Answer:
One reason, that water did not boil at 100 C could be the atmospheric pressure
Calculate the percent error for each sample from the theoretical 100°c
Percent error ( 102C) = (102-100/ 100) *100
= 2% error
In an atom electron travels outside the nucleus.
electrons travels around the nucleus in orbital but it is impossible to know both the speed and location of electron at the same time. Unlike the proton and neutrons which are located inside the nucleus electrons are located outside the nucleus.
Answer :
(A) The rate expression will be:
![Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
(B) The average rate of the reaction during this time interval is, 0.00176 M/s
(C) The amount of Br₂ (in moles) formed is, 0.0396 mol
Explanation :
Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.
The given rate of reaction is,

The expression for rate of reaction :
![\text{Rate of disappearance of }HBr=-\frac{1}{2}\frac{d[HBr]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DHBr%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D)
![\text{Rate of disappearance of }H_2=+\frac{d[H_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DH_2%3D%2B%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D)
![\text{Rate of formation of }Br_2=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DBr_2%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
<u>Part A:</u>
The rate expression will be:
![Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
<u>Part B:</u>
![\text{Average rate}=-\frac{1}{2}\frac{d[HBr]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20rate%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D)


The average rate of the reaction during this time interval is, 0.00176 M/s
<u>Part C:</u>
As we are given that the volume of the reaction vessel is 1.50 L.
![\frac{d[Br_2]}{dt}=0.00176M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D%3D0.00176M%2Fs)
![\frac{d[Br_2]}{15.0s}=0.00176M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BBr_2%5D%7D%7B15.0s%7D%3D0.00176M%2Fs)
![[Br_2]=0.00176M/s\times 15.0s](https://tex.z-dn.net/?f=%5BBr_2%5D%3D0.00176M%2Fs%5Ctimes%2015.0s)
![[Br_2]=0.0264M](https://tex.z-dn.net/?f=%5BBr_2%5D%3D0.0264M)
Now we have to determine the amount of Br₂ (in moles).



The amount of Br₂ (in moles) formed is, 0.0396 mol
Answer:
6 μL
Explanation:
To prepare a solution by a concentrated solution, we must use the equation:
C1xV1 = C2xV2, where <em>C</em> is the concentration, <em>V</em> the volume, 1 is the initial solution and 2 the final solution.
The final concentration must be 15 μg/mL with a volume of 0.1L or 100 mL. The concentration of the stock solution is 250 mg/mL. 1 mg has 1000 μg, so the concentration is 250000 μg/mL. Then:
250000xV1 = 15x100
V1 = 15/2500
V1 = 0.006 mL = 6 μL