Eezy real talk ma frend!
The fromula for sodium cyanide is NaCn.
P1/T1 = P2/T2
125⁰C = 398.15 k
182⁰C = 455.15 k
1.22/398.15 = p2/455.15
p2= 1.39atm
the pressure of the gas be after the temperature change is 1.39 atm
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: The scientific method is an empirical method of acquiring knowledge that has characterized the development of science since at least the 17th century. It involves careful observation, applying rigorous skepticism about what is observed, given that cognitive assumptions can distort how one interprets the observation.
Explanation:
Hope this helps. :)
Pressure can be defined as the force acting on a perpendicular surface per unit area.
Force exerted by a man of mass 100 kg wearing snow shoes = m.a
Where m = mass of the man = 100 kg
a = acceleration due to gravity= 9.8 
Force exerted by the man of mass 100 kg = 
Force exerted by woman of mass 60 kg = 
Force exerted by 100 kg man is greater than that exerted as 60 kg woman. The area on which this force is acting determines the pressure. Pressure is inversely proportional to the area on which the force acts. Therefore, the pressure exerted by 100 kg man wearing snow shoes is less than the pressure exerted by a 60 kg woman woman wearing high heels as the force acts over a larger area when the man wears snow shoes when compared to the force exerted over a smaller area in case of the woman wearing high heels.