It is energy associated with gravity or gravitational force. (Potential<span> </span>energy<span> held by an object because of its high position compared to a lower position).</span>
Answer:
a) 0,014 min⁻¹
b) 1452 seconds.
c) 0,02156 M
d) The faster rate is for the reaction A. The largest rate constant, k, is for A reaction
Explanation:
a) For a first order reaction the half-life is:
<em>(1) </em>As half.life is 49 min. Rate constant -k- is: 0,014 min⁻¹
b) The half-life is defined as the time that takes in descompose the half of the initial concentration of a compound.
As half-life is 726s. In 726s the concentration will be 0,300M, In the next 726s the concentration will be 0,150M. Thus, thae time it takes to decrease concentration until 0,150M is 726s×2= 1452 seconds
c) In the same way, 11072 years are 4 half-life for this reaction, thus:
0,345M/2⁴ = 0,02156 M
d) By definition of half-life, the less half-life, the faster rate. Thus, The faster rate is for the reaction A.
By (1) half-life is inversely proportional to rate constant. Thus, the less half-life, the largest rate constant. The largest rate constant, k, is for A reaction
I hope it helps!
Answer:
(a) The system does work on the surroundings.
(b) The surroundings do work on the system.
(c) The system does work on the surroundings.
(d) No work is done.
Explanation:
The work (W) done in a chemical reaction can be calculated using the following expression:
W = -R.T.Δn(g)
where,
R is the ideal gas constant
T is the absolute temperature
Δn(g) is the difference between the gaseous moles of products and the gaseous moles of reactants
R and T are always positive.
- If Δn(g) > 0, W < 0, which means that the system does work on the surroundings.
- If Δn(g) < 0, W > 0, which means that the surroundings do work on the system.
- If Δn(g) = 0, W = 0, which means that no work is done.
<em>(a) Hg(l) ⇒ Hg(g)</em>
Δn(g) = 1 - 0 = 1. W < 0. The system does work on the surroundings.
<em>(b) 3 O₂(g) ⇒ 2 O₃(g)
</em>
Δn(g) = 2 - 3 = -1. W > 0. The surroundings do work on the system.
<em>(c) CuSO₄.5H₂O(s) ⇒ CuSO₄(s) + 5H₅O(g)
</em>
Δn(g) = 5 - 0 = 5. W < 0. The system does work on the surroundings.
<em>(d) H₂(g) + F₂(g) ⇒ 2 HF(g)</em>
Δn(g) = 2 - 2 = 0. W = 0. No work is done.
Answer:
In a movable pulley system, where does the other effort come from? is carried by this end of the rope. This means that the structure (ceiling) does half the effort.
Explanation:
