Answer: A) More base is likely required to reach the endpoint for the diprotic acid than for the monoprotic acid under these conditions
Explanation:
The monoprotic acid (HA) has a valency of 1 and diprotic acid
has a valency of 2.
As the concentration and volume of the diprotic acid and the monoprotic acids are equal.
The neutralization reaction for monoprotic acid is:

The neutralization reaction for diprotic acid is:

Thus more number of moles of base are required for neutralization of diprotic acid and thus the volume required will be more as concentration and volume of the diprotic acid and the monoprotic acids are equal.
Answer:
the valence electrons of atoms in a pure metal can be modeled as a sea of electrons
<h3>
Answer:</h3>

<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
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2C + O₂ → 2CO₂
[Given] 0.25 moles O₂
[Solve] moles CO₂
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol O₂ → 2 mol CO₂
<u>Step 3: Stoichiometry</u>
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:
