Answer:
The correct answer is option D.
Explanation:
Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.
Rate of the reaction:
![R=-\frac{1}{1}\times \frac{d[NO_2]}{dt}=-\frac{1}{1}\times \frac{d[CO]}{dt}](https://tex.z-dn.net/?f=R%3D-%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BCO%5D%7D%7Bdt%7D)
Rate of decrease in nitrogen dioxide concentration is equal to the rate of decrease in carbon monoxide.
Given rate expression of the reaction:
![R = k[NO2]^2[CO]](https://tex.z-dn.net/?f=R%20%3D%20k%5BNO2%5D%5E2%5BCO%5D)
Rate of the reaction on doubling concentration of nitrogen dioxide and carbon monoxide : R'
![R'=k(2\times [NO_2])^2(2\times [CO])=8\times k[NO2]^2[CO]=8R](https://tex.z-dn.net/?f=R%27%3Dk%282%5Ctimes%20%5BNO_2%5D%29%5E2%282%5Ctimes%20%5BCO%5D%29%3D8%5Ctimes%20k%5BNO2%5D%5E2%5BCO%5D%3D8R)
Doubling the concentrations of nitrogen dioxide and carbon monoxide simultaneously will increase the rate of the reaction by a factor of eight.
Hence, none of the given statements are true.
Moles = mass / molar mass
molar mass of O2 is 32
therefore moles = 72/32
= 2.25 moles
Answer:
See Below
Explanation:
'an aerobic ' means it does not use oxygen
193.38 K was the initial temperature of the krypton.
Explanation:
Data given:
Initial volume of the krypton gas = 6 litres
initial pressure of the krypton gas = 0.960 atm
initial temperature of the krypton gas = ?
final volume of the krypton gas = 7.70 litres
final pressure of the Krypton gas = 1.25 atm
final temperature of the krypton gas = 55 degrees or 273.25+55 = 323.15 K
Applying the Combined Gas Laws:
Rearranging the equation:
T1 = 
Putting the value in the equation:
T1 = 
T1 = 193.38 K
Initial temperature of the krypton gas is 193.78 K
The correct answer is 3 moles of nitrogen
Hope this helped :)
