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The Virtual Child Project Paper

My Virtual Life Project PaperYour ID numberYour nameO O university※What are some behaviors or choices that you have a tendency to do (habits, perceptions, stress or other health-detrimental aspects of your life) that you might need to work on? How might these serve as obstacles to later life success?I think that this is very important that I feel satisfaction that I did something well. So If I think I did badly, I try again, again, again, until I feel satisfaction. I know my habit is not good for my health, because I have more stress, but I could not stop working. I need more time than other people, because I want to be a perfect person.※What parts of development do you predict might stay the same as you move into and through your adulthood years? What might influence this stability as you mature?I think that physical development might be the same as now late adulthood, because really I like exercise, so I think my body will stay strong. I understand exactly that health is very importaped my virtual children because when I was young, really I did not like to read books, but I tried to read many books to my babies. My children really like to read books. I gave them a lot of free time. This means I was not very strict with them. So they grew up to become very friendly and smart people.※Do you see yourself as the kind of person that will stay in the same job for a long time, perhaps into retirement, or as one of a job climber who will jump from one position to the next in order to climb the professional ladder? What factors would you be likely to prioritize as you make such decisions? What do theories of career development predict in terms of personal growth and achievement of career goals?I like challenge so I want to jump up from one position on the professional ladder to the next. Also I do not want to work in the same place for a long time. I would like to try many different jobs. I consider that what I want, what I need, and what I can do. Therefore, switching- an, family experiences, and goals, how do you think your work history will play into your transition into and through retirement, as you forecast into the later adulthood years? Perhaps you might consider financial factors (social security or retirement savings planning) or preferences to stay employed part-time in your response.I love my job, and I still enjoy my work. I know that I might reach my transition time in later adulthood, and at the time I will think about changing my job and work. I will never stop working. If I will get sick or when I get older, I will try to retire from my work. Hopefully I do not have to consider financial problems because I have enough money from working. I just love my work. My family understands me, because they know me how much I love my job. I am sure I will not regret what I did in my life.※What do you think might lead some to experience a full-on crisis, while others experience a mild crisis or change just one thing in an effort to accomplish a wore needs of two generations of family members along with your own needs as a working, married individual in your own home, with your own bills, relationships, goals, and plans?I will pay someone who helps to take care my family, so I never retire my job. I would reserve one night a week to go to dinner with my friends. Others night I would spend with my family.※How can involvement in civic or religious activity buffer you against stress effects? Give some examples from your personal life.Sometimes I have some stress from work, and people. But to relieve the stress I go to church and meet many friends. We talk about our daily experiences and share our feeling and news. Also I have a meeting every Wednesday with my university friends. If I have some stress, I just meet my friends and sing a song with my husband.※Why might your friends (or even you) continue to work past the age of retirement?The first reason is they really have pride for their work, because they like their job. Second, th spectators; (c) adventurers; (d) searchers; (e) easy gliders; (f) retreaters. Which of these paths seem most probable for you? Why?I think I am an adventurer because I like adventures. Also, if I retire from on job, I will find a new one. I like to learn new things, so I will try to find new ways. And sometimes I will go travel to other countries, and I will not follow someone, I will just follow my self. If I want to try something, I will do this.※Some of the best predictors of successful aging are an individual’s general negativity towards life, as well as the sense of control or one’s ability to adapting to life’s events (expected or not!). Looking back over your virtual life course, what sorts of experiences could contribute to successful aging, and which could have put you at risk for poor transition through the final stages of life?I spent a lot of time with my family, and sometimes I met my friends. This helps me keep more deep relationships with my family. We also went a lot o

교육학| 2012.12.14| 8페이지| 3,500원| 조회(130)

인간발달, human development, Pearson, virtual c..

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Spectrophotometry

Lab #4SpectrophotometryYour nameChemistry LabInstructor:Lab time:Date:Conclusion:In this lab knowledge was obtained by analyzing different substances using the MicroLAB program. Different colored substances were analyzed and then mixed together. Using selective absorption of light determined the concentration of the differing substances and applied that data to the different brands of tonic water. We also compared the fluorescence between the concentration of unknown substance and Canada Dry tonic water. We measured percent transmittance, convert that quantity to absorbance, and then use absorbance to find the concentration of unknown solution. Then, we should to use Beer’s law.What is Beer’s law? Beer explained about the relationship of the absorption of light to the properties of the material through which the light is traveling. Also he made the equation:A = a b CA is the solution’s absorbancea is a constant called the molar absorptivity constantb is the path lengthC is the solutionionSo we wanted to know that if beer’ law is correct or not in this experiment.Through this experiment, we found out Beer’s law is true. Because the absorbance’ linear goes up at the same time, concentration’s linear also goes up.The next part of the lab required mixing the different analyzed colors together. This first color mixed was blue with yellow. It was observed that when the substances were mixed, the resulting colors absorbed were violet, blue, and orange. The colors that were transmitted were that of red and green. The next colors mixed were purple and yellow. The most absorbed colors in this mixture were violet, blue, and a little green. The transmitted colors were yellow, orange and red. The next colors mixed were purple and blue. They absorbed green, yellow, and blue. The transmitted colors of this mix were observed as red, and indigo (dark blue). Through this experiment, we could know about the difference between absorbance and transmittance. Transmittance is the fractiont light at a specified Hyperlink "http://en.wikipedia.org/wiki/Wavelength" o "Wavelength" wavelength that passes through a sample. Absorbance is the fraction of radiation absorbed by a sample at a specified wavelength. (← what is the difference between absorbance and transmittance?)The next part of the lab was to obtain the red stock solution and take measurements based on diluting the solution to observe how the absorption and transmittance levels differ. As the concentration level decreased, it was observed that the transmittance level increased. Since more light is able to pass the diluted solution the absorption level decreases. The exact opposite occurs when the solution has more concentration. The worksheet that was turned in during the lab shows the mathematical equations used in determining the answers. From the calculations, it was observed that the optimum wavelength for the given color is 502 nanometers. So we found out the concentration of red flood color solution from the tool, because absorbance has the steepest slope than the other tools. (← how can we get the concentration of red flood color solution from this experiment?)What is fluorescence and how does it work? Fluorescence is the emission of light by a substance that has absorbed light or other Hyperlink "http://en.wikipedia.org/wiki/Electromagnetic_radiation" o "Electromagnetic radiation" electromagnetic radiation. We figure out the color according to emission spectrum of fluorescence, because fluorescence differs from emission spectra. The rotation and vibration in molecules broadens each energy level, and one observes a band of emitted color instead of a sharp line as in gaseous emission spectra.The next part of the lab was to create wavelength graphs for known tonic water and unknown tonic water samples, and diluting as done in the previous color experiment. The transmittance level of the known tonic water at 100% concentration was 55.7 and as it was diluted with the DI water the transmittadropped to .25. The unknown tonic water at 100% concentration had a transmittance level of 38.1 and absorbance level of .44. As the transmittance levels increased the absorption numbers dropped.It was concluded in this lab that as each of the tonic waters had similar transmittance and absorbance levels, the unknown sample decreased and increased at a faster rate than the known sample. The known tonic sample had a higher level of quinine than the unknown.What is difference between the colored light and colored pigment? Colored light can pass through every substance. But colored pigment reflects every light. This means colored pigment is unable to absorb the light.Work Cite“Transmittance.” Wikipedia. 2010. < http://en.wikipedia.org/wiki/Transmittance>“Beer–Lambert law.” Wikipedia. 2008.< Hyperlink "http://en.wikipedia.org/wiki/Beer%E2%80%93Lambert_law" l "Equations" http://en.wikipedia.org/wiki/Beer%E2%80%93Lambert_law#Equations>“Fluorescence.” Wikipedia. < http://en.wikipedia.org/wiki/e>

자연과학| 2012.12.14| 4페이지| 2,500원| 조회(84)

Spectrophotometry, Fluorescence, Quantitat..

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Spectroscopy

Lab #3SpectroscopyYour nameChemistry LabInstructor:Lab time:Date:Conclusion:In this lab knowledge of the property of waves was gained. Wavelength, frequency, and interference were observed with visually watching the metal spring move in different ways. When the spring was rapidly moved (high frequency), the wavelength produced a short wave. The exact opposite was observed when the spring was moved slower, creating a low frequency, long wavelength. Constructive interference was observed when the two waves of the spring began at opposite ends (starting at the same side of the spring) and the waves met in the middle. When the waves began at opposite ends of the spring, the waves crossed the middle creating constructive interference. The higher the frequency of the wave, the shorter the wavelength of the standing wave occurs.Bohr believed orbit has a constant line (fixed distances). The electrons only can go constant line. He thought the energy absorb when electrons jump up, and during electrons go down, the energy emits. Also Bohr’s Hydrogen atom model was accurately coincided.The next part of the lab was to use the spectroscope to view different light objects. Our eyes just can see “visible lights (rainbow)”. This mean is that our eyes cannot see most of lights, so spectroscope helps us to see other lights. As the frequency increased looking at the incandescent light bulb for example, the wavelength decreased. Starting from low to high frequency, the colors red, orange, yellow, green, blue, and violet were observed. The florescent light when observed through the spectroscope produced the colors, red, orange, green, blue, indigo, and violet. The reason the color are different is because of the different reaction the ions were creating.The next part of lab was to observe gas discharge tubes using the spectroscope. The compounds observed were He, Ar, Ne, and Hg bulbs. Each element created a different spectrum of light as noted in the attached worksheet. When the electrons are excited to different levels, it produces a varying light spectrum, which makes it possible to identify the particular element in question. Each element created its own unique spectrum of light. Looking again through the spectroscope at the florescent light, it was concluded that the element contained in the bulb was mercury. The color spectrum from the florescent light matched the mercury gas discharge tube.Observing a flame and introducing the compounds of nichrome wire, lithium ion, potassium ion, sodium acetate, and sodium chloride to the flame was the next step in the lab. Without the spectroscope, we can see different spectra of light. We were able to determine that the sodium samples identity was dependent on the ions of the particular compound. This was a challenging part of the lab. It was difficult to observe the compounds, because they burnt very rapidly, making if difficult to tell if the spectrum observed was the burning of the compound or light from the flame of the burner or the lighting in the room.When comparing the solid vs. gaseous spectra it was concluded that the flame test (sodium), fluorescent light and Bunsen burner flam only produced a gas spectra. The only light source that produced a solid spectra was the incandescent light bulb, because the light produced is created with a solid filament. The light source from the other sources is created from a gas. When the gas is excited through electricity it produces its own unique fingerprint and compound identification is possible. It was concluded that solids produce a band spectrum and gases produce a line spectrum.What is the difference of band spectrum and line spectrum? Band spectrums are part of optical spectra of polyatomic systems, including condensed materials, large molecules etc. Each line corresponds to one level in the atom splits in the molecules. When the number of atoms is large, one gets a continuum of energy levels, the so called "spectral bands." They are often labeled in the same way as the mono-atomic lines. Line spectrums appear the light like a line. Each atom emits a different wavelength.After setting up the atomic spectrum in MicroLAB with the unknown letter “W” and printing the graph, it was concluded that unknown compound was Mercury. By taking an average of the wavelengths and comparing it to the known mercury we were able to come to this conclusion. By comparing the differing wavelengths and spectrum of lights one is able to identify the same with mercury.Work Cite“Spectral bands.” Wikipedia. 2007.< http://en.wikipedia.org/wiki/Spectral_bands>

자연과학| 2012.12.14| 4페이지| 2,500원| 조회(69)

energy, spectroscopy, Waves, frequency, Ligh..

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The Behavior of Gases-Boyle`s and Charles` law

Lab #8The Behavior of Gases: Boyle’s and Charles’ LawsYour nameChemistry LabInstructor:Lab time:Date:Purpose:The purpose of the lab was to determine the effect of pressure on the volume of a gas, determine the effect of temperature on the volume of a gas, graphically deduce the relationship between P and V and T and V, and to compare the relationships to Boyle’s and Charles’ laws.Procedure & Observation:The first part of the lab was to compare pressure and volume. A syringe was obtained and connected to MicroLAB. The syringe was set to 30 mL. Pressure measurements were taken from 24 mL to 54 mL in 5 ml steps. Each of the steps measured were recorded in MicroLAB.The second part of lab was to observe the effect of temperature on a gas volume. A temperature sensor was obtained and connected to MicroLAB. A new program was opened in MicroLAB to measure temperature on the y variable. A 400 mL beaker was obtained and filled to about 2/3 with regular tap water. The 400 mL beaker was then placed on the hot plate, along with the capillary. A ring stand was obtained to assist in holding the temperature sensor in the water until 80 C was met.Holding the temperature sensor in place with the ring stand, the capillary tubes volume was recorded by reading the volume at the bottom of the liquid plug and the data entered in the MicroLAB program. Additional volume readings were taken every minute as the water cooled. Once the temperature of the water reached 50 C the beaker was taken off the hot plate and the additional volume reading were taken every minute. Once the waters temperature reached 35 C ice was added to aid in the cooling process. The readings were still taken every minute of the experiment. Readings were to be taken until the temperature of the water reached 5 C.Data:Table 1Boyle’s Law MicroLAB GraphTable 2Please see attached graph for Charles’ LawData Analysis & Calculations:Applying Boyle’s law means that gas pressure will change with volume change. If both P and V are kept at the same temperature, they are inversely proportional to each other. Applying Charles’ law means that at a constant pressure, the volume of a gas changes with the temperature. Celsius in MicroLAB must be converted to Kelvin using K = C + 273.15 calculation.Charles’s LawExperimentally determined absolute zero :-91.7 CY = 0.24x + 22.02-22.02 = 0.24x∴ - 91.7Percent error : 33.6 -91.7/-273 × 100∴ -33.6 % errorConclusion:The first part of lab measuring pressure and volume on a fixed sample of gas was successful. The data was recorded and plotted in MicroLAB. The pressure was plotted on the y-axis and 1/V on the x-axis of the program. Comparing both the tests showed a correlation of the relationship between the volume of gas and the pressure exerted on it.In this lab the second part of the experiment was unsuccessful because of a failure in the MicroLAB program. When recording the temperature changes ever minute the program froze around 22.02 C and would not record a lower temperature, even though ice was being added to the water. With the equipment failure my lab partner and myself were not able to determine the effects of pressure and temperature on the volume of gas in the capillary tube.Boyle’s law is P1V1=P2V2 which means that gas pressure changes with volume. If both P and V are kept at the same temperature, they are inversely proportional to each other. As the pressure on a gas is increased, the volume decreases. When pressure vs. 1/V is plotted there should be a straight line on a mathematical graph. Charles law, states that at a constant pressure, the volume of a gas changes with the temperature. Charles’ law states that at constant pressure, the volume occupied by a fixed amount of gas is directly proportional to the absolute temperature, measured in Kelvin. On a graph the line would start at absolute zero, or 0 Kelvin and go out in a straight line. Celsius readings must be converted to Kelvin before applying Charles’ law.If MicroLAB were operating the proper way, the experiment would have been successful, however with it not working it was difficult to obtain data points for calculations.

자연과학| 2012.12.14| 5페이지| 3,500원| 조회(107)

Boyle, The Behavior of Gase, Charles, gases, volume of a gas

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Quantitative Analysis

Lab #6Quantitative AnalysisYour nameChemistry LabInstructor:Lab time:Date :Purpose:The purpose of the lab is to identify the separate quantities of an unknown nickel-containing compound and weigh the insoluble reaction products. Quantitative analysis would then determine how much of the element or compound is in the sample. To have a chemical reaction, both the reactants and products must be in balance and coefficients are used to balance the reaction. Also the “mole method” had to be used to compute the amount and percentage of an element present in an unknown nickel sample.Procedure & Observation:The first step was to mass about the plastic “boat” empty. And determine the mass of the filter paper. About 0.214 g unknown nickel compound was measured. A clean 100mL beaker was prepared.The next step was to prepared 30 mL of deionized water, 0.214g unknown nickel compound. All of these were put into the beaker carefully and stirred until the nickel compound was completely dissolved. Then 5 drops of concentrated ammonium hydroxide solution (NH4NO) were added. Mixed were 25 mL of dimethylglyoxime (DMG). This mixture produced a red precipitate.A Buchner funnel with side-arm vacuum flask and filtration apparatus was needed for the next step of the lab. The vacuum system was used to filter the Ni(DMG)2. The Ni(DMG)2 was poured into the top of the flask slowly. Any remaining precipitate from the beaker was rinsed with deionized water. The filtrate was cloudy with red(pink) precipitate that had passed through the filter paper during the first filtration. This process repeated 3 times. After filtering, the precipitate was dried. And then the precipitate was weighed. The mass Ni(DMG)2 was 0.504 g without filter paper.The third step was to repeat the entire experiment one more time. It should have repeated this two more times, but the experimenters did not have time.Data:A: 20.18 % NiB: 14.86 % NiC: 22.33% Ni12Mass unknown Ni²0.214g0.200gMass filter paper0.275g0.286gMass Ni(DMG) + filter paper3.395g3.008gMass Ni(DMG)0.504g0.51gData Analysis & Calculations:1) Ni² + 2 C4 H 8 N2O2 + 2 OH → Ni (C4 H 7 N2O2)2 + 2 H 2O0.214g Ni(DMG) 2×1 mole Ni(DMG) 2/ 289g Ni(DMG)2 = 7.40×10¯⁴mole Ni(DMG)27.40×10¯⁴mole Ni(DMG)2 ×1 mole Ni² / 1 mole Ni(DMG) 2= 7.40×10¯⁴mole Ni²7.40×10¯⁴mole Ni² × 58.7g Ni² / 1 mole Ni² = 0.0434g Ni²0.0434 g nickel / 0.214 g sample × 100% = 20.28 % nickel2) 0.200 g Ni(DMG)2 × 1 mole Ni(DMG) 2/ 289g Ni(DMG) 2 = 6.92 × 10¯⁴mole Ni(DMG)26.92 ×10¯⁴mole Ni(DMG)2 ×1 mole Ni² / 1 mole Ni(DMG) 2= 6.92 × 10¯⁴mole Ni²6.92 ×10¯⁴mole Ni² × 58.7g Ni² / 1 mole Ni² = 0.0406g Ni²0.0406 g nickel / 0.200 g sample × 100% = 20.3 % nickelConclusion:In this lab we made some errors when we got an unknown nickel compound. We had to take a Nickel “B” but we did not check this. We thought all of these nickels were the same, so we were not aware of nickel’s A, B, and C. just we did the experiment. When we figured out the “mole method”, we realized that we had used Nickel A. First, we got an unknown nickel compound of 20.28% by the mole method. I think that this was Nickel “A”, because Nickel “A” is 20.18 %. These two numbers were very similar.Also we made an error in the second experiment. When we solved the reaction equation, we got an unknown nickel compound of 20.3%. I think that this was almost Nickel “A” too.I am sure if we had used nickel “B”, we would have gotten a good data. Both our data were different from Nickel “B.” The percentage of Nickel “B” was 14.86 %, but we got a 20.28% and 20.3%. Between our first data and Nickel “B” there was a 5.42% error. Our second data and Nickel “B” differed by 5.44%.

자연과학| 2012.12.14| 3페이지| 3,000원| 조회(77)

nickel, quantitative, Quantitative Analysi, quantitatively separ

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