Exercise 1

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National University College *

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104

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Chemistry

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Jan 9, 2024

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Mitchell Spangle PHS 104 LAB EXERCISE 1 -7 1.) What patterns do you observe based on the information in Table 4? a. Based on the table we can see that the fish count goes up along with the PPM of the dissolved oxygen until it reaches what appears to be the max number of fish that can be in the body of water regardless of the Dissolved oxygen. At a ratio of 12/15 O2 to fish 2.) Develop a hypothesis relating to the amount of dissolved oxygen measured in the water sample and the number of fish observed in the body of water. a. I hypothesis that the body of water will only allow a maximum of 15 fish regardless of the dissolved oxygen. 3.) What would your experimental approach be to test this hypothesis? a. Have a set body of water a 10 gallon fish tank everything being equal. 4.) What would be the independent and dependent variables? a. The dependent variable would be the type of fish, water temp, enclosure. b. The independent variable would be the adders of oxygen to the water in one and fish to the other. 5.) What would be your control? a. A 3 rd tank with 13 fish in it at 10(PPM) of dissolved oxygen based on the chart this is balanced before the fish start to fall of 6.) What type of graph would be appropriate for this data set? Why? a. A plot/line chart would be best to show a comparison of the independent variables added and results 7.) 0 2 4 6 8 10 12 0 2 4 6 8 10 12 Chart Title Dissolved oxygen Number of Fish --- EXERCISE 2 --- 1.) Testable a. This would be quantitative. b. Hypothesis: Placing a plant on a windowsill result in it growing three inches faster per day compared to when it is placed on a cofee table in the middle of the living room. Public

Mitchell Spangle PHS 104 LAB EXERCISE 1 -7 i. Null Hypothesis: There is no significant diference in the growth rate of a plant when placed on a windowsill compared to when it is placed on a cofee table in the middle of the living room. c. This I believe would be a simple experiment to test. Take the same plan in the same pot with the same soil adding the same amount of water equally one in the window and on on a cofee table and record the results. d. The Dependent variables would be the constants water, pot size, soil composition. i. The independent variables would be the amount of sunlight each one get, the amount of UV rays each one gets as the window “U factor” would have to be taken into account as well. Along with the weather, what time of year it is can be set but the weather itself is hard to predict. e. Data can be collected in a few ways but perhaps the most efficient way would be to set up two highspeed cameras with a vertical ruler or measuring tap stood separately but next to each plant as to interfere with plants light source. And record the plants growth. f. I would present data with a chart as follows. i. (Please note that these numbers are made up and do not represent actual data) ii. 1 2 3 4 5 6 7 8 9 10 0 5 10 15 20 Chart Title Days Plant A (window) growth Plant B (Table) growth g. I would analyze this data using the highspeed camera doing a time lapse to track and progress to growth of the plants over time and char that time like the example above 2.) Testable a. This would be quantitative. b. Hypothesis: Putting the teller at the bank with brown hair against the other tellers would show that the teller with the brown hair is taller i. Null Hypothesis: there is no measurable diference between the tellers based on high c. I would line up all the tellers against a measuring stick all the tellers would be without shoes or socks and the measurement would be taken from the top of the head not the top of the hair. d. The dependent variables would be the level floor and same messureing stick, i. The independent variable would be the height that the teller would be. e. We can collect this date simply by lining up the tellers and recording their heights f. A simple bar graph would be the best way to illustrate this. g. I would analyze the data collected via the measurements and record it. 3.) I don’t think this is testable, there are to many variables. Public

Mitchell Spangle PHS 104 LAB EXERCISE 1 -7 4.) I’m not sure if its testable or not technically it is but you would only need to use simple observation. a. This would be quantitative. b. Hypothesis: the Italian restaurant across the street will close at 9pm as opposed to the one two blocks away. c. Simple observation can test this by having one person stand out side each restaurant with synchronized watches and observe what time each one closes. Or you could call each restaurant and ask them when the close. d. The dependent variables would be the time zone, synchronized watches or a phone connection based on what method yours going to use to test. i. The independent variable would be if a special even would to be going on at one or both restaurants. e. We can collect this data with simple observation and record what is observed. f. I would use a simple bar graph i. Place A Place B 8:24 PM 8:38 PM 8:52 PM 9:07 PM 9:21 PM 9:36 PM 9:50 PM 10:04 PM 10:19 PM Close time g. I would analyze the data collected via the observations and record it. 5.) Not testable 6.) Not testable --- EXERCISE 3 --- 1.) Start with the given value: 46.0 g. Then divide the given value by 1000 to convert grams to kilograms: 46.0 g / 1000 = 0.046 kg. Therefore, 46.0 grams is equal to 0.046 kilograms. 2.) Start with the given value: 57 seconds. Then divide the given value by 60 to convert seconds to minutes: 57 seconds / 60 = 0.95 minutes. Therefore, 57 seconds is equal to 0.95 minutes. Public

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Mitchell Spangle PHS 104 LAB EXERCISE 1 -7 3.) Start with the given value: 13.5 cm. Then divide the given value by 2.54 to convert centimeters to inches: 13.5 cm / 2.54 = 5.31 inches (rounded to two decimal places). Therefore, 13.5 centimeters is approximately equal to 5.31 inches. 4.) Start with the given value: 47 °F. Then subtract 32 from the given value: 47 °F - 32 = 15 °F. Then divide the result by 1.8 to convert Fahrenheit to Celsius: 15 °F / 1.8 ≈ 8.33 °C (rounded to two decimal places). Therefore, 47 °F is approximately equal to 8.33 °C. --- EXERCISE 4 --- 1.) Both 2.) Accurate 3.) Accurate 4.) Accurate 5.) Neither --- EXERCISE 5 --- 1.) The number 405000 has five significant digits: 4, 0, 5, 0, and 0 2.) The number 0.0098 has two significant digits: 9 and 8. 3.) The number 39.999999 has eight significant digits: 3, 9, 9, 9, 9, 9, 9, and 9. 4.) The number 13.00 has four significant digits: 1, 3, 0, and 0. 5.) The number 80,000,089 has eight significant digits: 8, 0, 0, 0, 0, 0, 0, and 9. Public

Mitchell Spangle PHS 104 LAB EXERCISE 1 -7 6.) The number 55,430.00 has six significant digits: 5, 5, 4, 3, 0, and 0. 7.) The number 0.000033 has two significant digits: 3 and 3. 8.) The number 620.03080 has eight significant digits: 6, 2, 0, 0, 3, 0, 8, and 0 --- PART 2 --- 1.) The number 70,000,000,000 can be written in scientific notation as 7.0 x 10 ^10 2.) The number 0.000000048 can be written in scientific notation as 4.8 x 10 ^-8 3.) The number 67,890,000 can be written in scientific notation as 6.789 x 10 ^7 4.) The number 70,500 can be written in scientific notation as 7.05 x 10 ^4 5.) The number 450,900,800 can be written in scientific notation as 4.509008 x 10 ^8 6.) The number 0.009045 can be written in scientific notation as 9.045 x 10 ^-3 7.) The number 0.023 can be written in scientific notation as 2.3 x 10 ^-2 --- EXERSICE 6--- 1.) In this scenario, the son guessed that the dad was holding 81 cents, but the dad actually had 90 cents. First, we calculate the diference between the guessed value and the actual value: |81 - 90| = 9 Then, we divide the absolute diference by the actual value: 9 / 90 = 0.1 Next, we multiply the result by 100 to find the percent error: 0.1 x 100 = 10 The percent error is 10%. Public

Mitchell Spangle PHS 104 LAB EXERCISE 1 -7 Since the percent error is greater than the allowed 5% error, the son did not guess close enough to get the money. 2.) In this scenario, Jennifer and Johnny determined the velocity of the car to be 34.87 m/s, while the actual velocity is 34.15 m/s. First, we calculate the diference between the measured value and the actual value: |34.87 - 34.15| = 0.72 Then, we divide the absolute diference by the actual value: 0.72 / 34.15 = 0.0211 Next, we multiply the result by 100 to find the percent error: 0.0211 x 100 = 2.11 The percent error is 2.11%. Since the percent error is less than the allowed 2.5% error, Jennifer and Johnny will pass their final project. Their measured value is within the acceptable range of error set by the teacher. 3.) In this scenario, the expected trip duration is 3.15 hours, but the actual duration is 3.26 hours. First, we calculate the diference between the measured value and the actual value: |3.26 - 3.15| = 0.11 Then, we divide the absolute diference by the actual value: 0.11 / 3.15 = 0.0349 Next, we multiply the result by 100 to find the percent error: 0.0349 x 100 = 3.49 The percent error is 3.49%. Since the percent error is slightly higher than the allowed 3% error, the train company did not live up to its reputation on this trip. The actual trip duration exceeded the expected duration by more than the specified margin of error. 4.) In this scenario, Tommy ended the season with a batting average of 0.258, while the desired batting average, within a 7% error, is 0.275. First, we calculate the diference between the measured value and the actual value: |0.258 - 0.275| = 0.017 Then, we divide the absolute diference by the actual value: 0.017 / 0.275 = 0.0618 Next, we multiply the result by 100 to find the percent error: 0.0618 x 100 = 6.18 The percent error is 6.18%. Since the percent error is greater than the allowed 7% error, Tommy did not have a great season according to his coach's criteria. His batting average falls below the desired range, indicating that he did not meet the coach's expectation. --- EXERCISE 7 --- 1.) A. The independent variable in this experiment is the habitat space, which refers to the diferent sized aquariums used for the fish populations. B. The dependent variable in this experiment is the number of surviving fish. This variable is dependent on the habitat space and will be measured at the end of the experiment. Public

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Mitchell Spangle PHS 104 LAB EXERCISE 1 -7 C. The controls in this experiment include: - Same type and amount of food: By keeping the food consistent across all aquariums, the researchers ensure that the only diference among the habitats is the space available. - Equally maintained and cleaned aquariums: Maintaining and cleaning the aquariums equally ensures that the cleanliness and conditions within the habitats do not afect the fish populations diferently. - Constant water temperature: By keeping the water temperature constant, the researchers eliminate the variable of temperature as a potential factor influencing the fish populations. 2.) A. The independent variable in this experiment is the habitat space, which refers to the diferent sized aquariums used for the fish populations. B. The dependent variable in this experiment is the number of surviving fish. This variable is dependent on the habitat space and will be measured at the end of the experiment. C. The controls in this experiment include: - Same type and amount of food: By keeping the food consistent across all aquariums, the researchers ensure that the only diference among the habitats is the space available. - Equally maintained and cleaned aquariums: Maintaining and cleaning the aquariums equally ensures that the cleanliness and conditions within the habitats do not afect the fish populations diferently. - Constant water temperature: By keeping the water temperature constant, the researchers eliminate the variable of temperature as a potential factor influencing the fish populations. Public