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Here are the best resources to pass Chem108. Find Chem108 study guides, notes, assignments, and much more.
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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) • 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...
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
Last document update:
ago
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...
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...
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
Last document update:
ago
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...
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...
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
Last document update:
ago
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...
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
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
Last document update:
ago
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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