Reagent blanks gave the following values: Blank 1 2 3 4 5 6 7 Signal, nA 1.4 2.2 1.7 0.9 0.4 1.5 0.7 The slope of the calibration curve for higher concentrations is m = 0.229 nA/µM. (a) Find the signal detection limit and the minimum detectable concentration. (b) What is the concentration of analyte in a sample that gave a signal of 7.0 nA. [2.9; 7.3; 25.0] 8. Low concentrations of EDTA near the detection limit gave the following dimensionless instrument readings: Replicate Reading 1 2 3 4 5 6 7 9. 10 175 104 164 193 131 189 155 133 151 176 Ten blanks had a mean reading of 45.0. The slope of the calibration curve is 1.75 x 10° M-'. Estimate the signal and concentration detection limits and the lower limit of quantitation of EDTA. [129; 4.8 x 10-8; 1.6 x 10-"] 9. An analysis of the calibration data for the determination of lead based upon its flame emission spectrum yielded an equation: S = 1.12C + 0.312 where C is the Pb concn in ppm and S is a measure of the relative emission intensity. The following replicate data were obtained: Concentration, ppm Number of replicates Mean value of S Standard deviation 10.0 10 11.62 0.15 1.00 10 1.12 0.025 24 0.0296 0.0082 Calculate (a) the calibration sensitivity and (b) the detection limit. [1.12; 0.022]
Reagent blanks gave the following values: Blank 1 2 3 4 5 6 7 Signal, nA 1.4 2.2 1.7 0.9 0.4 1.5 0.7 The slope of the calibration curve for higher concentrations is m = 0.229 nA/µM. (a) Find the signal detection limit and the minimum detectable concentration. (b) What is the concentration of analyte in a sample that gave a signal of 7.0 nA. [2.9; 7.3; 25.0] 8. Low concentrations of EDTA near the detection limit gave the following dimensionless instrument readings: Replicate Reading 1 2 3 4 5 6 7 9. 10 175 104 164 193 131 189 155 133 151 176 Ten blanks had a mean reading of 45.0. The slope of the calibration curve is 1.75 x 10° M-'. Estimate the signal and concentration detection limits and the lower limit of quantitation of EDTA. [129; 4.8 x 10-8; 1.6 x 10-"] 9. An analysis of the calibration data for the determination of lead based upon its flame emission spectrum yielded an equation: S = 1.12C + 0.312 where C is the Pb concn in ppm and S is a measure of the relative emission intensity. The following replicate data were obtained: Concentration, ppm Number of replicates Mean value of S Standard deviation 10.0 10 11.62 0.15 1.00 10 1.12 0.025 24 0.0296 0.0082 Calculate (a) the calibration sensitivity and (b) the detection limit. [1.12; 0.022]
Principles of Instrumental Analysis
7th Edition
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Chapter17: Applications Of Infrared Spectrometry
Section: Chapter Questions
Problem 17.13QAP
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Question
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Kindly help me solve #8. The answers are on the bracket. thank you.
![1:30 PM Sun Oct 11
* 100%
a learn-ap-southeast-1-prod-fleet02-xythos.s3.ap-southeast-1.amazonaws.com
Diglidi, IA
J.0
5.0
4.2
4.0
0.0
Reagent blanks gave the following values:
Blank
1
3
4
5
7
Signal, nA
1.4
2.2
1.7
0.9
0.4
1.5
0.7
The slope of the calibration curve for higher concentrations is m = 0.229 nA/µM.
(a) Find the signal detection limit and the minimum detectable concentration.
(b) What is the concentration of analyte in a sample that gave a signal of 7.0 nA. [2.9; 7.3; 25.0]
8.
Low concentrations of EDTA near the detection limit gave the following dimensionless instrument readings:
Replicate
Reading
1
2
3
4
5
7
8
9.
10
175
104
164
193
131
189
155
133
151
176
Ten blanks had a mean reading of 45.0. The slope of the calibration curve is 1.75 x 10° M²'. Estimate the
signal and concentration detection limits and the lower limit of quantitation of EDTA. [129; 4.8 x 10-8; 1.6 x 10-7]
An analysis of the calibration data for the determination of lead based upon its flame emission spectrum yielded an
equation: S =1.12C + 0.312 where C is the Pb concn in ppm and S is a measure of the relative emission intensity.
The following replicate data were obtained:
9.
Concentration, pm
Number of replicates
Mean value of S
Standard deviation
10.0
10
11.62
0.15
1.00
10
1.12
0.025
24
0.0296
0.0082
Calculate (a) the calibration sensitivity and (b) the detection limit. [1.12; 0.022]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6edcf33c-bf36-493f-a581-e1351e207dd6%2F7450c215-c1c2-42bc-b2a6-47ce32aab8ce%2Faxqxml_processed.png&w=3840&q=75)
Transcribed Image Text:1:30 PM Sun Oct 11
* 100%
a learn-ap-southeast-1-prod-fleet02-xythos.s3.ap-southeast-1.amazonaws.com
Diglidi, IA
J.0
5.0
4.2
4.0
0.0
Reagent blanks gave the following values:
Blank
1
3
4
5
7
Signal, nA
1.4
2.2
1.7
0.9
0.4
1.5
0.7
The slope of the calibration curve for higher concentrations is m = 0.229 nA/µM.
(a) Find the signal detection limit and the minimum detectable concentration.
(b) What is the concentration of analyte in a sample that gave a signal of 7.0 nA. [2.9; 7.3; 25.0]
8.
Low concentrations of EDTA near the detection limit gave the following dimensionless instrument readings:
Replicate
Reading
1
2
3
4
5
7
8
9.
10
175
104
164
193
131
189
155
133
151
176
Ten blanks had a mean reading of 45.0. The slope of the calibration curve is 1.75 x 10° M²'. Estimate the
signal and concentration detection limits and the lower limit of quantitation of EDTA. [129; 4.8 x 10-8; 1.6 x 10-7]
An analysis of the calibration data for the determination of lead based upon its flame emission spectrum yielded an
equation: S =1.12C + 0.312 where C is the Pb concn in ppm and S is a measure of the relative emission intensity.
The following replicate data were obtained:
9.
Concentration, pm
Number of replicates
Mean value of S
Standard deviation
10.0
10
11.62
0.15
1.00
10
1.12
0.025
24
0.0296
0.0082
Calculate (a) the calibration sensitivity and (b) the detection limit. [1.12; 0.022]
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