125 mile *1gallon/35 mi = 135/35 = (27/7) gallon gasoline
27/7 gallon * 1 L/0.264 gallon = 14.6 L gasoline
14.6 L gasoline * 2.5kg CO2/1L gasoline= 36.5 kg CO2
36.5 kg CO2 * 1lb/0.454 kg = 80.4 lb
Answer: 80.4 lb CO2
There are five states of matter out of which we encounter three states of matter in our day today life
a) gas b) solid and c) liquid
the main difference between the three is of
a) Inter molecular forces of attraction
b) thermal energy
due to this
a) solid has high intermolecular forces and low thermal energy: thus they have fix shape and occupy fix volume
b) liquid has intermediate forces and medium themal energy. Thus they may have fixed volume and but no fix shape
c) gas has weak intermoelcular forces and high thermal energy. thus they have no fixed volume no fix shape
so in the given problem
the state of the substance D- Gas.
Answer:
k+
Ba2+
Al3+
K+
S2-
you may think that if it has a + it would be bigger, but it is actually the opposite.
Answer:
131.5 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
First, we will calculate the standard enthalpy of the reaction (ΔH°).
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)
) - 1 mol × ΔH°f(CaCO₃(s)
)
ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)
ΔH° = 179.2 kJ
Then, we calculate the standard entropy of the reaction (ΔS°).
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)
) - 1 mol × S°(CaCO₃(s)
)
ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)
ΔS° = 160.2 J/K = 0.1602 kJ/K
Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.
ΔG° = ΔH° - T × ΔS°
ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K
ΔG° = 131.5 kJ
