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 : The mass of the water in two significant figures is, 
Explanation :
In this case the heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of iron metal = 
= specific heat of water = 
= mass of iron metal = 32.3 g
= mass of water = ?
= final temperature of mixture = 
= initial temperature of iron metal = 
= initial temperature of water = 
Now put all the given values in the above formula, we get
Therefore, the mass of the water in two significant figures is,
Answer:
13.5 g
Explanation:
This question is solved easily if we remember that the number of moles is obtained by dividing the mass into the atomic weight or molar mass depending if we are referring to elements or molecules.
Therefore, the mass of aluminum in the reaction will the 0.050 mol Al times the atomic weight of aluminum.
number of moles = n = mass of Al / Atomic Weight Al
⇒ mass Al = n x Atomic Weight Al = 0.050 mol x 27 g mol⁻¹
= 13.5 g
We have three significant figures in 0.050 and therefore we should have three significant figures in our answer.
Answer:
1. Many different hypotheses will be tested.
Explanation:
Scientists test every hypothesis they make for their investigation, before moving onto bigger conclusions.
Hello!
The chemical reaction for the dissolving of calcium fluoride is the following:
CaF₂(s) ⇄ Ca⁺²(aq) + 2F⁻(aq)
In this reaction, and according to Le Chatelier's principle, the action that would shift this reaction away from solid calcium fluoride and towards the dissolved ions is the removing of fluoride ions.
Le Chatelier's principle states that in an equilibrium reaction, the system would shift in the opposite direction of the changes. If we remove fluoride ions from the system, it will shift towards the formation of more fluoride ions by dissolving more Calcium Fluoride to achieve equilibrium again.
Have a nice day!
