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zimovet [89]
3 years ago
10

Suppose you want to create a 6 ng/μL solution in a 25 mL volumetric flask. However, this concentration cannot really be accurate

ly prepared in a single step due to the small amount of mass that would need to be measured. Therefore, you decided to conduct this as a two step process. You will create a high concentration stock solution in one 10 mL flask. You then transfer 0.1 mL into a second 25 mL volumetric flask that will be the concentration of 6 ng/μL. What is the mass of chemical you need to add to the stock flask to achieve the desired concentration after dilution?
Chemistry
1 answer:
Over [174]3 years ago
4 0

Answer:

Mass of chemical = 1.5 mg

Explanation:

Step 1: First calculate the concentration of the stock solution required to make the final solution.

Using C1V1 = C2V2

C1 = concentration of the stock solution; V1 = volume of stock solution; C2 = concentration of final solution; V2 = volume of final solution

C1 = C2V2/V1

C1 = (6 * 25)/ 0.1

C1 = 1500 ng/μL = 1.5 μg/μL

Step 2: Mass of chemical added:

Mass of sample = concentration * volume

Concentration of stock = 1.5 μg/μL; volume of stock = 10 mL = 10^6 μL

Mass of stock = 1.5 μg/μL * 10^6 μL = 1.5 * 10^6 μg = 1.5 mg

Therefore, mass of sample = 1.5 mg

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8 0
4 years ago
When (R)-6-bromo-2,6-dimethylnonane is dissolved in CH3OH, nucleophilic substitution yields an optically inactive solution. When
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Answer:

See explanation

Explanation:

The nucleophile here is CH3OH. We know that CH3OH is a good nucleophile that promotes SN2 reanction. However, (R)-6-bromo-2,6-dimethylnonane is a tertiary alkyl halide so the reaction proceeds by SN1 mechanism. This means that a racemic mixture is obtained at the end of the reaction because the attack occurs at the stereogenic carbon atom (6R) hence the product is optically inactive.

On the other hand, when (5R)-2-bromo-2,5-dimethylnonane is reacted with CH3OH, an optically active product is obtained because; though a tertiary alkyl halide and reaction occurs by SN1 mechanism, the attack does not occur at the stereogenic carbon atom (5R). Therefore, an optically active product is obtained in this case.

7 0
3 years ago
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6 0
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I believe your answer would be A.
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