Answer:
5.95 L
Explanation:
V1/V2 = T1/T2
4.82 L/V2 = 243.1K/300.1K
V2 = 4.82*300.1/243.1 L = 5.95 L
The enthalpy change of the reaction is <u>-1347.8 kJ.</u>
<h3>What is the enthalpy change, ΔH, of the reaction?</h3>
The enthalpy change, ΔH, of the reaction is calculated from Hess's law of constant heat summation as follows:
Hess's law states that the enthalpy change of a reaction is the sum of the enthalpies of the intermediate reaction.
Given the reactions below and their enthalpy values;
1. X (s) + 12 O₂ (g)⟶ XO (s) ΔH₁ = −850.5 kJ
2. XCO₃ (s) ⟶ XO (s) + CO₂ (g) ΔH₂ = +497.3 kJ
The enthalpy change, ΔH, of the reaction whose equation is given below, will be:
X (s) + 12 O₂ (g) + CO₂ (g) ⟶ XCO₃ (s)
ΔH = ΔH₁ - ΔH₂
ΔH = − 850.5 kJ - (+497.3 kJ)
ΔH = -1347.8 kJ
Learn more about enthalpy change at: brainly.com/question/14047927
#SPJ1
To find the freezing point, you use the following formula
fp solution= fp solvent - Δfp solution
fp solvent= 0 C (since the solvent was not stated, you can assume is water which freezing point is 0 degree celsius)
but first, we need to find the depression in freezing point (Δfp solution)
Δfp solution= Kf x i x m, where Kf is constant for the solvent, i is the number of particles produced by the solute and m is the molality of the solution.
Kf= 1.86 for water
i= 4 , becausethe solute breaks into 4 ions, K3PO4---> 3K+ + PO4-3
m= 2.60
Δfp solution= (1.86) (4) (2.60)= 19.3 C
fp solution= 0 - 19.3C=<span> -19.3C
</span>
for boiling point, we use a similar formula and way to solve it.
Δbp solution= bp solvent + Δbp solution
bp solvent= 100 C for water
to find the elevation boiling point (Δbp soln) is equal to:
Δbp solution= Kb x i x m
Kb= 0.512
i= 4
m= 2.60
Δbp solution= (0.512) (4) (2.60)= 5.32 C
bp solution= 100C + 5.32C= <span>105.32 C</span>
The balanced equation for the combustion of CH₄ is as follows
CH₄ + 2O₂ ---> CO₂ + 2H₂O
stoichiometry of CH₄ to CO₂ is 1:1
therefore molar ratio of CH₄ to CO₂ is 1:1
the number of CH₄ moles combusted - 1.00 mol
number of CO₂ moles formed = number of CH₄ moles reacted
therefore number of CO₂ mol formed - 1.00 mol
mass of CO₂ produced - 1.00 mol x 44 g/mol = 44.0 g
mass of CO₂ produced is 44.0 g
It is so false....it dose not mater how big the orbital is.