Answer:
- <em><u>Step 2 (the slow step).</u></em>
Explanation:
The rate-determining step is always the slow step of a mechanism.
That is so, because it is the slow step which limits the reaction.
Imaging that for assembling a toy you have process of three steps:
- 1. order ten pieces, which you can do in 1 minute: meaning that you can order order the pieces for 60/1 = 60 toys in 1 hour.
- 2. glue the pieces and hold the toy until the glue hardens, which takes 1 hour: meaning finishingh 1 toy in 1 hour.
- 3. pack the toy, which takes 2 minutes: meaning that you can pack 60/2 = 30 toys in one hour.
The time to glue and hold one toy until the glue hardens determines that you can assemble 1 toy in 1 hour and not 60 toys or 30 toys.
Thus, the step that determines the rate at which the reaction happens is the slowest step: step 2.
Chimps feed on seeds more
2.49 x 10^46 is the answer
Answer: The given statement is true.
Explanation:
It is known that due to increase in temperature ice melts. Therefore, during ice age there is also melting of ice and solid state of water changes into liquid state of water.
Therefore, this water moves from its initial place and changes its position result in the change of land and specific areas.
Thus, we can conclude that the statement during an ice age, land can move and specific areas can be permanently changed is true.
Answer:
(a)
(b)
Explanation:
Hello,
(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:
![\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BNOBr%5D%7D%3Dkt%20%2B%5Cfrac%7B1%7D%7B%5BNOBr%5D_0%7D%5C%5C%5Cfrac%7B1%7D%7B%5BNOBr%5D%7D%3D%5Cfrac%7B0.8%7D%7BM%2As%7D%2A22s%2B%5Cfrac%7B1%7D%7B0.086M%7D%3D%5Cfrac%7B29.3%7D%7BM%7D%5C%5C)
![[NOBr]=\frac{1}{29.2/M}=0.0342M](https://tex.z-dn.net/?f=%5BNOBr%5D%3D%5Cfrac%7B1%7D%7B29.2%2FM%7D%3D0.0342M)
(b) Now, for a second-order reaction, the half-life is computed as shown below:
![t_{1/2}=\frac{1}{k[NOBr]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BNOBr%5D_0%7D)
Therefore, for the given initial concentrations one obtains:
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