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:
brainly.com/question/11653058
#SPJ4
Answer:
The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).
Explanation:
The reaction between an acid and a base is called neutralization, forming a salt and water.
Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.
When an acid is neutralized, the amount of base added must equal the amount of acid initially present. This base quantity is said to be the equivalent quantity. In other words, at the equivalence point the stoichiometry of the reaction is exactly fulfilled (there are no limiting or excess reagents), therefore the numbers of moles of both will be in stoichiometric relationship. So:
V acid *M acid = V base *M base
where V represents the volume of solution and M the molar concentration of said solution.
In this case:
- V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
- M acid= 0.129 M
- V base= ?
- M base= 0.135 M
Replacing:
0.0137 L* 0.129 M= V base* 0.135 M
Solving:
V base=0.0131 L = 13.1 mL
<u><em>
The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).</em></u>
It will belong to the metals because metals bond with nonmetals like chlorine to form ionic compounds
Answer:
The answer is 98.07848. We assume you are converting between grams H2SO4 and mole. You can view more details on each measurement unit: This compound is also known as Sulfuric Acid. The SI base unit for amount of substance is the mole. 1 grams H2SO4 is equal to 0.010195916576195 mole.
<u>Quick conversion chart of moles H2SO3 to grams</u>
1 moles H2SO3 to grams = 82.07908 grams
2 moles H2SO3 to grams = 164.15816 grams
3 moles H2SO3 to grams = 246.23724 grams
4 moles H2SO3 to grams = 328.31632 grams
5 moles H2SO3 to grams = 410.3954 grams
6 moles H2SO3 to grams = 492.47448 grams
7 moles H2SO3 to grams = 574.55356 grams
8 moles H2SO3 to grams = 656.63264 grams
9 moles H2SO3 to grams = 738.71172 grams
10 moles H2SO3 to grams = 820.7908 grams
