Answer:
Ksp = 8.8x10⁻⁵
Explanation:
<em>Full question is:</em>
<em>After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2+ is 2.8 × 10–2 M. What is Ksp for PbCl2?</em>
<em />
When an excess of PbCl₂ is added to water, Pb²⁺ and Cl⁻ ions are produced following Ksp equilibrium:
PbCl₂(s) ⇄ Pb²⁺ + 2Cl⁻
Ksp = [Pb²⁺] [Cl⁻]²
If an excess of PbCl₂ was added, an amount of Pb²⁺ is produced (X) and twice Pb²⁺ is produced as Cl⁻ (2X):
Ksp = [X] [2X]²
Ksp = 4X³
As X is the amount of Pb²⁺ = 2.8x10⁻²M:
Ksp = 4(2.8x10⁻²)³
<h3>Ksp = 8.8x10⁻⁵</h3>
<h3>
Answer:</h3>
7.4797 g AlF₃
<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>Stoichiometry</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃
[Given] 15.524 g KF
<u>Step 2: Identify Conversions</u>
[RxN] 6 mol KF = 2 mol AlF₃
Molar Mass of K - 39.10 g/mol
Molar Mass of F - 19.00 g/mol
Molar Mass of Al - 26.98 g/mol
Molar Mass of KF - 39.10 + 19.00 = 58.1 g/mol
Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 5 sig figs.</em>
7.47966 g AlF₃ ≈ 7.4797 g AlF₃
Answer:
wavelength = 0.56 × 10⁻² m
Explanation:
Given data:
Frequency of radiation = 5.4 × 10¹⁰ Hz
Wavelength = ?
Solution:
Formula:
speed of light = wavelength × frequency
wavelength = speed of light / frequency
wavelength = 3 × 10⁸ m/s / 5.4 × 10¹⁰ Hz ( Hz= 1/s)
wavelength = 0.56 × 10⁻² m