Q = m . C . Δ T
q : Heat absorbed
m : mass of the sample
C : The specific heat of the substance.
Δ T : The change in temperature (Final T - initial T)
So:
q = 15.5 x 4.18 x (50 - 25) = 1619.75 J
Answer:
Explanation:
Hello,
The law of mass action, allows us to know the required amounts, thus, for this chemical reaction it is:
![\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B-3%7D%20%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-1%7D%20%5Cfrac%7Bd%5BE%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-2%7D%20%5Cfrac%7Bd%5BF%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B5%7D%20%5Cfrac%7Bd%5BG%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B4%7D%20%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D)
Now, we answer:
(a)
![\frac{d[H]}{dt}=4*\frac{1}{-3} *(-0.12M/s)=0.16M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D%3D4%2A%5Cfrac%7B1%7D%7B-3%7D%20%2A%28-0.12M%2Fs%29%3D0.16M%2Fs)
(b)
![\frac{d[E]}{dt}=-1*\frac{1}{5} *(0.2M/s)=-0.04M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BE%5D%7D%7Bdt%7D%3D-1%2A%5Cfrac%7B1%7D%7B5%7D%20%2A%280.2M%2Fs%29%3D-0.04M%2Fs)
(c) Since no initial data is specified, we could establish the rate of the reaction as based of the law of mass action:
![r=\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B1%7D%7B-3%7D%20%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-1%7D%20%5Cfrac%7Bd%5BE%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-2%7D%20%5Cfrac%7Bd%5BF%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B5%7D%20%5Cfrac%7Bd%5BG%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B4%7D%20%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D)
Thus, any of the available expressions are suitable to quantify the rate of the reaction.
Best regards.
You take the 1.00 M and multiply it by 375L witch gives you 375mol and then you take the mol and multiply it by the molar mass giving you 23630.505g nitric acid or HNO3
It is.
An acid will be strong when its conjugated base is highly stable, and vice-versa.
That can occur for instance through electronic delocalization.