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CHEM 103 Module 1 Exam 2026–2027 | Portage Learning General Chemistry I Questions & Verified Answers

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Prepare for the Portage Learning CHEM 103 Module 1 Exam 2026–2027 with this comprehensive General Chemistry I resource featuring questions and verified answers covering Measurement, Matter & Atomic Theory, including units and conversions, dimensional analysis, significant figures, accuracy and precision, density, classification and properties of matter, temperature conversions, atomic structure, atomic theory, elements, isotopes, and foundational chemistry calculations. Designed for students enrolled in CHEM 103 General Chemistry I w/Lab, this resource provides focused preparation for Exam 1 and reinforces the high-yield concepts emphasized in Portage Learning's current Module 1 curriculum.

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Portage Learning CHEM 103 General Chemistry I | Module 1 Exam Questions & Verified Answers

1.1 EXPONENTIAL NUMBERS (지수) The speed of light is 186,000 miles per second. This is 1.86 x 105 (in exponential
form)
The diameter of an E. coli cell is 0.00000080 meters = 8.0 x 10-7 (in exponential
form)
4950000000. A number larger than one will have a positive exponent. Move the
decimal point nine places to the left to give 4.95 x 109
0.00000000000056 A number smaller than one will have a negative exponent.
Move the decimal point 13 places to the right to give 5.6 x 10-13
1.) 0.00000456 = smaller than 1 = negative exponent, move decimal 6 places =
4.56 x 10-6
2.) 2.63 x 107 = positive exponent = larger than 1, move decimal 7 places =
26300000.
3.) 7800000. = larger than 1 = positive exponent, move decimal 6 places = 7.8 x
106
4.) 8.26 x 10-5 = negative exponent = smaller than 1, move decimal 5 places =
0.0000826
5.) 5.38 x 10-3 = negative exponent = smaller than 1, move decimal 3 places =
0.00538
6.) 0.000673 = smaller than 1 = negative exponent, move decimal 4 places =
6.73 x 10-4
7.) 3.82 x 104 = positive exponent = larger than 1, move decimal 4 places =
38200.
8.) 623000000. = larger than 1 = positive exponent, move decimal 8 places =
6.23 X 108


1.2: METRIC SYSTEM - UNIT CONVERSIONS 1 foot (ft)=12 inches (in)
1 kilogram (kg)=1000 grams (g)
1 pound (lb)=16 ounces (oz)
1 gallon (gal)=4 quarts (qt)
1000 milliliters (ml)=1 liter (L)
100 centigrams (cg)=1 gram (g)
10 decimeters (dm)=1 meter (m)


The prefixes kilo (= 1000), milli (= 1/1000), centi (= are used with any metric unit, such as grams (weight), liters (volume), and
1/100), and deci (= 1/10) meters (distance).



The original quantity and unit (to be converted) are 48 ounces = ? pounds
multiplied by a conversion factor (which is a fraction 48 o z × 1 p o u n d / 16 o z = 3 p o u n d s
made up as follows):
New unit (on top) / Old unit (on bottom) 4 kilograms = ? grams
4 Kg X 1000 grams/1kg = 4000 grams

1.) 656 cm x 1 m / 100 cm = 6.56 m

2.) 20 gal x 4 qt / 1 gal = 80 qts

3.) 7820 ml x 1 liter / 1000 ml = 7.82 liters

4.) 36 g x 1000 mg / 1 g = 36000 mg

5.) 25.3 g x 10 dc / 1 g = 253 dg

6.) 6.2 m x 1000 mm / 1 m = 6200 mm

7.) 725 l x 1000 ml / 1 liter = 725,000 ml

8.) 56 ft x 12 in / 1 ft = 672 in


1.3: TEMPERATURE CONVERSIONS 1.) 132 oC + 273 = 405 oK
2.) 132 oF - 32 x 1.8 = 55.6 oC + 273 = 328.6 oK
3.) 285 oK - 273 = 12 oC
4.) 60 oC x 1.8 + 32 = 140 oF
5.) 85 oF - 32 x 1.8 = 29.4 oC
6.) 52 oC + 273 = 325 oK
7.) 32 oC x 1.8 + 32 = 89.6 oF
8.) 412 oK - 273 = 139 oC x 1.8 + 32 = 282.2 oF


K to °C(°C and K differ by 273) K (larger) - 273 = °C (smaller)




°C to K(°C and K differ by 273) °C (smaller) + 273 = K (larger)

, Portage Learning CHEM 103 General Chemistry I | Module 1 Exam Questions & Verified Answers
°F to °C(Subtract 32, and divide by 1.8) °F - .8 = °C




°C to °F(Multiply by 1.8, and add 32) °C X 1.8 + 32 = °F




1.4: DENSITY Density represents the mass of a substance in a unit volume of that substance.
The density of most solids or liquids is expressed in the unit grams per
milliliter (or g/ml).


The following equation is used to calculate density if you D (g/ml) = M(g) / V(ml)
know the mass (in grams) and the volume (in 1.) 35.6 ml = V 1.86 g/ml = D find M = D x V = 1.86 x 35.6 = 66.2 grams
milliliters).
2.) 13.6 g/ml = D 1000 g = M find V = M / D = .6 = 73.5 ml

3.) 200 ml = V 620 g = M find D = M / V = 620 g / 200 ml = 3.10 g/ml


1.5: SIGNIFICANT FIGURES Rules for determining significant figures in a number:
All non-zero digits are significant.
All zeros between or following non-zero numbers are significant.
Zeros to the left of a decimal or preceding other non-zero numbers are not
significant.
4326.7 - This number contains 5 significant figures (in bold).
400.70 - This number contains 5 significant figures (in bold).
0.3267 - This number contains 4 significant figures (in bold).
0.0070 - This number contains 2 significant figures (in bold).
0.5070 - This number contains 4 significant figures (in bold).


The appropriate number of significant figures must be Examples:
reported in any answer resulting from a calculation. The 8.612 + 4.51 + 0.20 + 3.9 = 17.222 = 17.2 (round off to tenths place because of
rules for reporting the correct number of significant 3.9)
figures in an answer depends on whether the calculation 47.60 - 23. = 24.60 = 25 (round off to the unit's place because of 23.)
involved addition/subtraction or multiplication/division.
In addition or subtraction calculations, the result
should be rounded off (more about that later) so that it
has the same number of decimal places as the
measurement having the fewest decimal places
(counting from left to right).


In multiplication or division calculations, the result Examples:
should be rounded off so as to contain the same number 128..20 = 2.51191 = 2.512 (round off to 4 significant figures because of
of significant figures as the measurement with the least 51.20)
number of significant figures. 54.06 x 17.0 = 919.02 = 919 (round off to 3 significant figures because of 17.0)


Rules for rounding off numbers: If the digit to be dropped is greater than 5, the last retained digit is increased by
Wait until all operations (in a multiple-step problem) one. For example, 12.6 is rounded to 13.
have been done before rounding off the final answer.
If the digit to be dropped is less than 5, the last remaining digit is left as it is. For
example, 12.4 is rounded to 12.

If the digit to be dropped is 5, and if any digit following it is not zero, the last
remaining digit is increased by one. For example, 12.51 is rounded to 13.

If the digit to be dropped is 5 and is followed only by zeroes, the last remaining
digit is increased by one if it is odd but left as it is if even. For example, 11.5 is
rounded to 12, 12.5 is rounded to 12.


How many significant figures are there in each of the 1.) 0.004035 contains 4 significant figures
following numbers? 2.) 306.2 contains 4 significant figures
3.) 20.70 contains 4 significant figures
4.) 0.3450 contains 4 significant figures
5.) 0.0231 contains 3 significant figures
6.) 45.670 contains 5 significant figures
7.) 0.00034 contains 2 significant figures


1.) (5.625 + 8.15) x 2.34 + 3.2 = 35.4 (to tenths place, 3.2)

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