Answer:
202 g/mol
Explanation:
Let's consider the neutralization between a generic monoprotic acid and KOH.
HA + KOH → KA + H₂O
The moles of KOH that reacted are:
0.0164 L × 0.08133 mol/L = 1.33 × 10⁻³ mol
The molar ratio of HA to KOH is 1:1. Then, the moles of HA that reacted are 1.33 × 10⁻³ moles.
1.33 × 10⁻³ moles of HA have a mass of 0.2688 g. The molar mass of the acid is:
0.2688 g/1.33 × 10⁻³ mol = 202 g/mol
Answer:
sugar and salt, as they can dissolve in water. Sugar and salt have a high solubility in water. So, sugar and salt can form a solution with water.
Explanation:
<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]:
Answer:
13.8 moles of water produced.
Explanation:
Given data:
Moles of KMnO₄ = 3.45 mol
Moles of water = ?
Solution:
Chemical equation:
16HCl + 2KMnO₄ → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O
Moler ratio of water and KMnO₄:
KMnO₄ : H₂O
2 : 8
3.45 : 8/2×3.45 = 13.8 mol
Hence, 3.45 moles of KMnO₄ will produced 13.8 mol of water.