A liquid has definite volume but no definite shape .
So what I know is that enzyme and substrate are like lock and key meaning that when the active site of the enzyme changes, the enzyme will not fit to the substrate which will lead the enzyme to denature. Hope this helps.
<h2>a)
The rate at which 
is formed is 0.066 M/s</h2><h2>b)
The rate at which molecular oxygen 
is reacting is 0.033 M/s</h2>
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Rate in terms of disappearance of 
= ![-\frac{1d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BNO%5D%7D%7B2dt%7D)
= 0.066 M/s
Rate in terms of disappearance of = ![-\frac{1d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_2%5D%7D%7Bdt%7D)
Rate in terms of appearance of = ![\frac{1d[NO_2]}{2dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1d%5BNO_2%5D%7D%7B2dt%7D)
1. The rate of formation of
![-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BNO_2%5D%7D%7B2dt%7D%3D%5Cfrac%7B1d%5BNO%5D%7D%7B2dt%7D)
![\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s](https://tex.z-dn.net/?f=%5Cfrac%7B1d%5BNO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B2%7D%7B2%7D%5Ctimes%200.066M%2Fs%3D0.066M%2Fs)
2. The rate of disappearance of
![-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO%5D%7D%7B2dt%7D)
![-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%200.066M%2Fs%3D0.033M%2Fs)
Learn more about rate law
brainly.com/question/13019661
https://brainly.in/question/1297322
1-H NMR spectroscopy tool will be used for distinguishing a sample of 1,2,2-tribromopropane from 1,1,2-tribromopropane.
The preferred method for determining or validating the structure of organic molecules or those containing protons is H NMR. When compared to other nuclei, a solution-state proton spectrum may be obtained relatively quickly, and it contains a wealth of knowledge regarding a compound's structure.
It can be calculated by simply counting the number of unique hydrogens on one side of the symmetry plane will give you the count of signals individual molecules emit in a 1H NMR spectrum.
Therefore, 1-H NMR spectroscopy tool will be used for distinguishing a sample of 1,2,2-tribromopropane from 1,1,2-tribromopropane.
To know more about 1-H NMR spectroscopy
brainly.com/question/20111886
#SPJ4
First find the oxidation states of the various atoms:
<span>in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 </span>
<span>Note that N gained electrons, ie, was reduced; Cr was oxidized </span>
<span>Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out </span>
<span>B.NH4+ → N2 </span>
<span>Which of the following is an oxidation half-reaction? </span>
<span>A.Sn 2+ →Sn 4+ + 2e- </span>
<span>Sn lost electrons so it got oxidized</span>
