Answer:
C is the reaction intermediate.
Explanation:
A reaction intermediate is a molecular structure that is formed during the reaction but then is converted in the final products.
Usually, these reaction intermediates are unestable and, for that reason, the lifetime of these structures is low.
In the reaction, you can see in the first step C is produced, but also, in the second step reacts producing D. As is produced and, immediately consumed,
<h3>C is the reaction intermediate.</h3>
<u>Answer:</u> The half life of the sample of silver-112 is 3.303 hours.
<u>Explanation:</u>
All radioactive decay processes undergoes first order reaction.
To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:
![k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%20%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = ?
t = time taken = 1.52 hrs
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= Initial concentration of reactant = 100 g
[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g
Putting values in above equation, we get:
To calculate the half life period of first order reaction, we use the equation:
where,
= half life period of first order reaction = ?
k = rate constant = 
Putting values in above equation, we get:
Hence, the half life of the sample of silver-112 is 3.303 hours.
In order to solve this, we need to make use of Hess' Law.
We are already given the equations and their corresponding deltaH. Using Hess' Law, we can generate this equation:
104 kJ = x - (-1182 kJ) - (-1144 kJ)
Among the choices, the answer is
<span>B.104 = x - [(-1182) + (-1144)]
</span>
Answer:
No, ΔE does not always equal zero because it refers to the systems internal energy, which is affected by heat and work
Explanation:
According to the first law of thermodynamics, energy is neither created nor destroyed. This implies that the total energy of a system is always a constant.
So, according to the first law of thermodynamics we have that ΔE = q + w. This means that the value of ΔE depends on q (heat) and w(work). Hence ΔE is not always zero since it depends on the respective values of q and w.
