In normal conditions, warm water does "pile up" in the" Western Pacific Ocean.
Answer:
2Ag⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + H₂O (l)
Explanation:
Step 1: RxN
2AgNO₃ + 2NaOH → Ag₂O + 2NaNO₃ + H₂O
Step 2: Define states of matter
2AgNO₃ (aq) + 2NaOH (aq) → Ag₂O (s) + 2NaNO₃ (aq) + H₂O (l)
Step 3: Total Ionic Equation
2Ag⁺ (aq) + 2NO₃⁻ (aq) + 2Na⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + 2Na⁺ (aq) + 2NO₃⁻ (aq) + H₂O (l)
Step 4: Cancel out spectator ions
2Ag⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + H₂O (l)
The answr to your question is 325.15
<h3>
Answer:</h3>
10.6 mol NO
<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>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 4NH₃ + 5O₂ → 4NO + 6H₂O
[Given] 13.2 mol O₂
<u>Step 2: Identify Conversions</u>
[RxN] 5 mol O₂ → 4 mol NO
<u>Step 3: Stoich</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
10.56 mol NO ≈ 10.6 mol NO