Answer: The energy required to vaporize 12.5 g of liquid water is 28.2 kJ
Explanation:
Latent heat of vaporization is the amount of heat required to convert 1 gram of liquid into its vapor state without change in its temperature.
Given : The enthalpy of vaporization of water is 40.65 kJ/mol.
n = number of moles =
Thus 1 mole of water requires heat = 40.65 kJ
0.694 moles of water requires heat =
Thus the energy required to vaporize 12.5 g of liquid water is 28.2 kJ
Answer:
Explanation:
Hello,
In this case, considering the Arrhenius law, the decreasing ratio is:
By considering the 115°C as 1 and 230°C as 2, we obtain:
And the increasing ratio:
Best regards.
Formula for water is H20 and there are 5 mol water. 2 mol of hydrogen per 1 mol water so 5 mol of water times 2 mol hydrogen will give you 10 mol of hydrogen.
Same process with oxygen except 1 mol of oxygen per mol of water. So 5 mol water times 1 mol of oxygen will give you 5 mol of oxygen.