Answer:
1.42 M
Explanation:
First calculate the amount of moles.
that's done by dividing the mass with the molecular mass so 660g / 310.18 g/mol = 2.13 mol
Then you can calculate the molarity by dividing the moles with the volume so 2.13 mol / 1.5 l = 1.42 M
(without rounding: 1.418531175 M)
Answer:
False
Explanation:
While we do know that A. Leeuwenhoek used a simple microscope that consisted of only 1 lens, Hooke used a compound microscope. Although, after trying a compound microscope, Hooke found out that it strained his eyes and continued to use a simple microscope for his <em>Micrographia</em>.
Thus, we can say that the (compound) microscopes used today are different than the (simple) microscope used by Hooke and Leeuwenhoek.
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
Answer:
'Oxidation states have changed.'
Explanation:
Redox Reactions can be identifiable by a change in oxidation number/states of two of the atoms in the reaction. Any reaction in which no oxidation numbers/state change is not a redox reaction.