Answer:
![[F^-]_{max}=4x10{-3}\frac{molF^-}{L}](https://tex.z-dn.net/?f=%5BF%5E-%5D_%7Bmax%7D%3D4x10%7B-3%7D%5Cfrac%7BmolF%5E-%7D%7BL%7D)
Explanation:
Hello,
In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:
![[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}} \\](https://tex.z-dn.net/?f=%5BF%5E-%5D_%7Bmax%7D%3D2.0x10%5E%7B-3%7D%5Cfrac%7BmolCa%5E%7B2%2B%7D%7D%7BL%7D%2A%5Cfrac%7B2molF%5E-%7D%7B1molCa%5E%7B2%2B%7D%7D%20%20%5C%5C)
![[F^-]_{max}=4x10{-3}\frac{molF^-}{L}](https://tex.z-dn.net/?f=%5BF%5E-%5D_%7Bmax%7D%3D4x10%7B-3%7D%5Cfrac%7BmolF%5E-%7D%7BL%7D)
Best regards.
Total atoms is 9 ( 2 carbon atoms, 5 hydrogen atoms, 1 oxygen atom and 1 hydrogen atom = 9 atoms)
Element is 3 ( Carbon, Hydrogen and Oxygen)
1 mole contains = 6.02x10^23 atoms. 0.31mole contains = 0.31x6.02x10^23 = 1.8662x10^23.
<h3>
Answer:</h3>
25.4 g CH₄
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
1.58 mol CH₄
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of C - 12.01 g/mol
[PT] Molar Mass of H - 1.01 g/mol
Molar Mass of CH₄ - 12.01 + 4(1.01) = 16.05 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
25.359 g CH₄ ≈ 25.4 g CH₄