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Chapter 1 TAVSS

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

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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

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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 μm/s B. 1.51 x 10 10 μm/s C. 4.18 x 10 12 μm/s D. 1.51 x 10 -2 μm/s E . 4.18 x 10 6 μm/s Plan: m → μm; min → sec 251 m/min x (1 μm/10 -6 m) x (1 min/60 sec) = 4.18 x 10 6 μm/s 3

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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. μ 10 -9 D . G 10 9 E. p 10 -15 m = milli = 10 -3 M = mega = 10 6 μ = 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 you’re trying to find. 2.8 mL x (1 cm 3 /1 mL) x (3.51 g/cm 3 ) x (1 carat/0.200 g) = 49.1 carats 25. Chem 161-2004 Exam I Zumdahl 6 th edition Chapter 1 5

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Unit Conversions Given that 1 inch = 2.54 cm and that 1 L = 1000 cm 3 convert 0.092 ft 3 to liters A. 3.2 x 10 -3 L B. 26 L C. 0.40 L D. 1.8 L E . 2.6 L Plan: ft 3 → in 3 → cm 3 → L 0.092 ft 3 x (12 in/1 ft) 3 x (2.54 cm/in) 3 x (1 L/1000 cm 3 ) = 2.605 L 2 Chem 161-2004 Exam I Zumdahl 6 th edition Chapter 1 Units/unit conversions/dimensional analysis Convert 55.6 ° F to kelvins.
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• Spring '10
• Kimmel
• International System of Units, Chemical substance, Chemical compound, Chemical Foundations, Chemical Foundations Unit

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

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