The heat released by the water when it cools down by a temperature difference AT
is Q = mC,AT
where
m=432 g is the mass of the water
C, = 4.18J/gºC
is the specific heat capacity of water
AT = 71°C -18°C = 530
is the decrease of temperature of the water
Plugging the numbers into the equation, we find
Q = (4329)(4.18J/9°C)(53°C) = 9.57. 104J
and this is the amount of heat released by the water.
Answer: acid dissociation constant Ka= 2.00×10^-7
Explanation:
For the reaction
HA + H20. ----> H3O+ A-
Initially: C. 0. 0
After : C-Cx. Cx. Cx
Ka= [H3O+][A-]/[HA]
Ka= Cx × Cx/C-Cx
Ka= C²X²/C(1-x)
Ka= Cx²/1-x
Where x is degree of dissociation = 0.1% = 0.001 and c is the concentration =0.2
Ka= 0.2(0.001²)/(1-0.001)
Ka= 2.00×10^-7
Therefore the dissociation constant is
2.00×10^-7
Answer:
hydrocarbons are the answer
Answer:
<h2>pH = 4.44 </h2>
Explanation:
The pH of a substance can be found by using the formula
![p H = - log[ H^{ + } ]](https://tex.z-dn.net/?f=p%20H%20%20%3D%20%20-%20%20%20log%5B%20H%5E%7B%20%2B%20%7D%20%20%5D)
where [ H+ ] is the hydrogen ion concentration of the solution
From the question
[ H + ] = 3.60 × 10^-5 M
So the pH is
We have the final answer as
<h3>pH = 4.44 </h3>
Hope this helps you
