Answer:
Half-reactions:
Cr³⁺ + 1e⁻ → Cr²⁺; Zn → Zn²⁺ + 2e⁻
Net ionic equation:
2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺
Explanation:
The Cr³⁺ is reduced to Cr²⁺:
<h3>
Cr³⁺ + 1e⁻ → Cr²⁺ -Half-reaction 1-</h3>
Zn is oxidized to Zn²⁺:
<h3>
Zn → Zn²⁺ + 2e⁻ -Half-reaction 2-</h3>
Twice the reduction of Cr:
2Cr³⁺ + 2e⁻ → 2Cr²⁺
Now this reaction + Oxidation of Zn:
2Cr³⁺ + 2e⁻ + Zn → 2Cr²⁺ + Zn²⁺ + 2e⁻
<h3>2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺ - Net ionic equation</h3>
<h3>
Answer:</h3>
5.00 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>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.<u>
</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.01 × 10²⁴ atoms O₂
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<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>
4.99834 mol O₂ ≈ 5.00 mol O₂
Answer:
The answer is (H30+) =3,55e-8M and (OH-)=2,82e-7M
Explanation:
We use the formulas:
pH= - log(H30+) and Kwater=(H30+)x(OH-)
pH= - log(H30+) ----< (H30+)= antilog- pH=antilog- 7,45=3,55E-8M
Kwater=(H30+)x(OH-)
(OH-)=Kwater/(H30+)= 1,00e-14/3,55e-8 = 2,82e-7
The numbers may be off by a few, if you worked it out, since my periodic table is for AP chemistry. Hope this helps!!!