D. Should be correct- the other ones don’t make sense
<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:
Answer:
pH = 11.60
Explanation:
When we add a base, we are increasing [OH⁻], so the pH will be totally basic.
pH of water = 7
Basic pH > 7
We are adding 2 M . 1 mL = 2 mmoles
2 mmoles of OH⁻ are contained in 501 mL of total volume.
[OH⁻] = 2 mmol / 501 mL = 3.99×10⁻³ M
- log 3.99×10⁻³ M = 2.39 → pOH
pH = 14 - pOH → 11.61
Water equilibrium
2H₂O ⇄ H₃O⁺ + OH⁻ Kw = 1×10⁻¹⁴
NaOH → Na⁺ + OH⁻
Unburned hydrocarbon on reacting with oxygen undergoes combustion reaction. However, the activation energy of this reaction is significantly high. When a catalyst like Pd is added to the reaction system, it provides active sites for the reaction to occur. It acts are a heterogeneous catalyst. It is pertinent of note that catalyst is refereed as heterogeneous, when it exist in different phase as compared to reactant and products. In present case, reactants and products are in gas phase, while catalyst is in solid phase. Due to availability of larger surface area at active site of Pd, activation energy of reaction decreases and decrease in activation energy favors higher reaction rates.
