Answer:c the correct technology cannot support this mission
Explanation:
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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The property of potential energy that distinguishes it from kinetic energy are Shape and position
Answer:
Mass= 2.77g
Explanation:
Applying
P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28
PV=nRT
NB
Moles(n) = m/M
PV=m/M×RT
m= PVM/RT
Substitute and Simplify
m= (2.09×1.13×28)/(0.082×291)
m= 2.77g
Answer:
25.42 atm
Explanation:
Data Given:
Volume of a gas ( V )= 2.00 L
temperature of a gas ( T ) = 310 K
number of moles (n) = 2 mol
Pressure of a gas ( P ) = to be find
Solution:
Formula to be used
PV= nRT
Rearrange the above formula
P = nRT / V . . . . . . . . . . (1)
Where R is ideal gas constant
R = 0.08205 L atm mol⁻¹ K⁻¹
Put values in equation 1
P = nRT / V
P = 2 mol x 0.08205 L atm mol⁻¹ K⁻¹ x 310 k / 2 L
P = 50.84 L atm / 2 L
P = 25.42 atm
P ressure of gas (P) will be = 25.42 atm
