Answer:
<em>C</em><em> </em><em>.</em><em> </em><em>the</em><em> </em><em>dramatic</em><em> </em><em>scenery</em><em> </em><em>created</em><em> </em><em>by</em><em> </em><em>volcanic</em><em> </em><em>eruptions</em><em> </em><em>attracts</em><em> </em><em>tourists</em><em>. </em>
9.096491 lbs/gal hope it helps
Answer:
lol I hate chemical but let me give some advice
Explanation:
Please go on Khan Academy or look at your notes and I promise you can figure out! Seriously, I am trying to be helpful not like the annoying teacher that says "figure it out"
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 .
The molecular weight of a given compound would simply the
sum of the molar weights of each component.
The molar masses of the elements are:
C = 12 amu
H = 1 amu
N = 14 amu
O = 16 amu
where 1 amu = 1 g / mol
Since there are 6 C, 5 H, 1 N and 2 O, therefore the
total molecular weight is:
molecular weight = 6 (12 amu) + 5 (1 amu) + 1 (14 amu) +
2 (16 amu)
molecular weight = 123 amu
Therefore the molecular weight of nitrobenzene is 123 amu
or which is exactly equivalent to 123 g / mol.
