Q=mc(change in temp)
Q is amount of heat, m is mass, c is specific heat
Water:
Q= (50.0g)(4.18 J/g°C)(63°C)
Q= 13167J
Aluminum:
Q= (200.0g)(0.900 J/g°C)(63°C)
Q= 11340J
Water requires more heat
The latent heat is correlated with energy as follows:
Q = mL
550 * 103 = 14 * 103 * L
L = 39.285 J /g
Thus, latent heat of the substance is 39.285 j /g
The type of bonding present in water (H2O) is hydrogen bonding.
Answer : The value of
of the weak acid is, 4.72
Explanation :
First we have to calculate the moles of KOH.


Now we have to calculate the value of
of the weak acid.
The equilibrium chemical reaction is:

Initial moles 0.25 0.03 0
At eqm. (0.25-0.03) 0.03 0.03
= 0.22
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[HK]}{[HA]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BHK%5D%7D%7B%5BHA%5D%7D)
Now put all the given values in this expression, we get:


Therefore, the value of
of the weak acid is, 4.72