<span>Due to limitations on typography, I will have to describe the equation instead of actually writing it.
Crude appearance.
18 18 0
F --> O + e
9 8 1
Detailed description. Each of the 3 components have both a left superscript and a left subscript which is a superscript and a subscript to the LEFT of the main figure unlike the usual right side that you see subscripts and superscripts.
The equation will be F with an 18 left superscript and a 9 left subscript to represent Florine with atomic weight of 18 and 9 protons.
Followed by a right arrow to indicate the direction the reaction is going.
Followed by the letter O with a left superscript of 18 and a left subscript of 8 to represent Oxygen with atomic weight of 18 and 8 protons.
Followed by a plus sign to indicate more.
Followed by either the lower case letter "e" or the upper case Greek character beta with a left superscript of 0 and a left subscript of 1 or +1 to represent the positron being emitted with a positive charge and an atomic weight of 0.</span>
Answer:
308 g
Explanation:
Data given:
mass of Fluorine (F₂) = 225 g
amount of N₂F₄ = ?
Solution:
First we look to the reaction in which Fluorine react with Nitrogen and make N₂F₄
Reaction:
2F₂ + N₂ -----------> N₂F₄
Now look at the reaction for mole ratio
2F₂ + N₂ -----------> N₂F₄
2 mole 1 mole
So it is 2:1 mole ratio of Fluorine to N₂F₄
As we Know
molar mass of F₂ = 2(19) = 38 g/mol
molar mass of N₂F₄ = 2(14) + 4(19) =
molar mass of N₂F₄ = 28 + 76 =104 g/mol
Now convert moles to gram
2F₂ + N₂ -----------> N₂F₄
2 mole (38 g/mol) 1 mole (104 g/mol)
76 g 104 g
So,
we come to know that 76 g of fluorine gives 104 g of N₂F₄ then how many grams of N₂F₄ will be produce by 225 grams of fluorine.
Apply unity formula
76 g of F₂ ≅ 104 g of N₂F₄
225 g of F₂ ≅ X of N₂F₄
Do cross multiplication
X of N₂F₄ = 104 g x 225 g / 76 g
X of N₂F₄ = 308 g
So,
308 g N₂F₄ can be produced from 225 g F₂
1) Calculate the number of mols,n, of the substance
n = mass/ molar mass = 326.0 g / 58.45 g / mol = 5.577 moles
2) Calculate the molar heat of fusion as the total heat released by the sample divided by the number of moles
hf = heat released / n = 4325.8 cal / 5.577 moles = 775. 59 cal /mol
