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Sidana [21]
3 years ago
8

To plot the calibration curve, you need to prepare iron solutions with known concentrations and measure their absorbance.

Chemistry
1 answer:
kap26 [50]3 years ago
8 0

To plot the calibration curve, you need to prepare iron solutions with known concentrations and measure their absorbance. You need to pipet 0 mL of the diluted solution to have 0.00 mg of iron.

In spectrophotometry, to plot the calibration curve, you need to prepare solutions with known concentrations and measure their absorbance.

We have a standard iron solution with a concentration of 0.2500g/L of pure iron (C₁). We pipet 25.00mL (V₁) of this standard iron solution into a 500mL (V₂) volumetric flask and dilute up to the mark with distilled water.

We can calculate the concentration of the diluted solution (C₂) using the dilution rule.

C_1 \times V_1 = C_2 \times V_2\\\\C_2 = \frac{C_1 \times V_1}{V_2}  = \frac{0.2500 g/L \times 25.00 mL}{500 mL} = 0.0125 g/L

Then, if we wanted to prepare the blank, that is, the solution that contains the same matrix but not the analyte, and whose concentration in iron is 0.00 mg/L, we wouldn't pipet any of the diluted solution.

To plot the calibration curve, you need to prepare iron solutions with known concentrations and measure their absorbance. You need to pipet 0 mL of the diluted solution to have 0.00 mg of iron.

Learn  more: brainly.com/question/24195565

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Heat = mC(T2-T1)

By substituting the given values, we can calculate for C which is the specific heat of the material.

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A pycnometer is a precisely weighted vessel that is used for highly accurate density determinations. Suppose that a pycnometer h
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Answer:

5.758  is the density of the metal ingot in grams per cubic centimeter.

Explanation:

1) Mass of pycnometer = M = 27.60 g

Mass of pycnometer with water ,m= 45.65 g

Density of water at 20 °C = d =998.2 kg/m^3

1 kg = 1000 g

1 m^3=10^6 cm^3

998.2 kg/m^3=\frac{998.2 \times 1000 g}{10^6 cm^3}=0.9982 g/cm^3

Mass of water ,m'= m - M = 45.65 g -  27.60 g =18.05 g

Volume of pycnometer = Volume of water present in it = V

Density=\frac{Mass}{Volume}

V=\frac{m'}{d}=\frac{18.05 g}{0.9982 g/cm^3}=18.08 cm^3

2) Mass of metal , water and pycnometer = 56.83 g

Mass of metal,M' =  9.5 g

Mass of water when metal and water are together ,m''= 56.83 g - M'- M

56.83 g - 9.5 g - 27.60 g = 19.7 g

Volume of water when metal and water are together = v

v=\frac{m''}{d}=\frac{19.7 g}{0.9982 g/cm^3}=19.73 cm^3

Density of metal = d'

Volume of metal = v' =\frac{M'}{d'}

Difference in volume will give volume of metal ingot.

v' = v - V

v'=19.73 cm^3-18.08 cm^3=

v'=1.65 cm^3

Since volume cannot be in negative .

Density of the metal =d'

=d'=\frac{M'}{v'}=\frac{9.5 g}{1.65 cm^3}=5.758 g/cm^3

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