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r-ruslan [8.4K]
3 years ago
6

Recovery standards are a necessary tool for determining exactly how much of a particular analyte you are able to extract from a

sample. You have a recovery standard that contains 4.0 mg Fe/1.0000 g sample, d.w. Your analysis determines that you extracted 3.5 mg Fe from 1.0101 g sample. A sample that you ran in parallel with the recovery standard returned a measurement of 1.7 mg Fe in 0.9582 g sample. How much iron (mg Fe/g sample) is actually present in your sample?
Chemistry
1 answer:
Dafna11 [192]3 years ago
5 0

Answer:

2.05mg Fe/ g sample

Explanation:

In all chemical extractions you lose analyte. Recovery standards are a way to know how many analyte you lose.

In the problem you recover 3.5mg Fe / 1.0101g sample: <em>3.465mg Fe / g sample. </em>As real concentration of the standard is 4.0 mg / g of sample the percent of recovery extraction is:

3.465 / 4×100 = <em>86,6%</em>

As the recovery of your sample was 1.7mg Fe / 0.9582g, the Iron present in your sample is:

1.7mg Fe / 0.9582g sample× (100/86.6) = <em>2.05mg Fe / g sample</em>

<em></em>

I hope it helps!

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emmainna [20.7K]

Answer:

27 min

Explanation:

The kinetics of an enzyme-catalyzed reaction can be determined by the equation of Michaelis-Menten:

v = \frac{vmax[S]}{Km + [S]}

Where v is the velocity in the equilibrium, vmax is the maximum velocity of the reaction (which is directed proportionally of the amount of the enzyme), Km is the equilibrium constant and [S] is the concentration of the substrate.

So, initially, the velocity of the formation of the substrate is 12μmol/9min = 1.33 μmol/min

If Km is a thousand times smaller then [S], then

v = vmax[S]/[S]

v = vmax

vmax = 1.33 μmol/min

For the new experiment, with one-third of the enzyme, the maximum velocity must be one third too, so:

vmax = 1.33/3 = 0.443 μmol/min

Km will still be much smaller then [S], so

v = vmax

v = 0.443 μmol/min

For 12 μmol formed:

0.443 = 12/t

t = 12/0.443

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3 years ago
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Shkiper50 [21]
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velikii [3]

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koban [17]

Answer:

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Explanation:

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A~+~BC->AB~+~C

With this in mind, the reaction would be:

Cu~+~PtCl_2->Pt~+~CuCl_2

I hope it helps!

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