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# I am not sure what to do for B because I'm not certain what mental model looks like src="/qa/attachment/10700136/" alt="image.jpg" /> Attachment 1 Attachment 2 ATTACHMENT PREVIEW Download attachment image.jpg cantinn II. Data Analysis A. Identify a relationship between two of the properties you measured for brass by graphing. Try to come up with an algebraic relationship that expresses the pattern you found. Graph paper is provided at the back of this manual. (13, 07 , 1:00 ) and ( 57.53, 6,00) 6.00- (57. 53 , 6-00 ) m = 6100 - 1.00 ( 41. 49 , 5.00) 57, 53- 13,07 =0 . 112 4.00- b = (6100 ) - (0. 112 ) ( 57, 53 ) 6 = - 0.44 3.00- 8( 26. 32 , 3.0 ) Algebraic Equation : 2(21.83, 2.20) 200 (18. 39,2.0) y= 0.112x - 0.44 1.00 ( 13. 07 , 1.00 ) 10,06 20.00 30.00 40100 mass ( a ) 50.00 60.00 B. On the same sheet of graph paper, graph the same variables for the aluminum. What are similarities and differences in the two graphs? Offer an explanation for this. 9.00 ( 24.90. 9. 70) (24, 50 , 9.15 ) 18.73, 2.50 ) and ( 24.50, 9.20) (23.84, 8.90 ) (21.20, 7. 80) m = 9. 20- 2.50 6.70 15.77 = 0.425 (19. 96, 7.201 24.50 - 8.73 6 = ( 2. 50 ) - (0. 425 ) (8,73 ) 8 5.00 6 = - 1.21 Algebraic Equation: 3.00 P (8. 73, 2.50) 200 y = 0. 425 x - 1. 21 1. 00 10.00 70 , 00 30. 00 mass (92 40,00 50.00 60.00 C. Using a different combination of variables repeat the data analysis outlined in parts II.A and II.B. BRASS ( length vs. mass ) Aluminum (length us . mass ) ( 7. 60, 24. 90 ) (5.00, 57.53 ) 1 125. ouL (7. 30 , 24:50 ) . 50-00 20.00- ( 7,00, 23. 86 ) mass (o 40.00- ( 2.40 , 26.82 ) . (3. 60 , 41.49 (5. 70 , 19.95 ) MO 30.00- 0.00 (2.80, 8.73) 20.00 ( 1 .90 , 21.83 ) ( 1.50 , 15. 34 ) 5.00 - (1.20 , 13.07 100 2,00 3.00 4.00 500 600 7.00 8.00 1.00 J.00 3:00 4.00 5.00 Length ( cm ) ( 1.20 , 13. 07 ) and (3.60, 41. 49 ) Length (cm ) (2. 80, 8.73 ) and (7. 60, 24.90 ) Algebraic m = 41.49- 13.07 Algebraic Equation : m = 24. 90- 8:73 - 28.42 - 11.84 7.60- 2.80 2 = 3.37 Equation: 3.60 - 1.20 2.4 y = 11. 84 x - 1, 13 b = (41. 49 )-(11,84) (3.40) = -1 134 x- 1. 13 6 = (24, 90 ) - (3. 37) (7,60 ) =-0.71 y = 3. 37 X- 0.71 ATTACHMENT PREVIEW Download attachment image.jpg III. Interpretation A. Using the graphs and algebraic relationships you generated in this activity, summarize the relationships you discovered. Generalize about the relationship between the graphs and their algebraic equations. The graphs with the relationship between mass and volume ( density), shows an increasing trend and are explained algebraically. This means that as the mass ( 9) of the sample increased, the volume ( cm3 ) of the sample increased as well. The algebraic equation for brass density is y= 0. 112x- 0.44 suggests that as mass increases, volume increases by 0.112. On the other hand , the algebraic equation for aluminum density is y = 0.425x- 1.21 suggests that as mass increases, volume increases by 0.425. same as the the other graph, the graphs with the relationship between length and mass and its algebraic expressions showed an increasing linear trend. This means that as the length ( cm ) of the sample increased, the mass ( g ) also increases as well. The algebraic equation for brass (length us. mass ) is y = 11.84x - 1.13 suggests that as length increases, volume increases by 11.84. The algebraic equation for aluminum ( length us. mass ) is y= 3.37x- 0.71 suggests that as length increases, mass increases by 3.37. B. Mental Model-Draw a picture(s) that explains, at the level of atoms and molecules, the pattern observed in any of the relationships you investigated. Explain how your picture(s) illustrates the pattern.

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