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Ronch [10]
3 years ago
10

the frog population in the Amazon basin varied. every 500 years there was a mutational change leading to speciation. in between

the population remained the same with no observable changes. What pattern of macroevlution determined this change?
Chemistry
1 answer:
Usimov [2.4K]3 years ago
4 0
The frog population in the Amazon basin varied. Every 500 years there was a mutational change leading to speciation. In between the population remainedthe same with no observable changes. The pattern of macroevolution is that there is a change in species orientation and adaptation to change
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You might have a flashlight that runs (2) d-cells of 1.5 volts each in series. the bulb is rated for 0.7 amps. what’s its resist
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A particle that orbits the nucleus in an atom is called a(n
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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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