Answer: Option (c) is the correct answer.
Explanation:
When an acid or base is added to a solution then any resistance by the solution in changing the pH of the solution is known as a buffer.
This is because a buffer has the ability to not get affected by the addition of small amounts of an acid or a base. As a result, it helps in maintaining the pH of the solution.
In the give case, when we add the HCl then more number of protons will dissociate. This causes the acetate to react with the protons and leads to the formation of acetic acid.
We know that acetic acid is a weak acid and it dissociates partially or feebly. Therefore, no change in pH will take place.
Thus, we can conclude that equation
represents the chemical reaction that accounts for the fact that acid was added but there was no detectable change in pH.
Answer:
5.158 mol/L
Explanation:
To find the molarity, you need to use the formula:
Molarity (M) = moles / volume (L)
You have been grams sodium carbonate. You need to (1) convert grams Na₂CO₃ to moles (via molar mass), then (2) convert moles Na₂CO₃ to moles HCl (via mole-to-mole ratio from equation), then (3) convert mL to L (by dividing by 1,000), and then (4) use the molarity equation.
<u>Steps 1 - 2:</u>
2 HCl + 1 Na₂CO₃ ----> 2 NaCl + H₂O + CO₂
6.5287 g Na₂CO₃ 1 mole 2 moles HCl
-------------------------- x ------------- x ------------------------- = 0.12318 mole HCl
106 g 1 mole Na₂CO₃
<u>Step 3:</u>
23.88 mL / 1,000 = 0.02388 L
<u>Step 4:</u>
Molarity = moles / volume
Molarity = 0.12318 mole / 0.02388 L
Molarity = 5.158 mole/L
**mole/L is equal to M**
Do not ionize in solutions
Poor conductors of electricity/heat
Low melting/boiling points
gases or liquids at room temperature
Answer:
colloidal system liquid dispersed in gas.
Answer:
0.50 M
Explanation:
Given data
- Mass of sodium sulfate (solute): 7.1 g
- Volume of solution: 100 mL
Step 1: Calculate the moles of the solute
The molar mass of sodium sulfate is 142.04 g/mol. The moles corresponding to 7.1 grams of sodium sulfate are:

Step 2: Convert the volume of solution to liters
We will use the relation 1 L = 1000 mL.

Step 3: Calculate the molarity of the solution
