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exis [7]
2 years ago
11

What foods of the aztec found their way into mayan cooking?.

Chemistry
1 answer:
Lera25 [3.4K]2 years ago
7 0

Avocados and tomatoes were mainly eaten by the Aztecs and Maya.

<h3>How did Mayans cook their food?</h3>

"Mayans cooking method includes digging a shallow pit, lining it with stones or clay balls.

Avocados and tomatoes were mainly eaten by the Aztecs and Maya, with different types of fruit. Corn was made into a kind of porridge, called atole in Mesoamerica and capita in Inca territory.

Avocados and tomatoes were mainly eaten by the Aztecs and Maya.

Learn more about Mayan cooking here:

brainly.com/question/1979880

#SPJ1

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Paraphin [41]

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This would be 'chemical' weathering.

Acid rain would react with the minerals in the rocks, causing them to break down.

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"An unsaturated solution is formed when 80 grams of salt is dissolved in 100 grams of water at 40° celsius. This salt could be..
olchik [2.2K]

Answer:

3. NaNO3

Explanation:

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1. Which type of telescope is most useful for detecting a star that emits very long
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Methane (CH4), ammonia (NH3), and oxygen can react to form hydrogen cyanide (HCN) and water according to this equation:
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7 0
3 years ago
2AlF3 + 3K2O → 6KF + Al2O3<br><br> How many grams of AlF3would it take to make 15.524 g of KF?
mr_godi [17]
<h3>Answer:</h3>

7.4797 g AlF₃

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Reading a Periodic Table
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃

[Given] 15.524 g KF

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol AlF₃ = 6 mol KF

Molar Mass of Al - 26.98 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of K - 39.10 g/mol

Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol

Molar Mass of KF - 39.10 + 19.00 = 58.10 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                              \displaystyle 15.524 \ g \ KF(\frac{1 \ mol \ KF}{58.10 \ g \ KF})(\frac{2 \ mol \ AlF_3}{6 \ mol \ KF})(\frac{83.98 \ g \ AlF_3}{1 \ mol \ AlF_3})
  2. Multiply/Divide:                                                                                                    \displaystyle 7.47966 \ g \ AlF_3

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 5 sig figs.</em>

7.47966 g AlF₃ ≈ 7.4797 g AlF₃

3 0
3 years ago
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