The question is incomplete, here is the complete question:
At elevated temperature, nitrogen dioxide decomposes to nitrogen oxide and oxygen gas
The reaction is second order for 
with a rate constant of 
at 300°C. If the initial [NO₂] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M
a) 1.01 b) 5.19 c) 0.299 d) 0.0880 e) 3.34
<u>Answer:</u> The time taken is 5.19 seconds
<u>Explanation:</u>
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant =
t = time taken = ?
[A] = concentration of substance after time 't' = 0.150 M
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 0.260 M
Putting values in above equation, we get:
Hence, the time taken is 5.19 seconds
Answer:
27.64 liters
Explanation:
From the balanced equation, 2 moles of K2Cr2O7 requires 3 moles of CH3OH.
Mole of CH3OH = 1.9/32.04 = 0.0593 mole
Mole of K2Cr2O7 that will require 0.0593 mole of CH3OH:
2 x 0.0593/3 = 0.0395 mole
mole = molarity x volume
Volume of K2Cr2O7 needed = 0.0395/0.00143
= 27.64 Liter
<em>Hence, 27.64 liters of 0.00143 M K2Cr2O7 will be required to titrate 1.90 g of CH3OH dissolved in 50.0 mL of solution</em>
Celsius: -11.7
Kelvin: 261.5
Hope it helps! Please mark Brainliest! :)
