C6H5-C2H4-COO-CH3-CH2-CH3
1 answer:
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<h3>Answer:</h3>
0.0253 mol H₂O
<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> </u>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Step 1: Define</u>
[Given] 0.456 g H₂O (water)
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.025305 mol H₂O ≈ 0.0253 mol H₂O
Answer:
Explanation:
From the given information:
TO start with the molarity of the solution:
= 0.601 mol/kg
= 0.601 m
At the freezing point, the depression of the solution is
Using the depression in freezing point, the molar depression constant of the solvent
The freezing point of the solution
The molality of the solution is:
Molar depression constant of solvent X,
Hence, using the elevation in boiling point;
the Vant'Hoff factor