<h2>Answer:</h2>
Arrangement of inter molecular forces from strongest to weakest.
- Hydrogen bonding
- Dipole-dipole interactions
- London dispersion forces.
<h3>Explanation:</h3>
Intermolecular forces are defined as the attractive forces between two molecules due to some polar sides of molecules. They can be between nonpolar molecules.
Hydrogen bonding is a type of dipole dipole interaction between the positive charge hydrogen ion and the slightly negative pole of a molecule. For example H---O bonding between water molecules.
Dipole dipole interactions are also attractive interactions between the slightly positive head of one molecule and the negative pole of other molecules.
But they are weaker than hydrogen bonding.
London dispersion forces are temporary interactions caused due to electronic dispersion in atoms of two molecules placed together. They are usually in nonpolar molecules like F2, I2. they are weakest interactions.
Answer:
Take 100 ml of a 18 molar solution. The total number of moles is (1 liter/1000 ml) 100 ml 18 moles is 1.8 moles.
1.5 moles in 1 liter so If 1.1 liters of water is added, the total volume is 1.2 liters and 1.8 moles are dissolves in it. 1.8 moles/ 1.2 liters is 1.5 moles per liter.
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Answer:
D
Explanation:
the answer is d white precipitate
Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K
