Answer:
Part 1: - 1.091 x 10⁴ J/mol.
Part 2: - 1.137 x 10⁴ J/mol.
Explanation:
Part 1: At standard conditions:
At standard conditions Kp= 81.9.
∵ ΔGrxn = -RTlnKp
∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(81.9)) = - 1.091 x 10⁴ J/mol.
Part 2: PICl = 2.63 atm; PI₂ = 0.324 atm; PCl₂ = 0.217 atm.
For the reaction:
I₂(g) + Cl₂(g) ⇌ 2ICl(g).
Kp = (PICl)²/(PI₂)(PCl₂) = (2.63 atm)²/(0.324 atm)(0.217 atm) = 98.38.
∵ ΔGrxn = -RTlnKp
∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(98.38)) = - 1.137 x 10⁴ J/mol.
37.8 g CH2Br2 X (1 mol CH2Br2 / 173.83 g) = 4.60X10^-3 mol CH2Br2
<span>4.60X10^-3 mol CH2Br2 X (2 mol Br / 1 mol CH2Br2) X 6.02X10^23 atoms/mol = 5.54X10^21 bromine atoms</span>
Answer:
all atoms must have 6 protons to be a carbon atom
Explanation:
brainliest?
Combustion can be defined as the reaction of a compound with oxygen. The enthalpy of combustion of octane is 
for 
.
<h3>What is the enthalpy of reaction?</h3>
The enthalpy of reaction is the amount of heat energy absorbed or lost by the molecules in the chemical reaction.
The enthalpy of combustion is the amount of heat energy released by the compound in the reaction with oxygen.
The reaction in which heat is liberated with the reaction of a compound with oxygen has an enthalpy of combustion, equivalent to the enthalpy of reaction.
The combustion of octane can be given as:
Thus, the reaction has combustion energy equivalent to the enthalpy of the reaction is . Thus, option B is correct.
Learn more about enthalpy of reaction, here:
brainly.com/question/1657608
