Answer:
The answer is Relative plenitude alludes to the amount of a specific isotope is available in a given measure of test.
Explanation:
The 'relative plenitude' of an isotope implies the level of that specific isotope that happens in nature. Most components are comprised of a blend of isotopes. The total of the rates of the particular isotopes must indicate 100%. The relative nuclear mass is the weighted normal of the isotopic masses. The percent plenitude of every sort of sweets reveals to you what number of every sort of Aufbau there are in each 100 CANDIES. Percent wealth is additionally relative plenitude. This is only a method for giving us a photo on which kind exists all the more every now and again.
The pressure of the gas is expected to increase in accordance to Boyle's law.
<h3>What is Boyle's law?</h3>
Boyle's law states that, the volume of a given mass of gas is inversely proportional to its pressure at constant temperature.
By implication, when the piston is lowered and the volume of the gas is decreased, the pressure of the gas is expected to increase in accordance to Boyle's law.
Learn more about Boyle's law: brainly.com/question/1437490
<h3>
Answer:</h3>
5.55 mol C₂H₅OH
<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>
- Reading a Periodic Tables
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
[Given] 500. g C₆H₁₂O₆ (Glucose)
[Solve] moles C₂H₅OH (Ethanol)
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH
[PT] Molar mass of C - 12.01 g/mol
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
<u>Step 3: Stoichiometry</u>
- [DA] Set up conversion:

- [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>
5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH
Molality is the number of moles of solutes in 1 kg of solvent.
the molality of solution to be prepared is 2.0 molal.
therefore 2 moles in 1 kg water.
the mass of Li₂S required is - 46 g/mol x 2.0 mol = 92 g
the mass in 1 kg of solvent is - 92 g
Therefore mass of Li₂S required in 1600.0 g is - 92 g/kg x 1.6 kg = 147.2 g
Any compound with multiple covalent bonds