<h3>
Answer:</h3>
0.35 M
<h3>
Explanation:</h3>
<u>We are given;</u>
- Initial volume as 35.0 mL or 0.035 L
- Initial molarity as 12.0 M
- Final volume is 1.20 L
We are required to determine the final molarity of the solution;
- Dilution involves adding solvent to a solution to make it more dilute which reduces the concentration and increases the solvent while maintaining solute constant.
- Using dilution formula we can determine the final molarity.
M1V1 = M2V2
M2 = M1V1 ÷ V2
= (12.0 M × 0.035 L) ÷ 1.2 L
= 0.35 M
Thus, the final concentration of the solution is 0.35 M
Things go sublime when changing automatically into vapor when it is heated, usually forming a solid figure.
<h3>
Answer:</h3>
0.0157 g Au
<h3>
General Formulas and Concepts:</h3>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.113 g Au
<u>Step 2: Identify Conversions</u>
Molar Mass of Au - 197.87 g/mol
<u>Step 3: Convert</u>
<u />
= 0.015733 g Au
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
0.015733 g Au ≈ 0.0157 g Au
Answer:
3MnO₄⁻ + 24H⁺ + 5Au → 5Au³⁺ + 3Mn²⁺ + 12H₂O.
Explanation:
- KMnO₄ is strong oxidizing agent which is reduced and oxidizes Au according to the two-half reactions:
The oxidation reaction: Au → Au³⁺ + 3e.
The reduction reaction: MnO₄⁻ + 8H⁺ + 5e → Mn²⁺ + 4H₂O.
- To obtain the net redox reaction, we multiply the oxidation reaction by 5 and the reduction reaction by 5 to equalize the no. of electrons in the two-half reactions.
<em>So, the net redox reaction will be:</em>
<em>3MnO₄⁻ + 24H⁺ + 5Au → 5Au³⁺ + 3Mn²⁺ + 12H₂O.</em>
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