Answer :
(a) The average rate will be:
![\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D%3D9.36%5Ctimes%2010%5E%7B-5%7DM%2Fs)
(b) The average rate will be:
![\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BH%5E%2B%5D%7D%7Bdt%7D%3D1.87%5Ctimes%2010%5E%7B-4%7DM%2Fs)
Explanation :
The general rate of reaction is,

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.
The expression for rate of reaction will be :
![\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20A%7D%3D-%5Cfrac%7B1%7D%7Ba%7D%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D)
![\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20B%7D%3D-%5Cfrac%7B1%7D%7Bb%7D%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
![\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20C%7D%3D%2B%5Cfrac%7B1%7D%7Bc%7D%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D)
![\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20D%7D%3D%2B%5Cfrac%7B1%7D%7Bd%7D%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D)
![Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1%7D%7Ba%7D%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7Bb%7D%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7Bc%7D%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7Bd%7D%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D)
From this we conclude that,
In the rate of reaction, A and B are the reactants and C and D are the products.
a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.
The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.
The given rate of reaction is,

The expression for rate of reaction :
![\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DBr%5E-%3D-%5Cfrac%7B1%7D%7B5%7D%5Cfrac%7Bd%5BBr%5E-%5D%7D%7Bdt%7D)
![\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DBrO_3%5E-%3D-%5Cfrac%7Bd%5BBrO_3%5E-%5D%7D%7Bdt%7D)
![\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DH%5E%2B%3D-%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7Bd%5BH%5E%2B%5D%7D%7Bdt%7D)
![\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DBr_2%3D%2B%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
![\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DH_2O%3D%2B%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7Bd%5BH_2O%5D%7D%7Bdt%7D)
Thus, the rate of reaction will be:
![\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20reaction%7D%3D-%5Cfrac%7B1%7D%7B5%7D%5Cfrac%7Bd%5BBr%5E-%5D%7D%7Bdt%7D%3D-%5Cfrac%7Bd%5BBrO_3%5E-%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7Bd%5BH%5E%2B%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7Bd%5BH_2O%5D%7D%7Bdt%7D)
<u>Part (a) :</u>
<u>Given:</u>
![\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D%5Cfrac%7Bd%5BBr%5E-%5D%7D%7Bdt%7D%3D1.56%5Ctimes%2010%5E%7B-4%7DM%2Fs)
As,
![-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B5%7D%5Cfrac%7Bd%5BBr%5E-%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
and,
![\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B3%7D%7B5%7D%5Cfrac%7Bd%5BBr%5E-%5D%7D%7Bdt%7D)
![\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B3%7D%7B5%7D%5Ctimes%201.56%5Ctimes%2010%5E%7B-4%7DM%2Fs)
![\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D%3D9.36%5Ctimes%2010%5E%7B-5%7DM%2Fs)
<u>Part (b) :</u>
<u>Given:</u>
![\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D%5Cfrac%7Bd%5BBr%5E-%5D%7D%7Bdt%7D%3D1.56%5Ctimes%2010%5E%7B-4%7DM%2Fs)
As,
![-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B5%7D%5Cfrac%7Bd%5BBr%5E-%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7Bd%5BH%5E%2B%5D%7D%7Bdt%7D)
and,
![-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7Bd%5BH%5E%2B%5D%7D%7Bdt%7D%3D%5Cfrac%7B3%7D%7B5%7D%5Cfrac%7Bd%5BBr%5E-%5D%7D%7Bdt%7D)
![\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BH%5E%2B%5D%7D%7Bdt%7D%3D%5Cfrac%7B6%7D%7B5%7D%5Ctimes%201.56%5Ctimes%2010%5E%7B-4%7DM%2Fs)
![\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BH%5E%2B%5D%7D%7Bdt%7D%3D1.87%5Ctimes%2010%5E%7B-4%7DM%2Fs)
Answer:
Alpha Particle, Beta Particles and Gamma Rays
Explanation:
I THINK that that is true. It's just weathering the rock down not change composition. Lemme know if you get it right please! (:
Answer:
Refer to your periodic table. Lewis dot structures are based off the number of valence electrons an atom has.
Looking at the compounds, we can see that Gallium has three valence electrons in its outer shell and oxygen has six. Oxygen and Gallium are going to share electrons with one another, making a V shape in their diagram.
One Oxygen would make a double bond with a Gallium, leaving one valence electron to another oxygen. That oxygen takes that Final electron. It now has 7 in its outer shell. The remaining Gallium and Oxygen do the same double bond as the one before, leaving the 7 valence electron oxygen with one more electron.
Answer:
The answer is
<h2>2 cm/year</h2>
Explanation:
To find the rate in cm/year we must first convert 200 m into cm
1 m = 100 cm
if 1 m = 100 cm
Then 200 m = 200 × 100 = 20 ,000 cm
So the rate is
<h2>

</h2>
<u>Reduce the fraction with 10,000</u>
We have the final answer as
<h3>2 cm/year</h3>
Hope this helps you