Answer = option D
0.668 mg/ml
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From the plotted calibation line graph the concentrations of
unknown
can be ascertained.
The regression equation fro the above graph is given as ---
It is of the form y = mx-c with negaive intercept on absobance axis.
y = absorbance axis ( data)
x = concentration of
in mg/ml
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given that y = 0.650
X = ?
Substitute for y and determine x from the regression equation ( this is usually done from calibated curves in instrumentation laboratories )
0.650 = 0.99 x- 0.011
x = 0.6676 mg/ml
= 0.668 mg/ml
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From a fine divisions graph the same can be ascertained from graph also.
attempt SOUD ork - A... P4 18 Review Homework StatCrunch Do Homework - A... his Course...
please do 1 through 4. Thank you.
0 1. Fe 1.SCN Procedure B: 1. Prepare ICE tables for beakers 2 - 6 using the example below as a guideline (you will have 5 different ICE tables). Table 3. Sample "ICE" table. Fe(aq)*3 + SCN(aq) = FeSCN2 Initial: **M calculated **M calculated using your Table 2 using your Table 2 volumes volumes Change: - 1x - 1x Equilibrium: M 1x 1x = concentration calculated from Procedure A slope-intercept equation +1x M...