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:
From the image the answer is 24.
Answer:
Digestion of food.
Explanation:
I hope my answer help you.
Calcium fluoride.
Ca is metal, F is non-metal, so they form ionic bond.
Ca as metal can form only positive ion. Ca in the second group, so the charge of Ca ion is 2+. Ca²⁺
F is in the 17th group, so it has 7 electrons on the last level. It is non-metal, non-metal, so it has negative charge -(8-7)=-1. "8" because on the last level cannot be more than 8 electrons. F-ion is F¹⁻.
Ca²⁺ F¹⁻
Number of positive charges should be equal to number of negative charges,
Formula of calcium fluoride
CaF2.
2 atoms Fluorine bond with Calcium.
This is a one-step unit analysis problem. Since we are staying in moles, grams of our compound, and thus molar mass, is not needed.
1 mole is equal to 6.022x10²³ particles as given, so:
<h3>
Answer:</h3>
2.49 mol
Let me know if you have any questions.
