Assuming the concentration of stock solution is 50% sodium phosphate buffer solution, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.
<h3>What volume of a stock Sodium phosphate buffer and water is needed to 12 mL of 25% sodium phosphate buffer of pH 4?</h3>
The process of preparing solutions from stock solutions of higher concentration is known as dilution.
Dilution is done with the aid of the dilution formula given below:
where
- C1 is the concentration of stock solution
- V1 is the volume of stock solution required to prepare a diluted solution
- C2 is the concentration of the diluted solution prepared
- V2 is the final volume of the diluted solution
From the data provided:
C1 is not given
V1 is unknown
C2 = 25%
V2 = 12 mL
- Assuming C1 is 50% solution
Volume of stock, V1, required is calculated as follows:
V1 = C2V2/C1
V1 = 25 × 12 /50
V1 = 6 mL
Therefore, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.
Learn more about dilution formula at: brainly.com/question/7208546
6.02×[10]1 ×64.8=3900.96 moles
Explanation:
It is given that,
Initial orbit of electrons, 
Final orbit of electrons, 
We need to find energy, wavelength and frequency of the wave.
When atom make transition from one orbit to another, the energy of wave is given by :

Putting all the values we get :

We know that : 
So,

Energy of wave in terms of frequency is given by :


Also, 
is wavelength
So,

Hence, this is the required solution.
The question is incomplete, here is the complete question:
A chemist adds 370.0 mL of a 2.25 M iron(III) bromide solution to a reaction flask. Calculate the mass in grams of iron(III) bromide the chemist has added to the flask. Round your answer to 3 significant digits
<u>Answer:</u> The mass of iron (III) bromide is 246. grams
<u>Explanation:</u>
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

We are given:
Molarity of solution = 2.25 M
Molar mass of iron (III) bromide = 295.6 g/mol
Volume of solution = 370.0 mL
Putting values in above equation, we get:

Hence, the mass of iron (III) bromide is 246. grams