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
40% solution of glucose is where the solution contains, by weight, 40% glucose and 60% water.
Therefore, if the total weight of the solution is 250 g,
mass of the glucose (C6H12O6) = 250 g * 40% = 100 g
mass of water (H2O) = 250 g * 60% = 150 g
Mass of water can also be calculated by subtracting the weight of glucose from the total weight of the solution:
mass of water = 250g-100g = 150g.
When it comes to physical changes like phase changes, there are two types of heat energy: sensible heat and latent heat. Sensible heat is the heat absorbed/released when you heat the substance but it doesn't change phase. An example would be heating lukewarm water. The substance is liquid all throughout. Latent heat, on the other hand, is the heat absorbed/released when there is a phase change. An example would be boiling water, because it changes liquid to vapor.
Hence, for freezing liquid, you use the latent heat, specifically the heat of fusion. The answer should be
2.5 g * (1 mol/18.02 g) * 6.03 kJ/mol = 0.84 kJ/mol
The answer is not in the choices. You only use Hvap if you boil water.
Chadwick, Thompson, Rutherford, Bohr
10. For example, decagon and decade.
