Tuesday, July 22, 2025

Fall 2025 Courses

South Heights Christian Classes
Chemistry - Syllabus
Physical Science - Syllabus

Truth Academy
General Science + Lab (middle school) - Syllabus
Jekyll, Hyde, Scrooge, and the 1800s (middle school) - Syllabus
Jekyll, Hyde, Scrooge, and the 1800s (high school) - Syllabus
Western Civilizations 1 with CLEP Option (high school) - Syllabus

Online Classes (Self-Paced)
High School Chemistry Crash Course (self-paced)
High School Physics Crash Course (self-paced)
Strategies for the ACT Test (self-paced)
Emma by Jane Austen - English Literature (self-paced)
Pride and Prejudice by Jane Austen - English Literature (self-paced)
Sense and Sensibility by Jane Austen - English Literature (self-paced)
A Christmas Carol by Charles Dickens - English Literature (self-paced)
The Scarlet Pimpernel by Emma Orczy - English Literature (self-paced)
Strange Case of Dr Jekyll and Mr Hyde by Robert Louis Stevenson - English Literature (self-paced)
The Hobbit by J. R. R. Tolkien - English Literature (self-paced)
The Fellowship of the Ring by J. R. R. Tolkien - English Literature (self-paced)
The Two Towers by J. R. R. Tolkien - English Literature (self-paced)
The Return of the King by J. R. R. Tolkien - English Literature (self-paced)
Beethoven's Ninth Symphony - A Musical Journey (self-paced)
Handel’s Water Music - A Musical Journey (self-paced)
Holst: The Planets - A Musical Journey (self-paced)
Sergei Prokofiev: Peter and the Wolf - A Musical Journey (self-paced)

Saturday, July 19, 2025

Books and Other Educational Items for Sale

Items are marked with individual prices. Local pickup can be arranged in the western and southern suburbs of Minneapolis. If you're interested, please contact me at the contact form here



The Old Curiosity Shop by Charles Dickens: $8


The House of Seven Gables by Nathaniel Hawthorne: $9


Mutiny on the Bounty by Charles Nordhoff and James Norman Hall (hardcover): $7


The Red Pony by John Steinbeck: $7


Julie of the Wolves by Jean Craighead George: $6


Essential ACT flashcards: $5 or best offer

5 Steps To A 5 AP World History Flashcards: $5 or best offer


Numbers & Math Symbols stamps: $5 or best offer
(Please note that the number 1 is missing and will not be sold with the set.)





Thursday, July 10, 2025

Stoichiometry Explanation

Think of stoichiometry as the "recipe math" of chemistry. Just as a cookie recipe tells you that you need exactly two cups of flour and one cup of sugar to make a specific number of cookies, a balanced chemical equation tells you the exact ratio of reactants needed to create a specific amount of product. Stoichiometry is simply the set of calculations chemists use to measure these amounts. It allows you to predict the future of a reaction: if you know how much of one ingredient you have, stoichiometry gives you the mathematical tools to figure out exactly how much of the other ingredients you need or how much "stuff" you will end up with when the reaction is over.

In practice, stoichiometry relies heavily on the "mole," which serves as the bridge between the atomic world and the grams we measure on a scale in the lab. By using the coefficients (the big numbers in front of formulas) from a balanced equation, you create "mole ratios." These ratios are the heart of stoichiometry; they let you convert from moles of one substance to moles of another. Whether you are figuring out which chemical will run out first (the limiting reactant) or calculating the maximum amount of product you can produce (theoretical yield), stoichiometry ensures that matter is accounted for and that you aren't wasting chemicals by guessing.

Here are two examples of stoichiometry: one using a real-world analogy to visualize the concept, and one using a chemical reaction to show how it is used in the lab.

Example 1: The S'mores Analogy (Conceptual)

Stoichiometry essentially follows a recipe to ensure you don't waste ingredients.

The Balanced Equation:

2 Graham Crackers + 1 Marshmallow + 1 Chocolate Piece -> 1 S'more

The Scenario:

Imagine you have 10 Graham Crackers, 10 Marshmallows, and 10 Chocolate Pieces. How many S'mores can you make?

The Stoichiometry:

Even though you have plenty of marshmallows and chocolate, the stoichiometry (the ratio) says you need 2 crackers for every 1 S'more.

10 Crackers / 2 Crackers per S'more = 5 S'mores.

The Result:

You can only make 5 S'mores. The graham crackers are your "limiting reactant"—once they run out, the reaction stops, leaving you with leftover marshmallows and chocolate ("excess reactants").

Example 2: Airbag Safety (Chemical Application)

In the real world, engineers use stoichiometry to save lives. An airbag needs to inflate with a specific volume of nitrogen gas (N2) in milliseconds during a crash.

The Balanced Equation:

2NaN3 (s) -> 2Na (s) + 3N2 (g)

The Scenario:

An engineer needs exactly 60 Liters of Nitrogen gas to fill the driver's side airbag. If they use too little, the bag won't cushion the driver. If they use too much, the bag could explode or be too hard.

The Stoichiometry:

The engineer uses the mole ratio (2 moles of NaN3 produce 3 moles of N2) to calculate the precise mass of Sodium Azide powder to put inside the steering wheel. They calculate backwards from the gas needed (N2) to the solid required (NaN3) to ensure the reaction is perfect every time.


Wednesday, July 9, 2025

Grams to Moles and Moles to Grams Practice Problems (with answer key)

Part 1: Moles to Grams (Elements)
Convert the given amount of moles into grams.
  1. 2.00 mol of Helium
  2. 0.75 mol of Copper 
  3. 10.0 mol of Sulfur 
  4. 0.15 mol of Silver 
  5. 5.2 mol of Calcium 

Part 2: Grams to Moles (Elements)
Convert the given mass in grams into moles.
  1. 24.31 g of Magnesium 
  2. 100.0 g of Iron 
  3. 5.00 g of Lithium 
  4. 250.0 g of Gold 
  5. 15.5 g of Potassium 

Part 3: Moles to Grams (Compounds)
Calculate the molar mass of the compound, then convert moles to grams.
  1. 3.00 mol of water 
  2. 0.50 mol of sodium chloride 
  3. 1.20 mol of carbon dioxide 
  4. 2.50 mol of CH4
  5. 0.10 mol of H2SO4 

Part 4: Grams to Moles (Compounds)
Calculate the molar mass of the compound, then convert grams to moles.
  1. 100.0 g of calcium carbonate 
  2. 36.0 g of C6H12O6 
  3. 50.0 g of NH3 
  4. 117.0 g of sodium chloride 
  5. 22.0 g of carbon dioxide 

Answer Key
Part 1: Moles to Grams (Elements)
  1. 8.00 g
  2. 47.66 g
  3. 320.60 g
  4. 16.18 g
  5. 208.42 g
Part 2: Grams to Moles (Elements) 
  1. 1.00 mol 
  2. 1.79 mol 
  3. 0.72 mol 
  4. 1.27 mol 
  5. 0.40 mol
Part 3: Moles to Grams (Compounds) 
  1. 54.05 g 
  2. 29.22 g 
  3. 52.81 g 
  4. 40.11 g 
  5. 9.81 g
Part 4: Grams to Moles (Compounds) 
  1. 1.00 mol 
  2. 0.20 mol 
  3. 2.94 mol 
  4. 2.00 mol 
  5. 0.50 mol

Tuesday, July 8, 2025

Predicting the Products of a Chemical Reaction

Part I: Formation Reactions

Predict the chemical formula of the product formed, then balance the entire equation.

(General Pattern: A + B -> AB)

  1. Na + Cl2 -> ______________________

  2. Mg + O2 -> ______________________

  3. Al + Br2 -> ______________________

  4. K + S -> ______________________

  5. Li + N2 -> ______________________

  6. H2 + O2 -> ______________________

Part II: Decomposition Reactions

Predict the products formed (break them into elements), then balance the entire equation. Remember your homonuclear diatomic elements!

(General Pattern: AB -> A + B)

  1. H2O -> ______________________

  2. HgO -> ______________________

  3. MgCl2 -> ______________________

  4. NI3 -> ______________________

  5. Al2O3 -> ______________________

  6. Ag2O -> ______________________

Part III: Combustion Reactions

Predict the products for the complete combustion of the following compounds, then balance the equation.

(General Pattern: carbohydrate + O2 -> CO2 + H2O)

  1. CH4 + O2 -> ______________________ (Methane)

  2. C3H8 + O2 -> ______________________ (Propane)

  3. C6H12O6 + O2 -> ______________________ (Glucose)

  4. C2H5OH + O2 -> ______________________ (Ethanol)

  5. C4H10 + O2 -> ______________________ (Butane)

Part IV: Mixed Practice

Identify the reaction type (Synthesis, Decomposition, or Combustion), predict the products, and balance.

  1. Ca + N2 -> ______________________

    Type: ______________________

  2. FeS -> ______________________

    Type: ______________________

  3. C8H18 + O2 -> ______________________

    Type: ______________________


ANSWER KEY

Part I: Synthesis

  1. 2Na + Cl2 -> 2NaCl

  2. 2Mg + O2 -> 2MgO

  3. 2Al + 3Br2 -> 2AlBr3

  4. 2K + S -> K2S

  5. 6Li + N2 -> 2Li3N

  6. 2H2 + O2 -> 2H2O

Part II: Decomposition

7. 2H2O -> 2H2 + O2

8. 2HgO -> 2Hg + O2

9. MgCl2 -> Mg + Cl2

10. 2NI3 -> N2 + 3I2

11. 2Al2O3 -> 4Al + 3O2

12. 2Ag2O -> 4Ag + O2

Part III: Combustion

13. CH4 + 2O2 -> CO2 + 2H2O

14. C3H8 + 5O2 -> 3CO2 + 4H2O

15. C6H12O6 + 6O2 -> 6CO2 + 6H2O

16. C2H5OH + 3O2 -> 2CO2 + 3H2O

17. 2C4H10 + 13O2 -> 8CO2 + 10H2O

Part IV: Mixed Practice

18. Synthesis: 3Ca + N2 -> Ca3N2

19. Decomposition: 8FeS -> 8Fe + S8 (or Fe + S)

20. Combustion: 2C8H18 + 25O2 -> 16CO2 + 18H2O