"The other halogens are not as electronegative and so other hydrogen halides cannot form hydrogen bonds between molecules. Only London Forces are formed. - Therefore more energy is required to break the intermolecular forces in HF than the other hydrogen halides and so it has a higher boiling point."
not a hack link, just stating where i got your answer from! -
https://www.mytutor.co.uk/answers/17558/A-Level/Chemistry/Explain-the-unusually-high-boiling-point-of-HF/
Answer:
It is equal to Avogadro's number (NA), namely 6.022 x1023. If we have one mole of water, then we know that it will have a mass of 2 grams (for 2 moles of H atoms) + 16 grams (for one mole O atom) = 18 grams.
Explanation:
The question is not very much clear.
If you are asking for molecules then 1 mole water= 6.023 * 10^23
If you are asking for atoms then 1 mole water= 6.023 * 10^23 * 3
If you are asking for particles then,
So, in your example you would have one mole of water molecules. If you dissociated those water molecules, than you would end up with 2 moles of hydrogen atoms, and one mole of oxygen atoms.
I hope that was helpful!
H=1 proton,1 electron
O=8 protons,8 neutrons and 8 electrons
total particles in one H2O molecule-28
total no. of particles in 1 mole of water- 6.023 * 10^23 * 28
ΔG° at 450. K is -198.86kJ/mol
The following is the relationship between ΔG°, ΔH, and ΔS°:
ΔH-T ΔS = ΔG
where ΔG represents the common Gibbs free energy.
the enthalpy change, ΔH
The temperature in kelvin is T.
Entropy change is ΔS.
ΔG° = -206 kJ/mol
ΔH° equals -220 kJ/mol
T = 298 K
Using the formula, we obtain:
-220kJ/mol -T ΔS° = -206kJ/mol
220 kJ/mol +206 kJ/mol =T ΔS°.
-T ΔS = 14 kJ/mol
for ΔS-14/298
ΔS=0.047 kJ/mol.K
450K for the temperature Completing a formula with values
ΔG° = (450K)(-0.047kJ/mol)-220kJ/mol
ΔG° = -220 kJ/mol + 21.14 kJ/mol.
ΔG°=198.86 kJ/mol
Learn more about ΔG° here:
brainly.com/question/17214066
#SPJ4
