ACT Science : How to find synthesis of data in chemistry

Study concepts, example questions & explanations for ACT Science

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Example Questions

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Example Question #831 : Act Science

An undergraduate biology student working in a lab was reading about the latest health epidemic in the United States concerning obesity. He recently learned in his biology class that an enzyme called fatty acid synthase (FAS) catalyzes the formation of fatty acids. Accumulation of these fatty acids then creates the adipose fatty tissue that individuals typically see in their belly or side-regions. He hypothesizes that by decreasing the rate that this enzyme produces fatty acids by administering an inhibitor will aid in reducing the storage of fats, helping to alleviate obesity. After gaining approval from the university's clinical trials and ethics committee, he gathered a group of 20 test subjects. With the help of his research director, he measured the baseline rate of FAS activity for all test subjects and the averaged rate after oral administrators of a placebo and three types of inhibitors at a concentration of 0.6M to four equally-sized groups. The data is shown in Table 1.                      

Questionpicture2

Table 1.

The student then wonders if a mixture of different inhibitors were to increase the effect in the subject. He again measured the averaged baseline FAS level of the subjects and then attempted administering a placebo and three different combinations of inhibitors to subjects in a second trial. The results are shown in Table 2.

Questionpicture3

Table 2. 

A colleague of the researcher interpreted the results from Table 1 and 2 and suggested a third trial which would track the averaged weight change of the subjects over a 6-month period. The results are shown in Table 3.

Questionpicture4

Table 3.

Suppose that the researcher improperly mixed the Inhibitor II + III pills in the third experiment with Inhibitor I + III pills. What would be a reasonable change in weight that would be expected?

Possible Answers:

Correct answer:

Explanation:

The FAS activity level of Inhibitor I + III closely mimic the activity of the placebo group, in which the average weight gain was 3.71 kg. The closest answer to this would be .

Example Question #51 : How To Find Synthesis Of Data In Chemistry

The Millikin oil drop experiment is among the most important experiments in the history of science. It was used to determine one of the fundamental constants of the universe, the charge on the electron. For his work, Robert Millikin won the Nobel Prize in Physics in 1923.

Millikin used an experimental setup as follows in Figure 1. He opened a chamber of oil into an adjacent uniform electric field. The oil droplets sank into the electric field once the trap door opened, but were then immediately suspended by the forces of electricity present in the field.

Figure 1:

Millikin

By determining how much force was needed to exactly counteract the gravity pulling the oil droplet down, Millikin was able to determine the force of electricity. This is depicted in Figure 2.

Using this information, he was able to calculate the exact charge on an electron. By changing some conditions, such as creating a vacuum in the apparatus, the experiment can be modified. 

Figure 2:

Millikin_drop

When the drop is suspended perfectly, the total forces up equal the total forces down. Because Millikin knew the electric field in the apparatus, the force of air resistance, the mass of the drop, and the acceleration due to gravity, he was able to solve the following equation: 

Table 1 summarizes the electric charge found on oil drops in suspension. Millikin correctly concluded that the calculated charges must all be multiples of the fundamental charge of the electron. A hypothetical oil drop contains some net charge due to lost electrons, and this net charge cannot be smaller than the charge on a single electron.

Table 1: 

Trial #

Electric Charge Calculated in Coulombs (C)

Vacuum Used?

1

1.602176487 x 10-8

No

2

1.602176487 x 10-2

Yes

3

1.602176487 x 10-6

No

4

1.602176487 x 10-4

Yes

 

The decision to use oil drops in Millikin's experiment was made after water drops were found to evaporate too quickly.  

Another scientist conducts an experiment and places two beakers into sealed chambers within a single larger, constant pressure chamber, as in the figure below. The pressure and temperature conditions in the chambers are all very similar to the conditions used in the Millikin experiment from the passage, without a vacuum.  

One beaker contains oil, and one beaker contains water. The volumes of both beakers are identical. Based on the information in the original passage, which of the following is most likely true:

Oil_water

Possible Answers:

There will be a greater amount of water vapor at point 2 than oil vapor at point 1.

We cannot predict which point will have a greater amount of vapor without knowing the absolute pressure in the chamber.

There will be a smaller amount of water vapor at point 2 than oil vapor at point 1.

There will be about equal amounts of water vapor at point 2 as oil vapor at point 1.

Correct answer:

There will be a greater amount of water vapor at point 2 than oil vapor at point 1.

Explanation:

The question specifies that water was found to evaporate too quickly. This suggests that water will evaporate more quickly in the similar conditions established in the question. Water vapor will, therefore, likely be more abundant than oil vapor above their respective beakers.

Example Question #51 : How To Find Synthesis Of Data In Chemistry

The Millikin oil drop experiment is among the most important experiments in the history of science. It was used to determine one of the fundamental constants of the universe, the charge on the electron. For his work, Robert Millikin won the Nobel Prize in Physics in 1923.

Millikin used an experimental setup like the one shown in Figure 1. He opened a chamber of oil into an adjacent uniform electric field. The oil droplets sank into the electric field once the trap door opened, but were then immediately suspended by the forces of electricity present in the field.

 

Figure 1

Millikin

 

 

By determining how much force was needed to exactly counteract the gravity pulling the oil droplet down, Millikin was able to determine the force of electricity. This is depicted in Figure 2.

Using this information, Millikin was able to calculate the exact charge on an electron. By changing some conditions, such as creating a vacuum in the apparatus, the experiment can be modified. 

 

Figure 2

Millikin_drop

When the drop is suspended perfectly, the total forces up equal the total forces down. Because Millikin knew the electric field in the apparatus, the force of air resistance, the mass of the drop, and the acceleration due to gravity, he was able to solve the following equation: 

Table 1 summarizes the electric charge found on oil drops in suspension. Millikin correctly concluded that the calculated charges must all be multiples of the fundamental charge of the electron. A hypothetical oil drop contains some net charge due to lost electrons, and this net charge cannot be smaller than the charge on a single electron.

Table 1

Trial #

Electric Charge Calculated in Coulombs (C)

Vacuum Used?

1

1.602176487 x 10-8

No

2

1.602176487 x 10-2

Yes

3

1.602176487 x 10-6

No

4

1.602176487 x 10-4

Yes

 

A stronger electric field is needed to keep the oil drop in the field when a vacuum is used. This is most likely due to:

Possible Answers:

the effect of air resistance holding the oil drop in suspension.

the effect of air resistance pulling the oil drop out of suspension.

both the effect of air resistance pulling the oil drop out of suspension and the effect of extra pressure in the vacuum pulling the oil droplet out of suspension.

the effect of extra pressure in the vacuum pulling the oil drop out of suspension.

Correct answer:

the effect of air resistance holding the oil drop in suspension.

Explanation:

The question implies that the use of a vacuum removes some force that "assists" the electric charge in holding the droplet in suspension. What force is removed when a vacuum is used? A vacuum removes air resistance, one of the concepts mentioned in the answer choices. Because air resistance is removed when a vacuum is used, it is likely that the air resistance helps the droplet stay in suspension.

Other choices specify the opposite. They claim that in the absence of a vacuum, the oil drops fall out of suspension more quickly.

Example Question #51 : How To Find Synthesis Of Data In Chemistry

A student performed the following procedures to study various photosynthetic pigments (light-absorbing chemicals) in tree leaves and the wavelengths of light they absorb.

Experiment 1:

The student obtained samples of leaves from oaks, maples, ashes, sycamores, and poplars. Each leaf sample was ground separately with a mortar and pestle to release the pigments, and then each sample was suspended in water to make a colored solution of the pigment. The student then measured the absorption spectrum (a graph of how much light is absorbed by a pigment at varying wavelengths of light) of each solution in a device called a spectrophotometer. The setup of a spectrophotometer is shown below in Diagram 1.

 

 Spectrophotometer

 

The light source emits white light, which is split into its various wavelengths by the prism. Next, a slit, which can be moved up or down to select a particular wavelength, is used to transmit just a single wavelength to the sample. The sample absorbs a fraction of this light that is characteristic to the pigment in the sample, and the rest is transmitted to the detector for a readout. Using the spectrophotometer, the student found the λmax (the wavelength of light in nanometers (nm) that the pigment absorbs most intensely, for each sample) and recorded the results in Table 1. Table 1 also shows the transmittance and absorbance values at λmax. Transmittance, T, is defined as the fraction of light, expressed as a decimal, which passes through the sample. Absorbance, A, is given by:

                                    A = –log(T)   or  10–A = T

 

 Spectrophotometer_table_1

 

Experiment 2:

A student is given a leaf from an unknown source. She crushes and extracts the pigment according to the procedure in Experiment 1. Measuring the absorbance spectrum in the spectrophotometer produces the following readout, shown in Diagram 2.

       Leaf_absorbance

                                                     Diagram 2

A lumberjack wants to plant trees that make the most efficient use of sunlight in a shady area of her land. Which trees should she plant?

Possible Answers:

Sycamore

Popular

Maple

Ash

Correct answer:

Maple

Explanation:

Efficient use of sunlight here can be interpreted as high absorbance. As Table 1 shows, maple leaves have the highest absorbance at the λmax value.

Example Question #831 : Act Science

A student performed the following procedures to study various photosynthetic pigments (light-absorbing chemicals) in tree leaves and the wavelengths of light they absorb.

Experiment 1:

The student obtained samples of leaves from oaks, maples, ashes, sycamores, and poplars. Each leaf sample was ground separately with a mortar and pestle to release the pigments, and then each sample was suspended in water to make a colored solution of the pigment. The student then measured the absorption spectrum (a graph of how much light is absorbed by a pigment at varying wavelengths of light) of each solution in a device called a spectrophotometer. The setup of a spectrophotometer is shown below in Diagram 1.

 

 Spectrophotometer

 

The light source emits white light, which is split into its various wavelengths by the prism. Next, a slit, which can be moved up or down to select a particular wavelength, is used to transmit just a single wavelength to the sample. The sample absorbs a fraction of this light that is characteristic to the pigment in the sample, and the rest is transmitted to the detector for a readout. Using the spectrophotometer, the student found the λmax (the wavelength of light in nanometers (nm) that the pigment absorbs most intensely, for each sample) and recorded the results in Table 1. Table 1 also shows the transmittance and absorbance values at λmax. Transmittance, T, is defined as the fraction of light, expressed as a decimal, which passes through the sample. Absorbance, A, is given by:

                                    A = –log(T)   or  10–A = T

 

 Spectrophotometer_table_1

 

Experiment 2:

A student is given a leaf from an unknown source. She crushes and extracts the pigment according to the procedure in Experiment 1. Measuring the absorbance spectrum in the spectrophotometer produces the following readout, shown in Diagram 2.

 Leaf_absorbance

                                                Diagram 2

In Experiment 2, at which of the following wavelengths, in nanometers, is the most amount of light transmitted through the sample?

Possible Answers:

450

400

350

500

Correct answer:

500

Explanation:

As the equations in the description of Experiment 1, as well as the patterns in the data in Table 1 indicate, a low absorbance provides a high transmittance and vice-versa. Thus, to find what wavelength gives the highest value of transmittance, we must look for the lowest absorbance value. As 500 nm gives an absorbance of 0, it will give a transmittance of 100%, the highest possible value.

                             A = 0,           10–0 = T = 1, or 100%

Example Question #51 : How To Find Synthesis Of Data In Chemistry

A student performed the following procedures to study various photosynthetic pigments (light-absorbing chemicals) in tree leaves and the wavelengths of light they absorb.

Experiment 1:

The student obtained samples of leaves from oaks, maples, ashes, sycamores, and poplars. Each leaf sample was ground separately with a mortar and pestle to release the pigments, and then each sample was suspended in water to make a colored solution of the pigment. The student then measured the absorption spectrum (a graph of how much light is absorbed by a pigment at varying wavelengths of light) of each solution in a device called a spectrophotometer. The setup of a spectrophotometer is shown below in Diagram 1.

 

 Spectrophotometer

 

The light source emits white light, which is split into its various wavelengths by the prism. Next, a slit, which can be moved up or down to select a particular wavelength, is used to transmit just a single wavelength to the sample. The sample absorbs a fraction of this light that is characteristic to the pigment in the sample, and the rest is transmitted to the detector for a readout. Using the spectrophotometer, the student found the λmax (the wavelength of light in nanometers (nm) that the pigment absorbs most intensely, for each sample) and recorded the results in Table 1. Table 1 also shows the transmittance and absorbance values at λmax. Transmittance, T, is defined as the fraction of light, expressed as a decimal, which passes through the sample. Absorbance, A, is given by:

                                    A = –log(T)   or  10–A = T

 

 Spectrophotometer_table_1

 

Experiment 2:

A student is given a leaf from an unknown source. She crushes and extracts the pigment according to the procedure in Experiment 1. Measuring the absorbance spectrum in the spectrophotometer produces the following readout, shown in Diagram 2.

       Leaf_absorbance

                                                     Diagram 2

A standardized sample of a pigment often found in plant leaves, called Chlorophyll A, was placed into the spectrophotometer and found to have a λmaxof 425 nm. The leaves of which tree most likely have a high concentration of Chlorophyll A?

Possible Answers:

Popular

Maple

Sycamore

Oak

Correct answer:

Maple

Explanation:

Table 1 shows that maple leaves have a λmaxat 425 nm. Thus, we can assume that this is most likely caused by the presence of Chlorophyll A as it has the same λmax, and the value of λmax is characteristic of a particular pigment.

Example Question #151 : Chemistry

A student wanted to study the kinetics, or rates of a chemical reaction based on the concentrations of its reactants and products, of the reaction shown below.

 

This reaction is easy to monitor using a spectrophotometer, which measures how much light of a particular wavelength is absorbed by a solution. The deep purple potassium permanganate, or , absorbs light of a 550 nm wavelength in proportion to its concentration in the reaction solution. Manganese sulfate, or , is pale pink and absorbs light of a 500 nm wavelength in proportion to its concentration in the reaction solution. All other reactants and products are colorless and do not absorb visible light and thus cannot be monitored using the spectrophotometer.

 

Experiment 1:

The student constructed a standard curve, or a graph of the absorbance of solutions of varying concentrations of potassium permanganate, to quantify the relationship between concentration and absorbance. To prepare five sample of increasing concentration, he labeled five test tubes A, B, C, D, and E, weighed out 0.1, 0.2, 0.3, 0.4, and 0.5 grams of potassium permanganate into each, respectively, and added 1 milliliter (mL) of water to each test tube to dissolve. Then, he used the spectrophotometer to determine the absorbance at 550 nm of each sample. The data is graphed in Figure 1 below.

Kinetics_figure_1 

                                                        Figure 1

  

Experiment 2: 

The student then studied potassium permanganate in the presence of oxalic acid, , to observe the reaction. Monitoring both the absorbances of potassium permanganate and manganese sulfate, he was able to determine the reaction rate using a special setting on the spectrophotometer. The reaction rate at various concentrations of reactants is shown below in Table 1.

Kinetics_table_2

The rate constant, , for a chemical reaction involving two reactants is given by the following equation: 

What is the rate constant for the reaction studied in Experiment 2?

Possible Answers:

Correct answer:

Explanation:

We can obtain the value of the rate constant by plugging in values to this equation. Using the top row of data of Table 1, we can see that a rate of 2.5 occurs when concentrations of both potassium permanganate and oxalic acid are 0.5 grams/ml. Plugging this into our equation, we see that

or,

.

By dividing, we see that  must be 10.

Example Question #831 : Act Science

A student wanted to study the kinetics, or rates of a chemical reaction based on the concentrations of its reactants and products, of the reaction shown below.

 

This reaction is easy to monitor using a spectrophotometer, which measures how much light of a particular wavelength is absorbed by a solution. The deep purple potassium permanganate, or , absorbs light of a 550 nm wavelength in proportion to its concentration in the reaction solution. Manganese sulfate, or , is pale pink and absorbs light of a 500 nm wavelength in proportion to its concentration in the reaction solution. All other reactants and products are colorless and do not absorb visible light and thus cannot be monitored using the spectrophotometer.

 

Experiment 1:

The student constructed a standard curve, or a graph of the absorbance of solutions of varying concentrations of potassium permanganate, to quantify the relationship between concentration and absorbance. To prepare five sample of increasing concentration, he labeled five test tubes A, B, C, D, and E, weighed out 0.1, 0.2, 0.3, 0.4, and 0.5 grams of potassium permanganate into each, respectively, and added 1 milliliter (mL) of water to each test tube to dissolve. Then, he used the spectrophotometer to determine the absorbance at 550 nm of each sample. The data is graphed in Figure 1 below.

Kinetics_figure_1 

                                                        Figure 1

  

Experiment 2: 

The student then studied potassium permanganate in the presence of oxalic acid, , to observe the reaction. Monitoring both the absorbances of potassium permanganate and manganese sulfate, he was able to determine the reaction rate using a special setting on the spectrophotometer. The reaction rate at various concentrations of reactants is shown below in Table 1.

Kinetics_table_2

Reaction rate is given by the following equation:

Imagine that  for this particular reaction. If a student uses concentrations of 2.0 grams/mL for both of the reactants, what should he expect the rate to be?

Possible Answers:

Correct answer:

Explanation:

As we saw in Question 3, the rate is given by:

We are told in the question stem to assume that k=24 for this particular reaction. If we plug in 2 for both reactant concentrations, we get:

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