Chapter 1 TAVSS

Chapter 1 TAVSS - CHEM 161-2007 CHAPTER 1 CHEMISTRY: MATTER...

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Unformatted text preview: CHEM 161-2007 CHAPTER 1 CHEMISTRY: MATTER AND MEASUREMENT PRACTICE PROBLEMS DR. ED TAVSS Units/unit conversions/dimensional analysis Classification of matter Significant Figures/Precision/Accuracy Miscellaneous 1 UNITS/UNIT CONVERSIONS/DIMENSIONAL ANALYSIS 12 Chem 161-2006 Exam I Hill, Petrucci et al., 4 th edition Chapter 1 Chemistry: Matter and Measurement Unit conversions The density of gold is 19.3 g/cm 3 . What is the density of gold in lb/in 3 ? A . 0.697 B. 19.3 C. 100 D. 1000 E. 1.20 10 3 453.59 g = 1 lb 2.54 cm = 1 in 19.3g x (1 lb/453.6 g) = 0.0425 lb 1 cm 3 x (1 in/2.54 cm) 3 = 0.06102 in 3 0.0425 lb/0.06102 in 3 = 0.696 lb/in 3 A 2 23 Chem 161-2005 Final exam Chapter 1 - Chemical Foundations Unit conversions/dimensional analysis The value of gold is $472 per troy ounce. What is the value of the cube of gold 2.00 cm on a side? Given: 1 troy ounce = 31.1 g density of gold: 19.3 A . $2340 B. $5460 C. $1520 D. $3410 E. $4250 The volume of a cube of gold = (2.00 cm) 3 = 8 cm 3 . Plan: Volume grams troy ounces $ 8 cm 3 x (19.3g/cm 3 ) x (1 troy ounce/31.1 g) x $472/troy ounce = $2343 21 Chem 161-2005 Final exam Chapter 1 - Chemical Foundations Unit conversions/dimensional analysis Convert 251 into . A. 4.18 x 10-6 m/s B. 1.51 x 10 10 m/s C. 4.18 x 10 12 m/s D. 1.51 x 10-2 m/s E . 4.18 x 10 6 m/s Plan: m m; min sec 251 m/min x (1 m/10-6 m) x (1 min/60 sec) = 4.18 x 10 6 m/s 3 15 Chem 161-2005 Exam I Zumdahl 6 th edition Chapter 1 Units/unit conversions/dimensional analysis Pressure is force/area. The SI unit for force is kg m s-2 . Express pressure in terms of base SI units. A . kg m-1 s-2 B. kg m 2 s-2 C. kg m s-2 D. kg m-2 s-2 E. kg m-3 s-2 Area would have units of m 2 . Pressure = (kg x m x s-2 )/m 2 = kgs-2 /m = kgm-1 s-2 11 Chem 161-2005 Exam I Zumdahl 6 th edition Chapter 1 Units/unit conversions/dimensional analysis Which prefix is correctly associated with its power of ten? A. m 10-6 B. M 10 3 C. 10-9 D . G 10 9 E. p 10-15 m = milli = 10-3 M = mega = 10 6 = micro = 10-6 G = giga = 10 9 p = pico = 10-12 4 2 Chem 161-2005 Exam I Zumdahl 6 th edition Chapter 1 Units/unit conversions/dimensional analysis Convert 55.6 F to kelvins. A. 13.1 K B . 286.3 K C. 315.7 K D. 341.3 K E. 68.1 K C o = (5/9)(F-32) = (5/9)(55.6-32) = 13.11 K = 273 + C o = 273 + 13.11 = 286.11 11. Chem 161-2004 Exam I Zumdahl 6 th edition Chapter 1 Unit Conversions The density of a diamond is 3.51 g/cm 3 and 1 carat = 0.200 g. The mass of a diamond whose volume is 2.8 mL is A. 4.0 carats B. 2.0 carats C . 49 carats D. choose this choice if none of the others is correct E. 170 carats 1 cm 3 = 1 mL Begin with what you know, and end with what youre trying to find....
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This note was uploaded on 07/17/2010 for the course 160 162 taught by Professor Kimmel during the Spring '10 term at Rutgers.

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Chapter 1 TAVSS - CHEM 161-2007 CHAPTER 1 CHEMISTRY: MATTER...

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