Dear friend,
How are you doing? I finally switched to my new school and it's been so hard adjusting. But I know I have a new chance at making new friends and memories. Also, I can learn new things, like interacting with new people and how to keep my anxiety low. I really miss us hanging out and how much we'd laugh and get in trouble at class. But i really hope to see you soon & that it hasn't gone terribly wrong for you since i left.
your friend,
name
<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
The answer is 174Ir→4He+170Re
Answer: I think the formula is PV=nRT and I divide both sides by RT, but this is as far as I can get in my equation before I get stumped: (751 mm Hg) (8.3 L)/ (309 K) Can you help?
Explanation: