The following are spectrometric signals for methane in H_2. CH_4 (Vol %) 0 0.062
ID: 481662 • Letter: T
Question
The following are spectrometric signals for methane in H_2. CH_4 (Vol %) 0 0.062 0.122 0.245 0.486 0.971 0.921 Signal (mV) 9.1 47.5 95.6 193.8 387.5 812.5 1, 671.9 Subtract the blank value (9.1) from all other values. Then use LINEST to find the slope and intercept and their uncertainties. Plot the data table and the curve for this question and attach it to this activity. Print the data table and the curve for this question and attach it to this activity. What mass of Fe(NH_4)_2(SO_4)_2 middot 2H_2O (solid) is required to prepare a 500 mL solution containing 100 ppm (m:v) in Fe? (show work below) Using this stock solution (from 2a), what aliquot must be used to prepare calibration solutions, 100-mL volume, of the following concentrations: 0.100 ppm, 0.500 ppm, 2.00 ppm, 4.00 ppm and 7.00 ppm. Show calculations in the space below and write answers in the table in 2c The atomic absorbance spectrometry (AAS) data is shown below. Construct a calibration curve based on the data above and use LINEST to find the slope and intercept and their uncertainties. Plot the data for this question, with the linear equation expression and R^2 inserted in the graph. Are there any outliers in the data? Remove the 5th data point from the graph and recalculate R^2. Has the correlation factor improved' If R^2 has improved to two-9's (or better), recalculate and plot this second graph with the new linear equation expression inserted. Recalculate LINEST and use this new curve for question 2d, below. If a blood sample contains (m/m) 0.335% iron determine how much blood (mu L) should be diluted in a 250 mL volumetric flask with the appropriate solvent to give an absorbance reading of 0.2500. Assume that the solution being analyzed has a density of 1000 g/cc but blood has a density of 1.070 g/cc.Explanation / Answer
2a. To prepare 500 ml of 100 ppm solution of Fe
320.14 g of Fe(NH4)2(SO4)2.2H2O has 55.845 g of Fe
So for 100 ppm (0.1 g) Fe in 500 ml we would need,
= 0.1 x 320.14/55.845 x 2
= 0.30 g Fe(NH4)2(SO4)2.2H2O
2c. Using,
V1 = C2V2/C1
with,
C1 and C2 are concentrations of stock and final solution
V1 and V2 are volumes of stock and final volume of solution taken
Feeding the values,
To prepare 100 ml of 0.1 ppm solution
Volume of stock needed V1 = 0.1 x 100/100
= 0.1 ml
To prepare 100 ml of 0.5 ppm solution
Volume of stock needed V1 = 0.5 x 100/100
= 0.5 ml
To prepare 100 ml of 2 ppm solution
Volume of stock needed V1 = 2 x 100/100
= 2 ml
To prepare 100 ml of 4 ppm solution
Volume of stock needed V1 = 4 x 100/100
= 4 ml
To prepare 100 ml of 7 ppm solution
Volume of stock needed V1 = 7 x 100/100
= 7 ml
the remainder in each case would be water used for making the solution to 100 ml total volume.
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