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
1. Change in color
2. Formation of bubbles
3. Formation of a precipitate
4. Begins to make an odor
Answer:
Cl2 + 2NaBr --> 2NaCl + Br2
Explanation:
This is a single displacement reaction where one side of the ionic compound switches with the other.
So, Cl2 + NaBr ---> NaCl + Br2
This isolates the Bromine and puts the Chlorine in it's place.
Then, balance out the equation like so and you should get
Cl2 + 2NaBr --> 2NaCl + Br2
There are 5.43 x 10²³ present in 187 grams of XeF₄
<h3>Further explanation </h3>
The mole is the number of particles(molecules, atoms, ions) contained in a substance
1 mol = 6.02.10²³ particles
Can be formulated
N=n x No
N = number of particles
n = mol
No = Avogadro's = 6.02.10²³
Moles can also be determined from the amount of substance mass and its molar mass
mass of XeF₄ = 187 g
mol XeF₄ (MW=207,2836 g/mol) :
Number of molecules :
Answer:
If More solute is added and it does not dissolve,then the original solution was saturated. if the added solute dissolves,then the original solution was unsaturated. A solution that has been allowed to reach equilibrium but which has extra undissolved solute at the bottom of the container must be saturated.
