HF and NaF - If the right concentrations of aqueous solutions are present, they can produce a buffer solution.
<h3>What are buffer solutions and how do they differ?</h3>
- The two main categories of buffers are acidic buffer solutions and alkaline buffer solutions.
- Acidic buffers are solutions that contain a weak acid and one of its salts and have a pH below 7.
- For instance, a buffer solution with a pH of roughly 4.75 is made of acetic acid and sodium acetate.
<h3>Describe buffer solution via an example.</h3>
- When a weak acid or a weak base is applied in modest amounts, buffer solutions withstand the pH shift.
- A buffer made of a weak acid and its salt is an example.
- It is a solution of acetic acid and sodium acetate CH3COOH + CH3COONa.
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1.66 M is the concentration of the chemist's working solution.
<h3>What is molarity?</h3>
Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.
In this case, we have a solution of Zn(NO₃)₂.
The chemist wants to prepare a dilute solution of this reactant.
The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.
We want to know the concentration of this diluted solution.
As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:
= 
(1)
and we also know that:
n = M x 
If we replace this expression in (1) we have:
x 
= 
x
Where 1, would be the stock solution and 2, the solution we want to prepare.
So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is :
4.93 x 210 = 620 x
= 1.66 M
This is the concentration of the solution prepared.
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Percentage by volume of solution is the percentage volume of solute in total volume of solution.
Volume percentage (v/v%) = volume of solute / total volume of solution x 100%
volume of solute - 16.0 mL
total volume of solution - 155 mL
v/v% = 16.0 / 155 x 100% = 10.32%
this means that in a volume of 100 mL solution, 10.32 mL is acetone.
Answer:
T2 = 135.1°C
Explanation:
Given data:
Mass of water = 96 g
Initial temperature = 113°C
Final temperature = ?
Amount of energy transfer = 1.9 Kj (1.9×1000 = 1900 j)
Specific heat capacity of aluminium = 0.897 j/g.°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
Now we will put the values in formula.
Q = m.c. ΔT
1900 j = 96 g × 0.897 j/g.°C × T2 - 113°C
1900 j = 86.112 j/°C × T2 - 113°C
1900 j / 86.112 j/°C = T2 - 113°C
22.1°C + 113°C = T2
T2 = 135.1°C
