Chapter Exercises (1-8)

# 5 c what is its final temperature density of aluminum

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temperature was 32.5 °C, what is its final temperature? (density of aluminum = 2.70 g/cm3). Mass = volume x density = 98.5 x 2.70 = 265.95 24.200 J•mol−1•K−1 [1] g^-1 = 24.20/26.98= 0.897 J g^-1 ^-deg 67.4 = 265.95 x 0.897 x ΔT ΔT = 67.4 / (265.95 x 0.897) = 0.28 °C T = 32.5 + 0.28 = 32.78 °C

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CHAPTER EXERCISES (1-8) 3 Chapter 4, Exercise 72 Fill in the blanks to complete the following table. Chemical Symbol Group Number Group Name Metal or Nonmetal Cl 7A Halogens Nonmetal Ca 2A Alkaline Earth Metals Metal Xe 8A Noble Gases Nonmetal Na 1A Alkali metal Metal F 7A Halogen Nonmetal Chapter 5, Exercise 78 Calculate the formula mass for each of the following compounds. (a) CS2 = 12.011 + 2 (32.065) = 76.141 (b) C6H12O6 = 6 (12.011) + 12 (1.008) + 6 (15.999) = 180.156 (c) Fe(NO3)3 = 55.845 + 3(14.007) + 9(15.999) = 251.847 (d) C7H16 = 7 (12.011) + 16 (1.008) = 100.205 Chapter 6, Exercise 78 Calculate the mass percent composition of carbon in each the following carbon compounds. (a) C2H2 = 2 x Molar mass C Molar mass C2H2 x 100% = 2 (12.011) + 2 (1.008) = 26.038 = 2 x 12.011 26.038 x 100% = 0.923 (b) C3H6 = 3 (12.011) + 6 (1.008) = 42.081 = (3 x 12.011 / 42.081) x 100% = 0.86 (c) C2H6 = 2 (12.011) + 6 (1.008) = 30.07 = (2 x 12.011 / 30.07) x 100% = 0.799 (d) C2H6O = 2 (12.011) + 6 (1.008) + 15.999 = 46.069 = (2 x 12.011 / 46.069) x 100% = 0.52
CHAPTER EXERCISES (1-8) 4 Chapter 7, Exercise 82 A beaker of nitric acid is neutralized with calcium hydroxide. Write a balanced molecular equation and a net ionic equation for this reaction. Molecular = 2HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2H2O(l) Net Ionic = H+ + OH- H2O(l) Chapter 8, Exercise 82 Many home barbecues are fueled with propane gas (C3H8). How much carbon dioxide in kilograms is produced upon the complete combustion of 18.9 L of propane (approximate contents of one 5-gal tank)? Assume that the density of the liquid propane in the tank is 0.621 g/mL. ( Hint: Begin by writing a balanced equation for the combustion reaction.) C3H8 + 5O2 -----> 3CO2 + 4H2O mass of propane in 18.9L = 0.621 x 18.9 = 11.737kg = 11737g amount of propane = mass/molar mass = 11737/44 = 266.75mol amount of CO2 produced = 3 x 266.75 = 800.25mol mass of CO2 = amount x molar mass = 800.25 x 44 = 35211g = 35.211kg
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