Justin Estrella Chem M1 Lab 2 (2)
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SUNY Empire State College *
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1
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Mathematics
Date
Feb 20, 2024
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docx
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Experimental Data
You should also submit the Excel file, as well as copying and pasting the individual graphs in this section. Graph 1: Paste below a copy of your graph after Step 9 from Activity 1.
PASTE HERE
Graph 2: Paste below a copy of your graph after Step 10 from Activity 1.
PASTE HERE
Graph 3: Paste below a copy of your graph after Step 29 in Activity 2.
PASTE HERE
Graph 4: Paste below a copy of the first graph and trendline obtained from Step 30 in Activity 3.
PASTE HERE
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Graph 5: Paste below a copy of the second graph and trendline (y-intercept = 0) obtained from Step 30 in Activity 3.
PASTE HERE
Graph 6: Paste below a copy of the first graph and trendline obtained from Step 31 in Activity 3.
PASTE HERE
Lab Questions
Compare the graphs from step 9 and 10. Compare also, the two graphs from step 30. How did the equations change after setting the y-intercept to zero. What do you think this means?
When the intercept was set to zero on these graphs, the slope of the graph was reduced
Using one of the trendlines obtained in step 30, calculate the temperature in Kelvin at which the pressure of gas A will reach 450 kPa. Indicate which trendline you used. Show all your work for full credit. You may type your answers using ‘Equation Editor’ or you can paste a picture of your hand-
written work. Make sure your work is clearly labeled, neat and legible.
I used the trendline that had a y intercept value of 0. The equation associated with this trendline was y=0.2078x. The question asked to find the temperature, which is x, if the y value was equal to 450 kpa. What I did is just simply the reverse of what I would do with any other value that we found on our chart.
Since usually we have the temperature already and are seeking the y, and the solution for finding y (when the y intercept=0) involves multiplying our known slope value of m=0.2078 to our temperature values (x) to find the pressure (kPa), I instead said y=450 and m=0.2078, while x is the unknown. Since we multiply to find y when we know the x value, I divided to find x while we knew the y value. 450 divided by 0.2078 equals 2,165.54. To double check, we can plug the temperature value of 2,165.54 into
our equation This would be y=0.2078 x 2,165.54 When multiplied they equal 450, therefore the value of y (kPa) = 450. In other words, this confirms that if the value of y, the pressure of gas A, was equal to 450 then the temperature of gas A would have to be
equal to 2,165.54 kelvin
Using the information obtained from step 31, how much mass of each solution would be expected in 43 mL of solution? Note, that you must calculate the mass for each solution. Please label your work clearly. Show all of your work for full credit. You may type your answers using ‘Equation Editor’ or you can paste a picture of your hand-written work. Make sure your work is clearly labeled, neat and legible.
Mass of solution 1 is equal to y=1.05x-0.29. The 43 mL solution refers to the volume of the solution, which is the value of x. In this instance we are seeking the mass of solution 1, which will be the value of y. If the equation is then y=1.05 x 43 – 0.29, then y=44.86 in this instance. In other words, if the solution has a volume of 43 mL, then the mass of solution 1 is equal to 44.86 grams
Mass of solution 2 is equal to y=0.996x-0.04. Plugging in the new known x value, which is the 43mL solution, we get a new equation, which is y=0.996 x 43 – 0.04. Solving the equation with the new value of x gives a solution of y=42.90. In other words, if the volume of solution 2 is 43mL, then the mass of solution 2 is equal to 42.90 grams. Based on the definition of a slope as described in the background information, what do the slopes from the graphs in step 31 represent? (Hint: consider the units compared in the graph).
The slope is the value that shows the rate of change in the dependent variable, while the independent variable changes. Personal Reflection
Going into this lab assignment I was an open book for relearning these concepts, as I have not touched Microsoft excel in a very long time. There were some instances that required trial and error because I was using microsoft excel 365 off of the Microsoft website, and the layout seemed to be updated and was different than what was portrayed in the lab activities. Seemingly half of the description in the activity was different than what I had in my version of excel, which was a challenge as it required that I mess around with the program and find the corresponding locations. Creating new trends, creating two y values simultaneously, labeling the graph with names and more, were a bit different than the directions. Nonetheless, I spent an adequate amount of time and effort to learn and figure it out. I feel confident that I replicated the graphs in the way that was expected. The challenge actually helped me learn more and familiarize myself with excel once again. There is more to learn on the program in the future. I also
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realized that I can benefit from refreshing on math, as I haven’t taken math courses in a long time. Math
will be necessary to better understand concepts in my courses in the future, and math will be apart of my masters degree program, so I definitely want to keep up to date.