PORTAGE LEARNING CHEM 103 MODULE 1 TO 6 EXAM WITH QUESTIONS AND 100% CORRECT ANSWERS|| LATEST AND COMPLETE VERSION WITH EXPERT VERIFIED SOLUTIONS|| ASSURED PASS

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PORTAGE LEARNING CHEM 103 MODULE 1 TO 6
EXAM WITH QUESTIONS AND 100% CORRECT
ANSWERS|| LATEST AND COMPLETE VERSION 2025-
2026 WITH EXPERT VERIFIED SOLUTIONS|| ASSURED
PASS
Question 1 . 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 . 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) 1/100) milli (= 1/1000) deci (= 1/10) centi (=




1. 24.6 grams = ? kg
2. 6.3 ft = ? inches

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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


. 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 oC → oK (make larger)
+273
2. 158oF - 32 ÷ 1.8 = 70 oC oF → oC (make smaller)
-32 ÷1.8
3. 343oK - 273 = 70 oC x 1.8 + 32 = 158 oF oK → oC → oF


Question 4 . This may be helpful throughout 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.

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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 . This may be helpful throughout 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 . This may be helpful throughout 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

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Question 7 . This may be helpful throughout 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


. 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 +3