Can you please help to answer the following ? thank you!!! \"State in words and
ID: 998012 • Letter: C
Question
Can you please help to answer the following ? thank you!!!
"State in words and full sentences the results of your calibration curve, %Mn in eac sample, and uncertainty each %Mn measurement."
P44 Standard Curve Data: Average Calibration Volume Student Dilution Concen systematic Final e after 0.9000 Stock Used Volume tration mL 0.7000 06000 05000 correction 2.00 3.00 5.00 8.00 9.00 50.00 4.00E-06 50.00 6.00E-06 50.00 1.00E-05 50.00 1.60E-05 50.0 .80E-05 0.1840 0.2640 0.4487 0 3000 0.7920 00E00 500E-06 1.00E-05 1.50E-05 2.00E-05 Concentration g/ml Insert or calculate linear regression data below: Unlocked cells below for your use. Be sure to label slope 43611.11 0.00766667 y-intercept 1 Mean Y 0.4787 Sml 357.219 04322 Sb R 0.9998 0.004357 Sy Devx2 1.49E-10 S(trial 2) 1.095E-07 Sample Data: Sample #: 18 Mn concentrat Dilution volume Aliquot dilution after Final volume 50) (mL) Absolute uncertainty in %Mn concentration in sample Concen- Mn Concen- ion in ration in tration insteel Mass used Steel digestion (10,20, or (usually Average cuvet used 100 mL sample flask (g/mL) (mass %) %Relatve uncertainty 0.7039 0.9639 0.6956 0.9753 25) (mL) Absorbance (g/mL) trial 1 0.8073 100.0000 10.000050.0000 0.50330.0000114 0.00005683 0.006785 trial 2 0.8071 100.0000 10.000050.0000 0.4973 0.0000112 0.006784 0.00005614 Avg % Mn in Sample mi Bormos absolute 0.700 uncertainty0.006785Explanation / Answer
"Estado de las palabras y frases completas los resultados de la curva de calibración,% de Mn en cada muestra, y la incertidumbre de medición% cada Mn."
the beer laws say than Abs = eCl where C is contration where e and l are constants how this law have a linear comportament we can graph as Y = mx + b where m = el x is C and Y is Abs
the result s say than 46311x + 0.0077 = Abs whit a error of asociate to cuve of calibration of 1x10-7
for a trial 1 have a absorvance Abs = 0.5033 as X = C g/ml of Mn only need cleared of equation and get
46311x + 0.0077 = Abs =) x = (Abs - 0.0077 )/46311 = C relacing for the absorvance of trial 1 we have
x = (0.5033 - 0.0077 )/46311 = 1.07x10-5g/ml as it come to varius dilution s
we have than
1.07x10-5g/ml x 50ml/10ml = C =) C = 5.35x10-5g/ml and the begining solution have a V = 100ml
mMn = 100mlx5.35x10-5g/ml = 5.35x10-3g and the mass used for the solution was of 0.8073g
%Mn = 100% x 5.35x10-3g / 0.8073g = 0.66%
for the trial 2 are equal
for a trial 2 have a absorvance Abs = 0.4973 as X = C g/ml of Mn only need cleared of equation and get
46311x + 0.0077 = Abs =) x = (Abs - 0.0077 )/46311 = C relacing for the absorvance of trial 1 we have
x = (0.4973 - 0.0077 )/46311 = 1.06x10-5 g/ml as it come to varius dilution s
we have than
1.06x10-5g/ml x 50ml/10ml = C =) C = 5.30x10-5g/ml and the begining solution have a V = 100ml
mMn = 100mlx5.3x10-5g/ml = 5.30x10-3g and the mass used for the solution was of 0.8073g
%Mn = 100% x 5.30x10-3g / 0.8073g = 0.66%
for the error
1x10-7g/ml x 50ml/10ml = C =) C = 5x10-7g/ml and the begining solution have a V = 100ml
mMn = 100mlx5x10-7g/ml = 5x10-5g and the mass used for the solution was of 0.8073g
%Mn = 100% x 5x10-5g / 0.8073g = 0.06% any Mn have a error of 0.06%