Answer:
# In a familiar high-school chemistry demonstration, an instructor first uses electricity to split water into its constituent gases, Hydrogen and Oxygen. Then, by combining the two gases and igniting them with a spark, the instructor changes the gases back into water with a loud pop (That means the energy is released in the process).
# There are new other ways to produce water in laboratory, however, the scientists can not produce water in large quantity for the masses, because of some reasons.
1- Theoretically, this is possible, but it would be an extremely dangerous process. Since Hydrogen is extremely flammable and Oxygen supports combustion, it wouldn’t take much to create this force, but we also have an explosion. That’s why this process can be a deadly one if our experiment is big enough.
2- Personally, I think that it makes no sense to produce water in a laboratory ( or in a large plant) for people to use as daily water. The much more important thing we need to do is to save our environment, our planet Earth. Because the daily water people drink contains not just water molecules but other minerals, the marine life is depend not just in water molecules but diferent factors, etc.
Explanation:
This is just my personal opinion. Hope that can help you a little. Have a nice day
Answer:
19.264×
atoms are present in 3.2 moles of carbon.
Explanation:
It is known that one mole of each element is composed of Avagadro's number of atoms. This is same for all the elements in the periodic table.
So, as 1 mole of any element = Avagadro's number of atoms = 6.02× atoms
It is as simple as understanding a dozen of anything is equal to 12 pieces of that object.
As here the moles of carbon is given as 3.20 moles, the number of atoms in this mole can be determined as below.
1 mole of carbon = 6.02 × atoms
Then, 3.20 moles of carbon = 3.20 × 6.02 × atoms
Thus, 19.264× atoms are present in 3.2 moles of carbon.
Answer:
ΔSv = 0.1075 KJ/mol.K
Explanation:
Binary solution:
∴ a: solvent
∴ b: solute
in equilibrium:
- μ*(g) = μ(l) = μ* +RTLnXa....chemical potential (μ)
⇒ Ln (1 - Xb) = ΔG/RT
∴ ΔG = ΔHv - TΔSv
⇒ Ln(1 -Xb) = ΔHv/RT - ΔSv/R
∴ Xb → 0:
⇒ Ln(1) = ΔHv/RT - ΔSv/R
∴ T = T*b....normal boiling point
⇒ 0 = ΔHv/RT*b - ΔSv/R
⇒ ΔSv = (R)(ΔHv/RT*b)
⇒ ΔSv = ΔHv/T*b
∴ T*b = 80°C ≅ 353 K
⇒ ΔSv = (38 KJ/mol)/(353 K)
⇒ ΔSv = 0.1075 KJ/mol.K
