<h3>
Answer:</h3>
1.83 × 10⁻⁷ mol Au
<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>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.60 × 10⁻⁵ g Au (Gold)
<u>Step 2: Identify Conversions</u>
Molar Mass of Au - 196.97 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au
Answer:
Neutrons and Protons, Electrons
Explanation:
Neutrons and protons are in the inside as electrons orbit around it in a circular pattern
Yeah no one is gonna read all that babe.
The element that gains electrons, becomes reduced.
While the one which loses electrons, becomes oxidized.
In this equation,
CH₃OH + Cr₂O₇²⁻---- --> CH₂O + Cr³⁺.
By balancing the equation, we will get:
3CH₃OH + Cr₂O₇²⁻ + 8H⁺ --> 3CH₂O + 2Cr³⁺ + 7H₂O
Here the oxidation state of Cr changes from +6 to +3 that is it is being reduced thus serving as a oxidizing agent while other element retain their charges.
Here Cr₂O₇²⁻ is reduced while CH₃OH is oxidized.
So Cr₂O₇²⁻ serves as a oxidizing agent, while CH₃OH serves as reducing agent .
Answer:i read it Explanation: