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Teaching Company - Chemistry, 2nd Ed.

Professor Frank Cardulla

This is a course targeted toward high school students. It can stand alone or be used to supplement classroom work.

Lecture 1 Lecture 2 Lecture 3 Lecture 4 Lecture 5 Lecture 6 Lecture 7

Lecture 1 - Introduction and Philosophy

Workbook Questions answers (2017-06-22):

  1. D (develop deep understanding)
  2. False. The techniques used to solve high school chemistry problems are not specific to chemistry, per se.
  3. Self fulfilling prophesy is believing something, thereby, we work toward making it true, even subconsciously perhaps. Because some students believe chemistry is hard, they "allow" it to be hard, and don't develop the necessary deep understanding that makes it easy.
  4. The students and rooms problem shows that a problem with understood amounts and relationships is easy to solve. Chemistry is the same, if we understand the terms, relationships and principles.
  5. Students struggle with chemistry because:
    1. terms are unfamiliar
    2. values are sometimes very large or very small, using fractions/decimals and scientific notation
  6. Once past lecture 4, lectures can be watched in any order.
  7. Upon encountering unfamiliar material, simply stop and view the lecture that explains that material, then pick up where you left off.
  8. Develop deep understanding of the how and why things work the way they do; then each and every problem is not completely new and unfamiliar, but rather has different terms or arrangements.
  9. This is a Fibonacci Series, and the rule to generate the next value in the sequence is to simply add the preceding two values.

Lecture 2 - Basic Concepts of Quantitative Reasoning

Workbook Questions answers (2017-06-22):

  1. Amounts, and This-per-That ratios.
    1. this-per-that
    2. this-per-that
    3. amount
    4. amount
    5. this-per-that
    6. neither, but simplifies to amount
    7. amount
    1. Formula
    2. Factor-label
    3. Proportion
  4. Using formulas alone can limit true understanding of the fundamentals in the problem. Using formulas is fine so long as the underlying concepts that the formula solves are understood. By having true understanding, the formula can be derived when needed, so no need to memorize a bunch of formulas.
  5. Proportions and Factor-label techniques, too, can limit true understanding.
    1. read the question to get the overall feel of the problem
    2. read through slowly, understanding each amount and the relationships of the values
    3. solve the problem using the most useful technique
  7. Review the answer, and ask yourself if it is 'reasonable'. Do a quick mental estimation of what the answer should be.
  8. They don't develop deep understanding of the concepts and relationships.

Lecture 3 - Quantitative Reasoning in Everyday Life

Workbook Questions answers (2017-06-22):

  1. This-per-that tells me how many of something goes into ONE of something else.
  2. 33 mi/h tells me that 33 miles is traveled in ONE hour.
  3. 238 mi/4.5h tells me that 52.9 miles is traveled in ONE hour.
  4. 48 mi/gal tells me that it takes 0.0021 gallons to travel on mile (1/48).
  5. Simply perform the division to get a '1' in the denominator.
  6. Read the problem carefully, visualize the quantities, always attach units to values.
  7. 8.5 mi/day for 6 months (30.5 days/mo) tells me that the total distance is 8.5 * 6 * 30.5 = 1555.5 miles
  8. 47 mi/h over 5 miles. How many seconds? 5mi / 47mi/h * 60m/h * 60s/m = 383.0 seconds
  9. $11.85 purchases how much $6.99/1b. turkey? $11.85 / $6.99/lb. = 1.70 lbs.
  10. 825 mi. traveled at 29.7 mi./gal. @ $4.28/gal. What cost for gas? 825 mi. / 29.7mi./gal. * $4.28/gal. = $118.89
  11. 120 M&M/student, 87 students, 0.89g/M&M, 100g/bag. How many bags? 87 students * 120 M&M * 0.89g/M&M * bag/100g = 92.9 -> 93 bags

Lecture 4 - Quantitative Reasoning in Chemistry - Density

Workbook Questions answers (2017-06-22):

  1. Mass is how much material/matter there is in that object.
  2. Volume is how much space is occupied by that object.
  3. Density is the how much material per unit of volume
  4. F (B & C)
  5. Density is the same whether I have one unit of volume or two or more.
  6. dime, nickel, small metal fork, orange, liter bottle of soda
  7. density of 6.25g occupying 7.34 cm^3. 6.25g/7.34cm^3 = 0.85 g/cm^3
  8. 8.29cm^3 at 1.44g/cm^3? 1.44g/cm^3 * 8.29cm^3 = 11.94g
  9. 9g at 13.59g/cm^3? 9g / 13.59g/cm^3 = 0.66 cm^3
  10. 28.73g occupies 44.8cm^3 - 35.2cm^3 volume, density? 28.73g/(44.8cm^3-35.2cm^3) = 2.99 g/cm^3
  11. hard candy at 6.00g vs. Nerf ball at 5.55g? Nerf ball has greater volume.
  12. 60.0g/cm^3 on Earth. Density on Moon? Same, no change, since mass is not changed by variations in gravity.
  13. 1.096g/cm^3 density at 5.40 x 10^9 cells/cm^3, volume of single cell? 1cell / 5.40 x 10^9 cells/cm^3 = 1.85 x 10^-10 cm^3

Lecture 5 - The SI (Metrc) System of Measurement

Workbook Questions answers (2017-06-22):

  1. Five base units used in our course
    1. Meter (length) m
    2. Second (time) s
    3. Kilogram (mass) kg
    4. Mole (amount of substance) mol
    5. Kelvin (temperature) K
    1. Mass = kilogram (gram g)
    2. Volume - meter^3 (cubic centimeter cm^3)
    3. Temperature -
    4. Speed - meter/second (kilometer per hour kph)
  3. kilo (x10^3), mega (x10^6), giga (x10^9)
  4. deci (x10^-1, centi (x10^-2), milli (x10^-3), micro (x10^-6)
  5. meter has been redifined over time in order to develop a more accurate, and reproducible value, one that is not changeable in nature.
  6. kilogram
    1. 2 kilo(gram) mockingbird
    2. 1 peta-bullS
    3. 1 mega-phone
    4. 1 micro-phone
    5. 1 atto-boy
    6. 1 deka-cards
    7. 1 micro-fish (microfiche)
    8. 1 deka-ration (decoration)
    9. 1 centipede
    10. 1 deka-dent (decadent)

Lecture 6 - Converting between Systems of Measurements

Workbook Questions answers (2017-06-27):

  1. 3.77 mm -> 0.377 cm
  2. 9.18 mm -> 9180 um
  3. (WRONG) 7.93 Gmol (7.93 x10^9) -> 7.93x10^27 nmol SHOULD BE (7.93x10^18)
  4. 5.85X10^-7 Mg -> 5.85x10^8 ng
  5. 7.22x10^6 dg -> 7.22x10^2 kg
  6. 1.81 Ym (10^24) -> 1.81x10^36 pm (10^-12)
  7. 167 cm -> in? 167 cm / 2.54cm/in = 65.75 in
  8. 1.33 lb -> kg? (1.33 lb * 453.6g/lb) / (1000g/kg) = 0.603 kg
  9. 1 cm -> mi? 1 cm / (2.54 cm/in) / (12 in/ft) / (5280 ft/mi) = 6.214x10^-6 mi OR 6.214 umi
  10. 1,000,000 s -> days? 1x10^6 s / (m/60s) / (h/60m) / (day/24h) = 11.6 days

Lecture 7 - Elements, Atoms and the Periodic Table
  1. Operation Definition: