Answer:
Ka = 6.02x10⁻⁶
Explanation:
The equilibrium that takes place is:
We <u>calculate [H⁺] from the pH</u>:
- [H⁺] =
Keep in mind that [H⁺]=[A⁻].
As for [HA], we know the acid is 0.66% dissociated, in other words:
We <u>calculate [HA]</u>:
Finally we <u>calculate the Ka</u>:
- Ka =
![\frac{[9.12x10^{-4}]*[9.12x10^{-4}]}{[0.138]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B9.12x10%5E%7B-4%7D%5D%2A%5B9.12x10%5E%7B-4%7D%5D%7D%7B%5B0.138%5D%7D)
= 6.02x10⁻⁶
109.5
tetrahedral shape:
number of electron pair = 4,
number of bonded pair = 4,
number of lone pair = 0.
Answer:
There was 450.068g of water in the pot.
Explanation:
Latent heat of vaporisation = 2260 kJ/kg = 2260 J/g = L
Specific Heat of Steam = 2.010 kJ/kg C = 2.010 J/g = s
Let m = x g be the weight of water in the pot.
Energy required to vaporise water = mL = 2260x
Energy required to raise the temperature of water from 100 C to 135 C = msΔT = 70.35x
Total energy required = 
Hence, there was 450.068g of water in the pot.
