The average bond energy of the Xe¬F bonds in each fluoride is 132kJ/mol.
Given:
ΔH° of xenon difluoride (XeF2) = -105 kJ/mol
ΔH° of xenon tetrafluoride (XeF4)= -284 kJ/mol
ΔH° of xenon hexafluoride (XeF6) = -402 kJ/mol
The bond energy of Xe-F in XeF2 can be calculated as follows,
As we know that
ΔH° = ΔH°(bond formed) + ΔH°(bond broken)
The chemical reaction for the formation of XeF2 can be written in such a way,
Xe (g) + F2 (g) → XeF2 (g)
= [1 mol F2 (159 kJ/mol)] + [2(-Xe-F)] - 105 kJ/mol
= 159 kJ/mol + 2(-Xe-F) - 264 kJ/mol
= 2(-Xe-F)
Xe-F = 132 kJ/mol
Thus, we concluded that the average bond energy of the Xe¬F bonds in each fluoride is 132kJ/mol.
learn more about bond energy:
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Answer:
Option (C). Write the complete balanced chemical equation.
Explanation:
Writing a balanced chemical equation is the key to arriving at the correct stoichiometric calculations because it gives a fore knowledge of the correct amount of reactants needed to generate a predicted products for a given reaction
Answer:
A. a new substance is being produced.
Explanation:
The bubbles most likely indicates that a new substance is being produced by this reaction. In essence, we describe this sort of change as chemical change.
In a chemical change, new substances are usually produced. They are accompanied by the evolution or absorption of energy.
The reaction of Zinc with a strong acid to produce bubbles on the surface of the metal indicates a chemical change and the formation of a new kind of substance.
Take for example, let zinc reacts with hydrocholoric acid, HCl;
Zn + 2HCl → ZnCl₂ + H₂
Since Zn is higher than Hydrogen in the activity series, it will displace it from HCl and liberate hydrogen gas as a product. This will cause the bubbles observed in the reaction.
This is a chemical change and new products have been formed.
B and D are wrong because they are both physical changes.
C is wrong because no information about such is provided by the problem statement.
So, when a piece of zinc metal combines with a strong acid, a new kind of substance is produced.
NH₃:
N = 8*10²²
NA = 6.02*10²³
n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol
O₂:
N=7*10²²
NA = 6.02*10²³
n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol
4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant
N₂:
n = 0.0665mol
M = 28g/mol
m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>
Like dissolves like
so water is polar
CCl4 is nonpolar
LiCl is polar
CH4 is nonpolar
PCl6 is nonpolar
so LiCl would dissolve
