Answer:
0.00915 M of 
remain after 5.16 seconds.
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 that:
The rate constant, k = 
s⁻¹
Initial concentration = 0.054 M
Final concentration = ? M
Time = 5.16 s
Applying in the above equation, we get that:-
![[A_t]=0.054e^{-0.344\times 5.16}\ M=\frac{1\times \:0.054}{e^{1.77504}}\ M=0.00915\ M](https://tex.z-dn.net/?f=%5BA_t%5D%3D0.054e%5E%7B-0.344%5Ctimes%205.16%7D%5C%20M%3D%5Cfrac%7B1%5Ctimes%20%5C%3A0.054%7D%7Be%5E%7B1.77504%7D%7D%5C%20M%3D0.00915%5C%20M)
<u>0.00915 M of remain after 5.16 seconds.</u>
Answer:
in this file you will find it https://xlbrands.com/wp-content/uploads/2014/06/DEW-POINT-CALCULATION-CHARTenglish.pdf
Explanation:
We have to solve this question using the stoichiometry of the reaction:
The equation of the reaction is;
According to the question;
Number of moles of CO2 released = 21.3 g/44 g/mol = 0.48 moles
From the stoichiometry of the reaction:
Since;
24 moles of CO2 released 15,026 KJ
0.48 moles of CO2 will release 0.48 * 15,026/24
= 301 KJ of heat.
brainly.com/question/6901180
Answer: This was because the experiment showed that a substance could emit radiation even while it was not exposed to light.
