His distance and displacement are the same, which was 400 m
<h3>Further explanation</h3>
Given
Distance = 400 m
time = 2 min
Required
Distance and displacement
Solution
Distance is a scalar quantity that indicates the length of the trajectory that is traveled by an object within a certain interval. Distance has no direction, only has magnitude
Can be simplified distance = totals traveled
Displacement is a vector quantity that shows changes in the position of objects in a certain interval of time. Displacement has magnitude and direction
Can be simplified displacement = distanced traveled from starting point to ending point
From the definition above shows that the displacement and the distance that he traveled have the same value (magnitude), which is equal to 400 m
The value of the two will be different if he starts and finishes at the same point, then the displacement value is zero while the distance he has traveled is still 0
Answer:
a. forces acting on the object
Answer:
1.14 M
Explanation:
grams/molar mass = ans. / volume
317/110.98=2.86/2.50=1.14 M
- Hope that helps! Please let me know if you need further explanation.
The best answer choice here would be 'Combination'
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
