The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
The bond energies data is given as follows:
BE for C≡O = 1072 kJ/mol
BE for Cl-Cl = 242 kJ/mol
BE for C-Cl = 328 kJ/mol
BE for C=O = 766 kJ/mol
The enthalpy change for the reaction is given as :
ΔHr×n = ∑H reactant bond - ∑H product bond
ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )
ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )
ΔHr×n = 1314 - 1422
ΔHr×n = - 108 kJ
Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
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Answer:
Head loss in turbulent flow is varying as square of velocity.
Explanation:
As we know that head loss in turbulent flow given as
Where
F is the friction factor.
L is the length of pipe
V is the flow velocity
D is the diameter of pipe.
So from above equation we can say that
It means that head loss in turbulent flow is varying as square of velocity.
We know that loss in flow are of two types
1.Major loss :Due to surface property of pipe
2.Minor loss :Due to change in momentum of fluid.
Assuming the conditions of the reaction are maintained and appropriate for the reaction to still occur, the reaction rate can be affected by increasing the concentration of the reagents used in a reaction. It will speed it up.
Answer:
True
Explanation:
Soluble substances dissolve.
Therefore they wouldn't be called soluble if they can't dissolve in solvents
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