Given that there is 48 liters of gasoline to be burned and that 45 kJ of energy is released per gram of gasoline burned, the amount of energy that the gasoline fuel produces can then be calculated, First, we convert 48 liters of gasoline to units of mass (grams) in order to use the given conversion of 45 kJ per gram of gasoline. To do this, we use the density of gasoline which is 0.77 g/mL. The following expression is then used:
48 L gasoline x 1000 mL/L x 0.77 g/mL x 45 kJ/g gasoline = 1663200 kJ
<span>The amount of energy produced by burning 48 L of gasoline was then determined to be 1663200 kJ. </span>
Answer: the equilibrium will be displaced to the right leading an increase on the quantities of y(g) and z(s).
Justification:
According to the rules of equilibrium, based on Le Chatellier's priciple, any change in a system in equilibrium will be tried to be compensated to restablish the equilibrium
The higher the amount, and so the concentration, of X(g), the more will the forward reaction proceed leading to an increase on the concentration of the products y(g) and z (s). Look that that will also be accompanied by a decreasing on the pressure, since 2 molecules of the gas X(g) are converted into 1 molecule of the gas y(g).
Lets assume the gas is acting Ideally, then According to Ideal Gas Equation the density is given as,
d = P M / R T ------- (1)
Where;
P = Pressure = 1.03 atm
M = Molar Mass = 146.06 g/mol
R = Gas Constant = 0.08206 atm.L.mol⁻¹.K⁻¹
T = Temperature = 297 K
Putting Values in eq. 1,
d = (1.03 atm × 146.06 g/mol) ÷ (0.08206 atm.L.mol⁻¹.K⁻¹ × 297 K)
density = 6.17 g/L
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Answer:
Option (c) is correct.
Explanation:
According to 2nd law of thermodynamics, every spontaneous process is associated with increase in entropy (
).
As sodium acetate crystallizes out spontaneously therefore it is associated with .
During crystallization, bond formation takes place which results evolution of heat. In other words, crystallization phenomenon is exothermic (
).
So, the given crystallization process renders the following thermodynamic quantities- J/K and J
Option (c) is correct.
