Atomic <span>mass He = 4.002 u.m.a
4.002 g --------------- 6.02x10</span>²³ atoms
50 g ------------------ ? atoms
50 x ( 6.02x10²³) / 4.002
= 3.01x10²⁵ / 4.002
= 7.52x10²⁴ atoms of He
1 half-life - 4.5 days
x half-lives - 27 days
x=(1*27)/4.5=6
6 <span>half-lives have elapsed after 27 days.</span>
Https://www.youtube.com/watch?v=7ZtLaCucV98
The answer will be 25.92 L for 17.0 c
<h3>
Answer:</h3>
3.0 × 10²³ molecules AgNO₃
<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
- Writing Compounds
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
85 g AgNO₃ (silver nitrate)
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[PT] Molar Mass of Ag - 107.87 g/mol
[PT] Molar Mass of N - 14.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
3.01313 × 10²³ molecules AgNO₃ ≈ 3.0 × 10²³ molecules AgNO₃