Answer:
Bond energy of carbon-fluorine bond is 485 kJ/mol
Explanation:
Enthalpy change for a reaction, is given as:
![\Delta H_{rxn}=\sum [n_{i}\times (E_{bond})_{i}]-\sum [n_{j}\times (E_{bond})_{j}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn_%7Bi%7D%5Ctimes%20%28E_%7Bbond%7D%29_%7Bi%7D%5D-%5Csum%20%5Bn_%7Bj%7D%5Ctimes%20%28E_%7Bbond%7D%29_%7Bj%7D%5D)
Where 
and 
represents average bond energy in breaking "i" th bond and forming "j" th bond respectively.
and 
are number of moles of bond break and form respectively.
In this reaction, one mol of C=C, four moles of C-H and one mol of F-F bonds are broken. One mol of C-C bond, four moles of C-H bonds and two moles of C-F bonds are formed
So, 
or, 
or, 
So bond energy of carbon-fluorine bond is 485 kJ/mol
Answer:
82.0 mL
Explanation:
Step 1: Given data
- Concentration of concentrated acid (C₁): 12.2 M
- Volume of concentrated acid (V₁): ?
- Concentration of dilute acid (C₂): 1.00 M
- Volume of dilute acid (V₂): 1.00 L
Step 2: Calculate the required volume of the concentrated acid
We want to prepare a dilute solution from a concentrated one. We can calculate the volume of the concentrated acid using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 1.00 M × 1.00 L / 12.2 M = 0.0820 L = 82.0 mL
A covalent bond is formed between two non-metals that have similar electronegativities.
An <em>i</em><em>o</em><em>n</em><em>i</em><em>c</em><em> </em><em>b</em><em>o</em><em>n</em><em>d</em> is formed between a metal and a non-metal. Non-metals(-ve ion) are "stronger" than the metal(+ve ion) and can get electrons very easily from the metal. These two opposite ions attract each other and form the ionic bond.
