Answer:
is there like a passage to answer this question?
Explanation:
Answer:
c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt
Explanation:
Hello,
In this case, the undergoing chemical reaction is:

Thus, the rate is given as:
![rate=-\frac{1}{2} \frac{\Delta [HBr]}{\Delta t}=\frac{\Delta [Br_2]}{\Delta t} =\frac{\Delta [H_2]}{\Delta t}](https://tex.z-dn.net/?f=rate%3D-%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B%5CDelta%20%5BHBr%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B%5CDelta%20%5BBr_2%5D%7D%7B%5CDelta%20t%7D%20%3D%5Cfrac%7B%5CDelta%20%5BH_2%5D%7D%7B%5CDelta%20t%7D)
It is necessary to remember that each concentration to time interval is divided into the stoichiometric coefficient, that is why HBr has a 1/2. Moreover, the concentration HBr is negative since it is a reactant and it has a negative rate due to its consumption.
Therefore, the answer is:
c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt
Best regards.
When you use exponents, think of it like this. 1 squared is 1 x 1. 1 cubed is 1 x 1 x 1. And 1 to the power of 4 is 1 x 1 x 1 x 1. And so on. You basically just multiply them by themselves. 0 to the power of any exponent is 1, well that's just a rule. It doesn't make much sense but it's easy to remember and I wouldn't worry about it.