In an acidic solution, the concentration of H+ is greater than the concentration of OH-. The pH will be less than 7.
In a basic solution, the concentration of OH- is greater than the concentration of H+. The pH will be greater than 7.
In a neutral solution, the concentration of H+ ions to OH-ions will be equal, and will therefore have a pH of 7. (This is due to water autoionization, which we usually ignore because it is small in other circumstances.)
<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
=
= 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)
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Physical. you didn't burn it or create a chemical reaction
Answer:
Concentration of nitrate in the new solution = 0.007 M
Explanation:
Given:
Concentration nitrate solution = 0.070 m
Volume of aliquote of the nitrate solution is add = 10.0 ml
Total volume = 100 ml
Find:
Concentration of nitrate in the new solution
Computation:
Number of M. mole = 0.070 m x 10.0 ml
Number of M. mole = 0.7 m-moles
Concentration of nitrate in the new solution = 0.7 m-moles / 100 ml
Concentration of nitrate in the new solution = 0.007 M