C becuase that is the one that you would have to do becuase that is the only option
Answer: a) The concentration after 8.8min is 0.17 M
b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.
Explanation:
Expression for rate law for first order kinetics is given by:
![t=\frac{2.303}{k}\log\frac{a}{a-x}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B2.303%7D%7Bk%7D%5Clog%5Cfrac%7Ba%7D%7Ba-x%7D)
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) concentration after 8.8 min:
![8.8\times 60s=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{a-x}](https://tex.z-dn.net/?f=8.8%5Ctimes%2060s%3D%5Cfrac%7B2.303%7D%7B6.7%5Ctimes%2010%5E%7B-4%7Ds%5E%7B-1%7D%7D%5Clog%5Cfrac%7B0.25%7D%7Ba-x%7D)
![\log\frac{0.25}{a-x}=0.15](https://tex.z-dn.net/?f=%5Clog%5Cfrac%7B0.25%7D%7Ba-x%7D%3D0.15)
![(a-x)=0.17M](https://tex.z-dn.net/?f=%28a-x%29%3D0.17M)
b) for concentration to decrease from 0.25M to 0.15M
![t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{0.15}\\\\t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\times 0.20](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B2.303%7D%7B6.7%5Ctimes%2010%5E%7B-4%7Ds%5E%7B-1%7D%7D%5Clog%5Cfrac%7B0.25%7D%7B0.15%7D%5C%5C%5C%5Ct%3D%5Cfrac%7B2.303%7D%7B6.7%5Ctimes%2010%5E%7B-4%7Ds%5E%7B-1%7D%7D%5Ctimes%200.20)
![t=687s](https://tex.z-dn.net/?f=t%3D687s)
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.
160.0g
Mass =volume x density = 200.0 mL x 0.8 g/mL= 160.0 g