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.

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
What is the volume of 1.2 moles of water vapor at STP?
The answer is 26.9
Answer:
4.767 grams of KCl are produced from 2.50 g of K and excess Cl2
Explanation:
The balanced equation is
2 K+ Cl2 --->2 KCI
Here the limiting agent is K. Hence, the amount of KCl will be calculated as per the mass of 2.50 gram of K
Mass of one atom/mole of potassium is 39.098 grams
Number of moles is 2.5 grams = 
So, 2 moles of K produces 2 moles of KCL
0.064 moles of K will produces 0.064 moles of KCl
Mass of one molecule of KCl is 74.5513 g/mol
Mass of 0.064 moles of KCl is 4.767 grams
Colligative
properties calculations are used for this type of problem. Calculations are as
follows:<span>
ΔT(freezing point) = (Kf)(molality)
ΔT(freezing point)
= 1.86 °C kg / mol (molality)
</span>Tf - 102.08 = 1.86m
Tf = 1.86m + 102.08
The concetration of the solution is needed in order to obtain a specific value.