Answer:
3 H1 NMR signals
Explanation:
NB: kindly check the diagram of the chemical compound in the attached picture.
This particular Question is based on the part of chemistry which is known as spectroscopy. Spectroscopy is used in the Determination or in identifying chemical compounds. H'NMR works on the principle of nuclear magnetic resonance.
In order to solve this question, one has to count the number of hydrogen in unique location. The diagram in the attached show how hydrogen is been counted.
The numbers of signals is the number of different chemical environments in which hydrogen atoms are located.
NB: signals is also the same as peak in H'NMR.
Hence, the number of H1 NMR signals in this chemical compound is 3.
H^+(aq) + OH^-(aq) ---> H2O(l)
<span>Na^+ and ClO4^- are the spectator ions.</span>
Answer:
b
Explanation:
it's b because I just went over that frome my class and got it correct
The overall reaction is given by:
The fast step reaction is given as:
The slow step reaction is given as:
(slow step 
)
Now, the expression for the rate of reaction of fast reaction is:
![r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]](https://tex.z-dn.net/?f=r_%7B1%7D%3Dk_%7B1%7D%5BNO%5D%5BBr_%7B2%7D%5D-k_%7B-1%7D%5BNOBr_%7B2%7D%5D)
The expression for the rate of reaction of slow reaction is:
![r_{2}=k_{2}[NOBr_{2}] [NO]](https://tex.z-dn.net/?f=r_%7B2%7D%3Dk_%7B2%7D%5BNOBr_%7B2%7D%5D%20%5BNO%5D)
Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as ![[NOBr_{2}]](https://tex.z-dn.net/?f=%5BNOBr_%7B2%7D%5D)
takes place in this reaction.
The expression of rate of formation is:
= ![k_{2}[NOBr_{2}][NO]](https://tex.z-dn.net/?f=k_%7B2%7D%5BNOBr_%7B2%7D%5D%5BNO%5D)
(1)
Now, consider that the fast step is always is in equilibrium. Therefore, 
![k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]](https://tex.z-dn.net/?f=k_%7B1%7D%5BNO%5D%5BBr_%7B2%7D%5D%3D%20k_%7B-1%7D%5BNOBr_%7B2%7D%5D)
![[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]](https://tex.z-dn.net/?f=%5BNOBr_%7B2%7D%5D%20%3D%20%5Cfrac%7Bk_%7B1%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5BBr_%7B2%7D%5D)
Substitute the value of in equation (1), we get:
![\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28NOBr%29%7D%7Bdt%7D%3Dk_%7B2%7D%5BNOBr_%7B2%7D%5D%5BNO%5D)
=![k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]](https://tex.z-dn.net/?f=k_%7B2%7D%20%5Cfrac%7Bk_%7B1%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5BBr_%7B2%7D%5D%5BNO%5D)
= ![\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]](https://tex.z-dn.net/?f=%5Cfrac%7Bk_%7B1%7Dk_%7B2%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5E%7B2%7D%5BBr_%7B2%7D%5D)
Thus, rate law of formation of 
in terms of reactants is given by .
Hybridisation influences the bond length and bond enthalpy strength in organic compounds. The sp hybrid orbital contains more s character and hence it is closer to its nucleus and forms shorter and stronger bonds than the sp3 hybrid orbital.
