Which of these is an example of a chemical reaction that occurs at a very slow rate
1 answer:
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Answer:
1,812 wt%
Explanation:
The reactions for this titration are:
2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻
I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻
The moles in the end point of S₂O₃⁻ are:
0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:
4,463x10⁻⁴ moles of S₂O₃⁻× = 2,2315x10⁻⁴ moles of I₃⁻
As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:
2,2315x10⁻⁴ moles of I₃⁻× = 4,463x10⁻⁴ moles of Ce(IV). These moles are:
4,463x10⁻⁴ moles of Ce(IV)× = <em>0,0625 g of Ce(IV)</em>
As the sample has a 3,452g, the weight percent is:
0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>
I hope it helps!
Answer:
9.8 × 10²⁴ molecules H₂O
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Organic</u>
- Naming carbons
<u>Stoichiometry</u>
- Analyzing reaction rxn
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[RxN - Unbalanced] CH₄ + O₂ → CO₂ + H₂O
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 130 g CH₄
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[RxN] 1 mol CH₄ → 2 mol H₂O
[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: Stoichiometry</u>
- [DA] Set up conversion:
- [DA] Divide/Multiply [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
9.75526 × 10²⁴ molecules H₂O ≈ 9.8 × 10²⁴ molecules H₂O