As far as I remember, the needed formula for squaric acid is C4H2O4.
According to this one mole should be 114.06 g., which means we have <span>0.015mol of this acid.
Then we can easly calculate : </span><span>4(0.015) = 0.06 mol for both for Carbon and Oxygen and </span><span>0.03 mol of Hydrogen.
</span><span>To get more clear answer, we multiply by avogadros :
</span><span>6.022 x 10^23. Hope everything is clear! regards</span>
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)