Answer:
b is the anwer
Explanation:
the option is the explanation
I don't know the options but usually a small strainer or a coffee thing u put over a cup and let the water seep down and the sugar stays.
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Beer's law is no longer valid.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:
Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.
Explanation:
The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:
![Q = \displaystyle \frac{[\mathrm{SO_2\, (g)}]}{[\mathrm{O_2\, (g)}]}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdisplaystyle%20%5Cfrac%7B%5B%5Cmathrm%7BSO_2%5C%2C%20%28g%29%7D%5D%7D%7B%5B%5Cmathrm%7BO_2%5C%2C%20%28g%29%7D%5D%7D)
.
Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both ![[\mathrm{SO_2\, (g)}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BSO_2%5C%2C%20%28g%29%7D%5D)
and ![[\mathrm{O_2\, (g)}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BO_2%5C%2C%20%28g%29%7D%5D)
will increase if the pressure is increased through compression. However, because 
and 
have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient 
.
As a result, the increase in pressure will have no impact on the value of 
. If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.
