What is the boiling point of methanol, if ∆H°for vaporization is 35.3 kJ/mol, and ∆S° for vaporization is 105 J/mol K?
1 answer:
Answer: 336.2K
Explanation:
The boiling point is obtained from the standard enthalpy and entropy of vaporization as shown in the image attached.
Standard enthalpy of vaporization=35.3KJmol-1=35.3×10^3 Jmol-1
Standard entropy of vaporization= 105 JMol-1K-1
The formula and details of the solution are shown in the image attached.
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<h3>Answer:</h3>
126 g Fe
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<u>Step 1: Define</u>
2.25 mol Fe
<u>Step 2: Identify Conversions</u>
Molar Mass of Fe - 55.85 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
125.663 g Fe ≈ 126 g Fe