<h3>
Answer:</h3>
1.1 × 10²² atoms Au
<h3>
General Formulas and Concepts:</h3>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<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>
3.7 g Au
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of Au - 196.97 g/mol
<u>Step 3: Convert</u>
<u />
= 1.13121 × 10²² atoms Au
<u>Step 4: Check</u>
<em>We are given 2 sig figs. Follow sig fig rules and round.</em>
1.13121 × 10²² atoms Au ≈ 1.1 × 10²² atoms Au
Hello
Explanation:
well I don't know if I can answer you well but I try .... I think it's HC2H202Cl... Thank you for your attention ✌️
Answer:
Anhydride, any chemical compound obtained, either in practice or in principle, by the elimination of water from another compound. Examples of inorganic anhydrides are sulfur trioxide, SO3, which is derived from sulfuric acid, and calcium oxide, CaO, derived from calcium hydroxide
Explanation:
<h3>
<em><u>examples</u></em><em><u>.</u></em></h3>
1)acid anhydride.
2)basic anhydrides.
<h3>
<em><u>reactions</u></em><em><u>. </u></em></h3>
1)reaction with water
(CH3CO)2O + H2O → 2 CH3CO2H.
<h3>
Answer:</h3>
121 mol CH₄
<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>Organic</u>
- Writing chemical compounds
- Writing organic structures
- Prefixes
- Alkanes, Alkenes, Alkynes
<u>Atomic Structure</u>
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
7.31 × 10²⁵ molecules CH₄
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
<u />
<u />
<u />
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
121.388 mol CH₄ ≈ 121 mol CH₄