Answer:Light bounces off of the mirror and then appears to come from behind the mirror.
Explanation:Plane mirrors form images that are virtual, upright and the same size and shape as the object it is reflecting.
When rays of light from the object hits a plane mirror they bounces off the mirror,that is they undergo reflection, and appear to originate from behind the mirror, resulting to the formation of a virtual image.
The image formed appears to be behind the plane in which the mirror lies. A virtual image is an image that is formed at a location from which the rays of light appear to come from. The image can not be formed on a screen..
A is true
B is false
C is true
Answer:
9.1 seconds
Explanation:
Given that for a second order reaction
1/[A]t = kt + 1/[A]o
Where [A]t= concentration at time = t= 0.340M
[A]o= initial concentration = 0.820M
k= rate constant for the reaction=0.190m-1s-1
t= time taken for the reaction (the unknown)
Hence;
(0.340)^-1 = 0.190×t + (0.820)^-1
t= (0.340)^-1 - (0.820)^-1/0.190
t= 9.1 seconds
Hence the time taken for the concentration to decrease from 0.840M to 0.340M is 9.1 seconds.
Answer:
Explanation:
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given:
75 % is decomposed which means that 0.75 of is decomposed. So,
![\frac {[A_t]}{[A_0]}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D)
= 1 - 0.75 = 0.25
t = 60 min
![\frac {[A_t]}{[A_0]}=e^{-k\times t}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%3De%5E%7B-k%5Ctimes%20t%7D)
k = 0.023 min⁻¹
Considering the expression for half life as:-
Where, k is rate constant
So,
