Answer:
21.86582KJ
Explanation:
The graphical form of the Arrhenius equation is shown on the image attached. Remember that in the Arrhenius equation, we plot the rate constant against the inverse of temperature. The slope of this graph is the activation energy and its y intercept is the frequency factor.
Applying the equation if a straight line, y=mx +c, and comparing the given equation with the graphical form of the Arrhenius equation shown in the image attached, we obtain the activation energy of the reaction as shown.
Answer : The concentration after 17.0 minutes will be, 
Explanation :
The expression for first order reaction is:
![[C_t]=[C_o]e^{-kt}](https://tex.z-dn.net/?f=%5BC_t%5D%3D%5BC_o%5De%5E%7B-kt%7D)
where,
= concentration at time 't' (final) = ?
= concentration at time '0' (initial) = 0.100 M
k = rate constant = 
t = time = 17.0 min = 1020 s (1 min = 60 s)
Now put all the given values in the above expression, we get:
![[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}](https://tex.z-dn.net/?f=%5BC_t%5D%3D%280.100%29%5Ctimes%20e%5E%7B-%285.40%5Ctimes%2010%5E%7B-3%7D%29%5Ctimes%20%281020%29%7D)
![[C_t]=4.05\times 10^{-4}M](https://tex.z-dn.net/?f=%5BC_t%5D%3D4.05%5Ctimes%2010%5E%7B-4%7DM)
Thus, the concentration after 17.0 minutes will be, 
Answer:
kilograms
Explanation:
hope this helps, pls mark brainliest :D
Answer: The value of the equilibrium constant Kc for this reaction is 3.72
Explanation:
Equilibrium concentration of
= 
Equilibrium concentration of
= 
Equilibrium concentration of
= 
Equilibrium concentration of
= 
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as
For the given chemical reaction:
The expression for
is written as:
Thus the value of the equilibrium constant Kc for this reaction is 3.72
I think the answer would be dependent variable. An unknown or changeable quantity is called a dependent variable. It <span>is what you measure in the experiment and what is affected during the experiment. Hope this answers the question. Have a nice day.</span>