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.
DANNNNNNNNNNNNGGGGGG hella hot, the hottest I seen in this world !!
I think it’s to long to fit in a period??
Answer:
[N₂] = 0.032 M
[O₂] = 0.0086 M
Explanation:
Ideal Gas Law → P . V = n . R . T
We assume that the mixture of air occupies a volume of 1 L
78% N₂ → Mole fraction of N₂ = 0.78
21% O₂ → Mole fraction of O₂ = 0.21
1% another gases → Mole fraction of another gases = 0.01
In a mixture, the total pressure of the system refers to total moles of the mixture
1 atm . 1L = n . 0.082L.atm/mol.K . 298K
n = 1 L.atm / 0.082L.atm/mol.K . 298K → 0.0409 moles
We apply the mole fraction to determine the moles
N₂ moles / Total moles = 0.78 → 0.78 . 0.0409 mol = 0.032 moles N₂
O₂ moles / Total moles = 0.21 → 0.21 . 0.0409 mol = 0.0086 moles O₂
