When a liquid evaporates, it releases heat energy into the surroundings because the molecules obtain energy ( when they are being evaporates), causing the molecules to vibrate fast and escape into the atmosphere!
Explanation:
It is known that relation between
,
, and pH is as follows.
![E_{cell} = E^{o}_{cell} - (\frac{0.0591}{n}) \times log[H^{+}] ](https://tex.z-dn.net/?f=E_%7Bcell%7D%20%3D%20E%5E%7Bo%7D_%7Bcell%7D%20-%20%28%5Cfrac%7B0.0591%7D%7Bn%7D%29%20%5Ctimes%20log%5BH%5E%7B%2B%7D%5D%0A)
Also, it is known that
for hydrogen is equal to zero.
Hence, substituting the given values into the above equation as follows.
![E_{cell} = E^{o}_{cell} - (\frac{0.0591}{n}) \times log[H^{+}] ](https://tex.z-dn.net/?f=E_%7Bcell%7D%20%3D%20E%5E%7Bo%7D_%7Bcell%7D%20-%20%28%5Cfrac%7B0.0591%7D%7Bn%7D%29%20%5Ctimes%20log%5BH%5E%7B%2B%7D%5D%0A)
0.238 V = 0 - (\frac{0.0591}{1}) \times log[H^{+}]
[/tex]
= 4.03
= antilog 4.03
= 3.5
As, pH =
.
Thus, we can conclude that pH of the given unknown solution at 298 K is 3.5.
Answer:
1 mol SO2 contains 6.0213*10^23 molecules
6.023*10^24 molecules = 10 mol SO2
Equation
S(s) + O2(g) → SO2(g)
1 mol S reacts with 1 mol O2 to prepare 1 mol SO2
To prepare 10 mol SO2 you require : 10 mol S plus 10 mol O2
And that is the answer to the question
If you want a mass :
Molar mass S = 32 g/mol You require 10 mol = 320 g
Molar mass O2 = 32 g/mol :You require 10 mol = 320 g
So for the first one It's 784.56 J (GPE)
For the second one it's KE is equal to 100 J. If you can give me a few more minutes I'll try to get the GPE