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laiz [17]
3 years ago
14

The following conditions must be met before a standard calibration curve equation can be used to solve for the concentration of

an unknown solution:
a. The same instrument must be used to measure the unknown solution as was used to measure the known (standard) solutions that were used to create the calibration curve.
b. The analyte in the unknown solution must be the same analyte (or type of analyte) that is present in the known (standard) solutions that were used to create the calibration curve.
c. The equation from the calibration curve can only be linear.
d. Domain and range restrictions must be observed.
Chemistry
1 answer:
zzz [600]3 years ago
6 0

Answer:

The same instrument must be used to measure the unknown solution as was used to measure the known (standard) solutions that were used to create the calibration curve.

The analyte in the unknown solution must be the same analyte (or type of analyte) that is present in the known (standard) solutions that were used to create the calibration curve.

Domain and range restrictions must be observed.

Explanation:

Calibration curves are tools necessary in understanding the instrumental response for any analyte.

A calibration curve is obtained by preparing a set of  standard solutions with known concentrations of the analyte. The instrument response for each concentration is measured  and plotted against the concentration of the standard solution. The linear portion of this plot may be used to determine the unknown concentration of a sample of the analyte.  

The equation of the best-fit line is  used to determine the concentration of the unknown sample.

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Calculate the standard potential for the following galvanic cell: Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s) which has the overall bala
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Answer:

1.06  V  

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The half-cell and cell reactions for Ni | Ni^2+ || Ag^+ | Ag are

Ni → Ni^2+ + 2e-                     E° = 0.257   V

<u>2Ag^+ 2e- → 2Ag               </u>    <u>E° = 0.7996 V </u>

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To three significant figures, the standard potential for the cell is 1.06 V .

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How many grams of solid nickel will be plated if an aqueous nickel(II) sulfate solution is electroplated over 10.5 min with a co
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Answer:

613 mg

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Number of fargday's  $=\frac{It}{96500}$

Here, I = 9.20 A

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So, $\frac{It}{96500}$

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0.0104 mol = 59 x0.0104 gm Ni

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                    = (0.613 x 1000 ) mg of Ni

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