Answer:
The answer to your question is the letter C) 5648 kJ/mol
Explanation:
Data
C₁₂H₂₂O₁₁ + 12 O₂ ⇒ 12 CO₂ + 11 H₂O
H° C₁₂H₂₂O₁₁ = -2221.8 kJ/mol
H° O₂ = 0 kJ / mol
H° CO₂ = -393.5 kJ/mol
H° H₂O = -285.8 kJ/mol
Formula
ΔH° = ∑H° products - ∑H° reactants
Substitution
ΔH° = 12(-393.5) + 11(-285.8) - (-2221.8) - (0)
ΔH° = -4722 - 3143.8 + 2221.8
Result
ΔH° = -5644 kJ/mol
Answer: No, a<span>t high pressures, volume of a real gas does not compare with the volume of an ideal gas under the same conditions.
Reason:
For an ideal gas, there should not be any intermolecular forces of interaction. However, for real gases there are intermolecular forces of interaction like dipole-dipole and dipole-induced dipole. Further, at high pressures, molecules are close by. Hence, extend of these intermolecular forces is expected to be high. This results in decreases in volume of real gas. Thus, </span>volume of a real gas does not compare with the volume of an ideal gas under the same conditions.
Answer:
D
Explanation:
D. V1P1 / T1=V2P2 / T2 is correct
This is not chemistry but
see it is a triangular prim on it's side
V=BH
are of base times height
the base is a triangle
height is 18.5
base=1/2bh
b=8.6
h=8.4
base=1/2(8.6)(8.4)
base=36.12
V=bh
V=36.12*18.5
V=668.22
round
V=668.2 ft^3
One thing that does not change is the chemical composition of water, which is still H2O. And maybe mass, if all of the particles remain inside the beaker, which was never mentioned in the question so I am not sure.
