<h3>
Answer:</h3>
4 cm³
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtract Property of Equality
<u>Chemistry</u>
<u>Gas Laws</u>
Density = Mass over Volume
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
D = 25 g/cm³
m = 100 g
<u>Step 2: Solve for </u><em><u>V</u></em>
- Substitute variables [D]:

- Multiply <em>V</em> on both sides:

- Isolate <em>V</em>:

<h3>
Answer:</h3>
16.7 g 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>Stoichiometry</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)
[Given] 1.85 mol NaOH
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol NaOH → 1 mol H₂O
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
16.6685 g H₂O ≈ 16.7 g H₂O
Answer: Option (b) is the correct answer.
Explanation:
It is known that in a gas, molecules are away from each other due to more kinetic energy between its particles. As a result, there are more number of collisions between them.
So, when we apply pressure on a gas then its molecules come closer to each other. Due to which there occurs decrease in its volume.
Thus, we can conclude that to investigate the compressibility of a gas increase the pressure on it.