Answer:
-100 kJ
Explanation:
We can solve this problem by applying the first law of thermodynamics, which states that:

where:
is the change in internal energy of a system
Q is the heat absorbed/released by the system (it is positive if absorbed by the system, negative if released by the system)
W is the work done by the system (it is positive if done by the system, negative if done on the system)
For the system in this problem we have:
W = +147 kJ is the work done by the system
Q = +47 kJ is the heat absorbed by the system
So , its change in internal energy is:

Answer is: concentration of products increases (ammonia nad water).
Chemical reaction: heat + NH₄⁺ + OH⁻ ⇄ NH₃ + H₂<span>O.
</span>According to Le
Chatelier's Principle, the position of equilibrium moves to counteract the
change, because heat is increased, system consume that heat, so equilibrium is shifted to right, by decreasing concentration of reaactants and increasing concentration of product.
Answer:
Positive: a and b
Negative: c
Explanation:
The entropy (S) is the measure of the randomness of the system, and it intends to increase. The randomness can be determined by the energy of the molecules, their velocity and how distance they are between the other molecules.
When the entropy increases, ΔS is positive, when the entropy decreases, ΔS is negative. So, when gasoline mix with air in a car engine, the process intends to continue, the randomness increases and ΔS is positive. When hot air expands, the distance between the molecules increases, so ΔS is positive.
But, when humidity condenses, the molecules stay closer, so there's a decrease in the randomness, then ΔS is negative.
If I’m right this should be the answer
Answer:
The rate of disappearance of C₂H₆O = 2.46 mol/min
Explanation:
The equation of the reaction is given below:
2 K₂Cr₂O₇ + 8 H₂SO₄ + 3 C₂H₆O → 2 Cr₂(SO₄)₃ + 2 K₂SO₄ + 11 H₂O
From the equation of the reaction, 3 moles of C₂H₆O is used when 2 moles of Cr₂(SO₄)₃ are produced, therefore, the mole ratio of C₂H₆O to Cr₂(SO₄)₃ is 3:2.
The rate of appearance of Cr₂(SO₄)₃ in that particular moment is given 1.64 mol/min. This would than means that C₂H₆O must be used up at a rate which is approximately equal to their mole ratios. Thus, the rate of of the disappearance of C₂H₆O can be calculated from the mole ratio of Cr₂(SO₄)₃ and C₂H₆O.
Rate of disappearance of C₂H₆O = 1.64 mol/min of Cr₂(SO₄)₃ * 3 moles of C₂H₆O / 2 moles of Cr₂(SO₄)₃
Rate of disappearance of C₂H₆O = 2.46 mol/min of C₂H₆O
Therefore, the rate of disappearance of C₂H₆O = 2.46 mol/min