Ionic Equilibrium
Chemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibrium is established between the ions and unionized species in a system. Understanding the concept of ionic equilibrium is very important to answer the questions related to certain chemical reactions in chemistry.
Arrhenius Acid
Arrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.
Bronsted Lowry Base In Inorganic Chemistry
Bronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.
![### Balancing Chemical Equations: Coefficient of \( \text{H}_2\text{O} \)
**Problem Statement:**
What is the coefficient of \( \text{H}_2\text{O} \) when the following equation is properly balanced with the smallest set of whole numbers?
\[ \text{___} \, \text{Al}_4\text{C}_3 + \text{___} \, \text{H}_2\text{O} \rightarrow \text{___} \, \text{Al(OH)}_3 + \text{___} \, \text{CH}_4 \]
**Steps to Balance the Equation:**
1. **Identify the number of atoms of each element on both sides:**
- Left side:
- \( \text{Al}: 4 \times 1 = 4 \)
- \( \text{C}: 3 \times 1 = 3 \)
- \( \text{H}: 2 \times \text{coefficient of } \text{H}_2\text{O} \)
- \( \text{O}: 1 \times \text{coefficient of } \text{H}_2\text{O} \)
- Right side:
- \( \text{Al}: 1 \times \text{coefficient of } \text{Al(OH)}_3 \)
- \( \text{C}: 1 \times \text{coefficient of } \text{CH}_4 \)
- \( \text{H}: (3 \times \text{coefficient of } \text{Al(OH)}_3) + (4 \times \text{coefficient of } \text{CH}_4) \)
- \( \text{O}: 3 \times \text{coefficient of } \text{Al(OH)}_3 \)
2. **Balance each element one by one:**
- Start with Al (Aluminum):
- There are 4 aluminum atoms on the left side in 1 molecule of \( \text{Al}_4\text{C}_3 \), so put 4 in front of \( \text{Al(OH)}_3 \) on the right.
\[ \text{Al}_4\text{C}_3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd321eb40-4702-45a2-842a-571d0c67a0bc%2F7731ec13-7f38-4327-b9a4-9a6130937ed1%2Fhe3tdqn_processed.jpeg&w=3840&q=75)
Step by step
Solved in 2 steps with 2 images
