To get the percent yield, we will use this formula:
((Actual Yield)/(Theoretical Yield)) * 100%
Values given: actual yield is 220.0 g
theoretical yield is 275.6 g
Now, let us substitute the values given.
(220.0 grams)/(275.6 grams) = 0.7983
Then, to get the percentage, multiply the quotient by 100.
0.7983 (100) = 79.83%
Among the choices, the most plausible answer is 79.8%
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ANSWER:
the answer is burning as it is a chemical change emaning that the substance changes chemically, the other choices are exp of physical changing meaning they stay the same substance but change physically
~batmans wife dun dun dun...aka ~serenitybella
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:
B
Explanation:
Having the same electronegativity means that the both elements will form a nonpolar bond. The shared electrons of the bind will be found equidistant from the nucleus of the two atoms because none has an unusual ability to attract the electrons of a bond towards itself.