# Unit Conversions Chapter 7 Summary

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em study guide99 Chapter 8 Unit Conversions  Review Skills 8.1 Unit Analysis  An Overview of the General Procedure  Metric-Metric Unit Conversions  English-Metric Unit Conversions 8.2 Rounding Off and Significant Figures  Measurements, Calculations, and Uncertainty  Rounding Off Answers Derived from Multiplication and Division  Rounding Off Answers Derived from Addition and Subtraction 8.3 Density and Density Calculations  Using Density as a Conversion Factor  Determination of Mass Density 8.4 Percentage and Percentage Calculations 8.5 A Summary of the Unit Analysis Process 8.6 Temperature Conversions  Chapter Glossary Internet: Glossary Quiz  Chapter Objectives Review Questions Key Ideas Chapter Problems Section Goals and…show more content…
kg C = 1.456 × 104 kJ    3  = 0.444 kg C  32.8 kJ   10 g  84. The average adult male needs about 58 g of protein in the diet each day. A can of vegetarian refried beans has 6.0 g of protein per serving. Each serving is 128 g of beans. If your only dietary source of protein were vegetarian refried beans, how many pounds of beans would you need to eat each day? (Obj 15) ? lb beans 58 g protein  1 serving   128 g beans  1 lb  =     day 1 day  6.0 g protein   1 serving  453.6 g  = 2.7 lb beans per day 86. About 6.0  105 tons of 30% by mass hydrochloric acid, HCl(aq), is used each year to remove metal oxides from metals to prepare them for painting or for the addition of a chrome covering. How many kilograms of pure HCl would be used to make this hydrochloric acid? (Assume that 30% has two significant figures. There are 2000 lb/ton.) (Obj 15)  30 ton HCl   2000 lb   1 kg  ? kg HCl = 6.0 × 105 ton HCl soln      100 ton HCl soln   1 ton   2.205 lb   30 ton HCl   2000 lb   453.6 g   1 kg  ? kg HCl = 6.0 × 105 ton soln     3   100 ton soln   1 ton   1 lb   10 g  = 1.6  108 kg HCl 88. A typical nonobese male has about 11 kg of fat. Each gram of fat can provide the body with about 38 kJ of energy. If this person requires 8.0  103 kJ of energy per day to survive, how many days could he survive on his fat alone? (Obj