Answer:
The final dilution is 1:400
Explanation:
Let's analyze what we are told: we have an initial 1:5 dilution of protein lysate. This means that the initial solution (stock solution) was diluted 5 times. Then, from this dilution the student prepared another dilution taking 2 mL of the first dilution in 8 mL of water. This is the same as saying we took 1 mL of first dilution in 4 mL of water (the ratio is the same), so we now have a second 1:4 dilution of the first dilution (1:5). Finally, the student made a third 1:20 dilution, this means that the second dilution was further diluted 20 times.
So, to calculate the final dilution of protein lysate, we have to multiply all the dilution factors of every dilution prepared: in this case we have a final dilution of 1:20, this means we have a factor dilution of 20. But it was previously diluted 4 times, so we have a factor dilution of 20×4 = 80. However, this dilution was also previously diluted 5 times, so the new dilution factor is 80 × 5 = 400
This means that the final dilution of the compound was diluted a total of 400 times compared to the initial concentration of stock solution.
The melting point of the sample if it is not dried completely after filtering the recrystallized product will have a broad range and will occur at lower range than the actual value.
What is melting point?
Melting point is the temperature at which the solid form of a given substance changes to the liquid form at atmospheric pressure. It is important because, by using the physical property of a substance the substance can be identified.
The sharp range melting point of the substance indicates the purity of the substance. If the sample is not dried completely after recrystallisation, the melting point will have a broad range.
Therefore, if the sample given is not dried completely, it will be impure and the decreases the melting point of the substance. So the actual melting point of the substance cannot be determined.
To learn more about the melting point click on the given link brainly.com/question/40140
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Answer:
7.74%
<em>Hope this helped! Have a good day.</em>
These are buffer systems (solutions of substances) with a steady concentration of hydrogenium ions. At addition to such solution of a small amount of acid or alkali does not change рН.
For example:
CH₃COOH + CH₃COONa acetate buffer
KH₂PO₄ + K₂HPO₄ potassium phosphate buffer
