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

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

MODULE 1 EXAM
Question 1
Click this link to access the Periodic Table. This may be helpful throughout
the exam.
1. Convert 845.3 to exponential form and explain your answer.
2. Convert 3.21 x 10-5 to ordinary form and explain your answer.


1. Convert 845.3 = larger than 1 = positive exponent, move decimal 2 places
= 8.453 x 102
2. Convert 3.21 x 10-5 = negative exponent = smaller than 1, move decimal 5
places = 0.0000321

Question 2
Click this link to access the Periodic Table. This may be helpful throughout
the exam.


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. 24.6 grams = ? kg

2. 6.3 ft = ? inches


1. 24.6 grams x 1 kg / 1000 g = 0.0246 kg
2. 6.3 ft x 12 in / 1 ft = 75.6 inches

please always use the correct units in your final answer

Question 3

,Click this link to access the Periodic Table. This may be helpful throughout
the exam.


Do the conversions shown below, showing all work:
1. 28oC = ? oK
2. 158oF = ? oC
3. 343oK = ? oF

1. 28oC + 273 = 301 oK o
C → oK (make larger)
+273
2. 158oF - 32 ÷ 1.8 = 70 oC o
F → oC (make smaller)
-32 ÷1.8
3. 343oK - 273 = 70 oC x 1.8 + 32 = 158 oF o
K → o C → oF

Question 4
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the exam.


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


1. Show the calculation of the mass of a 18.6 ml sample of freon with
density of 1.49 g/ml


2. Show the calculation of the density of crude oil if 26.3 g occupies 30.5
ml.

1. M = D x V = 1.49 x 18.6 = 27.7 g
2. D = M / V = 26..5 = 0.862 g/ml

Question 5
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the exam.


1. 3.0600 contains ? significant figures.
2. 0.0151 contains ? significant figures.

,3. 3.0600 ÷ 0.0151 = ? (give answer to correct number of significant
figures)

1. 3.0600 contains 5 significant figures.
2. 0.0151 contains 3 significant figures.
3. 3.0600 ÷ 0.0151 = 202.649 = 203 (to 3 significant figures for 0.0151)


Question 6
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the exam.
Classify each of the following as an element, compound, solution or
heterogeneous mixture and explain your answer.
1. Coca cola
2. Calcium
3. Chili


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

2. Calcium - is on periodic table = Element

3. Chili - is not on periodic table (not element) - no element names (not
compound)
appears as more than one substance (meat, beans,
sauce) = Hetero Mix

Question 7
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the exam.


Classify each of the following as a chemical change or a physical change


1. Charcoal burns
2. Mixing cake batter with water

, 3. Baking the batter to a cake


1. Charcoal burns - burning always = chemical change
2. Mixing cake batter with water - mixing = physical change
3. Baking the batter to a cake - baking converts batter to new material =
chemical change




Question 8
Click this link to access the Periodic Table. This may be helpful throughout
the exam.


Show the full Nuclear symbol including any + or - charge (n), the atomic
number (y), the mass number (x) and the correct element symbol (Z) for
each element for which the protons, neutrons and electrons are shown -
symbol should appear as follows: xZy+/- n


31 protons, 39 neutrons, 28 electrons

31 protons = Ga31, 39 neutrons = 70Ga31, 28 electrons = (+31 - 28 = +3)
= 70Ga 31+3

Question 9
Click this link to access the Periodic Table. This may be helpful throughout
the exam.


Name each of the following chemical compounds. Be sure to name all acids
as acids (NOT for instance as binary compounds)


1. PF5
2. Al2(CO3)3
3. H2CrO4

1. PF5 - binary molecular = phosphorus pentafluoride
2. Al2(CO3)3 - nonbinary ionic = aluminum carbonate