Answer:
This indicates that the enzyme is a type of protein.
Explanation:
It is important to remember that proteins are composed of vast numbers of amino acids. Because these amino acids are tiny units, they cannot function as a catalyst on their own.
However, when they form a polymer, the protein enzyme will possess varying shapes, sizes, and both physical and chemical attributes differing from a single monomer.
Additionally, for proteins to function actively, a specific number of amino acids must combine to create a distinct shape suited to interact with another molecule, thus accelerating the chemical reaction and functioning as an enzyme.
Problem 2
You begin with 216 micrograms of Fermium - 253. After three days, the quantity halves, resulting in 108 micrograms left.
Another three days pass. Beginning with 108 micrograms, this amount gets halved again, leaving 54 micrograms.
Finally, after another three-day span, starting from 54 micrograms, you again halve this amount to reach 27 micrograms.
#days Amount in micrograms
0 216
3 108
6 54
9 27
Problem One
Your example is Nitrogen. Begin by completing the table, then formulate some rules to help prepare for possible alternate elements in the test. This approach is quite useful.
Table
Bond Energy Kj/Mol Bond Length pico meters
N - N 167 145
N=N 418 125
N≡N 942 110
Rules
As the number of bonds INCREASES, the energy within the bond also INCREASES
As the number of bonds INCREASES, the distance of the bond DECREASES.
Answer:
The rate law for the decomposition reaction is:
![R=k[D]^2](https://tex.z-dn.net/?f=R%3Dk%5BD%5D%5E2)
The unit for the rate constant will be 
Explanation:
The rate law can be expressed as:
![R=k[D]^x](https://tex.z-dn.net/?f=R%3Dk%5BD%5D%5Ex)
..[1]
When the drug concentration is tripled, the decomposition rate rises by a factor of nine.
![[D]'=3[D]](https://tex.z-dn.net/?f=%5BD%5D%27%3D3%5BD%5D)
![R'=k[D]'^x](https://tex.z-dn.net/?f=R%27%3Dk%5BD%5D%27%5Ex)
...[2]
[1] ÷ [2]
![\frac{R}{R'}=\frac{k[D]^x}{k[D']^x}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7BR%27%7D%3D%5Cfrac%7Bk%5BD%5D%5Ex%7D%7Bk%5BD%27%5D%5Ex%7D)
![\frac{R}{9R}=\frac{k[D]^x}{k[3D]^x}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B9R%7D%3D%5Cfrac%7Bk%5BD%5D%5Ex%7D%7Bk%5B3D%5D%5Ex%7D)
Solving for x results in:
x = 2.
This indicates a second-order reaction.
The decomposition reaction's rate law is:
The unit for the rate constant will be:
![k=\frac{R}{[D]^2}=\frac{M/s}{(M)^2}=M^{-1}s^{-1}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7BR%7D%7B%5BD%5D%5E2%7D%3D%5Cfrac%7BM%2Fs%7D%7B%28M%29%5E2%7D%3DM%5E%7B-1%7Ds%5E%7B-1%7D)
The unit for the rate constant will be .
Answer: Rearrange the lone pairs of electrons from the outer atom(s) to create double or triple bonds with the central atom.
Explanation:
Response:
The pKa value is 13.0.
Clarification:
pKa + pKb = 14
For trimethylamine, Kb = 6.3 × 
Calculating pKb: pKb = - log (6.3 × )
= 1.0
Thus, pKa = 14 - pKb = 14 - 1.0
pKa = 13.0
Verification: The typical range for pKa in weak acids is from 2 to 13.
