<h3>Answer:</h3>
4 mol O₂
<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>
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 4 mol H₂O
[Solve] x mol O₂
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol H₂O → 2 mol O₂
<u>Step 3: Stoichiometry</u>
- Set up conversion:

- Multiply/Divide:

ok so i think it would be b because gravity would pull you down but your the same size
Answer: 5622.6g
Explanation:
Note: Kf for water is 1.86°C/m.
The simple calculation is in the attachment below.
Answer : The concentration of
needed is, 
Explanation :
First we have to calculate the mole of phosphate.
As we are given that, 1 mg P/L that means, 1 mg of phosphate present in 1 L of solution.

Molar mass of phosphate = 94.97 g/mole

Now we have to calculate the concentration of phosphate.


Now we have to calculate the concentration of
.
The second equilibrium reaction is,

The solubility constant expression for this reaction is:
![K_{sp}=[Fe^{3+}][PO_4^{3-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BFe%5E%7B3%2B%7D%5D%5BPO_4%5E%7B3-%7D%5D)
Given: 
![\frac{1}{4}=[Fe^{3+}]\times 1.053\times 10^{-5}mol/L](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D%3D%5BFe%5E%7B3%2B%7D%5D%5Ctimes%201.053%5Ctimes%2010%5E%7B-5%7Dmol%2FL)
![[Fe^{3+}]=2.37\times 10^4M](https://tex.z-dn.net/?f=%5BFe%5E%7B3%2B%7D%5D%3D2.37%5Ctimes%2010%5E4M)
Thus, the concentration of
needed is, 
Yes they do. But a very small kinetic energy. They vibrate in fixed positions