Answer:
Oi ima french Not really
Explanation:
7-7 = 0 andbecause it is so ima done
Answer : The rate law for formation of NOBr based on this mechanism is, ![\frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2](https://tex.z-dn.net/?f=%5Cfrac%7Bk_1%5Ctimes%20k_2%7D%7Bk_1%5E-%7D%5BBr_2%5D%5BNO%5D%5E2)
Explanation :
The overall reaction is:

Rate law = ![k[Br_2][NO]^2](https://tex.z-dn.net/?f=k%5BBr_2%5D%5BNO%5D%5E2)
The first step of the overall reaction is:


Rate law 1 = ![k_1[Br_2][NO]](https://tex.z-dn.net/?f=k_1%5BBr_2%5D%5BNO%5D)
Rate law 2 = ![k_1^-[NOBr_2]](https://tex.z-dn.net/?f=k_1%5E-%5BNOBr_2%5D)
The second step of the overall reaction is:

Rate law 3 = ![k_2[NOBr_2][NO]](https://tex.z-dn.net/?f=k_2%5BNOBr_2%5D%5BNO%5D)
Now rate law of overall reaction can be obtained as follows.
We are multiplying rate law 1 and rate law 3 and dividing by rate law 2, we get:
Rate law = ![\frac{[k_1[Br_2][NO]]\times [k_2[NOBr_2][NO]]}{[k_1^-[NOBr_2]]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bk_1%5BBr_2%5D%5BNO%5D%5D%5Ctimes%20%5Bk_2%5BNOBr_2%5D%5BNO%5D%5D%7D%7B%5Bk_1%5E-%5BNOBr_2%5D%5D%7D)
Rate law = ![\frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2](https://tex.z-dn.net/?f=%5Cfrac%7Bk_1%5Ctimes%20k_2%7D%7Bk_1%5E-%7D%5BBr_2%5D%5BNO%5D%5E2)
Rate law = ![k[Br_2][NO]^2](https://tex.z-dn.net/?f=k%5BBr_2%5D%5BNO%5D%5E2)
The rate law for formation of NOBr based on this mechanism is, ![\frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2](https://tex.z-dn.net/?f=%5Cfrac%7Bk_1%5Ctimes%20k_2%7D%7Bk_1%5E-%7D%5BBr_2%5D%5BNO%5D%5E2)
In chemistry, the molar mass M is a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by its amount of substance. The base SI unit for molar mass is kg/mol. However, for historical reasons, molar masses are almost always expressed in g/mol.
Hope this helped!
Good luck :p
~Emmy <3
Let's apply the conservation of energy through the equation below:
Q for lead + Q for water = 0
So,
Q for lead = -Q for water
where
Q = mass*specific heat*(T₂ - T₁)
The specific heat of liquid water is 4.187 J/g·°C
Substituting the values:
(m)(0.129 J/g·°C)(17.9 - 91.6°C) = -(200 g)(4.187 J/g·°C)(17.9 - 15.5°C)
Solving for m,
m = <em>211.39 grams of lead</em>