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.
SAMPLE A - <span>pure substance.
</span>SAMPLE B - <span>homogeneous mixture.
</span>SAMPLE C - <span>heterogeneous mixture.
</span>Pure substance - <span>constant composition and properties.</span>
Homogeneous mixture - same uniform appearance and composition.
Heterogeneous mixture - <span>not </span>uniform<span> in composition, two phases (liquid and dust).
</span>
Hello.
The answer is: A. keep a cork in the test tube so the solution cannot spill out
B and C both could be very deadly and it shouldnt be D because if u have an open flame you shouldnt open it even if its not towards anyone.
have a nice day