Answer:
The change in entropy is -1083.112 joules per kilogram-Kelvin.
Explanation:
If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (
), in joules per gram-Kelvin, by the following model:

(1)
Where:
- Mass, in kilograms.
- Specific heat of water, in joules per kilogram-Kelvin.
,
- Initial and final temperatures of water, in Kelvin.
If we know that
,
,
and
, then the change in entropy for the entire process is:


The change in entropy is -1083.112 joules per kilogram-Kelvin.
<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
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<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
From the information given, the total volume of rubbing alcohol is 88.2 ml
68.6 % of this volume is isopropanol.
We will assume 88.2 ml represents 100% volume, so the volume of water will be 31.4 %
The volume of isopropanol is
68.6/100 x 88.2 → 0.686 × 88.2 = 60.505 ml
The volume of isopropanol is 60.5 ml.
Volume of water will be 88.20 - 60.5 = 27.7 ml
(27.7 / 88.2 × 100 = 31.4% )
Adding 60.5 ml of isopropanol to 27.7 ml of water to make up 88.2 ml will give 68.6 % v/v isopropanol to water solution.
Answer 19.9g. I’ve took the test last week at my uncle randy’s house