Answer:
A) Dilute the unknown so that it will have an absorbance within the standard curve. Once the diluted unknown concentration is determined, the full strength concentration can be calculated if the dilution process is recorded. Beer's law only applies to dilute solutions, so diluting the unknown is better than making new standards.
Explanation:
Beer's law states that <em>absorbance is proportional to the concentrations of the absorbing species</em>. This is verified in the case of diluted solutions (0≤0.01 M) of most substances. <u>As a solution gets more concentrated, solute molecules interact between themselves because of their proximity. </u>When a molecule interacts with another, the change in their electric properties (including absorbance) is probable. That's why <u>the plot of absorbance versus concentration stops being a straight line</u>, and <u>Beer's law is no longer valid.</u>
Therefore, if the absorbance value is higher than the highest standard, dilutions should be made. Once this concentration is determined, the full strength concentration can be calculated with the inverse of the dilution.
Answer:
This question is incomplete
Explanation:
This question is incomplete because the result of the described experiment would have better determined the type of scientific explanation to profer. However, the type of material that will preserve the relative hotness or temperature of the hot coffee for the longest time will be a material than can resist heat transfer. These materials tend to keep hot substances hot by not allowing the heat of the coffee to be conducted or pass through it. These materials are mostly insulators or made by placing an insulator between two heat conductors.
Generally, heat is usually transferred from a region of higher concentration to a region of lower concentration, hence when the heat is denied of this transfer, the heat will remain trapped in the "heat-donor" substance (in this case the hot coffee). Thus, the material chosen (A, B or C) will be the material that resists heat transfer the most based on the explanation above.
If the moon did not rotate we would see all hemispheres of the moon as it revolves around the Earth and not just the phases.
Answer:
![[Pb^{2+}]=3.9 \times 10^{-2}M](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%3D3.9%20%5Ctimes%2010%5E%7B-2%7DM)
this is the concentration required to initiate precipitation
Explanation:
⇄
Precipitation starts when ionic product is greater than solubility product.
Ip>Ksp
Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.
This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.
![Ip=[Pb^{2}][2Cl^-]^2=Ksp](https://tex.z-dn.net/?f=Ip%3D%5BPb%5E%7B2%7D%5D%5B2Cl%5E-%5D%5E2%3DKsp)

lets solubility=S
![[Pb^{2+}] = S](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%20%3D%20S)
![[Cl^-]=2S](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D2S)
![Ksp=[Pb^{2+}]\times [Cl^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BPb%5E%7B2%2B%7D%5D%5Ctimes%20%5BCl%5E-%5D%5E2)


![S=\sqrt[3]{\frac{Ksp}{4} }](https://tex.z-dn.net/?f=S%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BKsp%7D%7B4%7D%20%7D)

this is the concentration required to initiate precipitation
Answer
I think it might be B
Explanation