Answer:
[NaOH} = 0.4 M
Explanation:
In a reaction of neutralization, we determine the equivalence point of the titration. In this case, we have a strong base and a strong acid.
(H₂SO₄, is considered strong, but the first deprotonation is weak)
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
As we have 2 protons in the acid, we need 2 OH⁻ from the base to form 2 molecules of water.
In the equivalence point we know mmoles of base = mmoles of acid
Let's finish the excersise with the formula
25 mL . M NaOH = 28.2 mL . 0.355M
M NaOH = (28.2 mL . 0.355M) / 25 mL → 0.400
Answer:
The electron stays in an excited state for a short time. When the electron transits from an excited state to its lower energy state, it will gice off the same amound of energy needed to raise to that level. This emitted energy is a photon.
<h3>
Answer:</h3>
5.2 mol 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>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 6HCl + Fe₂O₃ → 2FeCl₃ + 3H₂O
[Given] 10.4 mol HCl
<u>Step 2: Identify Conversions</u>
[RxN] 6 mol HCl = 3 mol H₂O
<u>Step 3: Stoichiometry</u>
- Set up:
![\displaystyle 10.4 \ mol \ HCl(\frac{3 \ mol \ H_2O}{6 \ mol \ HCl})](https://tex.z-dn.net/?f=%5Cdisplaystyle%2010.4%20%5C%20mol%20%5C%20HCl%28%5Cfrac%7B3%20%5C%20mol%20%5C%20H_2O%7D%7B6%20%5C%20mol%20%5C%20HCl%7D%29)
- Multiply/Divide:
![\displaystyle 5.2 \ mol \ H_2O](https://tex.z-dn.net/?f=%5Cdisplaystyle%205.2%20%5C%20mol%20%5C%20H_2O)