Respuesta :
Answer:
The answer is "212.5 ml"
Explanation:
Please find the complete question in the attached file.
In 2 ml of [tex]35\%[/tex] SDS it makes 120 ml of [tex]7\%[/tex] SDS.
To mix x ml of [tex]35\%[/tex] SDS
[tex]\to V_1S_1 = V_2S_2\\\\\to V_1 = x \ ml\\\\\to V_2 = 120\ ml\\\\\to S_1 = 35\% \\\\ \to S_2 = 7\%\\\\\to V_1 = \frac{V_2S_2}{S_1}\\\\[/tex]
[tex]= \frac{(120 \times 7)}{35}\\\\=24 \ ml\\\\[/tex]
added water [tex]= 120-24 = 96\ ml[/tex]
So we have to add 24ml from [tex]35\%[/tex] SDS and add 96ml water to male 120ml [tex]7\%[/tex] SDS.
1500ml of 40% Tris Buffer solution
Let us assume we have to mix y ml of 40% Tris buffer solution
[tex]\to V_1S_1 = V_2S_2\\\\\to V_1 = y \ ml\\\\\to V_2 = 250\ ml\\\\\to S_1 = 40\% \\\\ \to S_2 = 6\%\\\\\to V_1 = \frac{V_2S_2}{S_1}\\\\[/tex]
[tex]= \frac{(250 \times 6)}{40}\\\\= 37.5\ ml[/tex]
added water [tex]= 250-37.5 = 212.5\ ml[/tex]
So we have to add 37.5 ml from 40% Tris Buffer and add 212.5 ml water to male 250 ml 6% Tris Buffer.

Answer:
- We can prepare 150 mL of 8% TRIS-buffer by adding [tex]30 ml[/tex] from [tex]40\%[/tex] Tris Buffer and add [tex]120 ml[/tex] water
Explanation:
1500ml of 40% Tris Buffer solution
Let us assume we have to mix y ml of 40% Tris buffer solution
We know,
[tex]V_1S_1 = V_2S_2[/tex]
where,
[tex]V_1 = y ml\\\\V_2 = 150 ml\\\\S_1 = 40\% and S_2 = 8\%[/tex]
So,
[tex]V1 = \frac{V_2S_2}{S_1}\\\\V_1 = \frac{150*8}{40}\\\\V_1 = 30 ml[/tex]
So water we have to add
[tex]= 150-30\\\\= 120 ml[/tex]
So we have to add 30 ml from 40% Tris Buffer and add 120 ml water to male 250 ml 6% Tris Buffer.
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