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.
To learn more about enthalpy here
brainly.com/question/13981382
#SPJ1
Answer:
0.01931034
Explanation:
Steps:
ρ = m/v
=
28 gram
1.45 cubic meter
= 19.310344827586 gram/cubic meter
= 0.019310344827586 kilogram/cubic meter
When you heated the can with the bit of water inside and you boiled it over a flame, the water turned to vapor (gas) and the pressure in the inside of the can is different from the pressure on the outside of the can. When you placed the can into a ice water beaker or a container, the can shrunk it's size, decreasing it's mass and density. The can shrunk as a result of the inside pressure being equalized with the outside pressure.
The part where you placed it in the ice bath or container was when the water vapor was forced out of the can.
Yes, I agree.
Chemistry can be difficult.
