<h2>
Answer:</h2>
390 g KNO₃
<h2>
General Formulas and Concepts:</h2><h3><u>Chemistry</u></h3>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3><u>Math</u></h3>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<h2>
Explanation:</h2>
<u>Step 1: Define</u>
2.3 × 10²⁴ formula units KNO₃
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of K - 39.10 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g.mol
Molar Mass of KNO₃ - 39.10 + 14.01 + 3(16.00) = 101.11 g/mol
<u>Step 3: Convert</u>
<u />
= 386.172 g KNO₃
<u>Step 4: Check</u>
<em>We are given 2 sig figs. Follow sig fig rules and round.</em>
386.172 g KNO₃ ≈ 390 g KNO₃
We can calculate years by using the half-life equation. It is expressed as:
A = Ao e^-kt
<span>where A is the amount left at t years, Ao is the initial concentration, and k is a constant.
</span>From the half-life data, we can calculate for k.
1/2(Ao) = Ao e^-k(1620)
<span>k = 4.28 x 10^-4
</span>
0.125 = 1 e^-<span>4.28 x 10^-4 (</span>t)
t = 4259 years
pressure......................................