Answer:
The experiment that did not give the results described is e) The Rutherford experiment proved the Thomson "plum-pudding" model of the atom to be essentially correct.
Explanation:
- First of all, let's start with the experiment listed in <em>b) The Rutherford experiment was useful in determining the nuclear charge on the atom</em>. This is true, because Rutherford was able to tell through this experiment that the atoms consisted mostly of empty space, with a positively charged nucleus in its center (he was able to tell it was positive because of the deflections observed on the positive alpha particles he used).
- Next, we have experiment <em>c) The electric discharge tube proved that electrons have a negative charge</em>. This is also true, given that Thomson was able to correctly determine that there existed particles that were negatively charged, called electrons (he determined this by observing that the rays deviated from the negatively charged plate of the discharge tube in his experiment).
- Experiment d) <em>Milikan's oil-drop experiment showed that the charge on any particle was a simple multiple of the charge on the electron</em>, also states the correct results. He was able to determine the charge of many electrically charged droplets of oil, and found out that the charges were a simple multiple of a base value, which he proposed to be that of the negative charge of an electron.
- Finally, we take a look at experiment <em>e) The Rutherford experiment proved the Thomson "plum-pudding" model of the atom to be essentially correct</em>. This is the experiment that did not give the result described. In fact, the Rutherford experiment proved that Thomson's "plum-pudding" model was not correct. This is because he determined that atoms consist mostly of empty space, with a dense, positively charged nucleus in its center, surrounded by negatively charged electrons. If Thomson's model were true, Rutherford would have not observed deflected alpha particles in his experiment.
<u>Answer:</u> The concentration of reactant after the given time is 0.0205 M
<u>Explanation:</u>
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = 11.0 min = 660 s (Conversion factor: 1 min = 60 s)
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount of the reactant = 0.400 M
[A] = amount left after decay process = ?
Putting values in above equation, we get:
![4.50\times 10^{-3}s^{-1}=\frac{2.303}{660s}\log\frac{0.400}{[A]}](https://tex.z-dn.net/?f=4.50%5Ctimes%2010%5E%7B-3%7Ds%5E%7B-1%7D%3D%5Cfrac%7B2.303%7D%7B660s%7D%5Clog%5Cfrac%7B0.400%7D%7B%5BA%5D%7D)
![[A]=0.0205M](https://tex.z-dn.net/?f=%5BA%5D%3D0.0205M)
Hence, the concentration of reactant after the given time is 0.0205 M
I don’t understand this one sorry
It is being aware of things around you and the decisions you make. Being in control of your own actions.
