The balanced chemical reaction for the complete combustion of C4H10 is shown below:
C4H10 + (3/2)O2 --> 4CO2 + 5H2O
The enthalpy of formation are listed below:
C4H10: -2876.9 kJ/mol
O2: none (because it is pure substance)
CO2: -393.5 kJ/mol
H2O: -285.8 kJ/mol
The enthalpy of combustion is computed by subtracting the total enthalpy formation of the reactants from that of the products.
ΔHc = (4)(-393.5 kJ/mol) + (5)(-285.8 kJ/mol) - (-2876.9 kJ/mol)
= -<em>126.1 kJ</em>
Thus, the enthalpy of combustion of the carbon is -126.1 kJ.
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
Answer:
Na₂O will also form a brittle ionic compound.
Explanation:
Sodium is located in the same group as Potassium: Alkali Metals. This is because both elements have similar chemical properties. In term of their electronegativity, both elements are very close. Na has an electronegativity of 0.93, while K has an electronegativity of 0.82. This makes both ideal elements to create ionic compounds with Oxygen.
Anwer:
1000000000000000000000000000000000000000000000000000000.01
Answer:
<em><u>1</u></em> C10H16 + <em><u>8</u></em> Cl2 --> <em><u>10</u></em> C + <em><u>16</u></em> HCl
