Answer:
ΔHf0=ΔHsub+IE+ΔHdiss+EA+U
ΔHf0=108+496+122−349−788=411kJ/mol
The enthalpy change in the formation of an ionic lattice from the gaseous isolated sodium and chloride ions is −788 kJ/mole. That enthalpy change, which corresponds to the reaction Na(g)+Cl(g)→NaCl(s), is called the lattice energy of the ionic crystal. Although the lattice energy is not directly measurable, there are various ways to estimate it from theoretical considerations and some experimental values. For all known ionic crystals, the lattice energy has a large negative value. It is ultimately the lattice energy of an ionic crystal which is responsible for the formation and stability of ionic crystal structures.
For sodium chloride, the Born
Explanation:
i hope it helps
The second approach is correct. The other two approaches are not correct because they are incomplete; first approach would have been right IF the 6.0 ppm was MEASURED in hexane. Third approach cannot be right since it calculates moles and grams but not L.
Answer:
1.5e+8 atoms of Bismuth.
Explanation:
We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):
For this, it is necessary to know the values in meters for any of these diameters:
Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.
<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>
1 atom of Bismuth = 320pm in diameter.
<h3>Diameter of a biscuit in meters</h3>
<h3>Resulting Ratio</h3>
How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:
In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.
171 because 57x3=171.
BOOM!!
Shakalaka.