Answer:
the reactor was overheating
Proton, the gold foil experiment indicated that the nucleus has a positive charge.
If temperature were to double the pressure would likewise double. Increased temperature would increase the energy of the molecules and the number of collisions would also increase causing the increase in pressure.
Answer : The pH of the solution is, 2.67
Explanation :
The equilibrium chemical reaction is:

Initial conc. 0.450 0 0
At eqm. (0.450-x) x x
As we are given:

The expression for equilibrium constant is:

Now put all the given values in this expression, we get:


The concentration of
= x = 0.00212 M
Now we have to calculate the pH of solution.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)


Therefore, the pH of the solution is, 2.67
Answer:
The new pressure is 53.3 kPa
Explanation:
This problem can be solved by this law. when the volume remains constant, pressure changes directly proportional as the Aboslute T° is modified.
T° increase → Pressure increase
T° decrease → Pressure decrease
In this case, temperature was really decreased. So the pressure must be lower.
P₁ / T₁ = P₂ / T₂
80 kPa / 300K = P₂/200K
(80 kPa / 300K) . 200 K = P₂ → 53.3 kPa