Hey there!
We Know that:
2 Ag⁺(aq) + Zn(s) <-> Zn²⁺(aq)+2 Ag(s)
The equilibrium expression for the reaction is:
Kc = [ Zn⁺² ] / [Ag⁺ ]²
Hope that helps!
Answer:
19.8m/s
Explanation:
Given parameters:
Mass of the ball = 10kg
Height of the rail = 20m
Unknown:
Velocity at the bottom of the rail = ?
Solution:
The velocity at the bottom of the rail is its final velocity.
Using the appropriate motion equation, we can find this parameter;
V² = U² + 2gH
V is the final velocity
U is the initial velocity
g is the acceleration due to gravity
H is the height
the ball was rolled from rest, U = 0
V² = O² + 2 x 9.8 x 20
V = 19.8m/s
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.
Here we go, 4 attachments, you made me work LOL
Answer:
(i) 5-day BOD of the waste is 120 mg/l.
(ii) The ultimate carbonaceous BOD (Lo) is 20 mg/l.
Explanation:
The dilution factor D is 0.05.
The initial DO is 8.0 mg/L and the DO after 5 days is 2.0 mg/L.
The BOD of the waste for an unseeded mixture is
The ultimate carbonaceous BOD (Lo) can be calculated as
