Answer: 11.2 moles of 
are produced when 5.60 mol of ethane is burned in an excess of oxygen.
Explanation:
The combustion of ethane is represented using balanced chemical equation:
As oxygen is preset in excess, ethane acts as the limiting reagent as it limits the formation of product.
According to stoichiometry :
2 moles of propane produces 4 moles of carbon dioxide
Thus 5.60 moles of propane will produce=
moles of carbon dioxide
Thus 11.2 moles of are produced when 5.60 mol of ethane is burned in an excess of oxygen.
Answer : 0.026 moles of oxygen are in the lung
Explanation :
We can solve the given question using ideal gas law.
The equation is given below.
We have been given P = 21.1 kPa
Let us convert pressure from kPa to atm unit.
The conversion factor used here is 1 atm = 101.3 kPa.
V = 3.0 L
T = 295 K
R = 0.0821 L-atm/mol K
Let us rearrange the equation to solve for n.
0.026 moles of oxygen are in the lung
The answer is 2.138 g of H2
Answer:
The answer to your question is 8.2 g of TiF₄
Explanation:
Data
Theoretical yield = ?
mass of Ti = 5 g
mass of F₂ = 5 g
Balanced chemical reaction
Ti(s) + 2F₂ (g) ⇒ TiF₄(g)
Process
1.- Calculate the Molar mass of reactants and products
Ti = 48 g
2F₂ = 2( 19 x 2) = 76 g
TiF₄ = 48 + 76 = 124 g
2.- Calculate the limiting reactant
theoretical proportion Ti/F₂ = 48/76 = 0.63
experimental proportion Ti/F₂ = 5/5 = 1
Conclusion The limiting reactant is F₂ because the experimental proportion was lower than the theoretical proportion.
3.- Calculate the theoretical yield using the mass of F₂
76 g of F₂ --------------- 124 g of TiF₄
5 g of F₂ --------------- x
x = (5 x 124) / 76
x = 8.15 g of TiF₄
Answer:
σ*2pₓ, also called 
Explanation:
I have drawn the MO diagram for fluorine below.
Each F atom contributes seven valence electrons, so we fill the MOs of fluorine with 14 electrons.
We have filled the 
and 
MOs.
They are the highest occupied molecular orbitals (HOMOs).
The next unfilled level (the LUMO) is the σ*2pₓ orbital. If you use the symmetry notation, it is called the orbital.
This is the orbital that fluorine uses when it acts as an electron acceptor.
