How can miscible liquids that different boiling points be separated
1 answer:
Answer:
By boiling and further condensing the liquid with the lowest boiling point.
Explanation:
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In this case, according to the attached diagram, it turns out possible for us to infer that the mechanism whereby miscible liquids with different boiling points are separated is distillation, because the flask is heated until the boiling point of the liquid with the lowest value, in order to boil it and subsequently condense it, whereas the liquid with the highest boiling point remains in the flask; and therefore, the two liquids are separated.
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<h3>Answer:</h3>
1.827 × 10²⁴ molecules H₂S
<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>Compounds</u>
- Writing Compounds
- Acids/Bases
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Step 1: Define</u>
103.4 g H₂S (Sulfuric Acid)
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of H - 1.01 g/mol
Molar Mass of S - 32.07 g/mol
Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 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 4 sig figs.</em>
1.82656 × 10²⁴ molecules H₂S ≈ 1.827 × 10²⁴ molecules H₂S
Answer:
(a) The proportion of dry air bypassing the unit is 14.3%.
(b) The mass of water removed is 1.2 kg per 100 kg of dry air.
Explanation:
We can express the proportion of air that goes trough the air conditioning unit as and the proportion of air that is by-passed as
, being
.
The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:
Replacing the first equation in the second one we have
(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.
It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.
The water removed of every 100 kg of dry air is
It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).
100 * 0.012 = 1.2 kgW