CHEM108

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Exam (elaborations) CHEM 108 PRACTICE FINAL EXAM ANSWERS- PORTAGE LEARNING PRACTICE EXAM Using the following information, complete the two conversions shown below, showing all work: 1 ft = 12 inches 1 pound = 16 oz 1 gallon = 4 quarts 1 mile = 5280 feet 1
  • Exam (elaborations) CHEM 108 PRACTICE FINAL EXAM ANSWERS- PORTAGE LEARNING PRACTICE EXAM Using the following information, complete the two conversions shown below, showing all work: 1 ft = 12 inches 1 pound = 16 oz 1 gallon = 4 quarts 1 mile = 5280 feet 1

  • Exam (elaborations) • 5 pages • 2022
  • Exam (elaborations) CHEM 108 PRACTICE FINAL EXAM ANSWERS- PORTAGE LEARNING PRACTICE EXAM Using the following information, complete the two conversions shown below, showing all work: 1 ft = 12 inches 1 pound = 16 oz 1 gallon = 4 quarts 1 mile = 5280 feet 1 ton = 2000 pounds 1 quart = 2 pints kilo (= 1000) milli (= 1/1000) centi (= 1/100) deci (= 1/10) 25.6 mg = ? grams 25. 6 mg x 1 g / 1000 mg = 0.0256 grams 1.5 gal = ? qts 1.5 gal x 4 qts / 1 gal = 6.0 qts Write the formula for each of the fol...
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Exam (elaborations) CHEM 108 Module 6 Exam Answers- Portage Learning Problem Set 1 1. Identify each of the compounds below as ACID, BASE or SALT on the basis of their formula and explain your answer based on the definition given in the module: (a) HC6H5O2
  • Exam (elaborations) CHEM 108 Module 6 Exam Answers- Portage Learning Problem Set 1 1. Identify each of the compounds below as ACID, BASE or SALT on the basis of their formula and explain your answer based on the definition given in the module: (a) HC6H5O2

  • Exam (elaborations) • 12 pages • 2022
  • Exam (elaborations) CHEM 108 Module 6 Exam Answers- Portage Learning Problem Set 1 1. Identify each of the compounds below as ACID, BASE or SALT on the basis of their formula and explain your answer based on the definition given in the module: (a) HC6H5O2 Acid (b) Be(OH)2 Base (c) Mn(ClO2)2 Salts (d) AgCl Salts (e) Sn(OH)4 bases (f) H2CrO4 acids (g) PbO2 Salts (h) Ni(OH)4 Bases (i) H3AsO4 Acids (j) HC2H3O2 Acids 2. Identify the following unknown substances as ACID, BASE or SALT on the basis of...
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Exam (elaborations) CHEM 108 Module 5 Exam Answers- Portage Learning Exam Page 1 In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below the following data table is obtained: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) Data Table #1 Time (sec) [N2O
  • Exam (elaborations) CHEM 108 Module 5 Exam Answers- Portage Learning Exam Page 1 In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below the following data table is obtained: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) Data Table #1 Time (sec) [N2O

  • Exam (elaborations) • 11 pages • 2022
  • Exam (elaborations) CHEM 108 Module 5 Exam Answers- Portage Learning Exam Page 1 In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below the following data table is obtained: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) Data Table #1 Time (sec) [N2O5] [O2] 0 0.200 M 0 300 0.182 M 0.009 M 600 0.166 M 0.017 M 900 0.152 M 0.024 M 1200 0.140 M 0.030 M 1800 0.122 M 0.039 M 2400 0.112 M 0.044 M 3000 0.108 M 0.046 M Complete the following three problems: Using the [O2] data from the table sho...
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Exam (elaborations) CHEM 108 Module 3 Exam Answers- Portage Learning Module 3 exam Exam Page 1 Show the calculation of the new temperature of a gas sample which has an original volume of 700 ml when collected at 730 mm and 35oC when the volume becomes 480
  • Exam (elaborations) CHEM 108 Module 3 Exam Answers- Portage Learning Module 3 exam Exam Page 1 Show the calculation of the new temperature of a gas sample which has an original volume of 700 ml when collected at 730 mm and 35oC when the volume becomes 480

  • Exam (elaborations) • 10 pages • 2022
  • Exam (elaborations) CHEM 108 Module 3 Exam Answers- Portage Learning Module 3 exam Exam Page 1 Show the calculation of the new temperature of a gas sample which has an original volume of 700 ml when collected at 730 mm and 35oC when the volume becomes 480 ml at 1.15 atm. 700ml/1000= 0.7 liters= Vi 730mm/700= 1.042 atm=Pi 35 ºC + 273 = 308K = Ti 480ml/ 1000 = 0.04 liters = Vf 1.15 amt= Pf (1.042) ×(0.7)= (1.15) × (0.04) (308) Tf Tf= 12.50 = 17.15K 0.729
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