter electrode matches with serving as the site for the complementary reduction reaction. The silver/silver chloride electrode matches with providing a constant, known half-cell potential. The potentiostat matches with regulating the voltage difference while recording the flow of charge.

Each component is correctly matched based on the principles of three-electrode photoelectrochemical cells and optical solar simulation. The water-filled filter absorbs heat-generating infrared light to maintain temperature stability. The silver/silver chloride electrode provides a stable potential reference. The platinum counter electrode completes the circuit and hosts the reduction reaction. The potentiostat manages and measures the electrical potentials and current of the cell.

Step-by-Step Solution

1

Analyze the role of the optical water filter.

Water absorbs light strongly in the infrared region. Removing infrared wavelengths from the simulated solar light prevents the electrolyte from heating up during the experiment, maintaining a constant temperature and stable ionic conductivity.

This is crucial for isolating photoelectrochemical effects from thermal effects.

2

Analyze the role of the silver/silver chloride (

Ag/AgCl

) electrode.

In a three-electrode setup, the reference electrode must maintain a stable half-cell potential. The reference electrode serves as a stable reference point against which the working electrode potential is measured and controlled.

This prevents potential drift and ensures precise electrochemical measurements.

3

Analyze the role of the platinum counter electrode.

To avoid current passing through the reference electrode (which would alter its potential), a counter electrode is introduced. The platinum counter electrode completes the electrical circuit and provides the surface for the complementary reduction reaction (hydrogen evolution).

This maintains charge neutrality in the electrolyte and allows the photoanode current to flow.

4

Analyze the role of the potentiostat.

The potentiostat is the control instrument that maintains the potential of the working electrode at a constant level relative to the reference electrode by adjusting the current at the counter electrode.

This allows for precise control of the electrochemical driving force and measurement of the resulting photocurrent.

Key Concept

Function and operation of components in a three-electrode photoelectrochemical cell and optical filters in solar simulation.

Estimated Time:3m 0s

Question 32Question →

### Passage I

Researchers investigated the regulation of the electron transport chain (ETC) in isolated mammalian mitochondria by measuring the rate of oxygen ( $O_2$ ) consumption (in $\text{nmol } O_2/\text{min}/\text{mg protein}$ ) under different biochemical conditions.

In all trials, a fixed concentration of isolated mitochondria was suspended in a reaction chamber containing a physiological buffer solution at $37^\circ\text{C}$ . The baseline concentration of dissolved $O_2$ in the buffer was monitored. Respiration substrates, adenylates, and metabolic inhibitors were added sequentially or in combination to study their effects on the rate of mitochondrial $O_2$ consumption. The components added to each trial are shown in Table 1.

### Table 1

Trial	Mitochondria	Succinate (substrate)	ADP	Oligomycin (ATP synthase inhibitor)	FCCP (proton gradient uncoupler)
1	Yes	Yes	No	No	No
2	Yes	Yes	Yes	No	No
3	Yes	Yes	Yes	Yes	No
4	Yes	Yes	Yes	Yes	Yes
5	Yes	No	Yes	No	No
6	No	Yes	Yes	No	Yes

An investigator wants to establish a baseline condition to prove that the decrease in dissolved $O_2$ in Trials 1–5 is due to the biological activity of the electron transport chain in the mitochondria, rather than abiotic chemical reactions or gas leakage from the reaction chamber. Which of the trials listed in Table 1 serves as the most appropriate control group for this purpose?

Trial 1, because it contains mitochondria and succinate but lacks ADP, isolating resting state respiration.

Trial 5, because it contains mitochondria and ADP but lacks succinate, isolating the respiration fueled by endogenous substrates.

Trial 6, because it lacks mitochondria but contains all other components, isolating biological oxygen consumption from abiotic processes or gas leaks.

Trial 3, because it contains oligomycin to halt ATP synthase, serving as a baseline for the proton leak rate.

Show answer & explanation

Answer: Trial 6, because it lacks mitochondria but contains all other components, isolating biological oxygen consumption from abiotic processes or gas leaks.

Answer

Trial 6 serves as the most appropriate control group because it lacks the biological component (mitochondria) while retaining the chemical substrates, uncoupler, and physical conditions, thereby isolating any potential abiotic oxygen consumption or apparatus leakage.

The trial that lacks mitochondria but contains the other reaction components serves as the appropriate negative control. If any oxygen decrease occurs in this setup, it must be due to abiotic oxidation or chamber leakage. Comparing the rates of Trials 1–5 to this baseline allows the investigator to isolate the oxygen consumption directly caused by biological mitochondrial activity.

Step-by-Step Solution

1

Identify the confounding variables that need to be controlled based on the investigator's goal.

The investigator needs to prove that the oxygen consumption is due to biological mitochondrial respiration, rather than abiotic chemical reactions or gas leakage.

This requires identifying a baseline or negative control setup that removes the biological component (mitochondria) while keeping all other potential variables constant.

2

Examine the composition of the trials in Table 1 to find a setup that isolates mitochondrial biological activity.

Trial 6 contains the substrate (succinate), the phosphate acceptor (ADP), and the uncoupler (FCCP), but does not contain any mitochondria.

By omitting the mitochondria, any oxygen depletion measured in Trial 6 must be due to abiotic factors or leaks, establishing the non-biological baseline rate.

3

Compare the proposed trial to other potential baselines to confirm it is the correct control group.

Trial 6 is the only trial without mitochondria, whereas Trials 1 and 5 contain mitochondria but test physiological variables (lack of ADP and substrate, respectively).

Only a trial completely lacking mitochondria can isolate biological activity from abiotic oxygen loss.

Key Concept

Determining Control Groups and Baseline Conditions

Estimated Time:2m 0s

Question 33Question →

A student proposed the following hypothesis:
*Hypothesis*: As the temperature of water increases from $0^\circ\text{C}$ to $10^\circ\text{C}$ , the density of the water will decrease continuously, and this density decrease will be more pronounced in water with higher salinity.

To test this hypothesis, the student measured the density of freshwater ( $0\text{ ppt}$ salinity) and saline water ( $10\text{ ppt}$ salinity) at different temperatures. The results are shown in the table below.

Temperature ( $^\circ\text{C}$ )	Density at $0\text{ ppt}$ salinity ( $\text{g/cm}^3$ )	Density at $10\text{ ppt}$ salinity ( $\text{g/cm}^3$ )
$0$	$0.99984$	$1.00802$
$2$	$0.99994$	$1.00808$
$4$	$0.99997$	$1.00806$
$6$	$0.99994$	$1.00800$
$8$	$0.99985$	$1.00790$
$10$	$0.99970$	$1.00776$

Based on these results, how should the student modify the hypothesis regarding the relationship between temperature, salinity, and density?

Retain the hypothesis as proposed, because the density of both water samples at

10^\circ\text{C}

is lower than their density at

0^\circ\text{C}

, and the overall decrease in density is greater for the sample with higher salinity.

Modify the hypothesis to state that density decreases continuously for the sample with

0\text{ ppt}

salinity but first increases and then decreases for the sample with

10\text{ ppt}

salinity.

Modify the hypothesis to state that as temperature increases from

0^\circ\text{C}

to

10^\circ\text{C}

, density first increases and then decreases, and that the temperature at which maximum density occurs is lower for water with higher salinity.

Modify the hypothesis to state that as temperature increases from

0^\circ\text{C}

to

10^\circ\text{C}

, density first decreases and then increases, and that the temperature at which minimum density occurs is higher for water with higher salinity.

Show answer & explanation

Answer: Modify the hypothesis to state that as temperature increases from $0^\circ\text{C}$ to $10^\circ\text{C}$ , density first increases and then decreases, and that the temperature at which maximum density occurs is lower for water with higher salinity.

Answer

Modify the hypothesis to state that as temperature increases from

0^\circ\text{C}

to

10^\circ\text{C}

, density first increases and then decreases, and that the temperature at which maximum density occurs is lower for water with higher salinity.

The correct answer states that density first increases and then decreases, and that the temperature of maximum density is lower for water with higher salinity. This is supported by the data because for the

0\text{ ppt}

sample, the density increases from

0.99984 \text{ g/cm}^3

at

0^\circ\text{C}

to

0.99997 \text{ g/cm}^3

at

4^\circ\text{C}

before decreasing. Similarly, for the

10\text{ ppt}

sample, the density increases from

1.00802 \text{ g/cm}^3

at

0^\circ\text{C}

to

1.00808 \text{ g/cm}^3

at

2^\circ\text{C}

before decreasing. Comparing these peak temperatures (

2^\circ\text{C} < 4^\circ\text{C}

) confirms that the temperature of maximum density decreases with higher salinity.

Step-by-Step Solution

1

Examine the density values of both the

0\text{ ppt}

and

10\text{ ppt}

salinity samples as temperature increases from

0^\circ\text{C}

to

10^\circ\text{C}

.

For both samples, density increases to a peak value (

0.99997 \text{ g/cm}^3

at

4^\circ\text{C}

and

1.00808 \text{ g/cm}^3

at

2^\circ\text{C}

, respectively) and then decreases as temperature continues to rise to

10^\circ\text{C}

.

To evaluate whether the proposed hypothesis of a 'continuous decrease' matches the observed trend.

2

Determine the temperature of maximum density for each salinity concentration.

The

0\text{ ppt}

sample reaches its maximum density at

4^\circ\text{C}

, whereas the

10\text{ ppt}

sample reaches its maximum density at

2^\circ\text{C}

.

To identify the exact points of peak density and check if salinity shifts this temperature.

3

Compare the peak density temperatures between the two salinity groups (

0\text{ ppt}

and

10\text{ ppt}

).

The peak temperature for the higher salinity sample (

2^\circ\text{C}

) is lower than that of the lower salinity sample (

4^\circ\text{C}

).

To determine how salinity alters the temperature of maximum density.

4

Synthesize these observations to formulate a modified hypothesis.

A modified hypothesis must state that density first increases and then decreases, and that the temperature of maximum density decreases as salinity increases.

To construct a new hypothesis that is fully supported by the experimental data.

Key Concept

Formulating and Modifying Hypotheses

Estimated Time:2m 0s

Question 34Question →

A group of students studied the decomposition of sodium bicarbonate ( $NaHCO_3$ ) in aqueous solution. They proposed a hypothesis: *The rate of $NaHCO_3$ decomposition increases as the initial concentration of $NaHCO_3$ increases because a higher concentration of reactant particles leads to more frequent collisions.*

To test this, they performed two experiments. In Experiment 1, they measured the volume of carbon dioxide ( $CO_2$ ) gas produced over $10\text{ min}$ at a constant temperature of $25^\circ\text{C}$ using different initial concentrations of $NaHCO_3$ ( $0.1\text{ M}$ , $0.2\text{ M}$ , and $0.3\text{ M}$ ). In Experiment 2, they repeated the trials at a constant temperature of $50^\circ\text{C}$ . The results are shown in the table:

Trial	Temperature ( $^\circ\text{C}$ )	Initial $NaHCO_3$ Concentration ( $\text{M}$ )	Total $CO_2$ Produced in $10\text{ min}$ ( $\text{mL}$ )
1	25	0.1	12.4
2	25	0.2	12.5
3	25	0.3	12.3
4	50	0.1	24.8
5	50	0.2	25.1
6	50	0.3	24.9

Which of the following statements best describes how the students should modify their hypothesis?

The students should retain their original hypothesis because the total volume of

CO_2

gas produced increased when the temperature was raised from

25^\circ\text{C}

to

50^\circ\text{C}

.

The students should modify their hypothesis to state that the rate of

NaHCO_3

decomposition is independent of the initial

NaHCO_3

concentration but increases as temperature increases.

The students should modify their hypothesis to state that the rate of

NaHCO_3

decomposition increases as the initial

NaHCO_3

concentration increases, but only at a constant temperature of

50^\circ\text{C}

.

The students should modify their hypothesis to state that the rate of

NaHCO_3

decomposition decreases as the initial

NaHCO_3

concentration increases at higher temperatures.

Show answer & explanation

Answer: The students should modify their hypothesis to state that the rate of $NaHCO_3$ decomposition is independent of the initial $NaHCO_3$ concentration but increases as temperature increases.

Answer

The students should modify their hypothesis to state that the rate of

NaHCO_3

decomposition is independent of the initial

NaHCO_3

concentration but increases as temperature increases.

The experimental results demonstrate that changing the initial concentration of

NaHCO_3

from

0.1\text{ M}

to

0.3\text{ M}

has negligible effect on the volume of gas produced (remaining at approximately

12.4\text{ mL}

at

25^\circ\text{C}

and

25.0\text{ mL}

at

50^\circ\text{C}

). However, increasing the temperature from

25^\circ\text{C}

to

50^\circ\text{C}

significantly increases the rate of decomposition, doubling the gas output. Therefore, the hypothesis must be modified to state that the rate is independent of the initial concentration but increases with temperature.

Step-by-Step Solution

1

Analyze the effect of changing the initial

NaHCO_3

concentration at a constant temperature.

At

25^\circ\text{C}

(Trials 1-3), the

CO_2

volumes are

12.4\text{ mL}

,

12.5\text{ mL}

, and

12.3\text{ mL}

. At

50^\circ\text{C}

(Trials 4-6), the volumes are

24.8\text{ mL}

,

25.1\text{ mL}

, and

24.9\text{ mL}

.

To determine whether the data supports the students' hypothesis that a higher concentration increases the reaction rate.

2

Determine if there is a relationship between concentration and the rate of decomposition based on the step 1 results.

The volume of

CO_2

produced remains approximately constant regardless of the initial concentration, meaning concentration does not affect the rate.

To evaluate the validity of the original hypothesis.

3

Analyze the effect of temperature changes on the volume of gas produced at the same concentrations.

Comparing Trial 1 to Trial 4 (both

0.1\text{ M}

), the volume increases from

12.4\text{ mL}

to

24.8\text{ mL}

. Similar increases occur for other concentrations when temperature is raised.

To identify which variable (temperature) actually influences the rate of reaction.

4

Synthesize a modified hypothesis that aligns with both findings.

The modified hypothesis should state that the rate of decomposition is independent of the initial

NaHCO_3

concentration but increases as temperature increases.

To construct a hypothesis that accurately accounts for all observed trends in the experimental data.

Key Concept

Formulating and Modifying Hypotheses

Estimated Time:1m 30s

Question 35Question →

A student investigates the corrosion of iron in various acidic solutions. The student hypothesizes that the rate of corrosion, measured by the mass loss of an iron nail after $48\text{ hours}$ , increases linearly as the pH of the solution decreases from $7.0$ to $2.0$ . The student places identical iron nails in solutions of varying pH and records the mass loss in the table below:

pH of solution	Mass loss of iron nail (\text{mg})
$7.0$	$0.8$
$6.0$	$1.6$
$5.0$	$2.4$
$4.0$	$3.2$
$3.0$	$3.8$
$2.0$	$4.0$

Based on these results, which of the following statements best describes how the student should modify their hypothesis?

The rate of corrosion increases linearly as pH decreases from

7.0

to

4.0

, but the rate of increase begins to level off at a pH below

4.0

.

The rate of corrosion increases by a constant value of

0.8\text{ mg}

for every

1.0\text{ unit}

decrease in pH across the entire range of

7.0

to

2.0

.

The rate of corrosion decreases as the pH of the solution decreases from

7.0

to

4.0

, then increases at lower pH values.

The rate of corrosion remains constant at all pH values below

5.0

.

Show answer & explanation

Answer: The rate of corrosion increases linearly as pH decreases from $7.0$ to $4.0$ , but the rate of increase begins to level off at a pH below $4.0$ .

Answer

The rate of corrosion increases linearly as pH decreases from 7.0 to 4.0, but the rate of increase begins to level off at a pH below 4.0.

The correct option correctly identifies that from pH 7.0 to 4.0, the mass loss increases by exactly 0.8 mg for each 1.0-unit decrease in pH, showing a linear relationship. Below pH 4.0, the mass loss increases by 0.6 mg (from pH 4.0 to 3.0) and then by 0.2 mg (from pH 3.0 to 2.0), which shows that the rate of increase is leveling off.

Step-by-Step Solution

1

Analyze the student's initial hypothesis and the trend in the data.

The hypothesis predicts a constant, linear increase in mass loss (corrosion rate) as pH decreases from

7.0

to

2.0

. The data shows that as pH decreases from

7.0

to

4.0

, the mass loss increases by a constant

0.8\text{ mg}

per

1.0

pH unit (

0.8 \rightarrow 1.6 \rightarrow 2.4 \rightarrow 3.2

).

Establishing the baseline relationship helps identify where the data aligns with or deviates from the linear prediction.

2

Examine the data for pH values below

4.0

to determine if the linear trend continues.

From pH

4.0

to

3.0

, the mass loss increases by

0.6\text{ mg}

(

3.2 \rightarrow 3.8\text{ mg}

). From pH

3.0

to

2.0

, the mass loss increases by only

0.2\text{ mg}

(

3.8 \rightarrow 4.0\text{ mg}

).

Calculating the rate of change at lower pH values reveals whether the relationship remains strictly linear.

3

Formulate a modified hypothesis that accurately describes the entire dataset.

The rate of corrosion increases linearly as pH decreases from

7.0

to

4.0

, but the rate of increase begins to level off (increase by smaller amounts) at pH values below

4.0

.

A modified hypothesis must reflect both the initial linear range and the subsequent non-linear leveling off shown in the data.

Key Concept

Formulating and Modifying Hypotheses

Estimated Time:1m 30s

Question 36Question →

A student hypothesizes that as the angle of an inclined plane increases, the time it takes for a block to slide down the plane will decrease. The student conducts an experiment and measures the slide times at angles of $15^\circ$ , $30^\circ$ , and $45^\circ$ . If the student's hypothesis is correct, which of the following predictions is most likely true for the slide time of the block at an angle of $60^\circ$ ?

The slide time at

60^\circ

will be greater than the slide time at

45^\circ

.

The slide time at

60^\circ

will be less than the slide time at

45^\circ

.

The slide time at

60^\circ

will be equal to the slide time at

45^\circ

.

The slide time at

60^\circ

will be greater than the slide time at

15^\circ

.

Show answer & explanation

Answer: The slide time at $60^\circ$ will be less than the slide time at $45^\circ$ .

Answer

The slide time at

60^\circ

will be less than the slide time at

45^\circ

.

The student's hypothesis proposes that slide time decreases as the angle of inclination increases. Since

60^\circ

is greater than the largest tested angle of

45^\circ

, the slide time at

60^\circ

must be less than the slide time at

45^\circ

to remain consistent with the hypothesis.

Step-by-Step Solution

1

Identify the relationship proposed by the student's hypothesis.

The hypothesis states that as the angle of an inclined plane increases, the slide time decreases.

This establishes that there is an inverse relationship between the angle and the time taken.

2

Compare the test angle to the previously measured angles.

The target angle is

60^\circ

, which is greater than the previous angles of

15^\circ

,

30^\circ

, and

45^\circ

.

This determines how the independent variable is being modified relative to the existing data.

3

Apply the hypothesized relationship to make a prediction.

Since

60^\circ

is a larger angle than

45^\circ

, the slide time must be shorter than the slide time measured at

45^\circ

.

An increase in angle must result in a decrease in slide time to support the hypothesis.

Key Concept

Formulating predictions that align with a proposed hypothesis.

Estimated Time:45s

Question 37Question →

A student proposes the following hypothesis:

*Hypothesis*: Plants grown under blue light will grow taller than plants grown under green light or red light.

To test this hypothesis, the student grows $3$ identical pea plants under different wavelengths of light for $14$ days, keeping all other variables constant. The heights of the plants at the end of the experiment are shown in the table below:

Light Color	Final Plant Height ( $\text{cm}$ )
Blue	$12$
Green	$8$
Red	$15$

Based on these results, which of the following statements best describes how the student should modify the hypothesis?

Modify the hypothesis to state that plants grown under red light will grow taller than plants grown under blue light or green light.

Modify the hypothesis to state that plants grown under green light will grow taller than plants grown under blue light or red light.

Modify the hypothesis to state that plants grown under blue light and green light will grow to the same height.

Retain the hypothesis as originally proposed because the plant grown under blue light grew taller than the plant grown under green light.

Show answer & explanation

Answer: Modify the hypothesis to state that plants grown under red light will grow taller than plants grown under blue light or green light.

Answer

Modify the hypothesis to state that plants grown under red light will grow taller than plants grown under blue light or green light.

The experimental results demonstrate that the plant grown under red light achieved the greatest height (

15\text{ cm}

), followed by the plant grown under blue light (

12\text{ cm}

), and the plant grown under green light (

8\text{ cm}

). Because the original hypothesis predicted that blue light would yield the tallest plants, the finding that red light resulted in the tallest plants contradicts the prediction. Therefore, the hypothesis must be modified to reflect that red light produces the greatest height.

Step-by-Step Solution

1

Analyze the student's original hypothesis.

The hypothesis predicts that blue light results in the greatest plant height compared to green or red light.

To evaluate the hypothesis, we must first understand what it predicts.

2

Examine the experimental results in the table.

The final heights are: Red =

15\text{ cm}

, Blue =

12\text{ cm}

, Green =

8\text{ cm}

.

This provides the actual data to compare against the prediction.

3

Compare the data to the hypothesis and determine the correct modification.

Since red light resulted in the tallest growth (

15\text{ cm}

, which is greater than blue light's

12\text{ cm}

), the original hypothesis is incorrect. The hypothesis should be modified to state that red light leads to the tallest growth.

A scientific hypothesis must be updated when experimental evidence contradicts its initial prediction.

Key Concept

Evaluating and modifying a hypothesis based on experimental evidence.

Question 38Question →

Geophysicists study the rheology of magma to predict volcanic eruption styles. A volcanologist formulated a hypothesis regarding the combined effects of pressure ( $P$ ) and water content ( $W$ ) on the viscosity ( $\eta$ ) of rhyolitic magma at a constant temperature of $1000^\circ\text{C}$ :

*Hypothesis*: Increasing $W$ decreases $\eta$ by disrupting silicate network bonds. However, because higher $P$ compresses the melt and opposes this bond disruption, the viscosity-reducing effect of adding water (defined as the factor by which $\eta$ decreases when $W$ is increased from $0.1\text{ wt}\%$ to $3.0\text{ wt}\%$ ) will become weaker as $P$ increases.

To test this hypothesis, the volcanologist measured the viscosity of rhyolitic magma samples at $1000^\circ\text{C}$ under different combinations of $P$ and $W$ . The results are shown in the table below:

Trial	Pressure ( $P$ , $\text{MPa}$ )	Water content ( $W$ , $\text{wt}\%$ )	Viscosity ( $\eta$ , $\text{Pa}\cdot\text{s}$ )
1	50	0.1	$1.0 \times 10^8$
2	50	1.0	$5.0 \times 10^5$
3	50	3.0	$2.0 \times 10^3$
4	150	0.1	$8.0 \times 10^7$
5	150	1.0	$1.0 \times 10^5$
6	150	3.0	$9.0 \times 10^2$
7	300	0.1	$5.0 \times 10^7$
8	300	1.0	$3.0 \times 10^4$
9	300	3.0	$4.0 \times 10^2$

Which of the following statements best describes how the volcanologist should modify the hypothesis in light of these results?

Modify the hypothesis to state that the viscosity-reducing effect of water becomes stronger as pressure increases, because the factor by which viscosity decreases when water is added increases as pressure increases.

Modify the hypothesis to state that the viscosity-reducing effect of water becomes weaker as pressure increases, because the absolute decrease in viscosity when water is added is smallest at

300\text{ MPa}

.

Retain the hypothesis because at each pressure, increasing the water content from

0.1\text{ wt}\%

to

3.0\text{ wt}\%

decreases the magma viscosity.

Modify the hypothesis to state that increasing pressure increases magma viscosity, because compaction at higher pressures prevents water from effectively lubricating the silicate network.

Show answer & explanation

Answer: Modify the hypothesis to state that the viscosity-reducing effect of water becomes stronger as pressure increases, because the factor by which viscosity decreases when water is added increases as pressure increases.

Answer

Modify the hypothesis to state that the viscosity-reducing effect of water becomes stronger as pressure increases, because the factor by which viscosity decreases when water is added increases as pressure increases.

The correct option correctly identifies that the volcanologist's hypothesis must be modified to state that the viscosity-reducing effect of water becomes stronger at higher pressures. The hypothesis defines the effect as the factor by which viscosity decreases when water is increased from

0.1\text{ wt}\%

to

3.0\text{ wt}\%

. Calculating this factor (viscosity at

0.1\text{ wt}\%

divided by viscosity at

3.0\text{ wt}\%

) yields

50,000

at

50\text{ MPa}

, approximately

88,889

at

150\text{ MPa}

, and

125,000

at

300\text{ MPa}

. Since this factor increases with pressure, the effect becomes stronger, contradicting the hypothesis that it would become weaker.

Step-by-Step Solution

1

Identify the definition of the viscosity-reducing effect in the hypothesis.

The effect is defined as the factor (ratio) by which viscosity decreases when water content increases from

0.1\text{ wt}\%

to

3.0\text{ wt}\%

.

This sets the criteria for evaluating the strength of the water's effect.

2

Calculate the reduction factor at each pressure level using the formula:

\text{Factor} = \frac{\eta_{\text{at } 0.1\%}}{\eta_{\text{at } 3.0\%}}

.

At

50\text{ MPa}

:

\frac{1.0 \times 10^8}{2.0 \times 10^3} = 50,000

. At

150\text{ MPa}

:

\frac{8.0 \times 10^7}{9.0 \times 10^2} \approx 88,889

. At

300\text{ MPa}

:

\frac{5.0 \times 10^7}{4.0 \times 10^2} = 125,000

.

This quantifies the strength of the viscosity-reducing effect of water at each pressure.

3

Compare the calculated factors across different pressures to see if they increase or decrease.

As pressure increases from

50\text{ MPa}

to

300\text{ MPa}

, the factor of decrease increases from

50,000

to

125,000

.

This allows us to determine the trend in the viscosity-reducing effect as pressure increases.

4

Evaluate the hypothesis against this trend and determine the modification.

The hypothesis predicted the effect would become weaker (factor would decrease), but the data shows it becomes stronger (factor increases). Therefore, the hypothesis must be modified to state the effect becomes stronger as pressure increases.

This directly answers the question.

Key Concept

Evaluating and modifying a hypothesis based on experimental results by analyzing relative rates of change in a controlled experiment.

Estimated Time:2m 0s

Question 39Question →

A student proposed the following hypothesis regarding the growth of a bacterial biofilm in a flow chamber:

*Hypothesis*: The biofilm growth rate ( $G$ , in $\mu\text{m/day}$ ) is directly proportional to the bulk nutrient concentration ( $C$ , in $\text{mg/L}$ ) across all concentrations, and the rate of decrease in $G$ per unit increase in fluid shear stress ( $\tau$ , in $\text{Pa}$ ) is constant.

To test this hypothesis, the student conducted two experiments. In Experiment 1, the student varied $C$ while maintaining a constant $\tau$ of $0.10\text{ Pa}$ . In Experiment 2, the student varied $\tau$ while maintaining a constant $C$ of $10.0\text{ mg/L}$ . The results are shown in the tables below.

**Experiment 1 ( $\tau = 0.10\text{ Pa}$ )**

Bulk Nutrient Concentration ( $C$ , mg/L)	Biofilm Growth Rate ( $G$ , $\mu\text{m/day}$ )
2.0	0.5
4.0	1.0
8.0	2.0
16.0	2.0

**Experiment 2 ( $C = 10.0\text{ mg/L}$ )**

Fluid Shear Stress ( $\tau$ , Pa)	Biofilm Growth Rate ( $G$ , $\mu\text{m/day}$ )
0.05	2.4
0.10	2.0
0.20	1.5
0.40	1.0

Based on the results of Experiments 1 and 2, which of the following statements describes how the student should modify the hypothesis?

Biofilm growth rate (

G

) is directly proportional to bulk nutrient concentration (

C

) across all concentrations; and the rate of decrease in

G

per unit increase in shear stress (

\tau

) increases as

\tau

increases.

Biofilm growth rate (

G

) is directly proportional to bulk nutrient concentration (

C

) only up to a threshold concentration, after which

G

becomes independent of

C

; and the rate of decrease in

G

per unit increase in shear stress (

\tau

) decreases as

\tau

increases.

Biofilm growth rate (

G

) is independent of bulk nutrient concentration (

C

) at low concentrations but becomes directly proportional at high concentrations; and the rate of decrease in

G

per unit increase in shear stress (

\tau

) remains constant.

Biofilm growth rate (

G

) is directly proportional to bulk nutrient concentration (

C

) only up to a threshold concentration, after which

G

becomes independent of

C

; and biofilm growth rate (

G

) increases as shear stress (

\tau

) increases.

Show answer & explanation

Answer: Biofilm growth rate ( $G$ ) is directly proportional to bulk nutrient concentration ( $C$ ) only up to a threshold concentration, after which $G$ becomes independent of $C$ ; and the rate of decrease in $G$ per unit increase in shear stress ( $\tau$ ) decreases as $\tau$ increases.

Answer

Biofilm growth rate (

G

) is directly proportional to bulk nutrient concentration (

C

) only up to a threshold concentration, after which

G

becomes independent of

C

; and the rate of decrease in

G

per unit increase in shear stress (

\tau

) decreases as

\tau

increases.

The correct option states that the growth rate is directly proportional to the nutrient concentration up to a threshold, and that the rate of decrease per unit increase in shear stress decreases as shear stress increases. This is supported by Experiment 1, which shows a linear increase in growth rate from

0.5

to

2.0\text{ }\mu\text{m/day}

as nutrient concentration increases from

2.0

to

8.0\text{ mg/L}

, followed by a constant growth rate of

2.0\text{ }\mu\text{m/day}

at higher concentrations. Furthermore, Experiment 2 demonstrates that as shear stress increases, the growth rate decreases at a declining rate: the average rate of change decreases in magnitude from

-8.0

to

-5.0

, and then to

-2.5\text{ }\mu\text{m}/(\text{day}\cdot\text{Pa})

across successive intervals.

Step-by-Step Solution

1

Analyze Experiment 1 to determine the relationship between nutrient concentration (

C

) and growth rate (

G

).

G

increases linearly with

C

from

0.5

to

2.0\text{ }\mu\text{m/day}

as

C

goes from

2.0

to

8.0\text{ mg/L}

. However, at

16.0\text{ mg/L}

,

G

remains at

2.0\text{ }\mu\text{m/day}

, indicating that the relationship is directly proportional only up to a threshold concentration of

8.0\text{ mg/L}

.

This establishes how the first part of the hypothesis should be modified to account for nutrient saturation.

2

Analyze Experiment 2 to determine the general trend between shear stress (

\tau

) and growth rate (

G

).

As

\tau

increases from

0.05

to

0.40\text{ Pa}

,

G

decreases from

2.4

to

1.0\text{ }\mu\text{m/day}

, confirming that growth rate decreases as shear stress increases.

This rule-out step eliminates any modifications suggesting a positive relationship between growth rate and shear stress.

3

Calculate the rate of change (slope) of

G

with respect to

\tau

over successive intervals to test the linearity of the decrease.

From

0.05\text{ Pa}

to

0.10\text{ Pa}

, the rate of change is

\frac{2.0 - 2.4}{0.10 - 0.05} = -8.0\text{ }\mu\text{m}/(\text{day}\cdot\text{Pa})

. From

0.10\text{ Pa}

to

0.20\text{ Pa}

, it is

\frac{1.5 - 2.0}{0.20 - 0.10} = -5.0\text{ }\mu\text{m}/(\text{day}\cdot\text{Pa})

. From

0.20\text{ Pa}

to

0.40\text{ Pa}

, it is

\frac{1.0 - 1.5}{0.40 - 0.20} = -2.5\text{ }\mu\text{m}/(\text{day}\cdot\text{Pa})

.

Calculating the rates of change over different intervals evaluates whether the rate of decrease is constant, increasing, or decreasing.

4

Compare the calculated slopes to evaluate how the rate of decrease changes.

The magnitude of the rate of decrease (slopes of

-8.0

,

-5.0

, and

-2.5

) becomes smaller as shear stress increases. This means the rate of decrease in growth rate per unit increase in shear stress decreases.

This determines the correct modification for the second part of the hypothesis.

Key Concept

Evaluating and modifying a hypothesis based on experimental results showing non-linear relationships and saturation thresholds.

Estimated Time:3m 0s

Question 40Question →

A student proposed the following hypothesis regarding the behavior of gases:

*Hypothesis*: Under identical temperature and pressure conditions, gases with a larger molar mass will diffuse through a porous membrane at a faster rate than gases with a smaller molar mass.

To test this hypothesis, the student measured the diffusion time (the time required for a $1.0\text{ L}$ sample of gas to pass completely through a membrane) for four different gases. The results are shown in the table below:

Gas	Molar mass ( $\text{g/mol}$ )	Diffusion time ( $\text{s}$ )
Helium ( $\text{He}$ )	$4$	$12$
Neon ( $\text{Ne}$ )	$20$	$27$
Argon ( $\text{Ar}$ )	$40$	$38$
Krypton ( $\text{Kr}$ )	$84$	$55$

Based on these results, how should the student modify the hypothesis to accurately reflect the relationship between a gas's molar mass and its rate of diffusion?

The student should modify the hypothesis to state that as molar mass increases, the rate of diffusion increases, because the diffusion time decreases.

The student should retain the hypothesis because an increase in molar mass results in an increase in diffusion time, which represents an increase in the rate of diffusion.

The student should modify the hypothesis to state that as molar mass increases, the rate of diffusion decreases, because the diffusion time increases.

The student should retain the hypothesis because the gas with the largest molar mass had the shortest diffusion time, indicating the fastest rate of diffusion.

Show answer & explanation

Answer: The student should modify the hypothesis to state that as molar mass increases, the rate of diffusion decreases, because the diffusion time increases.

Answer

The student should modify the hypothesis to state that as molar mass increases, the rate of diffusion decreases, because the diffusion time increases.

The correct option states that the student should modify the hypothesis to show that as molar mass increases, the rate of diffusion decreases, because the diffusion time increases. A longer diffusion time means the gas takes more time to pass through the membrane, indicating a slower rate of movement. The data shows that as molar mass increases from

4\text{ g/mol}

to

84\text{ g/mol}

, the diffusion time increases from

12\text{ s}

to

55\text{ s}

, which supports this modification.

Step-by-Step Solution

1

Identify the independent variable (molar mass) and the dependent variable (diffusion time) from the data table, and determine their trend.

As the molar mass increases from

4\text{ g/mol}

(Helium) to

84\text{ g/mol}

(Krypton), the diffusion time increases from

12\text{ s}

to

55\text{ s}

.

This establishes the relationship between molar mass and diffusion time.

2

Relate the measured variable (diffusion time) to the concept in the hypothesis (rate of diffusion).

Since rate is inversely proportional to time, a longer diffusion time indicates a slower rate of diffusion.

To evaluate the hypothesis about the rate of diffusion, the student must translate diffusion times into diffusion rates.

3

Compare the trend in the rate of diffusion with the original hypothesis to decide how to modify it.

The original hypothesis states that larger molar mass leads to a faster diffusion rate, but the data shows that larger molar mass leads to a slower diffusion rate (longer time). Therefore, the student must modify the hypothesis to state that rate of diffusion decreases as molar mass increases.

This selects the option that correctly modifies the hypothesis according to the data.

Key Concept

Formulating and modifying hypotheses based on experimental observations and the relationships between measured variables.

Estimated Time:1m 0s

Flask	Temperature ( $^\circ\text{C}$ )	Added $NO_3^-$ ( $\text{mg/L}$ )	Added $PO_4^{3-}$ ( $\text{mg/L}$ )	Atrazine ( $\text{mg/L}$ )
1	20	0.0	0.0	0.0
2	20	5.0	0.0	0.0
3	20	5.0	1.0	0.0
4	20	5.0	1.0	0.1
5	25	5.0	1.0	0.1

Trial	Mitochondria	Succinate (substrate)	ADP	Oligomycin (ATP synthase inhibitor)	FCCP (proton gradient uncoupler)
1	Yes	Yes	No	No	No
2	Yes	Yes	Yes	No	No
3	Yes	Yes	Yes	Yes	No
4	Yes	Yes	Yes	Yes	Yes
5	Yes	No	Yes	No	No
6	No	Yes	Yes	No	Yes

Flask	Temperature ( $^\circ\text{C}$ )	Added $NO_3^-$ ( $\text{mg/L}$ )	Added $PO_4^{3-}$ ( $\text{mg/L}$ )	Atrazine ( $\text{mg/L}$ )
1	20	0.0	0.0	0.0
2	20	5.0	0.0	0.0
3	20	5.0	1.0	0.0
4	20	5.0	1.0	0.1
5	25	5.0	1.0	0.1

Trial	Mitochondria	Succinate (substrate)	ADP	Oligomycin (ATP synthase inhibitor)	FCCP (proton gradient uncoupler)
1	Yes	Yes	No	No	No
2	Yes	Yes	Yes	No	No
3	Yes	Yes	Yes	Yes	No
4	Yes	Yes	Yes	Yes	Yes
5	Yes	No	Yes	No	No
6	No	Yes	Yes	No	Yes

Experimental Design and Scientific Method

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Flask	Temperature ( $^\circ\text{C}$ )	Added $NO_3^-$ ( $\text{mg/L}$ )	Added $PO_4^{3-}$ ( $\text{mg/L}$ )	Atrazine ( $\text{mg/L}$ )
1	20	0.0	0.0	0.0
2	20	5.0	0.0	0.0
3	20	5.0	1.0	0.0
4	20	5.0	1.0	0.1
5	25	5.0	1.0	0.1

Trial	Mitochondria	Succinate (substrate)	ADP	Oligomycin (ATP synthase inhibitor)	FCCP (proton gradient uncoupler)
1	Yes	Yes	No	No	No
2	Yes	Yes	Yes	No	No
3	Yes	Yes	Yes	Yes	No
4	Yes	Yes	Yes	Yes	Yes
5	Yes	No	Yes	No	No
6	No	Yes	Yes	No	Yes