Answer:
The answer is indeed true
Explanation:
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
<span>The gaseous mixture contains 414.0 torr of h2(g), 345.7 torr of n2(g), and 80.1 torr of ar(g). Then, total pressure of the gas would be: 414 torr + 345.7 torr + 80.1 torr= 839.8 torr
The mole fraction of each gas:
H2= 414 torr/</span>839.8 torr= 0.49
N2= 345.7 torr/839.8 torr= 0.41
Ar= 80.1 torr/839.8 torr= 0.10
Answer:
See Explanation
Explanation:
Because you have to get through the d-block electron configurations for the rest of the p-block elements which is a hassle to do. You need to know how to account for electron stability, from which subshell to remove electrons, etc. because it is all weird for d-block.
