Exam (elaborations)

CHEM 103 Module 1, 2, 3, 4, 5, 6 Exam, CHEM 103 Final Exam (Latest update, ) / CHEM103 Module 1, 2, 3, 4, 5, 6 Exam, CHEM103 Final Exam: General Chemistry II: Portage Learning (QUESTIONS & ANSWERS)

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CHEM 103 Module 1, 2, 3, 4, 5, 6 Exam, CHEM 103 Final Exam (Latest update, ) / CHEM103 Module 1, 2, 3, 4, 5, 6 Exam, CHEM103 Final Exam: General Chemistry II: Portage Learning (QUESTIONS & ANSWERS) CHEM 103 General Chemistry Module 1, 2, 3, 4, 5, 6 Exam, CHEM 103 Final Exam (Latest update, ) / CHE...

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Contains Questions & answers

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CHEM 103 Module 1, 2, 3, 4, 5, 6 Exam and CHEM 103 Final Exam

EXAM 1
Question 1

1. Convert 1005.3 to exponential form and explain your answer.

2. Convert 4.87 x 10-6 to ordinary form and explain your answer.

Answer:

1. Convert 1005.3 = larger than 1 = positive exponent, move decimal 3 places = 1.0053
x 103

2. Convert 4.87 x 10-6 =negative exponent =smaller than 1, move decimal 6 places =
0.00000487

Question 2

Using the following information, do the 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)

1. 2.73 liters = ? ml 2. 8.6 pts = ? qts

Answer:

1. 2.73 liters x 1000 ml / 1 liter = 2730 ml

2. 8.6 pts x 1 qt / 2 pts = 4.3 qts

Question 3

Do the conversions shown below, showing all work:

1. 248oC = ? oK

2. 25oF = ? oC

3. 175oK = ? oF

Answer:

1. 248oC + 273 = 521 oK oC → oK (make larger) +273

,2. 25oF - 32 ÷ 1.8 = -3.88 oC oF → oC (make smaller) -32
÷1.8

3. 175oK - 273 = -98 oC x 1.8 + 32 = -144.4 oF oK → oC → oF

Question 4

Be sure to show the correct number of significant figures in each calculation.

1. Show the calculation of the mass of a 17.9 ml sample of chloroform with density
of 1.49 g/ml

2. Show the calculation of the volume of 19.4 grams of cresol with density of 1.02
g/ml

Answer:
1. M = D x V = 1.49 x 17.9 = 26.7 g

2. V = M / D = 19..02 = 19.0 ml
Question 5

1. 1.35601 contains ? significant figures.

2. 0.151 contains ? significant figures.

3. 1.35601 + 0.151 = ? (give answer to correct number of significant figures)

Answer:

, 1. 1.35601 contains 6 significant figures.

2. 0.151 contains 3 significant figures.

3. 1.35601 + 0.151 = 1.50701 = 1.507 (to the thousands place for 0.151)

Question 6

Classify each of the following as an element, compound, solution or heterogeneous
mixture and explain your answer.

1. Honey

2. Hydrogen iodide

3. Tea

Answer:

1. Honey - is not on periodic table (not element) - no element names (not
compound) appears to be one substance = Solution

2. Hydrogen iodide - is not on periodic table (not element) - has two element names
= Compound

3. Tea - is not on periodic table (not element) - no element names (not compound)

appears to be one substance = Solution

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Exam (elaborations) ·

Exam (elaborations)

CHEM 103 Module 1, 2, 3, 4, 5, 6 Exam, CHEM 103 Final Exam (Latest update, ) / CHEM103 Module 1, 2, 3, 4, 5, 6 Exam, CHEM103 Final Exam: General Chemistry II: Portage Learning (QUESTIONS & ANSWERS)

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