5 is 2 I’m not sure about 4 though....
Answer:
Explanation:
For the reaction
C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O
We can calculate the standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:
ΔHºc = 2ΔHºf CO2 (g) + 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )
( we were not give the water state but we know we are at standard conditions so it is in its liquid state )
The ΔHºfs can be found in appropiate reference or texts.
ΔHºc = 2ΔHºf CO2 (g)+ 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )
= [ 2 ( -393.52 ) + 3 ( -285.83 ) ] - [( -276.2 + 0 ) ] kJ
ΔHºc = -1368.33 kJ
Yes they are what are your options
Answer:
D.
Explanation:
The reaction is losing potential energy, which means that the reaction is losing that energy as heat. Exothermic is the loss of energy. Therefore it will be D.
Answer:
20.5torr
Explanation:
Given parameters:
V₁ = 15L
P₁ = 8.2 x 10⁴torr
V₂ = 6 x 10⁴L
Unknown:
P₂ = ?
Solution:
To solve this problem we have to apply the claims of Boyle's law.
Boyle's law is given mathematically as;
P₁ V₁ = P₂V₂
where P₁ is the initial pressure
V₁ is the initial volume
P₂ is final pressure
V₂ is final volume
8.2 x 10⁴ x 15 = P₂ x 6 x 10⁴
P₂ = 20.5torr
