I dont know that’s a hard one I’m in 7 th grade so google it bruh lol
In a non-flowering plant, the embryo is in spores found in the stem, and in a flowering plant, the embryo is in seeds found in the flower.
(don’t count on my answer but I think it might be this one and I apologize if you get it wrong)
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Beer's law is no longer valid.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.
The answer is 4.
Gases have low densities, because of the increased space between hight-energy particles.
Answer:
Explanation:
The result will be affected.
The mass of KHP weighed out was used to calculate the moles of KHP weighed out (moles = mass/molar mass).
Not all the sample is actually KHP if the KHP is a little moist, so when mass was used to determine the moles of KHP, a higher number of moles than what is actually present would be obtained (because some of that mass was not KHP but it was assumed to be so. Therefore, there is actually a less present number of moles than the certain number that was thought of.
During the titration, NaOH reacts in a 1:1 ratio with KHP. So it was determined that there was the same number of moles of NaOH was the volume used as there were KHP in the mass that was weighed out. Since there was an overestimation in the moles of KHP, then there also would be an overestimation in the number of moles of NaOH.
Thus, NaOH will appear at a higher concentration than it actually is.
